Download P(x, y)

Sections 1.4 Predicates
A predicate P(x) is a statement
Consider the statement x + 3 = 7.
1)
2) Quantify the variable(s)
a)
b)
Quantifying a statement containing two or more variables
Q(x, y) is the statement x + y = 5
xy Q(x, y)
xy Q(x, y)
xy Q(x, y)
xy Q(x, y)
domain
Definition (p. 43)
1. The symbol 
2. The symbol  means
3. The symbol  means
4. The symbols  and 
5. A counterexample is
P(x): x + 3 = 7
Predicate: x P(x)
Domain:
R(x): x3 ≥ 0
Predicates x R(x) and x R(x)
Domain
More examples
Translating Quantified Statements
Every student in EECS 210 has taken a programming class.
S(x):
Q(x):
P(x):
Some student in EECS 210 bought the textbook.
B(x):
When are the statements false?
Expressing Negations:
P(x): x is a professor
Q(x): x is ignorant
R(x): x is vain
a) "No professors are ignorant"
b) "All ignorant people are vain"
c) "No professors are vain"
Does statement c follow from a and b?
Negating Quantified Statements
Negation

(x P(x))
Equivalent
Statement
x (P(x))

(x P(x))
x (P(x))
When true
For some x,
P(x) is false
When false
P(x) is false for For some x,
all values of x P(x) is true
P(x) is true for
all x
Stmt: Every student in EECS 210 is a sophomore or junior.
Negation:
Q(n, r): For every nonzero integer n there is a rational number r
such that n = 1/r
There is a unique x such that for every y, xy = y.
R(x, y): xy = y
More Examples
D(x) is “x is a day”
S(x) is “x is sunny”
R(x) is “x is rainy”
M is “Monday”
Some days are not rainy.
No day is both sunny and rainy.
Monday was sunny; therefore every day will be sunny
Quantifications of Two Variables
Statement
When true
When false
xy P(x, y) P(x, y) true for all choices
yx P(x, y) of x and y
Some pair x, y makes
P(x, y) false
xy P(x, y)
Given x a y can be found
to make P(x, y) true
For some x, every choice
of y makes P(x, y) false
xy P(x, y)
For some value of x every
y makes P(x, y) true
For every x, there is some
y for which P(x, y) is false
xy P(x, y)
yx P(x, y)
There is an x, y pair that
makes P(x, y) true
P(x, y) is false for every
pair x, y
More Examples
Use the predicates given below to write each of the statements that
follow them in symbolic notation. The domain is the set of all
movies.
M(x): “x is a mystery”
D(x): “x is a drama”
C(x): “x is a comedy”
B(x, y): “x is better than y”
a. Some mysteries are dramas.
b. Some comedies are better than all dramas.
c. For every drama there is a comedy better than it.
d. Only comedies are better than mysteries.
e. Not every comedy is a mystery.
Negating Implications
The implication p  q is logically equivalent to p V q. Thus, the
negation of p  q is the statement p  q
p
q
T
T
T
F
F
T
F
F
F(x): x is a fruit
pq

p

pVq

q
p  q
V(x): x is a vegetable S(x, y): x is sweeter than y
Some vegetable is sweeter than all fruits
Every fruit is sweeter than some vegetable
Only fruits are sweeter than vegetables
Logical Equivalences for Conditional Statements

p  q  p V q
(p  q)  p  q
p  q  q  p
(p  q)  (p  r)  p  (q  r)
p V q  p  q
(p  q) V (p  r)  p  (q V r)
p  q  (p  q)
“No professor has been asked questions by all of his students.”
S(x): x is a student
P(x): x is a professor
A(x, y): x has asked y a question
What does each translation actually mean?
x y ((S(x)  P(y))  A(x, y)
x  y ((S(x)  A(x, y))  P(y)
x y (A(x, y))
x y (((S(x)  P(y) )A(x, y))
x y ((P(y)  S(x))  A(x, y))
x y(P(x)  A(x, y))
 x  y(P(x)  (S(y)  A(x, y) ))
A correct translation:
Section 1.6 Excursion—validity of arguments
A theorem is
A proof is
validity of an argument
Rules of inference
Example: If x ends in 0 then it is divisible by 10.
(p  (p  q))  q
modus ponens
p
T
T
F
F
|
|
|
|
|
q
T
F
T
F
| pq
| T
| F
| T
| T
Other Valid Arguments
Example:
If 6 divides x then 2 divides x
2 does not divide 5
Therefore, 6 does not divide 5
pq

q___
 p
modus tollens ((pq  q)  p)
p
T
T
F
F
|
|
|
|
|
q
T
F
T
F
| pq
| T
| F
| T
| T
Resolution: (p V q)  (p V r)  (q V r)
Example: (from Rosen #61 p. 76)
It is not raining or Sue has her umbrella
Sue does not have her umbrella or she does not get wet
It is raining or Sue does not get wet
p:
q:
r:
1st Statement
2nd Statement
3rd Statement
Table 1 Rules of Inference (p. 58 in Rosen)
Rule
Tautology
Name
__________________________________________________
p
p V q
p  (p V q)
Addition
__________________________________________________
pq
p
(p  q)  p
Simplification
__________________________________________________
p
q____
Conjunction
pq
__________________________________________________
p
pq
[p  (p  q)]  q
Modus ponens
q
__________________________________________________
q
pq
[q  (p  q)] p
Modus tollens
p
__________________________________________________
pq
qr
[(p  q)  (q  r)]  (p  r)
Hypothetical
p  r
Syllogism
__________________________________________________
pVq

p
[(p V q) p]  q
Disjunctive
q
Syllogism
_________________________________________________
pVq

pVr
Resolution
q V r
__________________________________________________
Fallacies
p
T
T
F
F
More Fallacies
Circular reasoning
|
|
|
|
|
q
T
F
T
F
| pq
| T
| F
| T
| T
Proving Validity of the Inference Rules
What we need to prove:
Proving the validity of disjunctive syllogism:
p
T
q
T
T
F
F
T
F
F
pVq

p
(p V q)  p [(p V q) p]  q
Validity of an argument
Example 1: (p. 76 # 2b) If this number is a perfect square then
the equation has a rational solution. The equation has a rational
solution. Therefore this number is a perfect square.
Example 2: If you fail the final you will fail the course. You failed
the final. Therefore, you will fail the course
Example 3: If it’s above 80 at noon then it will not rain. It rained.
Therefore it was not above
80 at noon.
Mathematical Notation and Terminology Used in EECS 210
Z
Z+
Q
R
N
Mathematical Functions
Floor function y
Ceiling function y
Mod function
Divides
Notation:
GCD
LCM
Prime number
Composite number
Relatively Prime
Chapter 2 A Primer of Mathematical Writing (Proofs)
Types of proofs
Give reasons for all steps in a proof
Proposition 1: (p. 87) Other than 3, 4 there is no pair of consecutive
integers where the first is a prime number and the second is a
perfect square.
Theorem 2: For all integers n > 4, if n is a perfect square, then n – 1
is not a prime number. (Basically a restatement of proposition 1.)
Tracing proofs
Typical Proofs about Numbers
An integer n is even if
An integer m is odd if
Closure Property of the Integers: Whenever the operation of
addition, subtraction or multiplication is applied to integers, the
result is an integer.
Proposition 3: The result of summing an odd integer and an even
integer is an odd integer.
Proof 1:
Proof 2:
Are both correct? Why or why not?
Proposition 5: If n is even then n2 is divisible by 4.
Proposition 6: For all integers n, if n2 is even then n is even.
Example: Prove that if n + 1 separate passwords are issued to n
students, then some student gets ≥ 2 passwords.
Section 2.2 Proofs About Numbers
An integer n is divisible by
Prove or disprove: If n is an odd integer, then n2 - 1 is divisible by 8.
If integers m and n are both divisible by 3 then
1. n + m is also divisible by 3.
2. m•n is divisible by 9.
3. n2 + 3n is divisible by 9.
Prop. 4: For any nonzero integer d, if integers m and n are both
divisible by d, then m + n is also divisible by d.
Definition: A real number r is rational if there exist integers a and
b (b 0) with r = a/b. Rational numbers (also called fractions) can
be expressed in many equivalent ways. (1/2 = 2/4 = 3/6 = …)It is
always possible to choose the integers a and b with no common
divisors greater than 1. Such numbers are called relatively prime.
2. A real number is irrational if it is not rational.
More Examples
Example 1: Prove that the sum of two rational numbers is rational
Example 2: Prove or disprove the product of two irrational
numbers is irrational.
Example 3: Prove that for every integer n, 3(n2 + 2n + 3) – 2n2 is a
perfect square.
Proposition 6: For any integer n, n2 + n is even.
Example 4: Prove n3 – n is divisible by 3 for all n ≥ 0
Example 5: The difference of two consecutive cubes is odd.
Proving by cases
Prove that for any two numbers x and y, |xy| = |x| • |y|
Case 1: x ≥ 0 and y ≥ 0
Case 2: x ≥ 0, y < 0
Case 3: y ≥ 0, x < 0
Case 4: x < 0, y < 0
Example 6: Prove or disprove there exist three consecutive odd
primes.
Example 7: Prove or disprove that given a positive integer n there
exist n consecutive odd positive integers that are prime.
The Division Theorem
Theorem 8: For all integers a and b with b > 0, there is an integer
q (the quotient) and an integer r (the remainder) such that
1. a = b•q + r and
2. 0  r < b
Proposition 9: If n is any integer not divisible by 5 then n has a
square that is either of the form 5k + 1 or 5k + 4.
Definition: We write a mod b = r to mean that r is the remainder
when a is divided by b. (i.e. a = b•q + r and 0  r < b)
Difference between the divides relation and division
Proving the equivalence of three or more statements
Prove that if n is an integer, then the following four statements
are equivalent:
(1) n is even
(2) n + 1 is odd
(3) 3n + 1 is odd
(4) 3n is even
Proof Technique
Exhaustive
Proof
Direct Proof
How to prove P  Q
Show P  Q for all
possible cases
Assume P, deduce Q
Comments
Only works for a finite
number of cases.
The standard
approach to try.
Contrapositive
Assume Q, deduce P Use if Q as a
(Indirect) Proof
hypothesis seems to
give more information
to work with.
Contradiction
Assume P Λ Q, deduce Try this approach
when Q says
a contradiction
something is not true.
Proof by Cases
Break the domain into
Try this for proving
two or more subsets
properties of numbers
and prove PQ for the where odd and even
elements in each such or positive and
negative numbers
subset.
require different
proofs.
Counterexample Prove statement false
Used to show that
by finding one value for PQ is false.
P such that Q is false.
Existence
Find one explicit
This is the only time
solution
you may "prove by
(e.g. there exists an
example"
even prime number)
Modification of Table 2.2 p. 91 Gersting