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WUCT121
Discrete Mathematics
Logic
1.
Logic
2.
Predicate Logic
3.
Proofs
4.
Set Theory
5.
Relations and Functions
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Logic
1
Section 1.
1.1.
Logic
Introduction.
In developing a mathematical theory, assertions or
statements are made. These statements are made in the
form of sentences using words and mathematical symbols.
When proving a theory, a mathematician uses a system of
logic. This is also the case when developing an algorithm
for a program or system of programs in computer science.
The system of logic is applied to decide if a statement
follows from, or is a logical consequence of, one or more
other statements.
You are familiar with using numbers in arithmetic and
symbols in algebra. You are also familiar with the ‘rules’ of
arithmetic and algebra.
Examples:
(3 + 4) + 6 = 3 + (4 + 6)
Associativity
= 3 + 10
•
= 13
•
3 x − 5 x = (3 − 5)x
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Distributivity
= −2 x
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2
In a similar way, Logic deals with statements or sentences
by defining symbols and establishing ‘rules’.
Roughly speaking, in arithmetic an operation is a rule for
producing new numbers from a pair of given numbers, like
addition (+) or multiplication (× ).
In logic, we form new statements by combining short
statements using connectives, like the words and, or.
Examples:
•
This room is hot and I am tired.
•
x < 1 or x > 7 .
1.2.
Statements
1.2.1.
Definition
Definition: Statement. A statement or proposition is an
assertion or declarative sentence which is true or false, but
not both.
The truth value of a mathematical statement can be
determined by application of known rules, axioms and laws
of mathematics.
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A statement which is true requires a proof.
Examples:
•
Is the following statement True or False?
For a real number x, if x 2 = 1, then x = 1 or x = −1.
The statement is TRUE. Therefore, we must prove it.
Consider x 2 = 1.
Adding − 1 to both sides gives x 2 − 1 = 0 .
Factorising this equation, we have ( x − 1)( x + 1) = 0 .
Therefore, x − 1 = 0 or x + 1 = 0 .
Case 1:
x −1 = 0.
Add 1 to both sides and we have x = 1.
Case 2:
x +1 = 0.
Add − 1 to both sides and we have x = −1.
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4
A statement which is false requires a demonstration.
Example:
•
Is the following statement True or False?
5 − (3 − 2) = (5 − 3) − 2
The statement is FALSE. Therefore, we must demonstrate
it.
5 − (3 − 2) = 5 − 1
=4
(5 − 3) − 2 = 2 − 2
=0
∴ 5 − (3 − 2) ≠ (5 − 3) − 2
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Exercise:
Determine which of the following sentences are statements.
For those which are statements, determine their truth value.
(i) 2 + 3 = 5
Statement
(ii) It is hot and sunny
Statement
True
outside.
(iii) 2 + 3 = 6
Statement
(iv) Is it raining?
Not a statement
(v) Go away!
Not a statement
(vi) There exists an
Statement
False
True
even prime
number.
(vii) There are six
Statement
people in this room.
(viii) For some real
Statement
True
number , x, x < 2
(ix) x < 2
See comment in notes
(x) x + y = y + x
See comment in notes
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Strictly speaking, as we don’t know what x or y are, in parts
(ix) and (x), these should not be statements. In
Mathematics, x and y usually represent real numbers and
we will assume this is the case here.
Therefore, (ix) is either true or false (even if we don’t know
which) and (x) is always true, so we will allow both.
1.2.2.
Simple Statements
Definition: Simple Statement. A simple or primitive
statement is a statement which cannot be broken down into
anything simpler.
A simple statement is denoted by use of letters p, q, r...
Examples:
•
p: There are seven days in a week
p is a simple statement
•
p:2+3= 6
p is a simple statement
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1.2.3.
Compound Statements
Definition: Compound Statement. A compound or
composite statement is a statement which is comprised of
simple statements and logical operations.
A compound statement is denoted by use of letters P, Q,
R...
Examples:
•
P: There are seven days in a week and twelve months
in a year.
Is a compound statement.
p: There are seven days in a week
q: There twelve months in a year
Operation: and
•
P: 2 + 3 = 6 or 5 − (3 − 2) = (5 − 3) − 2 .
Is a compound statement.
p: 2 + 3 = 6
q: 5 − (3 − 2) = (5 − 3) − 2
Operation: or
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•
P: If it is not raining then I will go outside and eat my
lunch.
Is a compound statement
p: It is raining
q: I will go outside
r: I will eat my lunch
Negation of p
Operations: If … then, and
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Exercises:
Determine which of the following are simple statements,
and which are not. For those which are not, identify the
simple statement(s) used.
Simple Statement
(i) 2 + 3 = 5
is a simple Statement
(ii) It is hot and
p: It is hot
sunny outside.
Operation
and
q: It is sunny outside
(iii) 2 + 3 ≠ 6
p: 2 + 3 = 6
negation
(iv) x ≤ 2
p: x < 2
or
q: x = 2
(v) − 5 < x < 2
p: − 5 < x
and
q: x < 2
(vi) If I study hard
then I will pass
p: I study hard
If..then
q: I will pass my exam
my exam
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1.3.
Truth Tables
A statement P can hold one of two truth values, true or
false. These are denoted “T” and “F” respectively.
Note: Some books may use “1” for true and “0” for false.
When determining the truth value of a compound statement
all possible combinations of the truth values of the
statements comprising it must be considered.
This is done systematically by the use of truth tables. Each
connective is defined by its own unique truth table.
There are five fundamental truth tables which will be
covered in the following sections.
1.3.1.
Truth Table Construction
To construct a truth table assign each statement a column.
The number of rows in the table is determined by the
number of statements. For n statements, 2 n rows will be
required.
Systematically assign truth vales to each of the statements,
beginning in the first column.
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11
Once all possible truth values for the simple statements are
inserted, determine the truth vales of the compound
statements following the rules for the operations.
Example:
•
Given three statements P, Q, R. The table setup is:
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P
Q
R
T
T
T
T
T
F
T
F
T
T
F
F
F
T
T
F
T
F
F
F
T
F
F
F
Compound Statement
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12
1.4.
Logical Operations
There are five main operations which when applied to a
statement will return a statement.
If P and Q are statements, the five primary operations used
are:
not P,
the negation of P.
P or Q,
the disjunction of P and Q.
P and Q,
the conjunction of P and Q.
P implies Q,
the conditional of P and Q.
P if and only if Q, the biconditional of P and Q.
1.4.1.
Negation, “not”
Definition: Statement Negation.
If P is a statement, the negation of P is “not P” or “it is not
the case that P” and is denoted ~P.
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Examples:
•
There are not seven days in a week
p: There are seven days in a week
•
P:
It is raining outside.
~P: ~(It is raining outside.)
It is not raining outside.
•
x > 2 or x < 2
Q:
~Q: ~( x > 2 or x < 2 )
Simplified: x = 2 .
Exercises:
For each statement P, write down ~P.
•
P:
Discrete Maths is interesting.
~P: ~( Discrete Maths is interesting)
Discrete Maths is not interesting.
•
P
~ P:
x2 −1 = 0
(
~ x 2 −1 = 0
)
x2 −1 ≠ 0
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1.4.1.1
Truth Table for Negation
The negation of P has the opposite truth value from P,
~P is false when P is true; ~P is true when P is false.
All possible
truth values
for P
P
~P
T
F
F
T
All possible
truth values for
~P depending
on the value of
P.
Example:
Write down the truth value of the following statements.
•
•
P
~P
2+5= 7
2+5≠ 7
T
F
This room is
empty
empty
F
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This room is not
T
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15
Exercise:
Write down the truth value of the following statements.
•
•
P
~P
1∈ 
T
1∉ 
F
Division is a closed
Division is not a closed
operation on 
T
operation on 
F
Note:
The truth table for negation tells us that for any
•
statement P, exactly one of P or ~P is true. So, to prove P
is true, we have two methods:
∗
Direct: Start with some facts and end up
proving P in a direct step-by-step manner.
∗
Indirect: Don’t prove P is true directly, but
prove that ~P is false.
•
Generally, brackets are left out around ‘ ~ P ’.
Thus, ~ P ∨ Q means (~ P) ∨ Q , and not ~ ( P ∨ Q) .
This is similar to arithmetic where − x + y means (− x ) + y
and not − ( x + y ).
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1.4.2.
Disjunction, “or”
Definition: Disjunction.
If P and Q are statements the disjunction of P and Q is “P
or Q”, denoted P ∨ Q .
Examples:
•
Given P : 2 + 3 = 5 , Q : 2 + 3 = 6 , write down P ∨ Q .
P ∨ Q : 2 + 3 = 5 or 2 + 3 = 6
alternatively : ( 2 + 3 = 5 ) ∨ ( 2 + 3 = 6)
simplified : 2 + 3 = 5 or 6
•
Write P : x ≤ 5 using “ ∨ ”.
( x < 5) ∨ ( x = 5)
Exercises:
•
Write the following statements using “ ∨ ”
∗
I am catching the bus or train home.
(I am catching the bus home) ∨ (I am catching the train
home)
∗
A month has 30 or 31 days.
(A month has 30 days) ∨ (A month has 31 days)
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•
For the statements P and Q, write down P ∨ Q .
∗
P: x >0
Q: x =0
P ∨ Q : ( x > 0) ∨ ( x = 0)
simplified :
x≥0
∗
P: x is the square of an integer, Q: x is prime
P ∨ Q : ( x is the square of an integer ) ∨ ( x is prime)
1.4.2.1
Truth Table for Disjunction
The disjunction of P and Q is true when either P is true, or
Q is true, or both P and Q are true; it is false only when
both P and Q are false.
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P
Q
P∨Q
T
T
T
T
F
T
F
T
T
F
F
F
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18
Example:
Write down the truth value of the following statements.
P
•
2+3=5
T
•
1∉ 
F
P∨Q
Q
2+3=6
F
T
0∈
F
F
Exercise:
Write down the truth value of the following statements.
P
•
•
•
2 >1
Q
( x + 1) 2 = x 2 + 2 x + 1
T
T
2 is odd
5 is odd
F
T
2 <1
T
T
This room is empty
F
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P∨Q
F
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F
19
1.4.3.
Conjunction, “and”
Definition: Conjunction.
If P and Q are statements the conjunction of P and Q is “P
and Q”, denoted P ∧ Q .
Examples:
•
Given P: It is hot, Q: It is sunny, write down P ∧ Q .
P ∧ Q : (It is hot) ∧ (It is sunny)
Simplified: It is hot and sunny
•
Write P : 0 < x < 5 using “ ∧ ”.
(0 < x ) ∧ ( x < 5)
Exercises:
•
Write the following statements using “ ∧ ”
∗
Snow is cold and wet.
(Snow is cold) ∧ (Snow is wet)
∗
Natural numbers are positive and whole
numbers.
(Natural numbers are positive numbers) ∧ (Natural
numbers are whole numbers)
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•
For the statements P and Q, write down P ∧ Q .
∗
P: x >0
Q : x <1
P ∧ Q : ( x > 0 ) ∧ ( x < 1)
simplified :
0 < x <1
∗
P: x is even, Q: x is a natural number
P ∧ Q : ( x is even ) ∧ ( x is a natural number )
1.4.3.1
Truth Table for Conjunction
The conjunction of P and Q is true when, and only when,
both P and Q are true.
If either P or Q are false, of if both are false, P ∧ Q is false.
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P
Q
P∧Q
T
T
T
T
F
F
F
T
F
F
F
F
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21
Example:
Write down the truth value of the following statements.
P
•
2+3=5
T
•
1∉ 
F
P∧Q
Q
2+3=6
F
F
0∈
F
F
Exercise:
Write down the truth value of the following statements.
P
•
•
•
Q
6 >π
2 >1
T
T
2 is odd
5 is odd
F
T
2 <1
T
F
4 = 23
F
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P∧Q
F
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F
22
1.4.4.
Conditional, “If … then”, “implies”
Definition: Conditional.
If P and Q are statements the conditional of P by Q is “If P
then Q” or “P implies Q”, and is denoted P ⇒ Q .
Examples:
•
Given P: It is raining, Q: I will go home, write down
P ⇒ Q.
P ⇒ Q : (It is raining) ⇒ (I will go home)
Simplified: If it raining then I will go home
•
Write “If x is even then x 2 is even” using “ ⇒ ”.
x is even ⇒ x 2 is even
Exercises:
•
Write the following statements using “ ⇒ ”
∗
If the snow is good then I will go skiing.
(The snow is good) ⇒ (I will go skiing)
∗
If x is a natural number then x is an integer.
(x is a natural number) ⇒ (x is an integer)
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•
For the statements P and Q, write down P ⇒ Q .
∗
P : x > −1
Q: x >0
P ⇒ Q : ( x > −1) ⇒ ( x > 0 )
∗
P: x is even, Q: x is a natural number
P ⇒ Q : ( x is even ) ⇒ ( x is a natural number )
If x is even then x is a natural number
1.4.4.1
Truth Table for Conditional
The conditional of P by Q is false when P is true and Q
false, otherwise it is true.
We call P the hypothesis (or antecedent) of the conditional
and Q the conclusion (or consequent).
In determining the truth values for conditional, consider the
following example.
Suppose your lecturer say to you:
“If you arrive for the lecture on time, then I will mark you
present.
Under what circumstances are you justified in saying the
lecturer lied? In other words under what circumstances is
the above statement false?
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It is false when you show up on time and are not marked
present.
The lecturers promise only says you will be marked present
if a certain condition (arriving on time) is met; it says
nothing about what will happen if the condition is not met.
So if the condition (arriving on time) is not met, you cannot
in fairness say the promise is false regardless of whether or
not you are marked present.
This example demonstrates that the only combination of
circumstances in which you have a conditional statement
false is when the hypothesis is true and the conclusion is
false.
Thus the truth table for conditional is:
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P
Q
P⇒Q
T
T
T
T
F
F
F
T
T
F
F
T
Logic
25
Example:
Write down the truth value of the following statements.
P
•
2+3=5
T
•
1∉ 
F
P⇒Q
Q
2+3=6
F
F
0∈
F
T
Exercise:
Write down the truth value of the following statements.
P
•
Q
2 >1
2 >1
T
•
2 is even
T
F
2 <1
F
4 <1
F
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T
5 is even
T
•
P⇒Q
F
Logic
T
26
Alternative wording for P ⇒ Q can be:
•
If P then Q.
•
P implies Q.
•
Q if P.
•
Q provided P.
•
Q whenever P.
•
P is a sufficient condition for Q.
•
Q is a necessary condition for P.
•
P only if Q.
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27
1.4.5.
Biconditional, “If and only if”
Definition: Biconditional.
If P and Q are statements the biconditional of P and Q is
“P if, and only if Q” and is denoted P ⇔ Q .
Examples:
•
Given P: Mark can study algebra, Q: Mark passes
pre-algebra, write down P ⇔ Q .
P ⇔ Q : (Mark can study algebra) ⇔ (Mark passes
pre-algebra)
Simplified: Mark can study algebra if, and only if, he
passes pre-algebra
•
Write “Water boils if, and only if, it’s temperature is
over 100 oC ” using “ ⇔ ”.
Water boils ⇔ Water temperature is over 100o C
Exercises:
•
Write the following statements using “ ⇔ ”
∗
I will go swimming if, and only if, the water is
warm.
(I will go swimming) ⇔ (The water is warm)
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28
∗
x is a natural number if, and only if, x is an
integer.
(x is a natural number) ⇔ (x is an integer)
•
For the statements P and Q, write down P ⇔ Q .
∗
P : x ∈
Q: x >0
P ⇔ Q : ( x ∈ ) ⇔ ( x > 0 )
∗
P: x is positive, Q: x is a natural number
P ⇔ Q : ( x is positive) ⇔ ( x is a natural number )
1.4.5.1
Truth Table for Biconditional
The biconditional of P and Q is true if both P and Q have
the same truth value, and is false if P and Q have opposite
truth values.
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P
Q
P⇔Q
T
T
T
T
F
F
F
T
F
F
F
T
Logic
29
Example:
Write down the truth value of the following statements.
P
•
2+3=5
T
•
1∉ 
F
P⇔Q
Q
2+3=6
F
F
0∈
F
T
Exercise:
Write down the truth value of the following statements.
P
•
•
•
Q
2 >1
2 >1
T
T
2 is odd
5 is odd
F
T
2 <1
T
F
4 <1
F
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P⇔Q
F
Logic
T
30
Alternative wording for P ⇔ Q can be:
•
P if, and only if Q.
•
P iff Q.
•
P implies and is implied by Q.
•
P is equivalent to Q.
•
P is a necessary and sufficient condition for Q.
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31
1.4.6.
Order of Operation for Logical Operators.
The order of operation for logical operators is as follows:
1.
Evaluate negations first
2.
Evaluate ∨ and ∧ second. When both are present,
parenthesis may be needed, otherwise work left to right.
Evaluate ⇒ and ⇔ third. When both are present,
3.
parenthesis may be needed, otherwise work left to right.
Note: Use of parenthesis will determine order of operations
which over ride the above order.
Examples:
Indicate the order of operations in the following:
•
~
{ p∧
{q
•
~
{ ( p∧
{q)
•
1
•
2
2
~
{ p∧
{( q∨
{r)
1
1
3
2
~
{ p⇒
{r
{q∧
1
3
2
Exercises:
Indicate the order of operations in the following:
•
(~
{ p⇒
{r
{ q )∧
•
~
{ ( p∨
{q)
•
1
•
2
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2
3
~
{ p⇒
{r
{ q∨
1
1
2
~
{ p⇔
{r
{ q∧
1
Logic
3
3
2
32
1.4.7.
Main Connective
Definition: Main Connective.
The main connective is the operation which “binds” the
statement together.
It is the final operation performed and is denoted with “*”.
Examples:
Indicate the main connective in the following:
•
~
{ p∧
{q
•
~
{ ( p∧
{q)
•
1
•
2*
2*
~
{ p∧
{ ( q∨
{r)
1
1
3*
2
~
{ p⇒
{r
{q∧
1
3*
2
Exercises:
Indicate the main connective in the following:
•
(~
{ p⇒
{r
{q)∧
•
~
{ ( p∨
{q)
•
1
•
2*
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2
3*
~
{ p⇒
{r
{ q∨
1
1
2
~
{ p⇔
{r
{ q∧
1
Logic
3*
3*
2
33
Example:
Construct a truth table for ~ ( p∧ ~ q ) , indicating order of
operations and the main connective
T
∧
F
F
F
F
T
T
F
T
T
F
F
F
F
T
F
T
3*
2
1
p
q
~
T
T
T
Step:
(p
~
q)
Exercises:
•
Construct a truth table for ~ p ⇒~ q ∧ p , indicating
order of operations and the main connective
⇒
~
F
T
F
∧
F
F
F
T
T
T
F
T
T
F
F
F
F
F
T
F
T
F
1
3*
1
2
p
q
~
T
T
T
Step:
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p
Logic
q
p
34
•
Construct a truth table for ( p ∨ q ) ∧ (r ∨ q ), indicating
order of operations and the main connective
p
T
T
T
T
F
F
F
F
Step:
•
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
(p
∨
T
T
T
T
T
T
F
F
1
q)
∧
T
T
T
F
T
T
F
F
2*
(r
∨
T
T
T
F
T
T
T
F
1
q)
Construct a truth table for (~ q ∧ r ) ∨ ~ ( p ∧ r ) ,
indicating order of operations and the main connective
p
T
T
T
T
F
F
F
F
Step:
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q
T
T
F
F
T
T
F
F
r (~ q
T F
F F
T T
F T
T F
F F
T T
F T
1
∧ r) ∨
F
F
F
T
T
T
F
T
F
T
F
T
T
T
F
T
2
3*
Logic
~ (p ∧ r)
F
T
T
F
F
T
T
F
T
F
T
F
T
F
T
F
2
1
35
1.5.
Tautologies and Contradictions
1.5.1.
Tautology
Definition: Tautology.
Any statement that is true regardless of the truth values of
the constituent parts is called a tautology or tautological
statement.
Examples:
Complete the truth table for the statement P ⇒ (Q ⇒ P )
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(Q
⇒
Q
T
T
T
T
T
F
T
T
F
T
T
F
F
F
T
T
2*
1
Step:
P
⇒
P
Logic
P)
36
Exercises:
•
Complete the truth table for the statement
(( P ⇒ Q ) ∧ P ) ⇒ Q to show it is a tautology.
⇒
Q
T
T
T
∧
T
T
F
F
F
T
F
T
T
F
T
F
F
T
F
T
1
2
3*
((P
Step:
•
⇒
P
Q)
P)
Q
T
Complete the truth table for the statement
(( P ⇒ Q ) ∧ ~ Q ) ⇒ ~ P to show it is a tautology.
Q
T
T
T
∧
F
T
F
F
F
T
F
F
Step:
WUCT121
⇒
P
((P
Q)
~Q) ⇒ ~ P
F
T
F
F
T
T
F
T
F
F
T
T
T
T
T
T
T
2
3
1
4*
1
Logic
37
1.5.1.1
Quick Method for Showing a
Tautology
In constructing a truth table for a compound statement
comprised of n statements, there will be 2n combinations of
truth values. This method can be long for large numbers of
statements.
We will consider a quicker method for determining if a
compound statement is a tautology. However, truth tables
are reliable (“safe”) and are highly recommended if the
“quick” method is confusing or leading nowhere!
The quick method relies on the fact that if a truth value of
“F” can occur under the main connective (for some
combination of truth values for the components), then the
statement is not a tautology. If this truth value is not
possible, then we have a tautology.
Therefore, to determine whether a statement is a tautology,
we place an “F” under the main connective and work
backwards.
WUCT121
Logic
38
Examples:
•
Determine if P ⇒ (Q ⇒ P ) is a tautology, using the
quick method
P
Step
1.Place “F” under main
⇒
(Q ⇒
2*
1
P)
F
connective
2. For “F” to occur under the
F
T
main connective, P must be
“T” and ⇒ must be “F”
3. For “F” to occur under ⇒ ,
T
F
Q must be “T” and P must be
“F”
P cannot be both “T” and “F”, thus P ⇒ (Q ⇒ P ) can only
ever be true and is a tautology.
WUCT121
Logic
39
•
Determine if ( P ∧ Q) ⇒ (R ∧ S ) is a tautology, using
the quick method
(P
Step
∧
1
Q)
⇒
(R
3*
1.Place “F” under main
∧
2
S)
F
connective
2. For “F” to occur under the
T
F
main connective, ( P ∧ Q ) must
be “T” and ( R ∧ S ) must be
“F”
T
3. For “T” to occur under
T
( P ∧ Q ) , P must be “T” and Q
must be “T”
3. For “F” to occur under
T
F
(R ∧ S ), R can be “T” and S
can be “F”
As there is a valid combination of truth values which gives
“F” under the main connective, ( P ∧ Q ) ⇒ (R ∧ S ) is not a
tautology.
WUCT121
Logic
40
Exercises:
•
Use the “quick” method for the statement
(( P ⇒ Q ) ∧ P ) ⇒ Q to determine if it is a tautology.
((P ⇒ Q)
Step
1
∧
2
P)
⇒
Q
3*
1. Place “F” under main
F
connective.
2. For “F” to occur under the
T
F
main connective, ∧ must be
“T” and Q must be “F”
3. For “T” to occur under ∧ , P
T
T
must be “T” and P ⇒ Q must
be “T”
T
4. For “T” to occur under
T
P ⇒ Q ,when P is “T” Q must
be “T”
Q cannot be both “T” and “F”, thus (( P ⇒ Q ) ∧ P ) ⇒ Q can only
ever be true and is a tautology.
WUCT121
Logic
41
•
Determine if the statement (( P ⇒ Q ) ∧ ~ Q ) ⇒ ~ P is
a tautology, using the “quick” method.
Step:
((P ⇒ Q) ∧ ~Q) ⇒
2
3 1 4*
1.Place “F” under main
~P
1
F
connective
2. For “F” to occur under
T
F
the main connective, ∧
must be “T” and ~P must
be “F”
T
3. For “T” to occur under
T
∧ , ~Q must be “T” and
P ⇒ Q must be “T”
4. For “T” to occur under
T
T
P ⇒ Q ,when P is “T”, Q
must be “T”
At step 2, ~P is “F”, thus P is “T”. Step 3 shows ~Q is “T”
thus Q is “F”and step 4 gives Q is “T”. Q cannot be both “T”
and “F”, thus (( P ⇒ Q ) ∧ ~ Q ) ⇒ ~ P can only ever be true
and is a tautology.
WUCT121
Logic
42
1.5.2.
Contradiction
Definition: Contradiction.
Any statement that is false regardless of the truth values of
the constituent parts is called a contradiction or
contradictory statement.
Examples:
Complete the truth table for the statement
~ ( P ∧ Q ) ⇔ (Q ∧ P )
P
Q
~
T
T
F
∧
T
T
F
T
F
T
F
F
Step:
WUCT121
⇔
F
∧
T
F
F
F
T
F
F
F
T
F
F
F
2
1
4*
3
(P
Logic
Q)
(Q
P)
43
Exercises:
•
Complete the truth table for the statement
~ ( P ∨ Q ) ∧ P to show it is a contradiction.
P
Q
~(P
T
T
F
∨
T
T
F
F
T
F
F
T
F
T
F
F
F
T
F
F
2
1
3*
Step:
•
Q)
∧
F
P
Complete the truth table for the statement
( P ∧ Q ) ∧ ~ Q to show it is a contradiction.
P
Q
T
T
∧
T
T
F
F
F
Step:
WUCT121
∧
F
~Q)
F
F
T
T
F
F
F
F
F
T
T
2
3*
1
(P
Logic
Q)
F
44
1.5.2.1
Quick Method for Showing a
Contradiction
The quick method for determining if a compound statement
is a tautology can be used similarly for showing a
contradiction.
The quick method relies on the fact that if a truth value of
“T” can occur under the main connective (for some
combination of truth values for the components), then the
statement is not a contradiction. If this truth value is not
possible, then we have a contradiction.
Therefore, to determine whether a statement is a
contradiction, we place a “T” under the main connective
and work backwards.
WUCT121
Logic
45
Example:
•
Use the “quick” method for the statement
~ ( P ∨ Q ) ∧ P to determine if it is a contradiction.
~
Step:
(P
2
∨
1
Q)
1.Place “T” under main
∧
3*
P
T
connective
2. For “T” to occur
T
T
under the main
connective, ~ must be
“T” and P must be “T”
F
3. For “T” to occur
under ~, P ∨ Q must be
“F”.
4. For “F” to occur
F
F
under P ∨ Q , P must be
“F” and Q must be “F”
P cannot be both “T” and “F”, thus ~ ( P ∨ Q ) ∧ P can only
ever be false and is a contradiction.
WUCT121
Logic
46
Exercise:
•
Use the “quick” method for the statement
( P ∧ Q ) ∧ ~ Q to determine if it is a contradiction.
(P
Step:
∧
2
Q)
1.Place “T” under main
∧
3*
~Q
1
T
connective.
T
2. For “T” to occur under
T
the main connective,
( P ∧ Q ) must be “T” and
~Q must be “T”
3. For “T” to occur under
T
T
( P ∧ Q ) , P must be “T”
and Q must be “T”.
At Step 2, ~Q is “T”, thus Q is “F”. Step 3 shows Q is
“T”. Q cannot be both “T” and “F”, thus ( P ∧ Q ) ∧ ~ Q
can only ever be false and is a contradiction.
WUCT121
Logic
47
1.5.3.
Contingent
Definition: Contingent.
Any statement that is neither a tautology nor a
contradiction is called a contingent or intermediate
statement.
Examples:
Complete the truth table for the statement Q ∨ (Q ⇒ P )
Q
T
T
∨
T
T
F
T
T
F
T
F
F
F
F
T
T
2*
1
Step:
WUCT121
⇒
P
Q
Logic
(Q
P)
T
48
Exercises:
•
Complete the truth table for the statement
( p ∨ r ) ⇒ ( p ∧ q ) to show it is contingent.
p
T
T
T
T
F
F
F
F
Step:
•
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
(p
∨
T
T
T
T
T
F
T
F
1
r)
⇒
T
T
F
F
F
T
F
T
3*
(p
∧
T
T
F
F
F
F
F
F
2
q)
Complete the truth table for the statement
~ (( p ∧ ~ q ) ∨ r ) ⇔ (r ⇒ q ) to show it is contingent.
p q
T T
T T
T F
T F
F T
F T
F F
F F
Step:
WUCT121
r ~( (p ∧
T F
F
F T
F
T F
T
F F
T
T F
F
F T
F
T F
F
F T
F
4
2
~ q) ∨
F
T
F
F
T
T
T
T
F
T
F
F
T
T
T
F
1
3
Logic
r) ⇔ (r ⇒ q)
F
T
T
T
T
F
F
T
F
T
T
T
T
F
T
T
6*
5
49
1.6.
Logical Equivalence
Definition: Logical Equivalence.
Two statements are logically equivalent if, and only if, they
have identical truth values for each possible substitution of
statements for their statements variables.
The logical equivalence of two statements P and Q is
denoted P ≡ Q .
If two statements P and Q are logically equivalent then
P ⇔ Q is a tautology
1.6.1.
Determining Logical Equivalence.
To determine if two statements P and Q are logically
equivalent, construct a full truth table for each statement. If
their truth values at the main connective are identical, the
statements are equivalent.
Alternatively show P ⇔ Q is a tautology and hence
conclude P ≡ Q .
WUCT121
Logic
50
Examples:
•
Determine if the following statements are logically
equivalent. P : p ⇒ q,
Q :~ p ∨ q
⇒
p
q
T
T
T
F
∨
T
T
F
F
F
F
F
T
T
T
T
F
F
T
T
T
1*
1
2*
p
Step:
q
~p
q
Since the main connectives * are identical, the statements P
and Q are equivalent. Thus P ≡ Q
•
i.e. p ⇒ q ≡ ~ p ∨ q
Determine if the following statements are logically
equivalent. P :~ ( p ∧ q ),
Q :~ p ∧ ~ q
p
q
~(
T
T
F
∧
T
T
F
T
F
T
F
F
Step:
~q
F
∧
F
F
F
F
T
T
F
T
F
F
T
F
T
T
T
2*
1
1
2*
1
p
q)
~p
F
Since the main connectives * are not identical, the
statements P and Q are not equivalent.
WUCT121
Logic
51
Exercises:
•
Determine if the following statements are logically
equivalent. P :~ ( p ∨ q ),
Q :~ p ∧ ~ q
p
q
~(
T
T
F
∨
T
T
F
F
F
T
F
F
Step:
~q
F
∧
F
T
F
F
T
F
T
T
F
F
T
F
T
T
T
2*
1
1
2*
1
p
q)
~p
F
Since the main connectives are identical, the statements P
and Q are equivalent. Thus P ≡ Q i.e. ~ ( p ∨ q ) ≡~ p ∧ ~ q
Determine if ~ ( p ∧ q ) ⇔ ~ p ∨ ~ q is a tautology, and
•
hence if ~ ( p ∧ q ) ≡~ p ∨ ~ q .
⇔
~p
T
F
T
p
q
~(
T
T
F
∧
T
T
F
T
F
T
F
F
Step:
~q
F
∨
F
T
F
T
T
F
T
T
T
F
T
F
T
T
T
T
2*
1
3*
1
2*
1
p
q)
F
Since the main connective is all T, the statement
~ ( p ∧ q ) ⇔ ~ p ∨ ~ q is a tautology, and hence
~ ( p ∧ q ) ≡~ p ∨ ~ q .
WUCT121
Logic
52
1.6.2.
Substitution
There are two different types of substitution into
statements.
Rule of Substitution: If in a tautology all occurrences of a
variable are replaced by a statement, the result is still a
tautology.
Examples:
•
We know P ∨ ~ P is a tautology.
Thus, by the rule of substitution, so too are:
∗
Q ∨ ~ Q , by letting Q = P .
∗
(( p ∧ q ) ⇒ r )∨ ~ (( p ∧ q ) ⇒ r ) , by letting
( p ∧ q) ⇒ r = P .
Note: We have simply replaced every occurrence of P in
the tautology P ∨ ~ P , by some other statement.
WUCT121
Logic
53
Rule of Substitution of Equivalence: If in a tautology we
replace any part of a statement by a statement equivalent to
that part, the result is still a tautology.
Example:
•
Determine if P ⇒ (~ Q ∨ P ) is a tautology.
We know: P ⇒ (Q ⇒ P ) is a tautology and
( P ⇒ Q ) ≡~ P ∨ Q
By the rule of substitution (Q ⇒ P ) ≡ ~ Q ∨ P
Thus, by the rule of substitution of equivalence,
P ⇒ (Q ⇒ P ) ≡ P ⇒ (~ Q ∨ P ) , and hence
P ⇒ (~ Q ∨ P ) is also a tautology.
Exercise:
•
~ T ∨ (~ S ∨ T ) a tautology?
Yes.
We know ( P ⇒ Q ) ≡~ P ∨ Q . So, ( S ⇒ T ) ≡~ S ∨ T and
T ⇒ (~ S ∨ T ) ≡ ~ T ∨ (~ S ∨ T ) (by RoS).
Hence, ~ T ∨ (~ S ∨ T ) ≡ T ⇒ ( S ⇒ T ) (by SoE).
P ⇒ (Q ⇒ P ) is a known tautology, thus (by (SoE)
T ⇒ ( S ⇒ T ) is a tautology, and since
~ T ∨ (~ S ∨ T ) ≡ T ⇒ ( S ⇒ T ) , ~ T ∨ (~ S ∨ T ) is a
tautology.
WUCT121
Logic
54
1.6.3.
Laws
The following logical equivalences hold:
Commutative Laws:
1.
•
•
•
( P ∨ Q ) ≡ (Q ∨ P )
( P ∧ Q ) ≡ (Q ∧ P )
( P ⇔ Q ) ≡ (Q ⇔ P )
Associative Laws:
2.
•
•
•
(( P ∨ Q ) ∨ R ) ≡ (P ∨ (Q ∨ R ))
(( P ∧ Q ) ∧ R ) ≡ (P ∧ (Q ∧ R ))
(( P ⇔ Q ) ⇔ R ) ≡ (P ⇔ (Q ⇔ R ))
Distributive Laws:
3.
•
•
(P ∨ (Q ∧ R) ) ≡ (( P ∨ Q) ∧ ( P ∨ R) )
(P ∧ (Q ∨ R) ) ≡ (( P ∧ Q) ∨ ( P ∧ R) )
4.
•
Double Negation (Involution) Law:
~~ P ≡ P
5.
•
•
De Morgan’s Laws:
~ ( P ∨ Q ) ≡ (~ P ∧ ~ Q )
~ ( P ∧ Q ) ≡ (~ P ∨ ~ Q )
WUCT121
Logic
55
6.
•
•
Implication Laws:
( P ⇒ Q ) ≡ (~ P ∨ Q )
( P ⇔ Q ) ≡ (( P ⇒ Q ) ∧ (Q ⇒ P ) ) ( Biconditional)
7.
•
•
Identity Laws:
(P ∨ F ) ≡ P
(P ∧ T ) ≡ P
8.
•
•
Negation (Complement) Laws:
( P∨ ~ P ) ≡ T
( P∧ ~ P) ≡ F
9.
•
•
Dominance Laws:
(P ∨ T ) ≡ T
(P ∧ F ) ≡ F
10.
•
•
( Implication )
Idempotent Laws:
( P ∨ P) ≡ P
( P ∧ P) ≡ P
11.
Absorption Laws:
•
P ∧ (P ∨ Q) ≡ P
•
P ∨ (P ∧ Q) ≡ P
12.
•
•
Property of Implication:
(P ⇒ (Q ∧ R) ) ≡ (( P ⇒ Q) ∧ ( P ⇒ R) )
(( P ∨ Q) ⇒ R ) ≡ (( P ⇒ R) ∧ (Q ⇒ R) )
WUCT121
Logic
56
Example:
Prove the first of De Morgan’s Laws using truth tables.
P
Q
~(
T
T
F
∨
T
T
F
F
F
T
F
F
Step:
~Q
F
∧
F
T
F
F
T
F
T
T
F
F
T
F
T
T
T
2*
1
1
2*
1
P
Q)
~P
F
Since the main connectives are identical, the statements are
equivalent., and first of De Morgan’s Laws is true.
Exercise:
Prove the second of De Morgan’s Laws using truth tables.
P
Q
~(
T
T
F
∧
T
T
F
T
F
T
F
F
Step:
~Q
F
∨
F
F
F
T
T
T
F
T
T
F
T
F
T
T
T
2*
1
1
2*
1
P
Q)
~P
F
Since the main connectives are identical, the statements are
equivalent, and second of De Morgan’s Laws is true.
WUCT121
Logic
57
Example:
Using logically equivalent statements, without the direct
use of truth tables, show: ~ (~ p ∧ q ) ∧ ( p ∨ q ) ≡ p
~ (~ p ∧ q ) ∧ ( p ∨ q ) ≡ (~ (~ p ) ∨ ~ q ) ∧ ( p ∨ q )
(De Morgan )
≡ ( p ∨ ~ q ) ∧ ( p ∨ q ) (Double Negation)
(Distributivity)
≡ p ∨ (~ q ∧ q )
(Commutativity)
≡ p ∨ (q ∧ ~ q )
(Negation)
≡ p∨F
(Identity)
≡ p
Exercises:
Using logically equivalent statements, without the direct
use of truth tables, show:
•
~ ( p ⇔ q ) ≡ ( p ∧ ~ q ) ∨ (q ∧ ~ p )
~ ( p ⇔ q ) ≡ ~ (( p ⇒ q ) ∧ (q ⇒ p ))
(Biconditional )
(De Morgan )
≡ ~ ( p ⇒ q ) ∨ ~ (q ⇒ p )
(Implication )
≡ ~ (~ p ∨ q ) ∨ ~ (~ q ∨ p )
≡ (~ ~ p ∧ ~ q ) ∨ (~ ~ q ∧ ~ p ) (De Morgan )
(Double Negation)
≡ ( p ∧ ~ q ) ∨ (q ∧ ~ p )
WUCT121
Logic
58
•
( p ⇒ q ) ≡ (~ q ⇒~ p )
( p ⇒ q ) ≡ (~ p ∨ q )
(Implication )
(Commutativity )
≡ (q ∨ ~ p )
(Double Negation)
≡ (~ (~ q )∨ ~ p )
(Implication )
≡ (~ q ⇒ ~ p )
•
p ⇒ (q ∧ r ) ≡ ( p ⇒ q ) ∧ ( p ⇒ r ) , without using the
property of implication
p ⇒ (q ∧ r ) ≡ ~ p ∨ (q ∧ r )
≡ (~ p ∨ q ) ∧ (~ p ∨ r )
≡ ( p ⇒ q) ∧ ( p ⇒ r )
WUCT121
Logic
(Implication )
(Distributive)
(Implication )
59
Section 2.
Predicate Logic
Discussion:
In Maths we use variables (usually ranging over numbers)
in various ways.
How does x differ in what it represents in the following
statements? x is real.
•
x2 = 0
•
x>2
•
x+0 = x
x represents all values
•
x2 +1 = 0
x represents no values
x represents one value, x = 0
x represents some, but not all values
Definition: Predicate
A predicate is a sentence that contains one or more
variables and becomes a statement when specific values are
substituted for the variables.
Definition: Domain
The domain of a predicate variable consists of all values
that may be substituted in place of the variable
WUCT121
Logic
60
Definition: Truth Set
If P(x) is a predicate and x has domain D, the truth set of
P(x) is the set of all elements of D that make P(x) true. The
truth set is denoted { x ∈ D : P( x )} and is read “the set of all
x in D such that P(x).”
Examples:
•
Let P(x) be the predicate “ x 2 > x ” with x ∈  i.e.
domain the set of real numbers  .
Write down P( 2), P(1), P( −2) and indicate which are true
and which are false.
Determine the truth set of P(x)
P( 2) : 2 2 > 2
or 4 > 2
True
P(1) :
(1)2 > 1
or 1 > 1
False
P( −2) : ( −2)2 > ( −2) or 4 > ( −2) True
{ x ∈  : x 2 > x} = { x ∈  : x < 0 ∨ x > 1}
•
Let Q(n) be the predicate “n is factor of 8”.
Determine the truth set of Q(n) if n ∈  +
8 = ±1 × ±8,
8 = ±2 × ±4
∴{n ∈  + :" n is a factor of 8"} = {1, 2, 4, 8}
WUCT121
Logic
61
Exercises:
•
3
Let P(x) be the predicate “ x > x ” with x ∈  i.e.
domain the set of integers,  .
Write down P(2), P(0), P( −2) and indicate which are true
and which are false.
Determine the truth set of P(x)
P( 2) :
23 > 2
or 8 > 2
True
P (0 ) :
(0 ) 3 > 0
or 0 > 0
False
P( −2) : ( −2) 2 > ( −2) or ( −8) > ( −2) False
{ x ∈  : x 3 > x} = { x ∈  : x > 1}
•
Let Q(n) be the predicate “n is factor of 6”.
Determine the truth set of Q(n) if n ∈ 
6 = ±1 × ±6, 6 = ±2 × ±3
∴ {n ∈  :" n is a factor of 6" } = {±1, ± 2, ± 3, ± 6}
WUCT121
Logic
62
2.1.
Quantifiers
A way to obtain statements from predicates is to add
quantifiers. Quantifiers are words that refer to quantities
such as “all”, “every”, or “some” and tell for how many
elements a given predicate is true.
2.1.1.
Universal Quantifier
The symbol ∀ denotes “for all” and is called the universal
quantifier.
Definition: Universal Statement
Let P(x) be a predicate and D the domain of x. A universal
statement is a statement of the form “ ∀x ∈ D, P( x ) ”. It is
defined to be true if, and only if, P(x) is true for every x in
D. It is defined to be false if, and only if, P(x) is false for at
least one x in D. A value of x for which P(x) is false is
called a counterexample to the universal statement.
Examples:
•
Write the sentence “All human beings are mortal”
using the universal quantifier.
Let H be the set of human beings.
∀h ∈ H ,h is mortal
WUCT121
Logic
63
•
Consider A = { x1 , x2 , x3 } . With ∀x ∈ A, P( x ) , the
following must hold: P( x1 ) ∧ P( x2 ) ∧ P( x3 )
Thus there will be 3 predicates which must hold.
Exercises:
Write the following statements using the universal
quantifier. Determine whether each statement is true or
false.
•
“All dogs are animals”
Let D be the set of dogs and A the set of animals
∀d ∈ D, d ∈ A . True
•
The square of any real number is positive.
∀x ∈ , x 2 > 0 .
False, consider x = 0 ∈ , x 2 = 0 2 u0.
Hence the statement is false by counterexample
•
Every integer is a rational number.
∀x ∈ , x ∈  . True
WUCT121
Logic
64
Exercises:
Write the following statements in words. Determine
whether each statement is true or false.
•
∀x ∈ ,
x ∈
The square root of any natural number is a natural number.
False. Consider x = 2 ∈ ,
x = 2 ∉  . Hence the
statement is false by counterexample.
•
∀x ∈ , x 2 ≠ −1.
The square of any real number does not equal –1. True.
WUCT121
Logic
65
2.1.2.
Existential Quantifier
The symbol ∃ denotes “there exists” and is called the
existential quantifier.
Definition: Existential Statement
Let P(x) be a predicate and D the domain of x.
An existential statement is a statement of the form
“ ∃x ∈ D, P( x ) ”.
It is defined to be true if, and only if, P(x) is true for at least
one x in D.
It is defined to be false if, and only if, P(x) is false for all x
in D.
Examples:
•
Write the sentence “Some people are vegetarians”
using the existential quantifier.
Let H be the set of human beings.
∃h ∈ H , h is a vegetarian
•
Consider A = { x1 , x2 , x3 }. With ∃x ∈ A, P( x ) , the
following must hold: P( x1 ) ∨ P( x2 ) ∨ P( x3 )
Thus there will be 1 predicate which must hold.
WUCT121
Logic
66
Exercises:
Write the following statements using the existential
quantifier. Determine whether each statement is true or
false.
•
“Some cats are black”
Let C be the set of cats.
∃c ∈ C ,c is black . True
•
There is a real number whose square is negative.
∃x ∈ , x 2 < 0 .False.
•
Some programs are structured.
Let P be the set of programs.
∃p ∈ P, p is structured . True
WUCT121
Logic
67
Exercises:
Write the following statements in words. Determine
whether each statement is true or false.
•
∃m ∈ , m 2 = m
There is an integer whose square is equal to itself.
True. Consider m = 1 ∈ , m 2 = 12 = 1 = m .
Hence the statement is true.
•
∃x ∈ , x 2 = −1.
There is a real number whose square is –1.
False.
•
1
∃ x ∈ , ∉ 
x
The reciprocal of some integer is not rational.
1 1
True. Consider x = 0 ∈ , = ∉  .
x 0
Hence the statement is true.
WUCT121
Logic
68
2.1.3.
Negation of Universal Statements
Let P(x) be a predicate and D the domain of x. The
negation of a universal statement of the form:
∀x ∈ D, P( x ) is logically equivalent to ∃x ∈ D , ~ P( x )
Symbolically ~ (∀x ∈ D , P( x )) ≡ ∃x ∈ D , ~ P( x )
Example:
•
Write down the negation of the following statement.
∀x ∈ , x 2 + 1 ≥ 2 x
Negation:
~ (∀x ∈ , x 2 + 1 ≥ 2 x )
≡ ∃x ∈ , ~ ( x 2 + 1 ≥ 2 x )
≡ ∃x ∈ , x 2 + 1 < 2 x
False.
WUCT121
Logic
69
Exercises:
•
Write down the negation of the following statement.
∀x ∈ , x 2 ≥ 0
Negation:
~ (∀x ∈ , x 2 ≥ 0)
≡ ∃x ∈ , ~ ( x 2 ≥ 0)
False.
≡ ∃x ∈ , x 2 < 0
•
Write down the negation of the following statement.
y +1 ⎞
⎛
∀y ∈ , ⎜ y ≠ 0 ⇒
< 1⎟
y
⎝
⎠
Negation:
⎛
y + 1 ⎞⎞
⎛
~ ⎜⎜ ∀y ∈ , ⎜ y ≠ 0 ⇒
< 1⎟ ⎟⎟
y
⎠⎠
⎝
⎝
y +1 ⎞
⎛
≡ ∃y ∈ , ~ ⎜ y ≠ 0 ⇒
< 1⎟
y
⎠
⎝
y +1 ⎞
⎛
≡ ∃y ∈ , ~ ⎜ ~ ( y ≠ 0) ∨
< 1⎟
y
⎠
⎝
⎛ y +1 ⎞
≡ ∃y ∈ , y ≠ 0∧ ~ ⎜
< 1⎟
⎝ y
⎠
y +1
≡ ∃y ∈ , y ≠ 0 ∧
≥1
y
True, choose y = 1.
WUCT121
Logic
70
Example:
•
Write the following statement using quantifiers. Find
its negation and determine whether the statement or its
negation is true, giving a brief reason..
“Every real number is either positive or negative.”
Statement:
∀x ∈ , x < 0 ∨ x > 0
Negation:
~ (∀x ∈ , x < 0 ∨ x > 0)
≡ ∃x ∈ , ~ ( x < 0 ∨ x > 0)
≡ ∃x ∈ , ~ ( x < 0)∧ ~ ( x > 0)
≡ ∃x ∈ , ( x ≥ 0) ∧ ( x ≤ 0)
≡ ∃x ∈ , x = 0
The true statement is the negation because x = 0 is neither
positive nor negative.
WUCT121
Logic
71
Exercises:
•
Write the following statement using quantifiers. Find
the negation.
“The square of any integer is positive.”
Statement:
∀x ∈  , x 2 > 0
Negation:
~ ( ∀x ∈  , x 2 > 0 )
≡ ∃ x ∈ , ~ ( x 2 > 0 )
≡ ∃ x ∈ , x 2 ≤ 0
There is an integer whose square is not positive. The
negation is true, choose x = 0.
•
Write the following statement using quantifiers. Find
the negation.
“All computer programs are finite.”
Let C be the set of computer programs
Statement:
∀x ∈ C , x is finite
Negation:
≡ ~ (∀x ∈ C , x is finite)
≡ ∃x ∈ C , x is not finite
Not all computer programs are finite.
Some computer programs are not finite. True?
WUCT121
Logic
72
2.1.4.
Negation of Existential Quantifiers
Let P(x) be a predicate and D the domain of x. The
negation of an existential statement of the form:
∃x ∈ D, P( x ) is logically equivalent to ∀x ∈ D , ~ P( x )
Symbolically ~ (∃x ∈ D, P( x )) ≡ ∀x ∈ D , ~ P( x )
Example:
•
Write down the negation of the following statement.
∃x ∈ , x 2 = 2
Negation:
~ (∃x ∈ , x 2 = 2)
≡ ∀x ∈ , ~ ( x 2 = 2)
≡ ∀x ∈ , x 2 ≠ 2
The negation is true.
WUCT121
Logic
73
Exercises:
•
Write down the negation of the following statement.
∃z ∈ , ( z is odd ) ∨ ( z is even)
Negation:
~ (∃z ∈ , ( z is odd ) ∨ ( z is even))
≡ ∀z ∈ , ~ (( z is odd ) ∨ ( z is even))
≡ ∀z ∈ , ~ ( z is odd ) ∧ ~ ( z is even)
≡ ∀z ∈ , ( z is not odd ) ∧ ( z is not even)
The negation is false
•
Write down the negation of the following statement.
∃n ∈ , ( n is even) ∧ ( n is prime)
Negation:
~ (∃n ∈ , ( n is even) ∧ ( n is prime))
≡ ∀n ∈ , ~ (( n is even) ∧ ( n is prime))
≡ ∀n ∈ , ~ ( n is even) ∨ ~ ( n is prime)
≡ ∀n ∈ , ( n is not even) ∨ ( n is not prime)
The negation is false.
WUCT121
Logic
74
Example:
•
Write the following statement using quantifiers. Find
its negation
“Some dogs are vegetarians.”
Let D be the set of dogs.
Statement:
∃d ∈ D , d is vegetarian
Negation:
~ (∃d ∈ D, d is vegetarian)
≡ ∀d ∈ D , ~ ( d is vegetarian)
≡ ∀d ∈ D , d is not vegetarian
All dogs are not vegetarian
Exercises:
•
Write the following statement using quantifiers. Find
the negation.
“There is a real number that is rational.”
Statement: ∃x ∈ , x ∈ 
Negation:
~ (∃x ∈ , x ∈ )
≡ ∀x ∈ , ~ ( x ∈ )
False
≡ ∀x ∈ , x ∉ 
All real numbers are not rational
WUCT121
Logic
75
•
Write the following statement using quantifiers. Find
the negation.
P(p): Some computer hackers are over 40.
Let C be the set of computer hackers.
P( p ) : ∃p ∈ C , p is over 40
~ P( p ) :
~ (∃p ∈ C , p is over 40)
≡ ∀p ∈ C , ~ ( p is over 40)
≡ ∀p ∈ C , p is not over 40
≡ ∀p ∈ C , p is 40 or under
All computer hackers are 40 or under
False
•
Write the following statement using quantifiers. Find
the negation.
“Some animals are dogs.”
Let A be the set of animals
Statement: ∃x ∈ A, x is a dog
Negation:
~ (∃x ∈ A, x is a dog )
≡ ∀x ∈ A, x is not a dog
All animals are not dogs. False
WUCT121
Logic
76
2.1.5.
Multiple Quantifiers
When a statement contains multiple quantifiers their order
must be applied as written and will produce different
results for the truth set.
Examples:
Write the following statements using quantifiers:
•
“Everybody loves somebody.”
Let H be the set of people.
Statement: ∀x ∈ H , ∃y ∈ H , x loves y.
•
“Somebody loves everyone.”
Let H be the set of people.
Statement: ∃x ∈ H , ∀y ∈ H , x loves y.
WUCT121
Logic
77
Exercises:
Write the following statements using quantifiers:
•
“Everybody loves everybody.”
Let H be the set of people.
Statement: ∀x ∈ H , ∀y ∈ H , x loves y.
•
The Commutative Law of Addition for 
Statement: ∀x ∈ ,, ∀y ∈ , x + y = y + x
•
“Everyone had a mother.”
Let H be the set of humans.
Statement: ∀x ∈ H, ∃y ∈ H , y was the mother of x
or x was the child of y.
•
“There is an oldest person.”
Let H be the set of humans.
Statement: ∃x ∈ H, ∀y ∈ H , x is older thany
WUCT121
Logic
78
Examples:
Write the following statements without using quantifiers:
•
∀x ∈ ,, ∃y ∈ , x + y = 0
Statement: Given any real number, you can find a real
number so that the sum of the two is zero. Alternatively:
Every real number has an additive inverse.
•
∃x ∈ ,, ∀y ∈ , x + y = y
Statement: There is a real number, which added to any
other real number results in the other number.
Alternatively: Every real number has an additive identity.
Exercises:
Write the following statements without using quantifiers:
•
∀c ∈ colours, ∃a ∈ animals, a is coloured c
Statement: For every colour, there is an animal of that
colour.
Alternatively: There are animals of every colour.
•
∃b ∈ books, ∀p ∈ people, p has read b
Statement: There is a book everyone has read.
WUCT121
Logic
79
2.1.6.
Interpreting Statements with Multiple
Quantifiers
To establish the truth of a statement with more than one
quantifier, take the action suggested by the quantifiers as
being performed in the order in which the quantifiers occur.
Consider A = { x1 , x 2 , x3 }, B = { y1 , y 2 } and the
predicate P( x, y ) .
There will be 6 possible predicates:
P( x1 , y1 ) , P( x1 , y 2 ),
P( x2 , y1 ), P( x2 , y 2 ),
P( x3 , y1 ) , P( x3 , y 2 ).
•
For ∀x ∈ A, ∀y ∈ B, P( x, y ) to be true the following
must hold:
P( x1 , y1 ) ∧ P( x1 , y 2 )
∧ P( x2 , y1 ) ∧ P( x2 , y 2 )
∧ P( x3 , y1 ) ∧ P( x3 , y 2 )
Thus there will be 6 predicates which must all be true. That
is for all pairs (x, y), P(x, y) must be true. It will be false if
there is one pair (x, y), for which P(x, y) is false.
WUCT121
Logic
80
•
For ∀x ∈ A, ∃y ∈ B, P( x, y ) to be true, the following
must hold:
P( x1 , y1 ) ∨ P( x1 , y 2 )
∧ P( x 2 , y1 ) ∨ P( x 2 , y 2 )
∧ P( x3 , y1 ) ∨ P( x3 , y 2 )
Thus there will be 3 predicates which must be true. That is
for every x there must be at least one y so that P(x, y) is
true. Given any element x in A you can find an element y in
B, so that P(x, y) is true. It will be false if there is one x in A
for which P(x, y) is false for every y in B.
•
For ∃x ∈ A, ∀y ∈ B, P( x, y ) to be true, the following
must hold:
P( x1 , y1 ) ∧ P( x1 , y 2 )
∨ P( x 2 , y1 ) ∧ P( x 2 , y 2 )
∨ P( x3 , y1 ) ∧ P( x3 , y 2 )
Thus there will be 2 predicates which must be true. That is
there is one x that when paired with any y, P(x, y) is true.
You can find one element x in A that with all elements y in
B, P(x, y) is true. It will be false if for every x in A, there is
a y in B for which P(x, y) is false.
WUCT121
Logic
81
•
For ∃x ∈ A, ∃y ∈ B, P( x, y ) to be true, the following
must hold:
P( x1 , y1 ) ∨ P( x1 , y 2 )
∨ P( x 2 , y1 ) ∨ P( x 2 , y 2 )
∨ P( x3 , y1 ) ∨ P( x3 , y 2 )
Thus there will be 1 predicate which must be true. That is
there is one x that when paired with one y, P(x, y) is true.
You can find one element x in A and one element y in B,
P(x, y) is true. It will be false if for all pairs (x, y), P(x, y) is
false.
Summary:
Statement
When true?
∀x, ∀y , P( x, y ) P(x, y) is true for
all pairs (x, y)
∀x, ∃y , P( x, y ) For every x, there
is a y for which
P(x, y) is true
∃x, ∀y , P( x, y ) There is an x such
that P(x, y) is true
for every y
∃x, ∃y , P( x, y ) There is a pair
(x, y) for which
P(x, y) is true
WUCT121
Logic
When false?
There is a pair
(x, y) for which P(x,
y) is false
There is an x such
that P(x, y) is false
for every y
For every x, there is
a y for which P(x, y)
is false
P(x, y) is false for
all pairs (x, y)
82
2.1.7.
Negation of Statements with Multiple
Quantifiers.
To negate statements with multiple quantifiers, each
quantifier is negated and the predicate must be negated.
•
To negate ∀x ∈ A, ∀y ∈ B, P( x, y )
~ (∀x ∈ A, ∀y ∈ B, P( x, y ) ) ≡ ∃x ∈ A, ∃y ∈ B, ~ P( x, y )
•
To negate ∀x ∈ A, ∃y ∈ B, P( x, y )
~ (∀x ∈ A, ∃y ∈ B, P( x, y ) ) ≡ ∃x ∈ A, ∀y ∈ B, ~ P( x, y )
•
To negate ∃x ∈ A, ∀y ∈ B, P( x, y )
~ (∃x ∈ A, ∀y ∈ B, P( x, y ) ) ≡ ∀x ∈ A, ∃y ∈ B, ~ P( x, y )
•
To negate ∃x ∈ A, ∃y ∈ B, P( x, y )
~ (∃x ∈ A, ∃y ∈ B, P( x, y ) ) ≡ ∀x ∈ A, ∀y ∈ B, ~ P( x, y )
Examples:
Write the negation of the following:
•
Statement: ∀x ∈ ,, ∃y ∈ , x + y = 0
Negation:
~ (∀x ∈ ,, ∃y ∈ , x + y = 0 )
≡ ∃x ∈ ,, ∀y ∈ , x + y ≠ 0
False:
Take y = − x , then x + y = x − x = 0
WUCT121
Logic
83
•
Statement: ∃x ∈ ,, ∀y ∈ , xy = 1
Negation:
~ (∃x ∈ ,, ∀y ∈ , xy = 1)
≡ ∀x ∈ ,, ∃y ∈ , xy ≠ 1
True :
Take y = − x, then xy = − x 2 ≠ 1
Exercises:
Write the negation of the following:
•
Statement: ∀c ∈ colours, ∃a ∈ animals, a is coloured c
Negation:
≡ ~ (∀c ∈ colours, ∃a ∈ animals, a is coloured c )
≡ ∃c ∈ colours, ∀a ∈ animals, a is not coloured c
There is a colour which every animal is not. True, my cat is
not purple.
•
Statement: ∃b ∈ books, ∀p ∈ people, p has read b
Negation:
≡ ~ (∃b ∈ books, ∀p ∈ people, p has read b )
≡ ∀b ∈ books, ∃p ∈ people, p has not read b
There is someone who has not read every book. True, me,
I’ve not read every book.
WUCT121
Logic
84
Section 3.
3.1.
Proofs
Introduction.
A proof is a carefully reasoned argument which establishes
that a given statement is true. Logic is a tool for the
analysis of proofs. Each statement within a proof is an
assumption, an axiom, a previously proven theorem, or
follows from previous statements in the proof by a
mathematical or logical rules and definitions.
3.1.1.
Assumptions.
Assumptions are the statements you assume to be true as
you try to prove the result.
Example:
If you want to prove:
“If x ∈  and n ∈  is even, then x n > 0 ”
Your proof should start with the assumptions that x ∈ 
and n ∈  is even. Further, you can use the “definition” of
an even natural number, and write the assumptions as
follows:
Let x ∈  , and n ∈  be even, that is, ∃p ∈ , n = 2 p .
WUCT121
Logic
85
Assumptions are often thought to be the “given
information” or information we “know” that can be used in
our proof. As in the example above, when you are proving
statements of the form P ⇒ Q , then the assumption is the
statement P.
Exercise:
Write the statement to be proven in the previous example
using logical notation:
[( x ∈ ) ∧ ( n ∈  : ∃p ∈ , n = 2 p )] ⇒ x n > 0
3.1.2.
Axioms.
Axioms are laws in Mathematics that hold true and require
no proof.
Examples:
•
x=x
•
x+0 = x
•
∀x, y , z ∈ , [ ( x = y ) ∧ ( y = z )] ⇒ ( x = z )
WUCT121
Logic
86
3.1.3.
Mathematical Rules.
Mathematical Rules are known rules that are often used.
Example:
∀x, y , z ∈ , ( x = y ) ⇒ ( x + z = y + z )
3.1.4.
Logical Rules.
Logical Rules are rules of logic such as Substitution and
Substitution of Equivalence using the laws introduced
earlier
3.2.
The Law of Syllogism
If P ⇒ Q and Q ⇒ R are both tautologies, then so is
P ⇒ R.
Exercise:
•
Write the Law of Syllogism using logical notation:
(( P ⇒ Q ) ∧ (Q ⇒ R )) ⇒ ( P ⇒ R )
WUCT121
Logic
87
•
Show, using the quick method that the Law of
Syllogism is a tautology.
((P
Step:
1.
2.
3.
4.
5.
⇒
1
Q)
∧
3
(Q
⇒
2
R))
⇒
5*
F
(P
T
⇒
4
F
T
T
T
F
T
T
F
F
1. Place “F” under main connective
2. For “F” to occur under the main connective, ∧ must be
“T” and ⇒ must be “F”
3. For “F” to occur under ⇒ , P must be “T” and R must
be “F”.
4. For “T” under ∧ , both ⇒ must be “T”
5. For the first ⇒ to be true, given P is “T”, Q must be
“T”. For the second ⇒ to be true, given R is “F”, Q must
be “F”.
Q cannot be both “T” and “F”, thus
(( P ⇒ Q ) ∧ (Q ⇒ R )) ⇒ ( P ⇒ R ) can only ever be true
and is a tautology.
WUCT121
R)
Logic
88
Examples:
•
s is a square ⇒ s is a rectangle
s is a rectangle ⇒ s is a parallelogram
s is a parallelogram ⇒ s is a quadrilateral
∴
s is a square ⇒ s is a quadrilateral.
•
x 2 ≥ 0 ⇒ ( x − 3) 2 ≥ 0
( x − 3) 2 ≥ 0 ⇒ ( x 2 − 6 x + 9) ≥ 0
( x 2 − 6 x + 9) ≥ 0 ⇒ x 2 − 6 x ≥ −9
∴ x 2 ≥ 0 ⇒ x 2 − 6 x ≥ −9
Exercise:
Complete the following using the Law of Syllogism:
•
t is studying WUCT121 ⇒ t is enrolled in a diploma
t is enrolled in a diploma ⇒ t is student at WCA.
∴
t is studying WUCT121 ⇒ t is student at WCA.
•
x ∈ ⇒ x ∈
x ∈ ⇒ x ∈
x ∈ ⇒ x ∈
∴ x ∈ ⇒ x ∈
WUCT121
Logic
89
Most results in Mathematics that require proofs are of the
form P ⇒ Q . The Law of Syllogism provides the most
common method of performing proofs of such statements.
The Law of Syllogism is a kind of transitivity that can
apply to ⇒ .
To use the Law of Syllogism, we set up a sequence of
statements, P ⇒ P1 , P1 ⇒ P2 , P2 ⇒ P3 ,K , Pn ⇒ Q .
Then, by successive applications of the law, we have
P ⇒ Q.
Example.
We wish to prove that for n ∈  , if n is even, then n 2 is
even.
In logic notation, we wish to prove:
n is even (P) ⇒ n 2 is even (Q).
This has the form P ⇒ Q and we note that our assumption
includes n ∈  and P: n is even.
WUCT121
Logic
90
Proof:
K( P ⇒ P1 )
n is even ⇒ ∃p ∈ , n = 2 p
∃p ∈ , n = 2 p ⇒ n 2 = 4 p 2 K( P1 ⇒ P2 )
K( P2 ⇒ P3 )
n 2 = 4 p 2 ⇒ n 2 = 2( 2 p 2 )
n 2 = 2( 2 p 2 ) ⇒ n 2 is even K( P3 ⇒ Q )
Completing the proof is simply a matter of applying the
Law of Syllogism three times to get n is even ⇒ n 2 is
even.
The previous proof can be simplified to:
n is even ⇒ ∃p ∈ , n = 2 p
⇒ n2 = 4 p 2
⇒ n 2 = 2( 2 p 2 ), 2 p 2 ∈ 
⇒ n 2 is even
The use of Law of Syllogism is a matter of common sense.
We shall use the Law of Syllogism without direct
reference.
Note. The use of the connective ⇒ in the previous proof
seems a little repetitive, albeit valid. For variety, the
connective can be replaced by words such as therefore,
thus, so we have, and hence.
WUCT121
Logic
91
3.3.
Modus Ponens
3.3.1.
Rule of Modus Ponens:
If P and P ⇒ Q are both tautologies, then so is Q.
In other words, Modus Ponens simply says that if we know
P to be true, and we know that P implies Q, then Q must
also be true.
Exercise:
•
Write the rule of Modus Ponens using logical notation:
( P ∧ ( P ⇒ Q )) ⇒ Q
WUCT121
Logic
92
•
Show, using the quick method that the rule of Modus
Ponens is a tautology.
(P
Step
∧
2
(P
⇒ Q)) ⇒
1
Q
3*
F
1. Place “F” under main
connective.
2. For “F” to occur under the
T
F
main connective, ∧ must be
“T” and Q must be “F”
3. For “T” to occur under ∧ , P
T
T
must be “T” and P ⇒ Q must
be “T”
4. For “T” to occur under
F
F
P ⇒ Q ,when P is “F” P must
be “F”
P cannot be both “T” and “F”, thus (( P ⇒ Q ) ∧ P ) ⇒ Q can only
ever be true and is a tautology.
WUCT121
Logic
93
Examples:
•
If Zak is a cheater, then Zak sits in the back row
Zak is a cheater
Therefore Zak sits in the back row.
•
If 2 = 3 then I will eat my hat
2=3
Therefore I will eat my hat
Exercise:
Complete the following using Modus Ponens
•
If Zeus is a God, then Zeus is immortal
Zeus is a God
Therefore Zeus is immortal.
•
If it is sunny then I will go to the beach
It is sunny
Therefore I will go to the beach
•
If I study hard then I will pass
I study hard
Therefore I will pass
WUCT121
Logic
94
3.3.2.
Universal Rule of Modus Ponens:
If P(x) and Q(x) are predicates, the universal rule of Modus
Ponens is (( P( x ) ⇒ Q ( x )) ∧ P( a )) ⇒ Q ( a ) .
This means Modus Ponens can be applied to predicates
using specific values for the variables in the domain.
Examples:
•
If x is even [P(x)], then x 2 is even [Q(x)]
x = 98374 [P(a)]
Therefore 98374 2 is even. [Q(a)]
•
The Principle of Mathematical Induction says that
when you have a statement, Claim(n), that concerns n ∈  ,
⎧Claim(1)
If P : ⎨
then Claim(n) is
⎩Claim( k ) ⇒ Claim( k + 1), ∀k ∈ 
true for all n ∈  (Q)
Thus we have P ⇒ Q .
WUCT121
Logic
95
Exercise:
According to Modus Ponens, what must we establish so we
can apply this principle to the following statement and be
able to say “Claim(n) is true for all n ∈  ”?
•
Claim(n): 4 n − 1 is a multiple of 3.
We must show that Claim(n) satisfies P.
So we need to establish two things:
1. Claim(1), i.e. 41 − 1 is a multiple of 3; AND
2. Claim( k ) ⇒ Claim( k + 1), ∀k ∈  , i.e. If 4 k − 1 is a
multiple of 3 for all k ∈  , then
4 (k +1) − 1 is a multiple
of 3.
WUCT121
Logic
96
3.4.
Modus Tollens
3.4.1.
Rule of Modus Tollens:
If ~Q and P ⇒ Q are both tautologies, then so is ~P.
In other words, Modus Ponens simply says that if we know
~Q to be true, and we know that P implies Q, then ~P must
also be true. Similarly if we know Q to be false, and we
know that P implies Q, then P must also be false
Exercise:
•
Write the rule of Modus Ponens using logical notation:
(( P ⇒ Q )∧ ~ Q ) ⇒ ~ P
WUCT121
Logic
97
•
Show, using the quick method that the rule of Modus
Tollens is a tautology.
((P ⇒ Q)
Step:
2
∧
3
~Q ⇒
1
1.Place “F” under main
4*
~P
5
F
connective
T
2. For “F” to occur under
F
the main connective, ∧
must be “T” and ~P must
be “F”
T
3. For “T” to occur under
T
∧ , ~Q must be “T” and
P ⇒ Q must be “T”
4. For “T” to occur under
T
T
P ⇒ Q ,when P is “T”, Q
must be “T”
At step 2, ~P is “F”, thus P is “T”. Step 3 shows ~Q is “T”
thus Q is “F” and step 4 gives Q is “T”. Q cannot be both “T”
and “F”, thus (( P ⇒ Q ) ∧ ~ Q ) ⇒ ~ P can only ever be true
and is a tautology.
WUCT121
Logic
98
Examples:
•
If Zak is a cheater, then Zak sits in the back row
Zak sits in the front row
Therefore Zak is not a cheater.
•
If 2 > 3 then Earth is flat
The Earth is not flat
Therefore 2 u 3
Exercise:
Complete the following using Modus Tollens
•
If Zeus is a God, then Zeus is immortal
Zeus is not immortal
Therefore Zeus is not a God.
•
If I go to the beach then it is sunny
It is not sunny
Therefore I don’t go to the beach
•
If I arrive on time then I will be marked present
I was marked absent
Therefore I did not arrive on time.
WUCT121
Logic
99
3.4.2.
Universal Rule of Modus Tollens:
If P(x) and Q(x) are predicates, the universal rule of Modus
Tollens is: (( P( x ) ⇒ Q ( x ))∧ ~ Q ( a )) ⇒ ~ P( a ) .
This means Modus Tollens can be applied to predicates
using specific values for the variables in the domain.
Example:
•
If x ∈  , then ∃a , b ∈ , b ≠ 0, x =
a
b
x= 2
Therefore
2 ∉ .
Exercise:
Complete the following using the universal rule of Modus
Tollens
•
If x ∈  , then x ≥ 1
x = −1
Therefore − 1 ∉  .
WUCT121
Logic
100
3.5.
Proving Quantified Statements
3.5.1.
Proving Existential Statements
A statement of the form ∃x ∈ D , P ( x ) is true if and only if
P ( x ) is true for at least one x ∈ D .
To prove this kind of statement, we need to find one x ∈ D
that makes P ( x ) true.
Examples:
•
Prove that there exists an even integer that can be
written two ways as the sum of two primes.
The statement is of the form ∃x ∈ D , P ( x ) , where D is the
set of even integers and P(x) is the statement “x can be
written as the sum of two primes”
Thus we need find only one even integer which satisfies
P(x).
Essentially, to find the appropriate number, we have to
“guess”.
Consider 14 = 7 + 7 (7 is prime); and 14 = 3 + 11 (3 and 11
are prime).
Therefore, there exists an even integer that can be written
two ways as the sum of two primes.
WUCT121
Logic
101
•
Let r , s ∈  . Prove ∃k ∈ , 22r + 18s = 2k
The statement is of the form ∃k ∈ D, P (k ), where D is the
set of integers and P(k) is the statement: “ 22r + 18s = 2k ”.
Thus we need find only one integer which satisfies P(k)
22r + 18s = 2(11r + 9 s )
= 2k where k = 11r + 9 s ∈ 
Exercises:
•
Prove ∃x ∈ , x + 5 = 0 .
Let x = −5 ∈  , then x + 5 = −5 + 5 = 0 .
•
Prove that if a , b ∈  ,then 10a + 8b is divisible by 2
(i.e., is even).
10a + 8b = 2(5a + 4b ) and 5a + 4b ∈  .
Thus, 2 (10a + 8b ).
WUCT121
Logic
102
3.5.2.
Proving Universal Statements
A statement of the form ∀x ∈ D , P ( x ) is true if and only if
P ( x ) is true for at every x ∈ D .
To prove this kind of statement, we need prove that for
every x ∈ D , P ( x ) is true.
In order to prove this kind of statement, there are two
methods:
Method 1: Method of Exhaustion.
The method of exhaustion is used when the domain is
finite.
Exhaustion cannot be used when the domain is infinite.
To perform the method of exhaustion, every member of the
domain is tested to determine if it satisfies the predicate.
WUCT121
Logic
103
Example:
•
Prove the following statement:
Every even number between 2 and 16 can be written as a
sum of two prime numbers.
The statement is of the form ∀x ∈ D, P ( x ), where
D = {4, 6, 8,10,12,14},and P(x) is the statement “x can be
written as the sum of two prime numbers”.
The domain D is finite so the method of exhaustion can be
used.
Thus we must test every number in D to show they can be
written as the sum of two primes.
4 = 2+2
6 = 3+3
8 = 3+5
10 = 5 + 5 12 = 5 + 7 14 = 7 + 7
Thus by the method of exhaustion every even number
between 2 and 16 can be written as a sum of two prime
numbers.
WUCT121
Logic
104
Exercise:
•
Prove for each integer n with 1 ≤ n ≤ 10 , n 2 − n + 11 is
prime.
The statement is of the form ∀n ∈ D, P (n ) , where
D = {1, 2, 3, 4, 5, 6, 7, 8, 9,10},and P(n) is the statement
“ n 2 − n + 11 is prime”. Thus we must show all numbers in
D satisfy P(n).
P(1) = 12 − 1 + 11
= 11 is prime
P( 2) = 2 2 − 2 + 11
= 13 is prime
P(3) = 32 − 3 + 11
= 17 is prime
P( 4) = 4 2 − 4 + 11
= 23 is prime
P(5) = 5 2 − 5 + 11
= 31 is prime
P(6) = 6 2 − 6 + 11
= 41 is prime
P(7) = 7 2 − 7 + 11
= 53 is prime
P(8) = 8 2 − 8 + 11
= 67 is prime
P(9) = 9 2 − 9 + 11 P(10) = 10 2 − 10 + 11
= 89 is prime
= 101 is prime
Thus by the method of exhaustion for each integer n with
1 ≤ n ≤ 10 , n 2 − n + 11 is prime.
WUCT121
Logic
105
Method 2: Generalised Proof.
The generalised proof method is used when the domain is
infinite.
It is called the method of generalizing from the generic
particular.
In order to show that every element of the domain satisfies
the predicate, a particular but arbitrary element of the
domain is chosen and shown to satisfy the predicate.
The method to show the predicate is satisfied will vary
depending on the form of the predicate.
Specific techniques of generalized proof will be outlined
later in this section.
WUCT121
Logic
106
Example:
•
Pick any number, add 3, multiply by 4, subtract 6,
divide by two and subtract twice the original. The result is
3.
Proof:
Choose a particular but arbitrary number, say x, and then
determine if it satisfies the statement.
Step
Result
Pick a number
x
Add 3
x+3
Multiply by 4
( x + 3) × 4 = 4 x + 12
Subtract 6
4 x + 12 − 6 = 4 x + 6
Divide by 2
( 4 x + 6) ÷ 2 = 2 x + 3
Subtract twice the original
2x + 3 = 2x = 3
In this example, x is particular in that it represents a single
quantity, but arbitrarily chosen as it can represent any
number.
WUCT121
Logic
107
3.6.
Disproving Quantified Statements
3.6.1.
Disproving Existential Statements
A statement of the form ∃x ∈ D , P ( x ) is false if and only
if P ( x ) is false for all x ∈ D .
To disprove this kind of statement, we need to show the for
all x ∈ D , P ( x ) is false.
That is we need to prove it’s negation:
~ (∃x ∈ D, P( x )) ≡ ∀x ∈ D , ~ P( x )
This is equivalent to proving a universal statement and so
the method of exhaustion or the generalized proof method
is used.
Example:
•
Disprove: There exists a positive number n such that
n 2 + 3n + 2 is prime.
Proving the given statement is false is equivalent to proving
its negation is true. That is proving that for all numbers n ,
n 2 + 3n + 2 is not prime. Since this statement is universal,
its proof requires the generalised proof method.
WUCT121
Logic
108
3.6.2.
Disproving Universal Statements
A statement of the form ∀x ∈ D , P ( x ) is false if and only
if P ( x ) is false for at least one x ∈ D .
To disprove this kind of statement, we need to find one
x ∈ D such that P ( x ) is false.
That is we need to prove it’s negation:
~ (∀x ∈ D , P( x )) ≡ ∃x ∈ D , ~ P( x )
This is known as finding a counterexample.
Example:
•
Disprove: ∀a , b ∈ , ( a 2 = b 2 ) ⇒ ( a = b ).
Let P( a , b ) : ( a 2 = b 2 ) ⇒ ( a = b ) .
We need to show ∃a , b ∈ , ~ P( a , b )
Counterexample:
Let a = 1, b = −1. Then a 2 = b 2 however a ≠ b .
Now true ⇒ false is false, thus P( a , b ) is false, and
~ P( a , b ) is true
So, we have shown, by counterexample ∃a , b ∈ , ~ P( a , b )
WUCT121
Logic
109
Exercises:
•
Disprove: ∀x ∈ , ( x > 0 ∨ x < 0).
Need to prove: ∃x ∈ , ~ ( x > 0 ∨ x < 0).
That is ∃x ∈ , ( x ≤ 0 ∧ x ≥ 0).
Let x = 0 .
•
Disprove ∀z ∈ , ( z is odd ) ⇒ (
Let z = 5 ∈ , ( z is odd ) , (
•
z −1
is odd ).
2
5 −1 4
= = 2 is even).
2
2
Prove or disprove: ∀x ∈ , ∃y ∈ , ( x + y = 0).
To prove the statement: Find a specific y for each “general”
x.
Consider an x ∈ . Let y = − x ∈ , then
x + y = x + (− x ) = 0 .
WUCT121
Logic
110
3.7.
Generalised Proof Methods
Before proving a statement, it is of great use to write the
statement using logic notation, including quantifiers, where
appropriate.
Doing this means you have clearly written the assumptions
you can make AND the conclusion you are aiming to reach.
Example:
•
Prove: For all integers n, if n is odd, then n 2 is odd.
Rewritten using logic notation:
∀n ∈ , n is odd ⇒ n 2 is odd
Here the domain is given as , , and the predicate involves
the statements: P(n) is n is odd, Q(n) is n 2 is odd.
The form of the predicate is P( n ) ⇒ Q ( n ) .
Thus the assumption that can be made is P(n) and the
conclusion to be reached is Q(n).
WUCT121
Logic
111
3.7.1.
Direct Proof
A direct proof is one in which we begin with the
assumptions and work in a straightforward fashion to the
conclusion.
The steps in the final proof must proceed in the correct
“direction” beginning with the initial assumption and
following known laws, rules, definitions etc. until the final
conclusion is reached. The proof must not start with what
you are trying to prove.
Example:
•
Prove that if 3 x − 9 = 15 then x = 8 .
Assuming the domain to be , then the statement is of the
form ∀x ∈ , P( x ) ⇒ Q( x ) .
Thus the assumption is P( x ) : 3 x − 9 = 15 , and the
conclusion Q( x ) : x = 8 .
3 x − 9 = 15 ⇒ 3 x − 9 + 9 = 15 + 9
⇒ 3 x = 24
3 x 24
⇒
=
3
3
⇒ x=8
WUCT121
Logic
112
Exercise:
•
Prove: For all integers n, if n is odd, then n 2 is odd.
Rewritten using logic notation:
∀n ∈ , n is odd ⇒ n 2 is odd
Here the domain is given as , , and the predicate involves
the statements: P(n) is n is odd, Q(n) is n 2 is odd.
The form of the predicate is P( n ) ⇒ Q ( n ) .
Thus the assumption that can be made is P(n) and the
conclusion to be reached is Q(n).
Proof:
n is odd ⇒ n = 2 p + 1
p ∈
definition of odd
⇒ n 2 = ( 2 p + 1) 2
⇒ n2 = 4 p2 + 4 p + 1
⇒ n 2 = 2( 2 p 2 + 2 p ) + 1
WUCT121
⇒ n 2 = 2q + 1
q = 2 p2 + 2 p ∈ 
⇒ n 2 is odd
definition of odd
Logic
113
When the statement to be proven is of the form:
P( x ) ⇒ Q( x ) , the assumption which begins the proof is
P( x ).
If the form is not P( x ) ⇒ Q( x ) , or P( x ) is not clear, it may
be necessary to examine what you are aiming to prove and
establish an assumption from which to begin.
Example:
•
(
)
Prove that for x ∈ , − x 2 + 2 x + 1 ≤ 2 .
(Do not start with this!)
In this case, the form is not P( x ) ⇒ Q( x ) .
By examining what we are aiming to prove, i.e.
(− x 2 + 2 x + 1) ≤ 2 a beginning to the proof can be found.
⎧− x 2 + 2 x + 1 ≤ 2 ⇒ x 2 − 2 x − 1 ≥ −2
⎪⎪
⇒ x2 − 2 x + 1 ≥ 0
working : ⎨
⎪
⇒ ( x − 1)2 ≥ 0 is true.
⎪⎩
We can now put the proof together:
We know that ( x − 1)2 ≥ 0 for any x ∈ .
Thus,
WUCT121
Logic
114
( x − 1)2 ≥ 0
what is known
⇒ x2 − 2x + 1 ≥ 0
expanding
⇒ x 2 − 2 x − 1 ≥ −2
subtracting 2 from both sides
⇒ − x2 + 2x + 1 ≥ 2
multiplying by − 1
Note. In the example, we did NOT start with the statement
(− x 2 + 2 x + 1) ≤ 2 , as we technically do not know whether
it is true or not. We started our proof with a statement we
know to be true.
Exercise:
•
Prove the following:
If the right angled triangle XYZ with sides of length x and y
z2
, then the triangle is
and hypotenuse z has an area of
4
X
isosceles.
z
y
Y
WUCT121
x
Z
Logic
115
The form of statement is P( x ) ⇒ Q( x ) .
What is known. i.e. the assumptions that can be made are:
•
z2
Area of the triangle: A =
,
4
•
Area of any triangle:
= 1 × (base) × (height ) = 1 xy
2
•
(1)
(2)
2
The sides are of length x and y and hypotenuse
z so by Pythagoras: z 2 = x 2 + y 2 .
(3)
What is to be proven: That triangle XYZ is isosceles. Thus
we must show two sides have equal length.
Proof:
Substituting (3) into (1) and equating with (2) gives:
x 2 + y 2 xy
=
4
2
x 2 + y 2 xy
=
⇒ x 2 + y 2 = 2 xy
4
2
⇒ x 2 − 2 xy + y 2 = 0
(multiplying both sides by 4)
⇒ (x − y ) = 0
⇒x= y
2
So two sides are equal and thus triangle XZY is isosceles.
WUCT121
Logic
116
3.8.
Indirect Proofs.
3.8.1.
Method of Proof by Contradiction
The method of proof by contradiction can be used when the
statement to be proven is not of the form P( x ) ⇒ Q( x ) .
The method is as follows:
Suppose the statement to be proven is false. That is,
1.
suppose that the negation of the statement is true.
2.
Show that this supposition leads to a contradiction
3.
Conclude that the statement to be proven is true.
Example:
•
Prove there is no greatest integer.
Suppose not, that is suppose there is a greatest integer. N.
Then N ≥ n for every integer n. Let M = N + 1 . Now M is
an integer since it is the sum of integers. Also M > N since
M = N +1
Thus M is an integer that is greater than N. So N is the
greatest integer and N is not the greatest integer, which is a
contradiction.
Thus the assumption that there is a greatest integer is false,
hence there is no greatest integer is true.
WUCT121
Logic
117
Exercise:
•
Prove there is no integer that is both even and odd.
Suppose not, that is suppose there is a greatest integer an
integer n, that is both even and odd.
By the definition of even n = 2k , k ∈  K (1) , and by the
definition of odd n = 2l + 1, l ∈ 
K ( 2) .
If n is both even and odd then equation (1) and (2) gives:
2k = 2l + 1
⇒ 2k − 2l = 1
⇒ 2( k − l ) = 1
1
⇒ k −l = ∉
2
Now since k and l are integers, the difference k – l must be
an integer. However k − l =
1
∉  . Thus k – l is an integer
2
and k – l is not an integer, which is a contradiction.
Thus the supposition is false and hence the statement
“There is no integer that is both even and odd” is true.
WUCT121
Logic
118
3.8.2.
Proof by Contraposition
Recall the following logical equivalence:
( P ⇒ Q ) ≡ (~ Q ⇒ ~ P ).
(~ Q ⇒ ~ P ) is known as the contrapositive of ( P ⇒ Q ) .
This equivalence indicates that if (~ Q ⇒ ~ P ) is a true
statement, then so too is ( P ⇒ Q ) .
Thus, in order to prove ( P ⇒ Q ) , we prove the
contrapositive, that is (~ Q ⇒ ~ P ) , is true.
The method of proof by contraposition can be used when
the statement to be proven is of the form P( x ) ⇒ Q( x ) .
The method is as follows:
1.
Express the statement to be proven in the form:
∀x ∈ D , P( x ) ⇒ Q ( x ).
2.
Rewrite the statement in the contrapositive
form: ∀x ∈ D, ~ Q( x ) ⇒ ~ P( x ).
3.
a.
Prove the contrapositive by a direct proof.
Suppose that x is a particular but arbitrary
element of D, such that Q(x) is false.
b.
WUCT121
Show that P(x) is false.
Logic
119
Example:
•
Prove that for all integers n if n 2 is even, n is even.
The statement can be expressed in the form:
∀n ∈ , n 2 is even ⇒ n is even.
Thus the contrapositive is
∀n ∈ , n is not even ⇒ n 2 is not even. That is
∀n ∈ , n is odd ⇒ n 2 is odd.
To prove the contrapositive:
Let n be any odd integer.
Then n = 2k + 1, k ∈ 
K(1)
Show n 2 is odd, i.e. show n 2 = 2l + 1, l ∈ 
n 2 = ( 2k + 1)2 by (1)
= 4k 2 + 4k + 1
= 2( 2k 2 + 2k ) + 1
= 2l + 1
l = 2k 2 + 2k ∈ 
So n 2 is odd, and the contrapositive is true.
Hence the statement “for all integers n if n 2 .is even, n is
even” is also true.
WUCT121
Logic
120
Exercise:
•
Prove that for all integers n if 5 /| n 2 .then 5 /| n .
The statement can be expressed in the form:
∀n ∈ , 5 /| n 2 ⇒ 5 /| n.
Thus the contrapositive is ∀n ∈ , 5 | n ⇒ 5 | n 2 .
To prove the contrapositive:
Let n be any odd integer.
Then 5 | n ⇒ n = 5k , k ∈ 
K (1)
Show 5 | n 2 , i.e. show n 2 = 5l , l ∈ 
n 2 = (5k ) 2
by (1)
= 25k 2
= 5(5k 2 )
= 5l l = 5k 2 ∈ 
So 5 | n 2 , and the contrapositive is true.
Hence the statement “for all integers n if 5 /| n 2 .then 5 /| n ”
is also true.
WUCT121
Logic
121
Exercise:
•
Prove if y is irrational, then y + 7 is irrational.
The statement can be expressed in the form:
∀y ∈ , y ∉  ⇒ y + 7 ∉ 
Thus the contrapositive is ∀y ∈ , y + 7 ∈  ⇒ y ∈ 
To prove the contrapositive:
Let. y ∈ , y + 7 ∈  , so y + 7 =
Show y ∈  , that is y =
c
,
d
a
,
b
a , b ∈ , b ≠ 0 .
c , d ∈ , d ≠ 0
a
a
⇒ y = −7
b
b
a − 7b
⇒y=
b
c
⇒ y = c = a − 7b, d = b ∈ , d ≠ 0
d
y +7 =
Therefore y ∈  , and the contrapositive is true.
Hence the statement “if y is irrational, then y + 7 is
irrational” is also true.
WUCT121
Logic
122
3.8.3.
Proof by Cases
When the statement to be proven is of the form, or can be
written in the form: ( P ∨ Q ) ⇒ R , the method of proof by
cases can be used.
It relies on the logical equivalence
(( P ⇒ R ) ∧ (Q ⇒ R )) ≡ (( P ∨ Q ) ⇒ R ).
The method is as follows:
1.
Prove P ⇒ R
2.
Prove Q ⇒ R.
3.
Conclude ( P ∨ Q ) ⇒ R.
If the statement is not written in the form ( P ∨ Q ) ⇒ R , it
is necessary to establish the particular cases by exhaustion.
WUCT121
Logic
123
Example:
•
Prove: If x ≠ 0 or y ≠ 0 , then x 2 + y 2 > 0 .
The statement can be expressed in the form:
( x ≠ 0) ∨ ( y ≠ 0) ⇒ x 2 + y 2 > 0
We assume x, y ∈  , thus x 2 ≥ 0, y 2 ≥ 0.
Proof:
Case 1:
Prove x ≠ 0 ⇒ x 2 + y 2 > 0
2
2
Let x ≠ 0 , then x > 0 and y ≥ 0 .
Thus x 2 + y 2 > 0 .
Case 2:
Prove y ≠ 0 ⇒ x 2 + y 2 > 0
Let y ≠ 0 , then y 2 > 0 and x 2 ≥ 0 .
Thus x 2 + y 2 > 0 .
Therefore If x ≠ 0 or y ≠ 0 , then x 2 + y 2 > 0 .
WUCT121
Logic
124
Exercise.
•
Prove: If x ≤ −2 or x ≥ 2 , then x 2 − 4 ≥ 0 .
The statement can be expressed in the form:
( x ≤ −2) ∨ ( x ≥ 2) ⇒ x 2 − 4 ≥ 0
Proof:
Case 1:
Prove x ≤ −2 ⇒ x 2 − 4 ≥ 0
x ≤ −2
⇒ x2 ≥ 4
⇒ x2 − 4 ≥ 0
Therefore x ≤ −2 ⇒ x 2 − 4 ≥ 0
Case 2:
Prove ( x ≥ 2 ⇒ x 2 − 4 ≥ 0
x≥2
⇒ x2 ≥ 4
⇒ x2 − 4 ≥ 0
Therefore x ≥ 2 ⇒ x 2 − 4 ≥ 0
Thus if x ≤ −2 or x ≥ 2 , then x 2 − 4 ≥ 0 .
WUCT121
Logic
125
If the statement is not written in the form ( P ∨ Q ) ⇒ R , it
is necessary to establish the particular cases by exhaustion.
Example.
•
Prove: ∀m ∈ , m 2 + m + 1 is odd.
The statement is not in the form ( P ∨ Q ) ⇒ R. However by
considering m ∈  ⇒ ( m is even) ∨ ( m is odd ) . Then the
statement can be expressed in the form:
( m is even) ∨ ( m is odd ) ⇒ m2 + m + 1 is odd
Case 1: Prove m is even ⇒ m 2 + m + 1 is odd
m is even ⇒ m = 2 p,
p∈
m 2 + m + 1 = (2 p )2 + 2 p + 1
K (1) .
by (1)
= 4 p2 + 2 p +1
(
)
= 2 2 p2 + p +1
(
)
= 2k + 1, where k = 2 p 2 + p ∈ 
Therefore, m is even ⇒ m 2 + m + 1 is odd .
WUCT121
Logic
126
Case 2: Prove m is odd ⇒ m 2 + m + 1 is odd
m is odd ⇒ m = 2q + 1, q ∈ 
K(2) .
m 2 + m + 1 = (2q + 1)2 + 2q + 1 + 1 by ( 2)
= 4q 2 + 4q + 1 + 2q + 1 + 1
= 4q 2 + 6q + 2 + 1
(
)
= 2 2q 2 + 3q + 1 + 1
(
)
= 2l + 1, where l = 2q 2 + 3q + 1 ∈ 
Therefore, m is odd ⇒ m 2 + m + 1 is odd .
Therefore, ( m is even) ∨ ( m is odd ) ⇒ m2 + m + 1 is odd .
Therefore, ∀m ∈ , m 2 + m + 1 is odd.
WUCT121
Logic
127
Exercise.
•
Prove: ∀n ∈ , n 2 − n + 3 is odd.
The statement is not in the form ( P ∨ Q ) ⇒ R. However by
considering n ∈  ⇒ ( n is even) ∨ ( n is odd ) . Then the
statement can be expressed in the form:
( n is even) ∨ ( n is odd ) ⇒ n 2 − n + 3 is odd
Case 1: Prove n is even ⇒ n 2 − n + 3 is odd
n is even ⇒ n = 2 p,
p ∈  K (1) .
n 2 − n + 3 = (2 p )2 − 2 p + 3
by (1)
= 4 p2 − 2 p + 2 +1
(
)
= 2 2 p2 − p +1 +1
(
)
= 2k + 1, where k = 2 p 2 − p + 1 ∈ 
Therefore, n is even ⇒ n 2 − n + 3 is odd .
WUCT121
Logic
128
Case 2: Prove n is odd ⇒ n 2 − n + 3 is odd
n is odd ⇒ n = 2q + 1, q ∈ 
K(2) .
n 2 − n + 3 = (2q + 1)2 − (2q + 1) + 3
by (2)
= 4q 2 + 4q + 1 − 2q − 1 + 3
= 4q 2 + 2q + 2 + 1
(
)
= 2 2q 2 + q + 1 + 1
(
)
= 2l + 1, where l = 2q 2 + q + 1 ∈ 
Therefore, n is odd ⇒ n 2 − n + 3 is odd .
Therefore, ( n is even) ∨ ( n is odd ) ⇒ n 2 − n + 3 is odd .
Therefore, ∀n ∈ , n 2 − n + 3 is odd.
WUCT121
Logic
129
Section 4.
4.1.
Set Theory
Definitions
A set may be viewed as any well defined collection of
objects, called elements or members of the set.
Sets are usually denoted with upper case letters, A, B, X,
Y,… while lower case letters are used to denote elements a,
b, x, y,…of a set.
Membership in a set is denoted as follows:
•
a ∈ S denotes that a is a member or element of a
set S. Similarly a , b ∈ S denotes that a and b are both
elements of a set S.
•
a ∉ S denotes that a is not an element of a set S.
Similarly a , b ∉ S denotes that neither of a and b are
elements of a set S.
In Set Theory, we work within a Universe, U, and consider
sets containing elements from U.
WUCT121
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130
A set may be specified in essentially two ways:
The elements of the set are listed within braces, {
1.
}, and separated by commas.
Technically, the listing of elements can be done only for
finite sets. However, if an infinite set is defined by a
“simple” rule, we sometimes write a few elements and then
use “…” to mean roughly “and so on” or “by the same
rule”.
Examples:
•
A = {1, 3, 5, 7, 9} . The set A is the finite collection of
odd integers, 1 to 9 inclusive
•
B = {K, − 4, − 2, 0, 2, 4,K}. The set B is the infinite
collection of even integers.
Exercises:
•
List a finite set, C, containing even integers between
10 and 20 inclusive.
C = {10,12,14,16,18, 20}
•
List an infinite set, D, containing natural numbers that
are divisible by 3
D = {0, 3, 6, 9,K}
WUCT121
Logic
131
2.
A statement defining the properties which
characterise the elements in the set is written within braces
Examples:
•
A = {z ∈  :( z is odd ∧ 1 ≤ z ≤ 9)}. The set A is the
finite collection of odd integers, 1 to 9 inclusive
•
B = {z ∈  : ∃k ∈ , z = 2k } . The set B is the infinite
collection of even integers.
Exercises:
•
Define a finite set, C, containing even integers
between 10 and 20 inclusive
C = {z ∈  : ( z is even ∧ 10 ≤ z ≤ 20)}
•
Define an infinite set, D, containing natural numbers
that are divisible by 3
D = {n ∈  : 3 | n}
WUCT121
Logic
132
4.1.2.
Axiom of Specification.
Given a Universe U and any statement P( x ) involving
x ∈ U , then there exists a set A such that
∀x ∈ U , ( x ∈ A ⇔ P( x )). Further, we write
A = { x ∈ U : P( x )}.
In other words, the Axiom of Specification says that we can
pick a set and a property and build a new set. This is why
the notation for A is sometimes referred to as set-builder
notation.
Example:
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and P( x ) be the
statement “x is odd”.
\ by the Axiom of Specification, A = { x ∈U : x is odd}.
Notes:
1.
We know an element x belongs to the set
A = { x ∈ U : P( x )} if x satisfies the condition P( x ) .
2.
This notation is more simply written
{ x ∈ D : P( x )} .
This is called set builder notation. In using this notation, the
elements of the domain, D, must belong to the Universe, U,
WUCT121
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133
and P ( x ) can be any predicate involving x. D could be all
of U.
Examples:
•
The interval [0, 1] can be written in set builder
notation as:
{x : x ∈  ∧ 0 ≤ x ≤ 1} = {x ∈  : 0 ≤ x ≤ 1}
•
The set of all rational numbers, – can be written as:
⎧a
⎫
 = ⎨ : a, b ∈  ∧ b ≠ 0⎬
⎩b
⎭
a
⎧
⎫
= ⎨ x : x = : a, b ∈  ∧ b ≠ 0⎬
b
⎩
⎭
•
{ x ∈  : x 3 = x} = {0, 1, − 1} .
Exercises:
Write down the following sets by listing their elements:
•
{ x ∈  : x 3 = x} = {1}
•
{ x ∈  : x 2 = 9} = {−3, 3}
•
{ x ∈  : x 2 = 7} = { }
WUCT121
Logic
134
4.2.
Venn Diagrams
Venn diagrams are a pictorial method of demonstrating the
relationship between set. The universal set, U, is
represented by a rectangle and sets within the universe are
depicted with circles.
While a Venn diagram may be used to demonstrate the
relationship between sets, it does not provide a method of
proving those relationships.
WUCT121
Logic
135
4.3.
Special Sets
4.3.1.
The Singleton Set
Sets having a single element are frequently called singleton
sets.
Example:
•
{1} is read “singleton 1”.
•
If a ∈U , then { x ∈U : x = a} = {a}
Note: The singleton set {a} is NOT the same as the element
a.
4.3.2.
The Empty Set
The empty set or null set is a set which contains no
elements.
It is denoted by the symbol ∆ or by empty braces { }.
Using set builder notation, one way of defining the empty
set is: ∆ = { x ∈  : x ≠ x}
WUCT121
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136
4.4.
Subsets
4.4.1.
Definition: Subset.
If A and B are sets, then A is called a subset of B, written
A ⊆ B , if and only if, every element in A is also in B.
Examples:
•
{1, 2} ⊆ {1, 2, 3}
•
The Venn diagram demonstrating A ⊆ B is:
Exercises:
•
Write the definition of subset using logic notation.
A ⊆ B ⇔ (∀x ∈ U , x ∈ A ⇒ x ∈ B )
•
Is {cat, dog} ⊆ {bird, fish, cat, dog}? Yes
WUCT121
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137
4.4.2.
Definition: Proper Subset.
If A and B are sets, then A is called a proper subset of B,
written A ⊂ B , if and only if, every element in A is also in
B but there is at least one element of B that is not in A.
A is a proper subset of B if A ⊆ B but A ≠ B .
Examples:
•
{1, 2} ⊂ {1, 2, 3}
Exercises:
•
Draw a Venn diagram demonstrating A ⊂ B , where
A = {1, 2} and B = {1, 2, 3, 4, 5}
•
Is {a , b, c} ⊂ {c, b, a}?
WUCT121
No
Logic
138
Notes.
1.
If A ⊆ B , then each element of A belongs to B, or
for each x ∈ A , it is true that x ∈ B .
2.
If A is a subset of B, then B is sometimes called a
superset of A.
3.
If A and B are sets, then to prove A ⊆ B , we need
to prove ∀x, x ∈ A ⇒ x ∈ B
4.
If A is a proper subset of B, there must be at least
one element in B that is not in A.
5.
If A and B are sets, to prove A is not a subset of B,
denoted A⊄B , we need to prove ~ ( A ⊆ B ) :
~ (∀x, x ∈ A ⇒ x ∈ B ) ≡ ∃x, ~ ( x ∈ A ⇒ x ∈ B )
≡ ∃x , ~ ( x ∉ A ∨ x ∈ B )
≡ ∃x , ( x ∈ A ∧ x ∉ B )
6.
The following relationships hold in the number
system: ∆ ⊆  ⊆  ⊆  ⊆ 
WUCT121
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139
4.4.3.
The null set as a subset.
For any set A in a Universe U, ∆ ⊆ A
Proof:
Suppose ~ ( ∆ ⊆ A) . Then, there exists x ∈ ∆ such that
x ∉ A . This, therefore, means that ∆ is not empty, which is
a contradiction. Therefore, ∆ ⊆ A .
4.4.4.
Distinction between elements and subsets
Examples:
•
2 ∈ {1, 2, 3} , 2⊄{1, 2, 3}
•
{2} ∉ {1, 2, 3} , {2} ⊆ {1, 2, 3}
•
1 ∈ { x ∈  : x 2 = 1}, {1} ⊆ { x ∈  : x 2 = 1}
Exercises:
Let S be a set in a Universe U. Determine whether the
following are true or false.
•
S ∈S
•
False
•
False
S ∈ {S }
•
True
WUCT121
S ⊆ {S }
∆ ⊆ {S }
True
Logic
140
•
•
∆ ∈ {S }
False
{∆ } ⊆ {S }
False
4.5.
Set Equality
4.5.1.
Definition: Set Equality.
If A and B are sets, then A equals B, written A = B , if and
only if, every element in A is also in B and every element in
B is also in A.
Equivalently, A = B if, and only if A ⊆ B and B ⊆ A .
Note: To prove that two sets are equal two things must be
shown:: A ⊆ B and B ⊆ A .
Examples:
•
The Venn diagram demonstrating A = B is:
WUCT121
Logic
141
Exercises:
•
Write the definition of set equality using logic
notation.
A = B ⇔ (∀x ∈U , x ∈ A ⇒ x ∈ B ∧ x ∈ B ⇒ x ∈ A)
⇔ (∀x ∈U , A ⊆ B ∧ B ⊆ A)
4.5.2.
Axiom of Extent.
If A and B are sets then A = B ⇔ (∀x ∈ U , x ∈ A ⇔ x ∈ B ) .
The Axiom of Extent says that a set is completely
determined by its elements, the order in which the elements
are listed is irrelevant, as is the fact that some members
may be listed more than once.
Examples:
•
{1, 2} = {1, 2}
•
{a , b, c} = {c, b, a}
Exercises:
•
Is {a , b, c, d } = {b, d , a , c}
Yes
•
Is {Ann, Bob, Cal} = {Bob, Cal, Ann, Cal}
Yes
WUCT121
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142
4.5.3.
Theorem: Equality by Specification
Let U be a universe and let P ( x ) be a statement.
If ∀x ∈U , ( P( x ) ⇔ Q( x )) , that is ∀x ∈U , ( P( x ) ≡ Q ( x ))
then { x ∈ U : P( x )} = { x ∈ U : Q ( x )}
The Theorem states that subsets of the same universe U
which are defined by equivalent statements are equal sets.
This theorem allows the use of tautologies of logic to prove
set theoretic statements, as will be outlined later.
Example:
•
We know that x 2 = 1 ⇔ ( x = 1 ∨ x = −1).
Therefore
{ x ∈  : x 2 = 1} = {( x = 1 ∨ x = −1)} = {1,−1}
•
If a1, a2 , a3 ,K an ∈ U , then we can write
A = {x ∈U : x = a1 ∨ x = a2 ∨ K ∨ x = an }
= {a1 , a2 , a3 ,K an }
In other words, if we know the elements of a set, we know
the set.
•
A = {x ∈  : x = 1 ∨ x = 2 ∨ x = 3} = {1, 2, 3}
WUCT121
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143
Exercise:
•
Are the following sets equal? Using logic, can you
prove your answer?
{1, 3, 1, 2}, {3, 2, 1}, {1, 2, 3} Yes
{1, 3,1, 2} = { x ∈  : x = 1 ∨ x = 3 ∨ x = 1 ∨ x = 2}
= { x ∈  : x = 1 ∨ x = 1 ∨ x = 3 ∨ x = 2}
= { x ∈  : x = 1 ∨ x = 3 ∨ x = 2}
= { x ∈  : x = 1 ∨ x = 2 ∨ x = 3}
= {1, 2, 3}
•
Are the following two sets equal? Give reasons.
E = {n ∈ :: n is even} and T = {n ∈ :: n 2 is even}.
Yes, previously is has been proven that
n is even ⇔ n 2 is even , thus by equivalence of statements,
the two sets are equal.
WUCT121
Logic
144
4.6.
Power Sets
4.6.1.
Definition: Power Set
If X is any set, then { A : A ⊆ X } is the power set of X.
The power set of X is often written as  ( X ) .
So  ( X ) = { A : A ⊆ X } .
A power set is a set whose elements are sets.
If the elements of X are in a universe U, those of  ( X ) are
in a universe (U ) .
Examples:
•
Let X = {1} and let S be the set of all subsets of X.
Write down the set S by listing its elements.
S = { A : A ⊆ X }.
∆ ⊆ {1} and {1} ⊆ {1}.
Thus S = {∆ , {1}}
•
Let X = {1, 2} and S = { A : A ⊆ X } . Write down the
set S by listing its elements.
∆ ⊆ {1, 2} , {1} ⊆ {1, 2}, {2} ⊆ {1, 2}, and {1, 2} ⊆ {1, 2}. Thus
S = {∆ , {1}, {2}, {1, 2}}
WUCT121
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145
Exercises:
•
Let X = {1, 2, 3} .
o
Write down the set  ( X ) by listing its elements.
 ( X ) = {∆ , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
o
How many elements are there in  ( X ) ?
o
Is ∆ ∈  ( X ) ?
Yes
o
Is ∆ ⊆  ( X ) ?
Yes
o
Is 1 ∈  ( X ) ?
No
o
Is {1} ∈  ( X ) ?
Yes
o
Is {2} ⊆  ( X ) ?
No
o
Is {{1, 2}} ⊆  ( X ) ?
WUCT121
8
Yes
Logic
146
4.7.
Hasse Diagrams
The elements of  ( X ) can be represented by diagrams
using the following procedure:
1.
An upward directed line between two sets indicates
that the “lower” set is a subset of the “upper” set.
2.
∆ is at the bottom and X is at the top.
3.
Each pair of sets is joined by an upward directed
line to the “smallest” set which contains each as a subset.
4.
Each pair of sets is joined by a downward directed
line to the “largest” set which is a subset of each.
Example:
Let X = {1, 2} , thus  ( X ) = {∆ , {1}, {2}, {1, 2}} and the
Hasse diagram is given by:
{1, 2}
{1}
{2}
ø
WUCT121
Logic
147
4.8.
Set Operations
There are five main set theoretic operations, one
corresponding to each of the logical connectives.
Set Operation Name
Logical
Name
Connective
A
Complement ~ P
Negation
A∪ B
Union
P∨Q
Disjunction
A∩ B
Intersection
P∧Q
Conjunction
A⊆ B
Subset
P⇒Q
Conditional
A=B
Equality
P⇔Q
Biconditional
P≡Q
Equivalence
The set operations can be defined in terms of the
corresponding logical operations. This means that each of
the tautologies proved by truth tables for the logical
connectives will have a corresponding theorem in set
theory.
WUCT121
Logic
148
We have seen how the logical conditional operator, P ⇒ Q
is related to subset, A ⊆ B and how the logical
biconditional operator, P ⇔ Q (or equivalence, P ≡ Q ) is
related to set equality, A = B .
The following sections will cover the three remaining set
operations: complement, union and intersection.
In our discussion of set theory, we will let U be a fixed set
and all other sets, whether denoted A, B, C, etc, will be
subsets of U. In other words, A, B, C ∈  (U ) . Thus, each
result should start with a statement similar to “Let A, B, C
be subsets of a universal set U” or “Let A, B, C ∈  (U ) ”.
4.8.1.
Definition: Compliment
Let U be a universal set, and let A ⊆ U . Then the
complement of A, denoted by A , is given by
A = {x ∈ U : ~ ( x ∈ A)} = {x ∈ U : x ∉ A}.
Notes.
1.
U \ A , A ′ and Ac are also used for A in some
books.
WUCT121
Logic
149
If the set U is fixed in a discussion, then A is
2.
sometimes written as A = {x : x ∉ A}
Example:
•
The shaded area in the following Venn diagram
depicts A :
Exercises:
Let U =  . Write down A for the following sets:
•
A = {1, 2, 3}
A = {x ∈  : x ≠ 1 ∧ x ≠ 2 ∧ x ≠ 3}
•
A = {x ∈  : x is even}
A = {x ∈  : x is odd}
•
A = {x ∈  : x > 0 ∨ x < 0}
A = {0}
WUCT121
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150
4.8.2.
Definition: Union
Let A and B be subsets of a universe U. Then the union of
A and B, denoted by A ∪ B , is given by
A ∪ B = {x ∈ U : x ∈ A ∨ x ∈ B}.
Example:
•
The shaded area in the following Venn diagram
depicts A ∪ B :
Exercises:
•
Let U =  . Write down A ∪ B for the following sets:
o
A = {1} and B = {2}.
A ∪ B = {1, 2}
o
A is the set of all even integers, B is the set of all odd
integers.
A ∪ B = .
WUCT121
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151
A = {x ∈  : 0 ≤ x ≤ 2} and B = {x ∈  : 1 ≤ x ≤ 3}
o
A ∪ B = { x ∈  : 0 ≤ x ≤ 3} = [0, 3]
•
If A ⊆ U and B ⊆ U , is it true that A ∪ B ⊆ U ?
Yes.
x ∈ A∪ B ⇒ x ∈ A ∨ x ∈ B
⇒ x ∈U ∨ x ∈U
⇒ x ∈U
4.8.3.
Definition: Intersection
Let A and B be subsets of a universe U. Then the
intersection of A and B, denoted by A ∩ B , is given by
A ∩ B = {x ∈U : x ∈ A ∧ x ∈ B}.
Example:
•
The shaded area in the following Venn diagram
depicts A ∩ B :
WUCT121
Logic
152
Exercises:
•
Let U =  . Write down A ∩ B for the following sets:
A = {1, 2, 3, 5} and B = {1, 4, 5, 6}.
o
A ∩ B = {1, 5}.
A is the set of all even integers, B is the set of all odd
o
integers.
A∩ B = .
A = {x ∈  : 0 ≤ x ≤ 2} and B = {x ∈  : 1 ≤ x ≤ 3}
o
A ∩ B = { x ∈  : 1 ≤ x ≤ 2} = [1, 2]
•
If A ⊆ U and B ⊆ U , is it true that A ∩ B ⊆ U ?
Yes.
x ∈ A∩ B ⇒ x ∈ A ∧ x ∈ B
⇒ x ∈U ∧ x ∈U
⇒ x ∈U
4.8.4.
Definition: Difference
Let A and B be subsets of a universe U. Then the
difference of A and B, denoted by A − B , is given by
A − B = {x ∈ U : x ∈ A ∧ x ∉ B}.
WUCT121
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153
Example:
•
The shaded area in the following Venn diagram
depicts A − B :
Notes.
1. The difference of A − B is sometimes called the
relative complement of B in A.
2. If we let A = U , then we have
U − B = {x ∈ U : x ∈ U ∧ x ∉ B}
= {x ∈ U : x ∈ B}
=B
3. Using Definitions for complement and intersection, we
can simplify the definition of difference as follows:
A − B = {x ∈ U : x ∈ A ∧ x ∉ B}
= {x ∈ U : x ∈ A ∧ x ∈ B}
= A∩ B
WUCT121
Logic
154
Exercises:
•
Let U =  . Write down A − B for the following sets:
A = {1, 2, 3, 5} and B = {1, 4, 5, 6}.
o
A − B = {2, 3}.
A is the set of all even integers, B is the set of all odd
o
integers.
A− B = A .
A = {x ∈  : 0 ≤ x ≤ 2} and B = {x ∈  : 1 ≤ x ≤ 3}
o
A − B = { x ∈  : 0 ≤ x < 1} = [0, 1)
•
If A ⊆ U and B ⊆ U , is it true that A − B ⊆ U ?
Yes.
x∈A− B ⇒ x∈A∧ x∉B
⇒ x ∈U ∧ x ∈U
⇒ x ∈U
•
Let U =  , A = {1, 2, 3}, B = {2}, C = {2, 3, 4} and
D = [0, 1] = {x ∈  : 0 ≤ x ≤ 1}.Write down:
o
A − C = {1}
o
B−C =
o
D−B=D
WUCT121
Logic
o
D − A = {x ∈  : 0 ≤ x < 1}
o
A − D = {2, 3}
155
4.8.5.
Definition: Disjoint sets
Let A and B be subsets of a universe U. Then A and B are
said to be disjoint if A ∩ B =  .
Example:
•
The following Venn diagram depicts disjoint sets A
and B:
Note. Disjoint sets have no elements in common.
Exercises:
•
Let U =  , A = {1, 2, 3}, B = {2}, C = {2, 3, 4} and
D = [0, 1] = {x ∈  : 0 ≤ x ≤ 1}. Which pairs of sets from A,
B, C, D are disjoint?
B and D are disjoint, as are C and D.
WUCT121
Logic
156
4.9.
Order of Operations for Set Operators.
The order of operation for set operators is as follows:
1. Evaluate complement first
2. Evaluate ∪ and ∩ second. When both are present,
parenthesis may be needed, otherwise work left to right.
3. Evaluate ⊆ and = third. When both are present,
parenthesis may be needed, otherwise work left to right.
Note: Use of parenthesis will determine order of operations
which over ride the above order.
Examples:
Indicate the order of operations in the following:
•
A∩ B
{{
•
( A∩
{ B)
13
12
•
12
•
A∩( B ∪
{{
{C )
13
2
A⊆
{
{C
{ B∩
13
2
2
WUCT121
Logic
157
Exercises:
Indicate the order of operations in the following:
•
({
A⊆
{C
{ B )∩
•
( A∪
{ B)
13
12
•
12
•
A⊆
{
{C
{ B∪
13
3
2
A{
= B∩
{
{C
1 3
2
2
Notes.
1. ∪ and ∩ are operations on sets, thus ∪ and ∩ can
only be put between two sets.
2. ∨ and ∧ are operations on statements, thus ∨ and ∧
can only be placed between statements.
Example:
•
If A, B, and C are sets then ( A ⊆ B ∧ B ⊆ C ) ⇒ A ⊆ C
is interpreted as (( A ⊆ B ) ∧ ( B ⊆ C )) ⇒ ( A ⊆ C )
•
( A ⊆ B ∧ B ⊆ C ) ≡/ ( A ⊆ ( B ∧ B ) ⊆ C )
WUCT121
Logic
158
4.10.
Set Laws
Let A, B, and C be subsets of a universal set U. That is
A, B, C ∈ (U ) . Then for all sets A, B, and C following set
laws hold:
1.
Commutative Laws:
•
•
•
( A ∪ B ) = ( B ∪ A)
( A ∩ B ) = ( B ∩ A)
( A = B ) = ( B = A)
2.
Associative Laws:
•
•
•
(( A ∪ B ) ∪ C ) = ( A ∪ ( B ∪ C ))
(( A ∩ B ) ∩ C ) = ( A ∩ ( B ∩ C ))
(( A = B ) = C ) = ( A = ( B = C ))
3.
Distributive Laws:
•
•
( A ∪ ( B ∩ C )) = (( A ∪ B ) ∩ ( A ∪ C ))
( A ∩ ( B ∪ C )) = (( A ∩ B ) ∪ ( A ∩ C ))
4.
Double Complement (Involution) Law:
•
( A) = A
5.
De Morgan’s Laws:
•
( A ∪ B) = ( A ∩ B)
•
( A ∩ B) = ( A ∪ B)
WUCT121
Logic
159
6.
Identity Laws:
•
•
( A ∪ ) = A
(A ∩U ) = A
7.
Negation (Complement) Laws:
•
( A ∪ A) = U
•
( A ∩ A) = 
•
U =
•
 =U
8.
Dominance Laws:
•
•
(A ∪U ) =U
( A ∩ ) = 
9.
Idempotent Laws:
•
•
( A ∪ A) = A
( A ∩ A) = A
10. Absorption Laws:
•
•
A ∩ ( A ∪ B) = A
A ∪ ( A ∩ B) = A
11. Set Difference
•
A− B = A∩ B
WUCT121
Logic
160
12. Subset properties of ∪ and ∩
•
•
( A ⊆ ( B ∩ C )) ⇔ (( A ⊆ B ) ∧ ( A ⊆ C ))
(( A ∪ B ) ⊆ C ) ⇔ (( A ⊆ C ) ∧ ( B ⊆ C ))
13. Subset property inclusion of intersection
•
A∩ B ⊆ A
•
A∩ B ⊆ B
14. Subset property inclusion in union
•
A⊆ A∪ B
•
B⊆ A∪ B
15. Transitive Property.
•
•
(( A ⊆ B ) ∧ ( B ⊆ C )) ⇒ ( A ⊆ C )
(( A = B ) ∧ ( B = C )) ⇒ ( A = C )
WUCT121
Logic
161
4.11.
Proving and Disproving Set Statements.
4.11.1.
Proof by Exhaustion
To prove set results for finite sets, the method of
exhaustion is used. That is every element in the set is tested
to ensure it satisfies the condition.
Example:
•
Let A = {1, 2} , B = {1, 2, 3, 4}. Prove A ⊆ A ∪ B .
To prove the statement, we must show every element in A
is in A ∪ B .
Now A ∪ B = {1, 2, 3, 4}
1 ∈ A, 1 ∈ A ∪ B
2 ∈ A, 2 ∈ A ∪ B
Thus all elements in A are in A ∪ B , and so by exhaustion
A ⊆ A ∪ B.
WUCT121
Logic
162
Exercise:
•
Let A = {1, 2} , B = {1, 2, 3, 4}. Prove A = A ∩ B .
To prove the statement, we must show every element in A
is in A ∩ B and every element in A ∩ B is in A.
Now A ∩ B = {1, 2}
1 ∈ A and 1 ∈ A ∩ B
2 ∈ A and 2 ∈ A ∩ B
Thus all elements in A are in A ∩ B and vice versa, and so
by exhaustion A = A ∩ B .
Exercise:
•
Give an example of three sets A, B and C such that
C ⊆ A∩ B.
Let A = {1, 2} , B = {1, 2, 3, 4}, C = {1} .
To prove C ⊆ A ∩ B , we must show every element in C is
in A ∩ B .
Now A ∩ B = {1, 2}
1 ∈ C and 1 ∈ A ∩ B
Thus all elements in C are in A ∩ B and so, for the given
sets A, B and C, C ⊆ A ∩ B .
WUCT121
Logic
163
4.11.2. Disproof by Counterexample.
A set result can be disproven by giving a counterexample.
To find a counterexample often creating a Venn diagram
will be of benefit.
Example:
•
Disprove A ⊆ A ∩ B .
To disprove the statement, we must give a counterexample.
Let A = {1, 2} , B = {3, 4}
Now A ∩ B = 
1 ∈ A, however 1 ∉ A ∩ B = 
Thus by counterexample A⊄A ∩ B .
Exercise:
•
Disprove A ⊆ A − B .
To disprove the statement, we must show a
counterexample.
Let A = {1, 2} , B = {1, 2, 3, 4}. Now A − B = 
1 ∈ A, however 1 ∉ A − B = 
Thus by counterexample A⊄A − B .
WUCT121
Logic
164
4.11.3. Proof by Typical Element.
To prove set results for infinite sets, generalised methods
must be used. The typical element method considers a
particular but arbitrary element of the set and by applying
knows laws, rules and definitions prove the result.
It is the method for proving subset relationships.
So prove that A ⊆ B , we must show that
∀x, ( x ∈ A ⇒ x ∈ B )
Begin by letting x ∈ A , that is, we take x to be a particular
but arbitrary element of A. Using the definitions, we prove
that x ∈ B . As long as we use no special properties of the
element x, we can conclude that (U ) , which is what we
wanted to prove.
This method can be used to prove set equalities. By using
the definition A = B ⇔ ( A ⊆ B ∧ B ⊆ A) and showing
A ⊆ B ∧ B ⊆ A , that is proving ∀x, ( x ∈ A ⇒ x ∈ B ) and
∀x, ( x ∈ B ⇒ x ∈ A) , the result A = B follows. Using this
definition is sometimes called a “double containment”
proof.
WUCT121
Logic
165
Examples:
•
Let U be a set and let A and B be elements of  (U ) .
Prove A ⊆ A ∪ B .
Need to prove ∀x, ( x ∈ A ⇒ x ∈ A ∪ B )
Let x ∈ A , then
x∈ A ⇒ x∈ A∨ x∈B
⇒ x∈ A∪ B
∴ A ⊆ A∪ B
see note
definition of ∪
Note: Appling rules of logic, we know P ⇒ P ∨ Q is a
tautology. Let P( x ) : x ∈ A, Q( x ) : x ∈ B . Thus
x ∈ A ⇒ x ∈ A ∨ x ∈ B is a tautology in the proof above.
•
Let U be a set and let A and B be elements of  (U ) .
Prove A ⊆ B ⇔ A ∪ B = B .
Need to prove two parts:
1. A ⊆ B ⇒ A ∪ B = B
2. A ∪ B = B ⇒ A ⊆ B
WUCT121
Logic
166
•
Proof of 1:
KNOW: A ⊆ B , that is ∀x, ( x ∈ A ⇒ x ∈ B )
K (1)
PROVE: A ∪ B = B .
Need to prove two parts:
i. A ∪ B ⊆ B
ii. B ⊆ A ∪ B
Proof of i.:
Let x ∈ A ∪ B then
x∈ A∪ B ⇒
⇒
⇒
∴ A∪ B ⊆ B
Proof of ii.:
x∈ A∨ x∈B
x∈B∨ x∈B
x∈B∨ x∈B
definition of ∪
by (1)
Logic rule P ∨ P ≡ P
Let x ∈ B then
x∈B ⇒ x∈ A∨ x∈B
⇒ x∈ A∪ B
∴B ⊆ A∪ B
see previous example
definition of ∪
Since A ∪ B ⊆ B and B ⊆ A ∪ B , A ∪ B = B
Thus A ⊆ B ⇒ A ∪ B = B .
WUCT121
Logic
167
Proof of 2:
KNOW: A ∪ B = B , that is
∀x, ( x ∈ A ∪ B ⇔ x ∈ B )
K( 2)
PROVE: A ⊆ B .
Let x ∈ A then
x∈ A ⇒
⇒
⇒
∴A⊆ B
x∈ A∨ x∈B
x∈ A∪ B
x∈B
see previous example
definition of ∪
by (2)
Thus A ∪ B = B ⇒ A ⊆ B
Since A ⊆ B ⇒ A ∪ B = B and A ∪ B = B ⇒ A ⊆ B it is
proven that A ⊆ B ⇔ A ∪ B = B .
Exercise:
•
Let U be a set and let A and B be elements of  (U ) .
Prove A ∩ B ⊆ A .
that is, prove ∀x, ( x ∈ A ∩ B ⇒ x ∈ A)
Let x ∈ A ∩ B , then
x∈ A∩ B ⇒
⇒
∴ A∩ B ⊆ A
WUCT121
x∈ A∧ x∈B
x∈A
Logic
definition of ∩
168
4.11.4. Proof by Equivalence of Statements.
If A can be written as A = { x ∈ U : P( x )} and
B = { x ∈ U : Q ( x )}, the equality of specification theorem to
show that A = B by showing that P( x ) ≡ Q( x ) , that is, by
showing that P( x ) ⇔ Q ( x ) is a tautology.
Examples:
•
Let A = { x ∈  : x 2 ≤ 1} and B = { x ∈  : −1 ≤ x ≤ 1} .
Prove A = B
Let P( x ) : x 2 ≤ 1 and Q( x ) : −1 ≤ x ≤ 1 . Now
x 2 ≤ 1 ⇔ −1 ≤ x ≤ 1
∴ P( x ) ⇔ Q( x )
∴A= B
•
Let U be a set and let A and B be elements of  (U ) .
Prove that ( A ∩ B ) = A ∪ B .
We need to show that the statements defining the sets
( A ∩ B ) and A ∪ B are equivalent.
( A ∩ B ) = { x ∈ U :~ ( x ∈ A ∩ B )} definition of A
A ∪ B = { x ∈ U : x ∈ A ∪ B}
WUCT121
Logic
169
Let P( x ) :~ ( x ∈ A ∩ B ) , and Q( x ) : x ∈ A ∪ B
x∈ A∪ B ≡ x∈ A ∨ x∈B
by definition of ∪
≡ ~ ( x ∈ A) ∨ ~ ( x ∈ B ) definition of A
≡ ~ (x ∈ A ∧ x ∈ B )
≡ ~ (x ∈ A ∩ B )
by Logic De Morgan' s
by definition of ∩
∴ Q( x ) ≡ P( x )
∴ (A ∩ B) = A ∪ B
Exercise:
•
Let U be a set and let A and B be elements of  (U ) .
Prove that ( A ∪ B ) = A ∩ B .
We need to show that the statements defining the sets
( A ∪ B ) and A ∩ B are equivalent.
( A ∪ B ) = { x ∈ U :~ ( x ∈ A ∪ B )} axiom of specification
A ∩ B = { x ∈ U : x ∈ A ∩ B} axiom of specification
Let P( x ) :~ ( x ∈ A ∪ B ) , and Q( x ) : x ∈ A ∩ B
x∈ A∩ B ≡ x∈ A ∧ x∈B
by definition of ∩
≡ ~ ( x ∈ A) ∧ ~ ( x ∈ B ) definition of A
≡ ~ (x ∈ A ∨ x ∈ B )
≡ ~ (x ∈ A ∪ B )
by Logic De Morgan' s
by definition of ∩
∴ Q( x ) ≡ P( x )
∴ (A ∪ B) = A ∩ B
WUCT121
Logic
170
•
Let U be a set and let A, B and C be elements of P(U).
Prove that ( A − B ) − C = ( A − C ) − B .
Let P( x ) : x ∈ ( A − B ) − C , and Q( x ) : x ∈ ( A − C ) − B
x ∈ (A − B) − C ⇔ (x ∈ A − B) ∧ x ∉ C
⇔ (x ∈ A ∧ x ∉ B ) ∧ x ∉ C
⇔ x ∈ A ∧ (x ∉ B ∧ x ∉ C
)
⇔ x ∈ A ∧ (x ∉ C ∧ x ∉ B )
⇔ (x ∈ A ∧ x ∉ C ) ∧ x ∉ B
⇔ (x ∈ A − C ) ∧ x ∉ B
⇔ x ∈ (A − C ) − B
∴ P( x ) ⇔ Q( x )
∴ (A − B) − C = (A − C ) − B
•
Let U be a set and let X and Y be elements of (U ) .
Prove that X − Y = X ∩ Y .
Let P( x ) : x ∈ X − Y , and Q( x ) : x ∈ X ∩ Y
x ∈ X − Y ≡ x ∈ X ∧ x ∉ Y Definition of set difference
≡ x ∈ X ∧ x ∈Y
Definition of complement
≡ x∈ X ∩Y
Definition of intersection
∴ P( x ) ≡ Q( x )
∴X −Y = X ∩Y
WUCT121
Logic
171
4.11.5. Proof by Set Laws.
Set equalities can be proven by using known set laws
Examples:
•
Let U be a set and let A, B and C be elements of P(U).
Prove ( A − B ) − C = ( A − C ) − B
(A − B) − C = (A − B) ∩ C
set difference
= (A ∩ B ) ∩ C
set difference
= A ∩ (B ∩ C )
associativity
= A ∩ (C ∩ B )
commutativity
= (A ∩ C ) ∩ B
associativity
= (A − C ) ∩ B
= (A − C ) − B
WUCT121
set difference
set difference
Logic
172
4.11.6. Further Examples.
Examples:
•
Let U be a set and let A, B and B be elements of P(U).
Using the following:
(i) A ⊆ B ⇔ A ∪ B = B ,
(ii): A ⊆ B ⇔ A ∩ B = A ,
(iii): A ∪ B = ( A ∩ B ) .
Prove that A ⊆ B ⇔ B ⊆ A .
Proof:
A ⊆ B ⇔ A∩ B = A
•
by part (ii)
⇔ (A ∩ B) = A
by taking complements
⇔ A∪ B = A
by part (iii)
⇔B⊆A
by part (i)
Let U be a set and let A, B and C be elements of P(U).
Disprove that A − (B − C ) = ( A − B ) − C .
Let A = {1, 2, 3}, B = {2, 3}, C = {3}.
A − (B − C ) = A − {2} = {1, 3}
( A − B ) − C = {1} − C = {1} ≠ A − (B − C )
WUCT121
Logic
173
•
Let U be a set and let X and Y be elements of  (U ) .
Use a typical element argument to prove X − Y = X ∩ Y .
Need to prove two parts:
1. X − Y ⊆ X ∩ Y
2. X ∩ Y ⊆ X − Y
Proof of 1:Let x ∈ X − Y be a typical element.
X − Y ≡ {x ∈ U : x ∈ X − Y }
⇒ {x ∈ U : x ∈ X ∧ x ∉ Y }
Axiom of Specification
Def of set difference
⇒ {x ∈ U : x ∈ X ∧ x ∈ Y }
Def of complement
⇒ {x ∈ U : x ∈ X ∩ Y }
Def of intersection
∴ ∀x ∈ U (x ∈ X − Y ⇒ x ∈ X ∩ Y )
∴
X −Y ⊆ X ∩Y
Proof 2: Let x ∈ X ∩ Y be a typical element.
X ∩ Y ≡ {x ∈ U : x ∈ X ∩ Y }
⇒ {x ∈ U : x ∈ X ∧ x ∈ Y }
Axiom of Specification
⇒ {x ∈ U : x ∈ X ∧ x ∉ Y }
Def of intersection
Def of complement
⇒ {x ∈ U : x ∈ X − Y }
Def of set difference
∴ ∀x ∈ U (x ∈ X ∩ Y ⇒ x ∈ X − Y )
∴
X ∩Y ⊆ X −Y
∴ ∀x ∈ U (x ∈ X − Y ⇔ x ∈ X ∩ Y ),
i.e. X − Y ⊆ X ∩ Y ∧ X ∩ Y ⊆ X − Y ,
∴ X −Y = X ∩Y
WUCT121
Axiom of extent
Logic
174
Section 5.
Relations and Functions
5.1.
Cartesian Product
5.1.1.
Definition: Ordered Pair
Let A and B be sets and let a ∈ A and b ∈ B .
An ordered pair ( a , b ) is a pair of elements with the
property that:
( a , b ) = ( c, d ) ⇔ ( a = c ) ∧ (b = d ) .
Notes:
∗
A pair set {a , b} is NOT an ordered pair, since
{a , b} = {b, a} .
∗
It should be clear from the context when ( a , b ) is an
ordered pair, and when ( a , b ) = { x ∈  : a < x < b} is an
open interval of real numbers.
WUCT121
Logic
175
Examples:
•
Points in the plane  2 are represented as ordered
pairs.
y
3
2
(1, 2)
1
(2, 1)
x
-4
-3
-2
-1
0
1
2
3
4
(-2, -1) -1
-2
(-1, -2)
-3
From the graph it can be seen (1, 2) ≠ (2,1) and
( −1, − 2) ≠ ( −2, − 1) .
•
Complex numbers a + ib where i = − 1 and a, b ∈  ,
are ordered pairs in the sense that,
a + ib = c + id ⇔ ( a = c ) ∧ (b = d ) .
WUCT121
Logic
176
5.1.2.
Definition: Cartesian Product
Let A and B be sets, then the Cartesian Product of A and B,
denoted A × B , is defined by
A × B = {(a , b ) : a ∈ A ∧ b ∈ B}.
Example:
•
 ×  = {( x, y ) : x ∈  : ∧ y ∈ }.
Sketch a graph of  ×  , otherwise known as  2 .
 2 is the usual Cartesian plane with the usual graph.
y
4
2
x
-6
-4
-2
0
2
4
6
-2
-4
WUCT121
Logic
177
Exercises:
•
Let A = {3} and B = {2, 3} . Write down A × B . Sketch
a graph of A × B in  2 .
A × B = {(3, 2), (3, 3)}
y
6
4
(3, 3)
2
(3, 2)
x
-6
-4
-2
0
2
4
6
-2
-4
-6
•
Let C = { x ∈  : −1 ≤ x ≤ 1} and D = {1, 2} . Write
down C × D . Sketch a graph of C × D in  2 .
C × D = {( x, y ) : −1 ≤ x ≤ 1 ∧ y ∈ (1, 2)}
y
4
2
x
-4
-3
-2
-1
0
1
2
3
4
-2
-4
WUCT121
Logic
178
5.2.
Relations
5.2.1.
Definition: Binary Relation
Let A and B be sets. We say that R is a (binary) relation
from A to B if R ⊆ A × B .
Notes:
∗
If R ⊆ A × A , we say that R is a relation on A.
∗
If ( a , b ) ∈ R , we will frequently write aRb and say
that “a is in the relation R to b”.
∗
Every relation is a subset of a Cartesian product
Examples:
•
Let A be the set of all male human beings and let B be
the set of all human beings. The relation T from A to B is
given by T = {( x, y ) : x is the father of y}.
•
W = {(1, 2 ), (2, 1), (5, π )}.
Note: W cannot be defined by a “rule”. Sometimes relations
are simply defined by a listing of elements.
WUCT121
Logic
179
•
Let R be the relation on  , defined by
R = {( x, y ) : x 2 + y 2 = 1} . Sketch the graph of R in  2
y
x
-2
-1
0
1
2
Exercise:
•
Let S be the relation on  , defined by
S = {( x, y ) : 4 x + y = 4} . Sketch the graph of S in  2
y
4
2
x
-2
WUCT121
-1
0
Logic
1
2
180
Example:
Consider the relation R on  given by R = {( x, y ) : x = y}
•
Sketch the graph of R in  2
y
4
2
x
-4
-3
-2
-1
0
1
2
3
4
-2
-4
•
•
Are the following true or false?
o 1R1
True
o 1R2.2
False
o ( −3, 3) ∈ R
False
If aR100, what is the value of a?
100
Note. The relation R in this example is called the identity
relation on  and is usually written R = {( x, x ) : x ∈ }.
WUCT121
Logic
181
Exercises:
•
Let X = {0, 1, 2, 3}, and let the relation R on X be
given by R = {( x, y ) : ∃z ∈ , x + z = y}.
o What is an easier way of expressing the relation R?
R = {( x, y ) : x, y ∈ X ∧ x < y}
o List all the elements of R.
R = {(0, 1), (0, 2 ), (0, 3), (1, 2 ), (1, 3), (2, 3)}
o Sketch X × X , and circle the elements of R.
y
2
x
0
•
1
2
3
4
Let S be the relation on  − {0} given by
S = {( x, y ) : ∃z ∈ , xz = y}
o Describe the relation S.
x is a factor of y, or x | y .
WUCT121
Logic
182
o Are the following true or false?
•
ƒ
(2,−4) ∈ S
ƒ
−3S0
ƒ
(3, 5) ∈ S
True, since 2 | −4 .
False, since 0 ∉  − {0}.
False, since 3F5.
Let R be the relation on  given by
R = {( x, y ) : y = x 2 } and let S be the relation on  given
by S = {( x, y ) : y = x 2 }.
o Sketch each relation. What difference does the
input” set make to the elements in each relation.
y
8
6
4
2
x
-4
-3
-2
-1
0
1
2
3
4
R is a set of isolated points.
WUCT121
Logic
183
y
3
2
1
x
-3
-2
-1
0
1
2
3
-1
S is a continuous curve.
Note. Care must be taken when writing relations. As can be
seen from this example, it must be very clear the sets a
relation is from and to.
•
Let A = {0, 1} and B = {− 1, 0, 1}. Let two relations
from A to B be given by R1 = {(0,−1), (1,−1), (1, 0 )}, and
R2 = {(0, 0), (1,1), (1, − 1)}.
Determine:
o
R 1 ∩ R 2 = {(1,−1)}.
o
R 1 ∪ R 2 = {(0,−1), (0, 0 ), (1,−1), (1, 0 ), (1, 1)}
WUCT121
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184
•
Let R3 and R4 be relations on  defined by
R3 = {( x, y ) : x = y}, and R4 = {( x, y ) : x = − y}.
Determine:
o
R3 ∪ R4 = {( x, y ) : x = y ∨ x = − y}
= {( x, y ) : x = ± y}
= {( x, y ) : x = y }
o
WUCT121
R3 ∩ R4 = {(0, 0 )}
Logic
185
5.2.2.
Definition: Domain
Let R be a relation from A to B.
Then the domain of R, denoted Dom R, is given by
Dom R = {x : ∃y , xRy}.
Notes:
∗
Let R be a relation from A to B, then Dom R ⊆ A .
∗
Dom R is the set of all first elements in the ordered
pairs that belong to R.
5.2.3.
Definition: Range
Let R be a relation from A to B.
Then the range of R, denoted Range R, is given by
Range R = {y : ∃x, xRy}.
Notes:
∗
Let R be a relation from A to B, then Range R ⊆ B .
∗
Range R is the set of all second elements in the
ordered pairs that belong to R.
WUCT121
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186
Examples:
•
Let A = {0, 1, 2, 3} and let R1 be the relation on A
given by R1 = {(0, 0), (0,1), (0, 2), (3, 0)}.
Determine:
o
Dom R1 = {0, 3}
o
Range R1 = {0, 1, 2}
Exercises:
•
Let R2 be the relation on  given by
R2 = {( x, y ) : xy ≠ 0}.
Determine:
•
o
Dom R2 =  − {0}
o
Range R2 =  − {0}.
Let R3 be the relation from  to  given by
{
}
R3 = ( x, y ) : x ≠ 0 ∧ y = 1 .
x
Determine:
o
Dom R3 =  − {0}
o
Range R3 = 1 : n ∈  ∧ n ≠ 0
WUCT121
{
}
n
Logic
187
5.2.4.
Definition: Inverse Relations
Let R be a relation from A to B. The inverse relation,
denoted R −1 , from B to A is defined as
R −1 = {( y , x ) : ( x, y ) ∈ R}.
Notes:
∗
For a relation R from A to B, the inverse relation
R −1 can be defined by interchanging the elements of all the
ordered pairs of R. This turns out to be easier for a finite
(listed) relation than an infinite (given by formula) relation.
Dom R −1 = Range R ⊆ B and
∗
Range R −1 = Dom R ⊆ A .
Examples:
•
Define a relation R on  as R = {( x, y ) : y = 2 x}.
o Write down 3 elements of R.
(1, 2 ), (2, 4 ), (3, 6)
o Write down 3 elements of R −1
(2, 1), (4, 2), (6, 3)
WUCT121
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188
o Sketch a graph of R and R −1 on coordinate axis,
circle elements of R −1 .
y
6
5
4
3
2
1
x
-1
0
1
2
3
4
5
6
7
-1
o Write down a simple definition for R −1 .
R −1 = {( y , x ) : y = 2 x}
= {( x, y ) : x = 2 y}
{
= ( x, y ) : y = 12 x
}
Exercise:
•
Let S be the identity relation on the set of reals. What
is S −1?
S = {( x, x ) : x ∈ }
S −1 = {( x, x ) : x ∈ }
=S
WUCT121
Logic
189
5.2.5.
Directed Graph of a Relation
When a relation R is defined on a set A, we can represent it
with a directed graph. This is a graph in which an arrow is
drawn from each point in A to each related point.
∀x, y ∈ A , there is an arrow from x to y ⇔ xRy ,
⇔ ( x, y ) ∈ R
If a point is related to itself, a loop is drawn that extends
out from the point and goes back to it.
Example:
•
Let A = {0, 1, 2, 3} and let R1 be the relation on A
given by R1 = {(0, 0), (0,1), (0, 2), (3, 0)}. Draw the directed
graph of R1 .
0
1
2
3
WUCT121
Logic
190
Exercise:
•
Let A = {0, 1, 2, 3} and let R2 be the relation on A
given by R2 = {(0, 0), (1, 2), ( 2, 2)}.
Draw the directed graph of R2 .
1
0
2
WUCT121
Logic
191
5.2.6.
Properties of Relations
Let R be a relation on the set A.
Reflexivity:
R is reflexive on A if and only if ∀x ∈ A, ( x, x ) ∈ R .
Example:
•
Let R1 be the relation on  defined by
R1 = {( x, y ) : x is a factor of y}.
For each x ∈  , we know that x is a factor of itself. Thus,
( x, x ) ∈ R1 , and so R1 is reflexive
Symmetry:
R is symmetric on A if and only if
∀x, y ∈ A, (( x, y ) ∈ R ⇒ ( y , x ) ∈ R ) .
Example:
•
Let R2 be the identity relation on  .
For x, y ∈  , if x = y , then y = x , that is, if ( x, y ) ∈ R2 ,
then ( y , x ) ∈ R2 and so R2 is symmetric
WUCT121
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192
Transitivity:
R is transitive on A if and only if
∀x, y , z ∈ A, ((( x, y ) ∈ R ∧ ( y , z ) ∈ R ) ⇒ ( x, z ) ∈ R ) .
Example:
•
Let R3 be the relation on  defined by
R3 = {( x, y ) : x < y}.
For x, y , z ∈  , if x < y and y < z , then x < z , that is, if
( x, y ) ∈ R3 and ( y, z ) ∈ R3 , then ( x, z ) ∈ R3and so R3 is
transitive.
Notes:
∗
A relation R on a set A is reflexive if each element
in A is in relation to itself.
∗
A relation R on a set A is symmetric if you can
“swap” the ordered pairs around and still get elements of R.
∗
A relation R on a set A is transitive if pairs of
elements are “related via” a third element (x and z related
via y).
WUCT121
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193
Exercises:
Which of the three properties do the following relations
satisfy? Give reasons why or why not.
•
R1 on  , given by R1 = {( x, y ) : x | y}.
∀x ∈ , x | x. ∴ ( x, x ) ∈ R1 Thus R1 is reflexive.
Consider ( 2, 4) ∈ R1 , since 2 | 4 , however ( 4, 2) ∉ R1 , as
4 F 2, so R1 is not symmetric.
∀x, y , z ∈ , x | y ∧ y | z ⇒ x | z.
∴ ( x, y ) ∈ R1 ∧ ∴ ( y , z ) ∈ R1 ⇒∴ ( x, z ) ∈ R1, thus R1 is
transitive.
•
R2 , the identity relation on 
Reflexive:
Yes
Symmetric:
Yes
Transitive:
Yes
•
R3 on  given by R3 = {( x, y ) : x < y}
Reflexive:
No
Symmetric:
No
Transitive:
Yes
WUCT121
Logic
194
•
R4 on  given by R4 = {( x, y ) : y = x 2 }
Reflexive:
No
Symmetric:
No
Transitive:
No
•
R5 on A where A is the set of all people.
R5 = {( x, y ) : x is in the family of y}
Reflexive:
Yes
Symmetric:
Yes
Transitive:
Yes
•
R6 on A where A is the set of all people.
R6 = {( x, y ) : x loves y}
Reflexive:
?
Symmetric:
?
Transitive:
?
WUCT121
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195
5.2.7.
Definition: Equivalence Relation
Let R be a relation on the set A. R is an equivalence relation
on A if and only if R is reflexive, symmetric and transitive
on A.
Example:
From the previous exercises, R2 and R5 are equivalence
relations.
Notes:
∗
If R is a relation on a set A, you must be able to
either prove or disprove the statement
“R is an equivalence relation.”
∗
To prove a relation R is an equivalence relation, it is
necessary to prove all three properties hold.
∗
To disprove that a relation R is an equivalence
relation, it is sufficient to show that one of the three
properties does not hold. This can usually be shown by
counterexample.
WUCT121
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196
Example:
•
Let R1 be the identity relation on  .
Prove or disprove R1 is an equivalence relation.
Proof:
Reflexive:
∀a ∈ , a = a , that is (a , a ) ∈ R1 . Thus R1 is reflexive.
Symmetric:
∀a , b ∈ , if a = b, then b = a , that is,
(a , b ) ∈ R1 ⇒ (b, a ) ∈ R1. Thus R1 is symmetric.
Transitive:
∀a , b, c ∈ , if a = b and b = c, then a = c , that is
(( a , b ) ∈ R1 ∧ (b, c ) ∈ R1 ) ⇒ ( a , c ) ∈ R1 . Thus R1 is
transitive.
Since R1 is reflexive, symmetric and transitive, R1 is an
equivalence relation.
WUCT121
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197
Exercises:
•
Let n ∈  . Consider the relation R2 on  given by
R2 = {(a , b ) : a ≡ b(mod n )}.
Prove or disprove R2 is an equivalence relation.
Recall: a ≡ b(mod n ) ⇔ ∃k ∈ , a − b = nk .
Proof:
Reflexive:
∀a ∈ , a − a = 0 = n × 0 , which implies that
a ≡ a(mod n ) , ∴ (a , a ) ∈ R2 Thus R2 is reflexive.
Symmetric:
∀a , b ∈ , if a ≡ b(mod n ), then a − b = nk
∴ b − a = − nk = n(− k ), giving b ≡ a (mod n ). Thus
(a , b ) ∈ R2 ⇒ (b, a ) ∈ R2 . So R2 is symmetric.
Transitive:
∀a, b, c ∈ , if a ≡ b(mod n ) and b ≡ c(mod n ), then
( a − b = nk ) ∧ (b − c = nl ) , ∴ a − c = n( k + l ) = np ,
so a ≡ c (mod n ) . That is
(( a , b ) ∈ R2 ∧ (b, c ) ∈ R2 ) ⇒ ( a , c ) ∈ R2 . Thus R2 is
transitive.
Since R2 is reflexive, symmetric and transitive, R2 is an
equivalence relation.
WUCT121
Logic
198
•
Let R3 be the relation on  given by
R3 = {(a , b ) : ab ≠ 0}.
o Prove or disprove R3 is an equivalence relation.
Disprove:
Reflexive:
We must show ∀a ∈ , a × a ≠ 0 .
However 0 ∈ , and 0 × 0 = 0 .
Thus R3 = {(a , b ) : ab ≠ 0} and so R3 is not
reflexive.
Therefore R3 is not an equivalence relation.
o Is R3 symmetric or transitive?
ab ≠ 0 ⇒ ba ≠ 0 ∴ symmetric
ab ≠ 0 ∧ bc ≠ 0 ⇒ ac ≠ 0 ∴ transitive
o How can we adjust the relation so it becomes an
equivalence relation?
R3 on  − {0}.
WUCT121
Logic
199
•
Let A = {0, 1, 2} and let R be the relation on A given by
R = {(0, 0 ), (1, 1), (2, 2 ), (0, 1), (1, 0 )}.
Prove or disprove R is an equivalence relation on A.
Reflexive:
For a = 0 :
(0, 0 ) ∈ R . For a = 1 : (1, 1) ∈ R .
For a = 2 :
(2, 2 ) ∈ R .
So, ∀a ∈ A, (a , a ) ∈ R . Thus R is reflexive.
Symmetric:
For (0, 0 ), (1, 1) and (2, 2 ) symmetry obviously holds.
(0, 1) ∈ R ⇒ (1, 0 ) ∈ R , (1, 0 ) ∈ R ⇒ (1, 0 ) ∈ R ,
So, ∀(a , b ) ∈ R ⇒ (b, a ) ∈ R , thus R is symmetric.
Transitive:
(0, 0 ), (0, 1) ∈ R ⇒ (0, 1) ∈ R , (1, 1), (1, 0 ) ∈ R ⇒ (1, 0 ) ∈ R ,
(0, 1), (1, 1) ∈ R ⇒ (0, 1) ∈ R , (0, 1), (1, 0 ) ∈ R ⇒ (0, 0 ) ∈ R
(1, 0 ), (0, 1) ∈ R ⇒ (1, 1) ∈ R , (1, 0 ), (0, 0 ) ∈ R ⇒ (1, 0 ) ∈ R ,
So ∀(a , b ) ∧ (b, c ) ∈ R ⇒ (a , c ) ∈ R , thus R is transitive.
Therefore, since R is reflexive, symmetric and transitive, R
is an equivalence relation.
WUCT121
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200
5.2.8.
Directed Graphs of Equivalence Relations
The directed graph of an equivalence relation on A has the
following properties:
∗
Each point of the graph has an arrow looping
around from it back to itself. (Reflexivity)
∗
In each case where there is an arrow going from one
point to a second, there is an arrow going from the second
point back to the first. (Symmetry)
∗
In each case where there is an arrow going from one
point to a second and from a second point to a third, there is
an arrow going from the first point to the third.
(Transitivity)
WUCT121
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201
Example:
•
Let A = {0, 1, 2} and let R be the relation on A given
by R = {(0, 0 ), (1, 1), (2, 2 ), (0, 1), (1, 0 )}.
Draw the directed graph for R.
Previously R was shown to be an equivalence relation on A.
The directed graph is then :
0
1
2
WUCT121
Logic
202
Exercise:
Let A = {2, 3, 4, 6, 7, 9}, and define a relation R on A
•
by R = {(a , b ) : a ≡ b(mod 3)}.
Draw the directed graph for R.
Solution:
R = {(2, 2 ), (3, 3), (4, 4 ), (6, 6 ), (7, 7 ), (9, 9 ), (3, 6 ),
(6, 3), (3, 9 ), (9, 3), (6, 9 ), (9, 6 ), (4, 7 ), (7, 4)}
It can be shown that R is an equivalence relation, and thus
the directed graph is:
4
3
2
9
7
6
WUCT121
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203
5.2.9.
Equivalence Class
The fundamental property of equivalence relations which
makes them important is that each one determines a
partition of the set A into a family of disjoint sets.
Definition:
Let R be an equivalence relation on the set A. Then for each
a ∈ A , we define the equivalence class of a as
class( a ) = {b ∈ A : (a , b ) ∈ R}.
Example:
•
Let A = {0, 1, 2} and let R be the relation on A given
by R = {(0, 0 ), (1, 1), (2, 2 ), (0, 1), (1, 0 )}. For each element in
A, we define equivalence classes as follows:
class(0 ) = {b ∈ A : (0, b ) ∈ R} = {0, 1}
class(1) = {b ∈ A : (1, b ) ∈ R} = {1, 0} = class(0 )
class(2 ) = {b ∈ A : (2, b ) ∈ R} = {2}
WUCT121
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204
Exercises:
•
Let R1 be the identity relation on  . Write down the
following equivalence classes:
o class(1) = {1}
o class(π ) = {π }
( ) {}
o class 12 = 12
o For any x ∈  , class( x ) = {x}.
•
Consider the relation R2 on  given by
R2 = {(a , b ) : a ≡ b(mod 3)}.
What kind of numbers are in class(2) (otherwise written as
[2])?
class(2 ) = {K − 4,−1, 2, 5, 8, 11, K ,3k + 2,K}.
•
Let R3 on A, the set of all people, be given by
R3 = {(a , b ) : a is in the family of b}.
Who is in your equivalence class?
WUCT121
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205
5.3.
Functions
5.3.1.
Definition
If F is a relation from A to B, then we say F is a function
from A to B, if and only if the domain of F is all of A and
for each element x ∈ A , there is only one value y ∈ B such
that ( x, y ) ∈ F .
Note:
A relation from A to B becomes a function if the domain is
all of A and if every first element is related to only one
second element. This last property is sometimes known as
the vertical line test.
Examples:
•
Is R1 on  , R1 = {( x, y ) : y = x 2 } a function?
y
3
2
1
x
-3
-2
-1
0
1
2
3
-1
Dom R1 =  , vertical line test holds, thus R1 is a function.
WUCT121
Logic
206
•
Is R2 on  , R2 = {( x, y ) : x = y 2 } a function?
y
3
2
1
x
-1
0
1
2
3
-1
-2
-3
Dom R2 =  , vertical line test fails, thus R2 is not a
function.
Exercises:
•
Is R3 on A = {x ∈  : x ≥ 0}, R3 = {( x, y ) : x = y 2 }
a function?
y
3
2
1
x
-1
0
1
2
3
-1
Dom R3 = A , vertical line test holds, thus R3 is a function.
WUCT121
Logic
207
•
Is R4 on  , R4 = {( x, y ) : y = x } a function?
y
3
2
1
x
-1
0
1
2
3
-1
Dom R4 = [0, ∞ ) ≠  , thus R4 is not a function.
Notes:
When determining if a relation is a function:
∗
Infinite Case: Is the domain is the entire set A.
Finite Case: Is every element of A a first element in
an ordered pair?
∗
Infinite Case: Graph the relation and apply the
vertical line test.
Finite Case: List the ordered pairs and check each
first element appears only once.
WUCT121
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208
Exercises:
Let A = {2, 4, 6} and let B = {1, 3, 5}. Which of the
•
following relations from A to B are functions?
o
R1 = {( x, y ) : x + 1 = y} = {(2, 3), (4, 5)}
Dom R1 = {2, 4} ≠ A .
Thus R1 is not a function.
o
R2 = {(2, 5), (4, 1), (4, 5), (6, 5)}.
Dom R2 = {2, 4, 6} = A .
However, (4, 1) ∈ R2 ∧ (4, 5) ∈ R2
Thus R2 is not a function.
o
R3 = {(2, 5), (4, 1), (6, 5)}.
Dom R3 = A , and each first element only appears
once.
Thus R3 is a function.
WUCT121
Logic
209
•
Which of the following are functions?
o
F1 , the identity relation on A = {1, 5, 10}.
F1 = {(1,1), (5, 5), (10,10 )}
Dom F1 = A , and each first element only appears
once.
Thus F1 is a function
o
F2 on  , F2 = {( x, y ) : y = 1}.
y
2
1
x
-3
-2
-1
0
1
2
3
-1
Dom F2 =  , vertical line test holds, thus F2 is a
function.
WUCT121
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210
o
F3 on  , F3 = {( x, y ) : y = x + 1}.
y
4
3
2
1
x
-3
-2
-1
0
1
2
3
-1
-2
Dom F3 =  , vertical line test holds, thus F3 is a
function.
WUCT121
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211
5.3.2.
One-to-one
Let F be a function from A to B. F is one-to-one if and only
if ∀x1 , x 2 ∈ A, (( x1 , y ) = ( x 2 , y ) ⇒ x1 = x 2 ).
For one-to-one functions, any given element from the
Range is related to only one element from the Domain.
That is each element in both the domain and the range is
related to just one element.
Notes:
∗
Only functions can be one-to-one.
∗
It is often the case that if a function F is one-to-one,
it satisfies a horizontal line test.
∗
To establish if a relation is one-to-one show if the
relation is, in fact, a function. Then determine if it is
one-to-one.
∗
To show a function is one-to-one, show each
element in the range occurs once in an ordered pair.
∗
To show a function is not one-to-one, give a
counterexample, that is, find an element of the
range that is related to two elements in the domain.
WUCT121
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212
Examples:
•
Consider the relation F1 on  given by
F1 = {( x, y ) : y = x 2 } . Is F1 a one-to-one function?
y
3
2
1
x
-3
-2
-1
0
1
2
3
-1
Dom F1 =  , vertical line test holds, thus F1 is a function.
Horizontal line test fails: ( −1,1) ∈ F1 ∧ (1,1) ∈ F , therefore
F1 is not a one-to-one function
WUCT121
Logic
213
•
Consider the relation F2 on  + = {x ∈  : x ≥ 0}
given by F2 = {( x, y ) : y = x 2 }. Is F2 a one-to-one
function?
y
4
3
2
1
x
-3
-2
-1
0
1
2
3
-1
Dom F2 =  + , vertical line test holds, thus F2 is a
function. Horizontal line test holds, therefore F2 is a oneto-one function
•
Let X = {0, 1, 2, 3}.
Consider the function F3 from  ( X ) to  given by
F3 = {( A, n ) : n is the number of elements in the set A}.
Is F3 a one-to-one function?
Consider A = {0, 1}∈ ( X ) and B = {1, 2} ∈ ( X ) .
Then ( A, 2 ) ∈ F3 and (B, 2 ) ∈ F3 , that is, 2 ∈  appears
twice.
Thus, F3 is not a one-to-one function.
WUCT121
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214
Exercises:
Which of the following relations are one-to-one functions?
•
F1 on A = {1, 2, 3}, F1 = {(1, 2 ), (2, 3), (3, 1)}.
5 y
4
3
2
1
x
0
1
2
3
-1
-2
Dom F1 = A , vertical line test holds, thus F1 is a function.
Horizontal line test holds, therefore F1 is a one-to-one
function.
•
F2 on A = {1, 2, 3}, F2 = {(1, 2 ), (2, 1), (3, 1)}.
5 y
4
3
2
1
x
0
1
2
3
-1
-2
Dom F2 = A , vertical line test holds, thus F1 is a function.
Horizontal line test fails: ( 2, 1) ∈ F2 ∧ (3, 1) ∈ F2 , therefore
F2 is not a one-to-one function.
WUCT121
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215
•
F3 on  , F3 = {( x, y ) : y = 2 x}.
y
4
3
2
1
x
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
Dom F3 =  , vertical line test holds, thus F3 is a function.
Horizontal line test holds, therefore F3 is a one-to-one
function
•
{
F4 from  − {0} to  , F4 = ( x, y ) : y =
}
x2 −1 .
y
4
3
2
1
x
-3
-2
-1
0
1
2
3
-1
Dom F4 =  − {0}, vertical line test holds, thus F4 is a
function.
Horizontal line test fails: (1, 0) ∈ F4 ∧ ( −1, 0) ∈ F4 ,
therefore F4 is not a one-to-one function.
WUCT121
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216
5.3.3.
Onto
Let F be a function from A to B. F is onto if and only if
Range F = B , that is,
∀y ∈ B, ∃x ∈ A, ( x, y ) ∈ F .
For a function to be onto, every given element from the
range must be related to at least one element from the
domain.
Notes:
∗
Only functions can be onto.
∗
To establish if a relation is onto show if the relation
is, in fact, a function. Then determine if it is onto.
∗
To show a function F from A to B is onto, show that
Range F = B , that is every element in the range
occurs at least once in an ordered pair.
∗
To show a function is not onto, give a
counterexample, that is, find an element of the
range that is not related to an element in the
domain.
WUCT121
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217
Example:
•
Consider the relation F1 from
A = { x ∈  : −1 ≤ x ≤ 1} to  given by
F1 = {( x, y ) : y = 1 − x 2 }. Is F1 an onto function?
y
x
-2
-1
0
1
2
Dom F1 = { x ∈  : −1 ≤ x ≤ 1} = A , vertical line test holds,
thus F1 is a function.
Range F1 = { y ∈  : 0 ≤ y ≤ 1} ≠  , thus F1 is not an onto
function.
By defining the function to F2 from
A = { x ∈  : −1 ≤ x ≤ 1} to B = { x ∈  : 0 ≤ x ≤ 1} given by
F2 = {( x, y ) : y = 1 − x 2 }.
Now Range F2 = { y ∈  : 0 ≤ y ≤ 1} = B , thus the function
F2 is an onto function
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Exercises:
Which of the following functions are onto?
•
F1 from A = {1, 2, 3, 4, 5} to B = {a , b, c, d } ,
F1 = {(1, a ), ( 2, c ), (3, c ), ( 4, d ), (5, d )}
Range F1 = {a , c, d } ≠ B . Therefore F1 is not an onto
function.
•
F2 from A = {1, 2, 3, 4, 5} to B = {a , b, c, d } ,
F2 = {(1, a ), ( 2, b ), (3, c ), ( 4, d ), (5, a )}.
Range F2 = {a , b, c, d } = B . Therefore F2 is an onto
function.
•
F3 on  , F3 = {( x, y ) : y = 4 x − 1}.
For each y ∈  , let x =
y +1
∈  , then
4
∀y ∈ , ∃x ∈ , ( x, y ) ∈ F3 . Thus Range F3 =  . Therefore
F3 is an onto function.
•
F4 on  , F4 = {( x, y ) : y = 4 x − 1} .
Consider y = 0 ∈  , then for ( x, 0) ∈ F4 requires
x=
0 +1
∉  . Thus Range F4 ≠  . Therefore F4 is not an
4
onto function
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5.3.4.
Inverse
Every relation has an inverse and this holds for functions
also.
For any function, there is an inverse relation; however, this
inverse relation is not always a function.
The inverse of a function F will also be a function when F
is one-to-one and onto.
Example:
Consider the relation F on the interval
[ −1,1] = { x ∈  : −1 ≤ x ≤ 1} , given by
F = {( x, y ) : y = 1 − x 2 } .
•
Sketch F. Is F a one-to-one and onto function?
y
x
-2
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220
Dom F = [−1,1] , and the vertical line test holds. Thus F is
a function. Horizontal line test fails, thus F is not a one-toone function. Range F = [0,1] ≠ [ −1,1] , thus F is not an
onto function.
−1
−1
Sketch F . Is F a function?
•
Since F is function, and thus a relation, there is an inverse
relation F −1 on [−1,1] given by
F −1 = {( x, y ) : x = 1 − y 2 } .
y
x
-2
-1
0
1
2
Dom F −1 = [0,1] ≠ [ −1,1] , and the vertical line test fails.
Thus F −1 is not a function.
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Exercises:
Consider the relation F on A = { x ∈  : x ≥ 0} given by
F = {( x, y ) : y = x 2 } .
•
Sketch F. Is F a one-to-one and onto function?
y
8
6
4
2
x
0
1
2
Dom F = A , vertical line test holds, horizontal line test
holds, Range F = A , thus F is one-to-one and onto
function.
•
Sketch F −1 . Is F −1 a function?
y
2
x
0
1
2
Dom F −1 = A , vertical line test holds, thus F −1 a function.
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5.4.
Permutations
5.4.1.
Definition
Let A be a set and let F be a function on A. Then F is a
permutation of A if F is one-to-one and onto.
Example:
Let A = {0,1, 2, 3}. Define F = {(0,1), (1, 2), ( 2, 3), (3, 0)}.
F is a one-to-one and onto function on A and thus is a
permutation of the elements of A.
Using conventional function notation each ordered pair in F
can be written as:
F (0) = 1, F (1) = 2, F (2) = 3, F (3) = 0
“Matrix” representation can also be used for permutations.
The function F can be written as
⎛ 0 1 2 3⎞
F =⎜
⎟
1
2
3
0
⎝
⎠
F is one possible permutation of the set A.
Other permutations are:
⎛ 0 1 2 3⎞
⎛0 1 2 3⎞
⎛ 0 1 2 3⎞
I =⎜
⎟, H = ⎜
⎟
⎟ ,G = ⎜
0
1
2
3
1
0
3
2
1
3
2
0
⎝
⎠
⎝
⎠
⎝
⎠
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There will be 4! = 4 × 3 × 2 × 1 total different permutations
of the set A.
I is known as the identity permutation, where each element
in A is mapped to itself.
Notes:
∗
In general, if A is a set with n elements, there are n!
different permutations of A.
∗
The set of all permutations on a set A with n
elements is often denoted by S n .
Exercises:
⎛0 1 2 3⎞
Let A = {0,1, 2, 3}and let G = ⎜
⎟ and
1
0
3
2
⎝
⎠
⎛ 0 1 2 3⎞
H =⎜
⎟ be permutation on A.
⎝1 3 2 0⎠
Write down the following.
•
G (1) = 0
•
H (1) = 3
•
G (3) = 2
•
G ( H (0)) = 0
•
H (0 ) = 1
•
G ( H (1)) = 2
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5.4.2.
Cycle notation
Obviously, the matrix notation for permutations can be
confusing when we start to combine permutations.
This notation can be mistaken for “normal” matrix
multiplication. Therefore, we introduce what is called cycle
notation for permutations.
Example:
Let A = {1, 2, 3, 4, 5} and let F be a permutation on A given
⎛ 1 2 3 4 5⎞
by F = ⎜
⎟
2
3
4
5
1
⎝
⎠
we note that:
1 “goes to” 2
2 “goes to” 3
3 “goes to” 4
4 “goes to” 5
5 “goes to” 1.
This can be written as a cycle: (1 2 3 4 5) .
Diagrammatically, this can be represented as
(1 2 3 4 5)
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If an element is mapped onto itself, then it is left out of the
cycle.
Examples:
Write the following permutations using cycle notation.
•
Let A = {0,1, 2, 3}
⎛ 0 1 2 3⎞
F =⎜
⎟ = (0 2 )
2
1
0
3
⎝
⎠
•
A = {1, 2, 3, 4, 5}
⎛1 2 3 4 5⎞
G=⎜
⎟ = (1 2 )(4 5)
2
1
3
5
4
⎝
⎠
•
A = {1, 2, 3}
⎛1 2 3 ⎞
I =⎜
⎟ = (1) or (2 ) or (3) or (1)(2 )(3)
⎝1 2 3 ⎠
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Exercises:
Write down the following permutations on
A = {0,1, 2, 3}, using cycle notation.
•
⎛ 0 1 2 3⎞
F =⎜
⎟ = (1 3)
0
3
2
1
⎝
⎠
•
⎛0 1 2 3⎞
G =⎜
⎟ = (0 1 3 2 )
1
3
0
2
⎝
⎠
•
⎛0 1 2 3⎞
H =⎜
⎟ = (0 1)(2 3)
1
0
3
2
⎝
⎠
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5.4.3.
Composition
In traditional Calculus, composition of functions is defined
to be ( g o f )( x ) = g ( f ( x )) .
The same idea is used when considering composition of
permutations.
Examples:
Let A = {1, 2, 3, 4} and let F = (1 2 3 4) ,
G = (1 2)(3 4) be permutations on A.
Write down the following:
•
G ( F (1)) = G (2) = 3
•
G ( F (2)) = G (3) = 4
•
G ( F (3)) = G (4) = 3
•
G ( F (4)) = G (1) = 2
What is G o F written using cyclic notation?
G o F = (2 4)
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This could be calculated by writing each function in cyclic
notation in the appropriate order, then determining the
resultant permutation.
G o F = FG = (1 2 3 4)(1 2)(3 4)
1 “goes to” 2 in the first cycle, then 2 “goes to” 1 in the
second. Thus, 1 “goes to” 1 overall.
2 “goes to” 3 in the first cycle, then 3 “goes to” 4 in the
third. Thus, 2 “goes to” 4 overall.
3 “goes to” 4 in the first cycle, then 4 “goes to” 3 in the
third. Thus, 3 “goes to” 3 overall.
4 “goes to” 1 in the first cycle, then 1 “goes to” 2 in the
second. Thus, 4 “goes to” 2 overall.
These calculations give G o F = FG = ( 2 4) .
(1
2 3 4 )(1 2 )(3 4 ) = ⎛⎜11 24 33 42 ⎟⎞ = (2 4 )
⎝
⎠
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Exercises:
Calculate the following compositions of permutations on
A = {0,1, 2, 3}.
•
⎛ 0 1 2 3⎞
(1 2)(1 0 2) = ⎜
⎟ = (0 2 )
2
1
0
3
⎝
⎠
•
⎛ 0 1 2 3⎞
(0 1)(2 3)(0 1 2 3) = ⎜
⎟ = (0 2 )
2
1
0
3
⎝
⎠
•
(1 2 3)(3 2) = ⎜
5.4.4.
⎛ 0 1 2 3⎞
⎟ = (1 3)
⎝ 0 3 2 1⎠
Inverse Permutations
Permutations are one-to-one and onto functions, thus their
inverses are also functions which are one-to-one and onto.
Thus, the inverse of a permutation is also a permutation.
Recall that to find the inverse of a relation or function, we
simply reverse the ordered pairs. For permutations, the
process is identical.
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Examples:
Let A = {1, 2, 3, 4} and let F = (1 2 4 3) .
In F:
1 “goes to” 2. Thus, in F −1 , 2 “goes to” 1.
2 “goes to” 4. Thus, in F −1 , 4 “goes to” 2.
3 “goes to” 1. Thus, in F −1 , 1 “goes to” 3.
4 “goes to” 3. Thus, in F −1 , 3 “goes to” 4.
Putting all these calculations together, we have
F −1 = (1 2 4 3) −1
= (1 3 4 2)
= (3 4 2 1)
Note that F −1 is just F written in the reverse order.
Exercises:
Let A = {0,1, 2, 3} Write down the following.
•
(1 2 3)−1 = (3 2 1)
•
(0 3 1)−1 = (1 3 0 )
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