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Mathematical Logic
The development of formal logic and its implementation is
essential in computer science.
It is mainly used for deriving a conclusion based on what one
already knows.
Logic is the study of correct reasoning. It provides rules to
determine whether a given argument is valid or not.
Proposition
A proposition is a statement that can be either ‘true’ or ‘false’.
Examples:
1) It rained yesterday.
2) India is a state.
3) Himachal Pradesh is a country.
Proposition
It is possible to determine whether any given sentence is a
proposition by prefixing it with:
It is true that ……
Or
It is false that ……
and check whether the result makes any grammatical sense.
Check whether the following sentences are propositions or not
1) New Delhi is the capital of Sri Lanka.
2) 1 + 1 = 10.
3) Two is less than five.
4) Man will reach Mars by 2050.
5) Close the door.
Proposition
6) This is true.
7) What time is it?
8) 6+5
9) x is greater than 5.
10) What a beautiful day!
Propositional Logic
It is the study of propositions (true or false statements) and
ways of combining them (logical operators) to get new
propositions. It is effectively algebra of propositions. In this
algebra, the variables stand for propositions and the
operators (connectives) are and, or, not, implies(if then), and
if and only if.
Propositional Logic
p and q
p or q
not p
Implies
(if then )


 or ~
→
Example
(p and q are simple
statements)
pq
pq
p or ~p
p→q
if and only if (iff)
↔
p ↔q
Connective
Symbol
Consider a statement “Raju will eat fruit-salad if the fruitsalad contains mangoes in it”. The statement is equivalent to
the statement “If the fruit-salad contains mangoes, then Raju
will eat it”. The statement is a complex statement constructed
from two simple statements say p and q, where
Propositional Logic
p: Fruit-slalad contains mangoes.
q: Raju will eat fruit-salad containing mangoes.
If p then q, when p and q are propositions can be written as
p → q.
The above sentence (p → q) states only that Raju will eat fruitsalad containing mangoes. It does not, however, rule out the
possibility that Raju will eat fruit-salad containing apples.
Whenever there is a statement p ↔ q (if and only if),its
meaning is different from the previous one. This is equivalent to
the statement “If the fruit-salad contains mangoes, then Raju will
eat it AND If Raju is eating fruit-salad, then it must be containing
mangoes”.
Propositional Logic
The values of the complex statement varies according to the
values of its constituent propositions.
p
q
pq
pq
p→q
(p ↔ q)
(q ↔ p)
F
F
T
T
F
T
F
T
F
F
F
T
F
T
T
T
T
T
F
T
T
F
F
T
Converse and Contrapositive
For a proposition p → q, the proposition q → p is called its
converse and the proposition q → p is called contrapositive.
Propositional Logic
Truth table for converse and contrapositive
p→q
q→p
q → p
F
T
T
T
F
T
T
F
T
T
F
F
T
F
T
T
T
T
T
p
q
F
Propositional Logic
Some of the logically equivalent propositions are listed below.
They are also called identities. The symbol  shows the logical
equivalence.
1. p  (p  p) ( idempotence of )
2. p  (p  p) ( idempotence of )
3. (p  q)  (q  p) ( commutativity of )
4. (p  q)  (q  p) ( commutativity of )
5. (p  q)  r  p  (q  r) (associativity of )
6. (p  q)  r  p  (q  r) (associativity of )
7. (p  q)  (p  q) (DeMorgan's Law)
8. (p  q)  (p  q) (DeMorgan's Law)
Propositional Logic
9. p  (q  r)  (p  q)  (p  r) (Distributive)
10. p  (q  r)  (p  q)  (p  r) (Distributive)
11. (p  True)  True
12. (p  False)  p
13. (p  True)  p
14. (p  False)  False
15. (p  p)  True
16. (p  p)  False
17. p  (p)
18. (p → q)  (p  q)
19. (p ↔ q)  [(p → q)  (q → p)]
20. (p → q)  (q → p)
Propositional Logic
Q1)Using the following statements
p:Raju is tall.
q:Raju is strong.
What is the symbolic form of the following statement?
“Raju is tall but week.”
Answer: The statement given is equivalent to “Raju is tall and
Raju is not strong”. So the corresponding symbolic
representation is (p  ¬q).
Q2) Using p and q of the previous question, what is the symbolic
representation of the following statement?
“ It is false that Raju is short or strong”.
Propositional Logic
Answer: The statement given is equivalent to the negation of
“Raju is not tall or Raju is strong”. So it is the negation of (¬p  q).
That is ¬(¬p  q).
Q3) Prove that (p  p)  (p → (q  q)) is equivalent to p  q
Answer: (p  p)  (p → (q  q))  p  (p → q)
 p  (¬p  q)
 (p  ¬p) ( p  q)
 F ( p  q)
 (p  q)
Q4) p → (q → r)  (p  q) → r
Propositional Logic
Answer: Let us explore the L.H.S first
p → (q → r)
 p  (q  r)
 (p  q)  r
 (p  q)  r
 (p  q) → r
Q5) Prove that (p  q)  (p  r)  p  (q  r)
Propositional Logic
Dual of a Proposition
Let X be a proposition involving  and  as connectives, and X*
be a proposition obtained from X by replacing  with ,  with
, T with F and F with T. Then X* is called the dual of X.
Eg: The dual of [(p  q)  r] is [(p  q)  r]
If two propositions X and Y are equivalent, then their duals X*
and Y* are also equivalent.
Tautology & Contradiction
A proposition whose truth value is always true is called a
tautology and one whose truth value is always false is called a
contradiction. The negation of a tautology is a contradiction
and that of a contradiction is a tautology.
Propositional Logic
Q1) Which of the following is true about the proposition
p  (¬p  q) ?
a)Tautology
b)Contradiction
c)Logically equivalent to p  q
d)None of these
Answer: The proposition can be written as
(p  ¬p)  (p  q)  F  (p  q)
 (p  q).So the answer is (c)
Q2) What is the dual value of (p  q)  T ?
Answer: The dual value of any expression is created by replacing
‘ with ’,‘ with ’, ‘T with F’ and ‘F with T’.
Thus the dual value of the given expression is (p  q)  F.
Propositional Logic
1) Which of the following proposition is a tautology?
a)(p  q) → p
b) p  (q → p)
c) p  (p → q)
d) p →(p → q)
Answer: There are two ways to solve the problem. The first
method is by drawing the Truth-table for each option given and
thereby finding the tautology. This approach sometimes
becomes a time and space consuming approach.
The second approach is by making the given expression of the
form T  (an expression). Where T is a known tautology
(like p  ¬p , q  ¬q etc..)
Propositional Logic
Option (a) can be written as ¬(p  q)  p.(There is no
tautology).
Option (b) can be written as:
p  (¬q  p)  (p  ¬q)  (p  p)
(p  ¬q)  p
(there is no tautology present in this also)
Option (c) can be written as:
p  (p → q)  p  (¬p  q)
 (p  ¬p)  (p  q)
 T  (expression)
So the truth value of the statement is always true.
So it is a tautology.
Propositional Logic
2) Let X denotes (p  q) → r and Y denotes (p → r)  (q → r).
Which of the following is a tautology?
a) X ↔ Y
b) X → Y
c) Y → X
d) ¬Y → X
Answer: We need to draw truth tables for all the options given.
p
F
F
F
T
T
T
F
T
q
F
F
T
T
F
T
T
F
r
F
T
T
T
F
F
F
T
pq
F
F
T
T
T
T
T
T
p→r
T
T
T
T
F
F
T
T
q→r
T
T
T
T
F
F
T
T
X
T
T
T
T
F
F
F
T
Y
T
T
T
T
T
F
T
T
¬Y
F
F
F
F
F
T
F
F
X → Y Y → X ¬Y → X
T
T
T
T
T
T
T
T
T
T
T
T
T
F
T
T
T
F
T
F
T
T
T
T
Propositional Logic
3) Let a, b, c and d be propositions. Assume that equivalences
a  (b  ¬b) and b  c hold. What is the truth value of the
formula (a  b) → ((a  c)  d) ?
Answer: The first equivalence a  (b  ¬b) shows that a is a
tautology as (b  ¬b)is a tautology.
b  c shows that b is false when c is false and b is true when c
is true.
(a  b) → ((a  c)  d)  (T  b) → ((T  c)  d)
 b → (c  d)
 ¬b  (c  d)
 (¬b  c)  d
Td
This is in the form of (Tautology  (any expression)).So the
truth value is always true.
Propositional Function (Predicates)
A propositional function defined on a set A is an expression
p(x),which has the property that p(a) is true or false for each
(a A).The set A is called the domain of p(x). The set containing
elements of A for which p(a) is true is called the truth set Tp.
Tp={x: x A, where p(x) is true}.
Eg: Consider a propositional function p(x) defined on the set of
positive numbers N.
1. Let p(x) be (2 + x > 9),its truth set is {8,9,…}
2. Let p(x) be (x < 0),its truth set is { }.
Propositional Function (Predicates)
Quantifiers
Quantifiers are symbols used with propositional functions.
There are two types of quantifiers as shown in the table below.
Name
Universal
Quantifier
Existential
Quantifier
Symbol
Meaning

“ for all”

“ there exists at
least one”
Eg: If N is a set of all positive numbers, then the following
statements are true.
x  N, (x + 3 > 2).
 x  N, (x + 2 < 7).
Propositional Function (Predicates)
Negation of Quantified Statements
Consider the statement “All cities are clean”. Its negation can
be written in two ways.
1.It is not the case that all cities are clean.
2.There exists at least one city which is not clean.
Symbolically we can write,
(x  M)(x is clean)  (x  M)(x is not clean)
or
(x  M) p(x)  (x  M) p(x)
Where M is the set of all cities.
According to Demorgan,
1) (x  A) p(x)  (x  A) p(x)
2) (x  A) p(x)  (x  A) p(x)
Propositional Function (Predicates)
Propositional Functions with more than one variable
A propositional function with more than one variable can be
written as p(x1,x2,…xn) over a product set A1A2…An.
Eg: Let A={1,2,3,4} and p(x,y) denotes x + y = 5.
So we can write a true statement as xy p(x,y).
But if we change the order of the quantifiers the statement
becomes false. i.e. if we write yx p(x,y) the meaning of the
statement becomes “There exists a y such that for all x we have
x+y=5”.(We know that no such y exists.)
Negating Quantified Statements
When we negate statement with more than one variable, each 
will be replaced by a  and each  will be changed to .
Eg: (xyz p(x,y,z))  xyz p(x,y,z)
Propositional Function (Predicates)
Q1) What is the predicate calculus statement equivalent to the
following?
“Every teacher is liked by some student”
a)x [ teacher(x) → y[student(y) → likes(y,x)]]
b)x [ teacher(x) → y[student(y)  likes(y,x)]]
c)yx[ teacher(x) → [student(y)  likes(y,x)]]
d)x [ teacher(x)  y[student(y) → likes(y,x)]]
Answer: The statement given can also be written as
“ For all x, if x is a teacher, then there exists a student y who loves
x”, which can also be represented using the quantifiers as follows.
x [ teacher(x) → y[student(y)  likes(y,x)]].
It is important to note that implication (→) is almost used in
conjunction with the quantifier . Mostly the quantifier  is
associated with .
Propositional Function (Predicates)
Q2)
P(x): x is a human being.
F(x, y): x is father of y.
M(x, y): x is mother of y.
Write the predicate corresponding to
“x is the father of the mother of y”
Answer: If we try to interpret the predicate using three
variables x, y, z we can write “z is a human being and x is the
father of z and z is the mother of y”. This can be represented
as ( z)(P(z)  F(x,z)  M(z,y)).
Propositional Function (Predicates)
Normal Forms
Some important points to remember here are,
1.An atomic proposition is a proposition containing no logical
connectives. Eg: p, q, r etc.
2.A literal is either an atomic proposition or a negation of an
atomic proposition. Eg:p, q, r etc.
3.A conjunctive clause is a proposition that contains only literals
and the connective . Eg: (p  q  r).
4. A disjunctive clause is a proposition that contains only literals
and the connective . Eg: (p  q  r)
The problem of finding whether a given statement is a tautology
or contradiction in a finite number of steps is called a decision
problem. Constructing truth tables is not a practical way.
Propositional Function (Predicates)
We can therefore consider alternate procedure known as
reduction to normal forms. Two such normal forms are:
1.Disjunctive Normal form(DNF)
2.Conjunctive Normal form(CNF)
1.Disjunctive Normal form(DNF)
A proposition in said to be in disjunctive normal form (DNF) if it is
a disjunction of conjunctive clauses and literals.
Eg: (p  q  r)  q  (q  r).
A proposition in said to be in principal disjunctive normal form if
it is a disjunction of conjunctive clauses only.
Eg: (p  q  r)  (q  r).
Propositional Function (Predicates)
2.Conjunctive Normal form(CNF)
A proposition in said to be in conjunctive normal form (CNF) if it is
a conjunction of disjunctive clauses and literals.
Eg: (p  q  r)  r  (q  r).
A proposition in said to be in principal conjunctive normal form if
it is a conjunction of disjunctive clauses only.
Eg: (p  q  r)  (q  r).
Q1) What is the disjunctive normal form of p  (p → q) ?
Answer: (p  p)  (p  q)
Q2) What is the principal disjunctive normal form of p  q?
Propositional Function (Predicates)
Answer: A proposition in said to be in principal disjunctive
normal form if it is a disjunction of conjunctive clauses only.
We can write,
p  q
 (p  T)  (q  T)
(p  (q  q))  (q  (p  p))
(p  q)  ( p  q)  (p  q)  ( p  q)
(p  q)  ( p  q)  (p  q)
Mathematical Reasoning
We need mathematical reasoning to determine whether a
mathematical argument is correct or incorrect
Mathematical reasoning is important for artificial intelligence
systems to reach a conclusion from knowledge and facts.
We can use a proof to demonstrate that a particular statement
is true. A proof consists of a sequence of statements that form
an argument.
Rules of Inference:
Inference is the act or process of deriving a conclusion based
solely on what one already knows. It uses hypotheses, axioms,
definitions etc. to reach a conclusion.
Mathematical Reasoning
The general form of a rule of inference is:
Where p1, p2,.., pnare known as the
hypotheses and q is known as the
conclusion and ‘∴’ means ‘therefore’.
The rule states that if p1and p2 and …

andpn are all true,
then q is true as well.
Some valid arguments:
Mathematical Reasoning
We say that an argument is valid, if whenever all its hypotheses
are true, its conclusion is also true.
Q1) check whether the following argument is valid or not.
“If it rains today, then we will not have a barbeque today. If we
do not have a barbeque today, then we will have a barbeque
tomorrow.
Therefore, if it rains today, then we will have a barbeque
tomorrow.”
p: “It is raining today.”
q: “We will not have a barbeque today.”
r: “We will have a barbeque tomorrow.”
So the argument is of the following form:
Mathematical Reasoning
Q2) check whether the following argument is valid or not.
Gary is either intelligent or a good actor.
If Gary is intelligent, then he can count from 1 to 10.
Gary can only count from 1 to 3.
Therefore, Gary is a good actor.
Answer: Yes
i: “Gary is intelligent.”
a: “Gary is a good actor.”
c: “Gary can count from 1 to 10.”
Step 1:  c
Step 2: i  c
Step 3:  i
Step 4: a  i
Step 5: a
Conclusion: a (“Gary is a good actor.”)
Proving Theorems
Direct proof:
An implication p  q can be proved by showing that if p is true,
then q is also true.
Example: Prove that “If n is odd, then n2 is odd.”
Assume that n is odd.
Then it can be written as n = 2k + 1, where k is an integer.
Consequently, n2 = (2k + 1)2.
= 4k2 + 4k + 1
= 2(2k2 + 2k) + 1
Since n2 can be written in this form, it is odd
Q1) Prove that “If n is even, then n2 is even.”
Proving Theorems
Indirect proof:
An implication p  q is equivalent to its contra-positive
 q   p. Therefore, we can prove p  q by showing that
whenever q is false, then p is also false.
Q2) Give an indirect proof of the theorem
“If 3n + 2 is odd, then n is odd”.
Assume that n is even.
Then n = 2k, where k is an integer.
It follows that 3n + 2 = 3(2k) + 2
= 6k + 2
= 2(3k + 1)
Therefore, 3n + 2 is even.
We have shown that the contrapositive of the implication is true,
so the implication itself is also true (If 3n + 2 is odd, then n is
odd).
Mathematical Induction
If we have a propositional function P(n), and we want to prove
that P(n) is true for any natural number n, we do the following:
• Show that P(0) is true.
(basis step)
• Show that if P(n) true then P(n + 1) is also true for any nN.
(inductive step)
• Then P(n) must be true for any nN.
(conclusion)
Example: Show that n < 2n for all positive integers n.
Let P(n) be the proposition “n < 2n.”
1. Show that P(1) is true. (basis step)
P(1) is true, because 1 < 21 = 2.
Mathematical Induction
2. Show that if P(n) is true, then P(n + 1) is true.
(inductive step)
Assume that n < 2n is true.
We need to show that P(n + 1) is true, i.e. n + 1 < 2n+1
We start from n < 2n:
n + 1 < 2n + 1  (2n + 2n = 2n+1)
Therefore, if n < 2n then n + 1 < 2n+1
3. Then P(n) must be true for any positive integer.
(conclusion)
n < 2n is true for any positive integer.
End of proof.
Mathematical Induction
Q1) Using mathematical induction, prove that
1 + 2 + … + n = n (n + 1)/2
Answer: Show that P(0) is true. (basis step)
For n = 0 we get 0 = 0. True.
Show that if P(n) then P(n + 1) (inductive step)
1 + 2 + … + n = n (n + 1)/2
1 + 2 + … + n + (n + 1) = n (n + 1)/2 + (n + 1)
= (2n + 2 + n (n + 1))/2
= (2n + 2 + n2 + n)/2
= (2 + 3n + n2 )/2
= (n + 1) (n + 2)/2
= (n + 1) ((n + 1) + 1)/2