Mathematical foundations of computer science

Mathematical Foundations of
Computer Science
 Mathematical logic
 Set theory
 Algebraic Structures
 Elementary Combinatorics
 Recurrence relations
 Graph theory

Mathematical logic
 Statement (Proposition) : A declarative sentence to which it is
meaningful to assign one and only one of the truth values “true”
or “false”. We call such sentences Propositions (Statements).
 Ex. London is a city. Ex. 2+3 = 4
The following sentences are not statements.
 What is your name?
 Close the door
For definiteness let us list our assumptions about propositions.
 The law of excluded middle: For every proposition p,
either p is true or p is false.
 The law of contradiction: For every proposition p, it is not the
case that p is both true and false.

Atomic and Compound statements
 Atomic statement : A statement which can not be divided
further, is called atomic statement (Simple statement or
primary statement).
These statements are denoted by p,q,r,s,……
Ex. Milk is white Ex. 2+3 = 5
 Compound Statement : Two or more simple statements can be
combined to form a new statement. These new statements are
called Compound statements or Molecular Statements or
Propositional function or Statement formulas.
Ex. It is raining today and there are 20 tables in this room.
 Compound statements can be formed from atomic statements
through the use of following sentential connectives.
not, and , or , if …then and if and only if .

Connectives
 Negation: If p is a statement, then the negation of p, written as
~p and read as “ not p ” is a statement.
Ex. p : London is a city.
~p : London is not a city.
 The truth table for not p is given below.
p ~p
T
F

Conjunction (and) pq
 If p and q are two propositions, then the conjunction of p and q
is the statement p  q which is read as “ p and q ”.
 The statement p  q has the truth value T whenever both p and
q have truth value T; otherwise it has the truth value false.
Conjunctive syllogism: If p  q is false and p is true, then q is
false.
p q p  q
F
F
T
T
F
T
F
T

Disjunction ( Or ) pq
 If p and q are two propositions, then the disjunction of p and q
is the statement p  q which is read as “ p or q”.
 The statement pq has the truth value F only when both p and
q have truth value F; otherwise it has the truth value T.
Disjunctive syllogism: If p  q is true and p is false, then q is
true.
p q p  q
F
F
T
T
F
T
F
T

Implication (Conditional) pq
 If p and q are two propositions, then the statement pq
which is read as “ if p, then q ” or “ p implies q “.
 The statement pq has truth value F only when p is true
and q is false; otherwise it has a truth value T.
 A false antecedent p implies any proposition q.
 A true consequent q is implied by any proposition q
p q pq
F
F
T
T
F
T
F
T

Biconditional (if and only if) pq
 Biconditional : If p and q are two propositions, then the
statement pq, which is read as “p if and only if q” is called a
biconditional statement.
 The statement pq has the truth value T whenever both p and
q have identical truth values.
p q pq
F
F
T
T
F
T
F
T

More on Implication
 The opposite of pq is p  q
 The converse of pq is q  p
 The contra positive of pq is q  p
 Note : pq is logically equivalent to q  p
i.e., pq  q  p
or pq  q  p
* Ex. p: Today is Sunday
q: Today is Holiday
p  q : If today is Sunday, then today is
Holiday
q  p : If today is not Holiday, then today is not
Sunday

Well formed formulas
 A well formed formula can be generated by the following rules.
1. A statement variable standing alone is a well formed formula.
2. If P is a well formed formula, then ~P is a well formed
formula.
3. If P and Q are well formed formulas, then (PQ) , (PQ) ,
(PQ) and (PQ) are well formed formulas.
4. A string of symbols containing the statement
variables,connectives and parenthesis is a well formed formula,
iff it can be obtained by finitely many applications of the rules
1,2 and 3.
 Ex. (PQ) , (PQ) , (P (PQ) ) , (P (Q R)) and
(PQ) (PQ) are well formed formulas.
 Ex. PQ , (PQ )Q ) and (P Q )  (Q) are not well
formed formulas.

Truth tables
 Our basic concern is to determine the truth value of a statement
formula for each possible combination of the truth values of the
component statements.
 A table showing all such truth values is called the truth table of
the formula.
 Ex.1 Construct truth table for the statement formula P  Q
P Q Q P  Q
F
F
T
T
F
T
F
T
T
F
T
F

Truth tables - Examples
Ex : 2 Construct the truth table for (PQ)  P
P Q PQ P (PQ)  P
F
F
T
T
F
T
F
T
F
T
T
T
T
T
F
F
T
T
T
T

Truth tables - Examples
Ex.3 Construct the truth table for (PQ)  (QP)
P Q PQ QP (PQ)  (QP)
F
F
T
T
F
T
F
T
T
T
F
T
T
F
T
T
T
F
F
T
Note:
(PQ)  {(PQ)  (QP)}

Tautology and Contradiction
 Tautology : A propositional function (Statement formula) whose
value is true all all possible values of the propositional variables
is called a Tautology ( A Universally valid formula or a logical
truth).
Ex: P  P is a tautology.
Ex. ( P  P )  Q is a tautology.
 Contradiction (Absurdity): A propositional function whose truth
value is always false is called a Contradiction
Ex. P  P is a Contradiction .
Ex. ( P  P )  Q is a Contradiction
 Contingency: A propositional function that is neither a tautology
nor a contradiction is called a Contingency.
Ex. P  Q , P  Q , P Q, ….

Logical Equivalence & Tautological Implication
 Two propositional functions P and Q are logically equivalent, if
they have same truth tables. Then we write
P  Q or P  Q
Ex: (P )  P
Ex: ( P  Q )  ( P  Q ).
Note : The symbol  is not a connective
 A Statement P is said to tautologically imply a Statement Q if
and only if PQ is a tautology.We shall denote this as P  Q.
 Here, P and Q are related to the extent that, Whenever P has
the truth value T then so does Q.
 Every logical implication is an implication, but all implications
are not logical implications.

More on Implications
 If P  Q and Q  P , then PQ.
 If PQ then PQ is a tautology.
 Ex: Show that ( P Q )  ( P  Q )
 Since columns 3 and 5 are identical, The result follows
P Q PQ P PQ
F
F
T
T
F
T
F
T
T
T
F
T
T
T
F
F
T
T
F
T

Ex.Construct truth table for [(pq) (r)]  p]
 The truth table is given below
p q r pq r (pq) (r) [(pq) (r)] p]
F F F
F F T
F T F
F T T
T F F
T F T
T T F
T T T
F
F
F
F
F
F
T
T
T
F
T
F
T
F
T
F
T
F
T
F
T
F
T
T
F
T
F
T
T
F
T
T

Ex. Show that (PQ)  (Q  P)
 Let us prove the result using truth table.
P Q PQ Q P (Q P)
F
F
T
T
F
T
F
T
T
T
F
T
T
F
T
F
T
T
F
F
T
T
F
T

Ex. Using truth tables, show that ( P  Q )  (Q)
is a tautology
 The truth table is given below.
P Q P  Q ( P  Q ) Q ( P  Q )  (Q)
F
F
T
T
F
T
F
T
T
T
F
T
F
F
T
F
T
F
T
F
T
T
T
T

Equivalences
Commutative laws:
 P  Q  Q  P
 P  Q  Q  P
Asociative laws:
 ( P  Q )  R  P  ( Q  R )
 ( P  Q )  R  P  ( Q  R )
Distributive laws:
 P  ( Q  R )  ( P  Q )  ( P  R )
 P  ( Q  R )  ( P  Q )  ( P  R )
Demorgan’s laws:
  ( P  Q)   P   Q
  ( P  Q)   P   Q

More Equivalences
  ( P )  P (Double negation)
 P  P  P
 P  P  P
 P   P  T
 P   P  F
 R  ( P   P )  R
 R  ( P   P )  R
 R  ( P   P )  T
 R  ( P   P )  F
 P  Q  ( P  Q)
 ( P  Q )  (P   Q)
 P  Q  ( Q   P )

More Equivalences
• P  F  P
• P  T  T
• P  F  F
• P  T  P
• P  ( Q  R)  ( P  Q )  R
  ( P  Q )  (P   Q)
• (P  Q )  [( P  Q)  ( Q  P )]
• ( P  Q )  [( P  Q)  ( P   Q )]
• Absorption laws
• P  ( P  Q )  P
• P  ( P  Q )  P

Ex. Without using truth tables, Show that
P  ( Q  R)  ( P  Q )  R
 Proof:
L.H.S = P  (Q  R)
 P  (Q  R) (Since A  B  ( A  B))
 P  (Q  R)
 (P  Q)  R (By associative property)
 ( P  Q )  R (By demorgan’s law)
 ( P  Q )  R
= R.H.S

Ex. Without using truth tables, Show that
( P  Q )  P is a tautology.
Proof:
Consider, ( P  Q )  P
 ( Q  P )  P ( By commutative law )
 Q  (P  P ) ( By associative property)
 Q  T
 T
 ( P  Q )  P is a tautology.

Ex. Show that the Statement formula
( P  Q )  (PQ)  P is a tautology.
 Proof : Consider,
 {( P  Q )  (PQ)}  P (Associative law)
 {(P  Q )  (PQ)}  P ( Demorgan’s law)
  {P  (Q  Q)}  P (Distributive law)
  {P  T }  P
  {P }  P
  T
  ( P  Q )  (PQ)  P is a tautology

Ex. Show that [{( P  Q )  ( P  Q )}  R ]  R
 Proof: L.H.S = {( P  Q )  ( P  Q )}  R
  { T }  R (Since P  Q  ( P  Q))
  R
 = R.H.S
 Ex. Show that {( P  Q )  ( P  Q )} is a Contradiction.
 Proof : Let P  Q = R
 Consider, {( P  Q )  ( P  Q )}
  { R  R }
  F
  { ( P  Q )  ( P  Q )} is a contradiction.

Ex. Show that (P  (Q  R))  ( Q  R )  (P  R)  R
 Proof : Consider,
 {P  (Q  R)}  ( Q  R )  (P  R)
  {(P  Q)  R}  {( Q  R )  (P  R)}, By associative law
  { (P  Q)  R}  {(Q  P )  R} , By distributive law
  {(P Q)  R}  {(Q  P )  R} , By Demorgan’s law
  {(P Q)  (Q  P ) } R, By distributive law
  {T } R (Since, A  A  T)
  R

Ex. S.T. ((P  Q)  (P  (Q  R)))  ( P  Q)  (P   R)
is a tautology.
 Consider,
 [(P  Q)  {P  (Q  R)}]  {(P  Q)  (P  R)}
 [(P  Q)  {P  (Q  R)}]  {(P  Q)  (P  R)}
(By De morgan’s laws)
 [(P  Q)  {P  (Q  R)}]  {(P  Q)  (P  R)}
(By De morgan’s laws)
 [(P  Q)  {P  Q) (P  R)}]  {(P  Q)  (P  R)}
(By Distributive law)
  {(P  Q)  (P  R)}  {(P  Q)  (P  R)}
(Since A  A  A)
  T ( Since A  A T)

Normal forms
 Elementary product:A product of the variables and their
negations in a formula is called an Elementary product.
Ex: P, PQ, PQ, PQ R
 Elementary Sum: A Sum of the variables and their negations in
a formula is called an Elementary Sum.
Ex: P, P  Q, P  Q, P Q R
 Disjunctive normal form: A formula which is equivalent to a
given formula and which consists of a sum of elementary
products is called a disjunctive normal form.
 Ex: (P )  ( PQ )  (PQ).
 Ex: ( PQ )  (PQ)  (PQ R ).

Normal forms (contd.,)
 Conjunctive normal form: A formula which is equivalent to a
given formula and which consists of a product of elementary
sums is called a conjunctive normal form.
Ex: (P )  ( P  Q ) (P  Q).
Ex: ( P  Q )  (P  Q) (P  Q  R ).
 Min terms: Let P and Q are two statement variables. Let us
construct all possible formulas which consist of conjunctions of
P or its negation and conjunctions of Q or its negation.
For two variables P and Q, there are 22 such formulas given by
PQ, PQ, PQ, PQ
These formulas are called ‘min terms’.

Normal forms (contd.,)
 For three variables P,Q and R, there are 23 such formulas given
by
PQ R, PQ R, PQ R, PQ R,
PQ R, PQ R, PQ R, PQ R
These min terms are denoted by m0, m1 , …, m7 respectively.
 In general, there are 2n min terms for n variables.
 Principal Disjunctive normal form (Sum of products canonical
form) : For a given formula, an equivalent formula consisting of
disjunctions of min terms only is known as Principal Disjunctive
normal form .

Ex. Obtain the Principal Disjunctive normal forms of the following
PQ , P  Q, (PQ)
 Solution:
 PQ  (PQ)  (PQ)  (PQ)
 P  Q  (PQ)  (PQ)  (PQ)
 (PQ)  (PQ)  (PQ)  (PQ)
P Q PQ P  Q PQ (PQ)
F
F
T
T
F
T
F
T
T
T
F
T
F
T
T
T
F
F
F
T
T
T
T
F

Ex. Obtain the Principal Disjunctive normal form of the following
P  {(PQ)  (P  Q)}
 Given formula is, [ P  {(PQ)  (P  Q)} ] = A (say)
The truth table for A is given below.
 A  (PQ)  (PQ)  (PQ)
 Which is the PDNF for A .
P Q PQ P  Q {(PQ) 
(P  Q)}
A
F
F
T
T
F
T
F
T
T
T
F
T
F
F
F
T
F
F
F
T
T
T
F
T

(P  Q)  (Q  R)  (P  R )
 Solution: Consider, (P  Q)  (Q  R)  (P  R )
  {(P  Q)  (R  R)} 
{(P  P)  (Q  R) } 
{(P  R )  (Q  Q)}
(PQ R)  (PQ R)  (PQ R) (PQ R)
Which is the PDNF for the given formula.

(P  Q)  (P  R )
= A (say)
 A  (PQ R)  (PQ R)  (PQ R) =  (m1, m6, m7)
P Q R P  Q P (P  R) A
F
F
F
F
T
T
T
T
F
F
T
T
F
F
T
T
F
T
F
T
F
T
F
T
T
T
F
F
F
F
T
T
T
T
T
T
F
F
F
F
F
T
F
T
T
T
T
T
F
T
F
F
F
F
T
T

Principal Conjunctive normal forms
(Product of Sums canonical forms)
 Max terms: For a given number of variables, the max term
consists of disjunctions in which each variable or its negation,
but not both, appears only once.
For two variables P and Q, there are 22 such formulas given by
(P  Q), (P  Q), (P  Q), (P  Q).
These formulas are called ‘max terms’.
 For three variables P,Q and R, there are 23 such formulas given
by
P  Q  R , P  Q  R, P  Q  R, P  Q  R,
P  Q  R, P  Q  R, P  Q  R, P  Q  R
These max terms are denoted by M0, M1 , …, M7 respectively.
 In general, there are 2n Max terms for n variables.

PCNF (Contd.,)
 Mi =  mi
 M0 =  m0
= (PQ R) = (P  Q  R)
 M1 =  m1
= (PQ R) = (P  Q   R)
 M2 =  m2
= (PQ R) = (P   Q  R)
 Principal Conjunctive normal form (Product of Sums canonical
form) : For a given formula, an equivalent formula
consisting of conjunctions of max terms only is known as
Principal Conjunctive normal form.

Ex. Obtain the Principal Conjunctive normal forms of the following
PQ , P  Q, (PQ)
 The PCNF’s are
 PQ  (P  Q)
 P  Q  (P  Q)  (P  Q)  (P  Q)
 (PQ)  (P  Q)  (P  Q)
P Q PQ P  Q PQ (PQ)
F
F
T
T
F
T
F
T
T
T
F
T
F
F
F
T
T
F
F
T
F
T
T
F

EX. Obtain the Principal Conjunctive normal form of the formula
given by (P  R)  (Q  P)
 Solution: (P  R)  (Q  P)
 (P  R)  {(PQ)  (QP)}
 (P  R)  (P  Q)  (Q  P)
 { (P  R)  (Q  Q) } 
{ (P  Q)  (R  R) } 
{ (Q  P)  (R  R) }
 (P  Q  R)  (P  Q  R)  (P  Q  R) 
( P  Q  R)  (P  Q  R)
= (0,2,3,4,5)
Which is the required PCNF.

Max terms and Min terms
 *
P Q R Min terms mi Max terms Mi
F
F
F
F
T
T
T
T
F
F
T
T
F
F
T
T
F
T
F
T
F
T
F
T
m0 : PQ R
m1 : PQ R
m2 : PQ R
m3 : PQ R
m4 : PQ R
m5 : PQ R
m6 : PQ R
m7 : PQ R
M0 : P  Q  R
M1 : P  Q  R
M2 : P  Q  R
M3 : P  Q  R
M4 : P  Q  R
M5 : P  Q  R
M6 : P  Q  R
M7 : P  Q  R

Ex. Obtain the Principal Conjunctive normal form and Principal
disjunctive normal form of A, where A = (P  Q) (P  R )
The PCNF of A = (0,2,4,5)
A  (P  Q  R)  (P  Q  R)  (P  Q  R) (P  Q  R)
P Q R P  Q P P  R A
F
F
F
F
T
T
T
T
F
F
T
T
F
F
T
T
F
T
F
T
F
T
F
T
F
F
F
F
F
F
T
T
T
T
T
T
F
F
F
F
F
T
F
T
F
F
F
F
F
T
F
T
F
F
T
T

Contd.,
The PDNF of A = (1,3,6,7)
A  (PQ R)  (PQ R)  (PQ R)  (PQ R)

Implications ,Arguments,Inferences
 Inference (Argument): From a set of premises (called
Hypotheses) {H1, H2, …., Hn } a conclusion C follows logically
iff H1  H2  ….  Hn  C.
• The rules of inference are criteria for determining the validity of
an argument.
• Any conclusion which is arrived at by following these rules is
called a valid conclusion, and the argument is called a valid
argument.
• The following statements are equivalent.
• 1. {H1 , H2 , …. , Hn }  C is a logical implication.
• 2. ( H1  H2  ….  Hn) C is a tautology.
• 3. {H1 , H2 ,…. , Hn }  C is a valid argument.

Rules of Inference
There are two rules of Inference
1) Rule P: A premise may be introduced at any point in the
derivation.
2) Rule T: A formula S may be introduced in a derivation if S is
tautologically implied by and/or equivalent to any one or more
of the preceding formulas in thederivation.

Rules of Inference (Logical Implications)
List of Implications

Rules of Inference (contd.,)
List of Equalence

Simplification rules:
 (P  Q)  P
(P  Q)  P is a tautology.
P logically follows from (P  Q)
 (P  Q)  Q
(P  Q)  Q is a tautology.
Q logically follows from (P  Q)
Addition rules:
• P  (P  Q)
P  (P  Q) is a tautology
(P  Q) logically follows from P

 Q  ( P  Q )
Q  (P  Q) is a tautology
(P  Q) logically follows from Q
 P (P  Q)
P (P  Q) is a tautology
(P  Q) logically follows from P
 Q  ( P  Q)
Q (P  Q) is a tautology
(P  Q) logically follows from Q
 (P  Q)  P
(P  Q)  P is a tautology (or) P follows from (P  Q)

Rules of Inference (Contd.,)
 (P  Q )  (Q)
(P  Q )  (Q) is a tautology
Q logically follows from (P  Q)
 Disjunctive syllogism
{P, P  Q}  Q
{P  ( P  Q)}  Q is a tautology.
The inference P  Q
 P
----------------
 Q is valid

Modus ponens (Rule of detachment)
 {P, PQ}  Q
 { P  (PQ) }  Q is a Tautology
 The argument
PQ
P
------------
 Q is valid
 Ex: The following argument is valid.
A) If today is a Sunday then today is a Holiday
B) Today is Sunday
C : Hence, Today is Holiday

Modus tollens
 { PQ, Q }  P
 { (PQ)  Q}  (P) is a Tautology
 The argument
PQ
Q
------------
 P is valid
A) If today is a Sunday then today is a Holiday
B) Today is not Holiday
C : Hence, Today is not Sunday

Rule of Transitivity (Hypothetical Syllogism)
 { PQ, QR }  (PR)
 { (PQ)  (QR}  (PR) is a Tautology
 The argument
PQ
QR
------------
 PR is valid
A) If I Study well, then I will get distinction.
B) If I get distinction, then I will get a Good Job.
C:  If I Study well, then I will get a good job

Dilemma
 The Inference
P  Q
P  R
Q  R
------------
 R is a valid Inference.
 {P  Q, PR, QR }  R is a logical implication.
 {(PQ)  (PR}  (QR) }  R is a Tautology

Constructive dilemma
 The Inference
P  Q
P  R
Q  S
------------
 R  S is a valid Inference.
 {P  Q, PR, QS }  ( R  S ) is a logical implication.
 {(PQ)  (PR}  (QS) }  (R  S) is a Tautology

Destructive Dilemma
 The Inference
P  R
Q  S
R  S
----------------
 P  Q is a valid Inference.
 { PR, QS, R  S }  (P  Q ) is a logical implication.
 {(PR)  (QS)  (R  S )}  (P  Q) is a Tautology

Conjunction and Conjunctive Syllogism
 Conjunction
P, Q
----------
 (PQ)
 Conjunctive Syllogism:
 {(PQ), P }  Q
 {(PQ) P } Q is a tautology.
 (PQ)
P
--------
Q

Fallacies
 1. The fallacy of affirming the Consequent (or affirming the
converse):
PQ
Q
_________
 P Fallacy
Ex: Consider, the following argument
If today is Mahatma Gandhi’s Birth day, then today is October 2nd.
Today is October 2nd.
 Today is Mahatma Gandhi’s Birth day.
The argument is not valid

2. Fallacy of denying the antecedent
( Or Assuming the opposite)
 Consider the following
PQ
P
_________
Q Fallacy
 Ex: Consider the following argument:
H1 : If today is Sunday, then today is Holiday
H2 : Today is not Sunday
C :  Today is not Holiday
The argument is not Valid.This is the fallacy of assuming the
opposite.

The non sequitur fallacy
 P , Q
---------
 R is a fallacy.
Ex: Consider the following argument:
1. India’s Capital is New Delhi
2. Milk is White
C:  Sun rises in the East.
The conclusion does not follow from the premises.
Hence, the argument is invalid.

Ex: Show that R follows logically from the premises
PQ, QR, P
 Proof: Consider the premises,
PQ -----(1)
QR -----(2)
P ------(3)
{1} (1) P  Q Rule P
{2} (2) P Rule P
{1, 2} (3) Q Rule T, (1), (2), and I11.
{4} (4) Q  R Rule P
{1, 2, 4} (5) R Rule T, (3), (4) and I11.

PQ, QR, P
PQ -----(1)
QR -----(2)
P ------(3)
From (1) and (2), By the rule of transitivity,we have
PR --------(4)
From (3) and (4), By the rule of Modus ponens,
R follows.
 R logically follows from the given premises

Ex: Show that P follows logically from the premises
PQ, QR, R
PQ -----(1)
QR -----(2)
R ------(3)
From (1) and (2), By the rule of transitivity,we have
PR --------(4)
From (3) and (4), By the rule of Modus tollens,
P follows.
 P logically follows from the given premises

PQ, QR, PM, M
P  Q -----(1)
Q  R -----(2)
P  M -----(3)
M ------(4)
From (3) and (4), By the rule of Modus tollens, we have
P --------(5)
From (1) and (5), By the rule of Disjunctive Syllogism,we have
Q --------(4)
R follows.

Ex: Show that (R  S) follows logically from the premises
C  D, (C  D) H, H (A B), (A B)  (R  S )
(C  D) -----(1)
(C  D)  H -----(2)
H  (A B) -----(3)
(A B)  (R  S ) ------(4)
From (2),(3) and (4), By the rule of Transitivity, we have
(C  D)  (R  S ) --------(5)
(R  S) follows.

Ex: Show that S follows logically from the premises
P  (R S), RP, P
P  (R S) -----(1)
R  P -----(2)
P -----(3)
From (1) and (3), By the rule of Modus ponens, we have
(R S) ------(4)
From (2), By Contra positive equivalence, we have
P  R -------(5)
(3) and (5), By the rule of Modus ponens, we have
R --------(6)
From (4) and (6), By the rule of Modus ponens, S follows.

Ex: Show that W follows logically from the premises
TR, S, T  W, R  S.
T  R ------(1)
S -----(2)
T  W -----(3)
R  S -----(4)
From (1), By Contra positive equivalence, we have
R  T -------(5)
From, (5) and (3), By the rule of Transitivity, we have
R W --------(6)
From (4) and (2), By the rule of Disjunctive syllogism,we have
R ---------- (7)
 From(6)and (7), By the rule of Modus ponens, W follows

Ex: Show that TP follows logically from the premises
R(ST), R  W, P S, W
R(ST) ------(1)
R  W -----(2)
P S -----(3)
W -----(4)
R ---------(5)
From(1)and (5), By the rule of Modus ponens, we have
S T ---------(6)
From, (3) and (6), By the rule of Transitivity, we have
P  T ---------(7)
 ( T P ) (By Contra positive equivalence)

Conditional Proof (CP rule)
 Theorem: If {H1, H2, …., Hn } and P imply Q, then
{H1, H2, …., Hn } imply (PQ).
 Proof: From our assumption we have,
(H1 H2  ….  Hn  P)  Q
This assumption means (H1 H2  ….  Hn  P)  Q is a tautology.
Using the equivalence P (Q R)  (P  Q)  R
We can say that (H1 H2  ….  Hn)  ( PQ ) is a tautology.
Hence the theorem.
 Rule CP : If we can derive Q from P and a set of premises,then
we can derive PQ from the set of premises alone

Ex:Show that RS can be derived from the premises
p (Q S), RP, Q
 Solution: Instead of deriving RS, we shall include R as an
additional premise and show S first.
p (Q S) …..(1)
RP …..(2)
Q ……(3)
R …….(4)
P ---------(5)
From(1)and (5), By the rule of Modus ponens, we have
Q S ………….(6)
From(3)and (6), By the rule of Modus ponens, S follows
 By CP rule, RS follows from the given premises.

Consistency, Inconsistency and Proof by Contradiction
 A set of formulas {H1, H2, …., Hn} is said to be consistent, if
their conjunction has truth value T for some assignment of the
truth values to the atomic variables appearing in H1, H2, …., Hn .
 A set of formulas {H1, H2, …., Hn} is said to be inconsistent, if
their conjunction implies a contradiction. that is
(H1 H2  ….  Hn )  (R  R) where R is any formula.
 Proof by Contradiction :
In order to show that,a conclusion C logically follows from the
premises H1, H2, …., Hn ,We assume that C is false and Consider
C as additional premise.
If the new set of premises is inconsistent, then our assumption
is wrong. Hence C follows.

Ex: Show that (PQ) follows from (P  Q)
 Solution: Let us introduce (PQ) as an additional premise and
show that this leads to contradiction.
(PQ) ….(1)
Which is equivalent to
(PQ) ….(2)
From (2), P follows
Given that, (P  Q) …..(3)
From (3), P follows
But, P and P cannot be simultaneously true (Contradiction).
 Our assumption is false.
Hence (PQ) follows from (P  Q)

Ex: Show that P follows from the premises PQ, (P  Q)
 Solution: Let us introduce P as an additional premise and show
that this leads to contradiction.
P ….(1)
PQ …..(2)
(PQ) ….(3)
From (1) and (2), By the rule of Moden ponens, we have
Q …….(4)
From (1) and (4), We have
( PQ) …….(5)
But, (3) and (5) cannot be simultaneously true (Contradiction).
 Our assumption is false.
Hence, P follows from the premises PQ, (P  Q)

Ex: Show that the following set of premises are inconsistent.
P Q, P R, Q  R, P
P  Q ------(1)
P  R -----(2)
Q R -----(3)
P ----(4)
From (1) and (3), By the rule of transitivity, we have
P R …….(5)
From(2)and (4), By the rule of Modus ponens, R follows
From(4)and (5), By the rule of Modus ponens, R follows
But, R and R cannot be simultaneously true (Contradiction).
Hence, the given premises are inconsistent.

Ex: Show that the following set of premises are inconsistent.
R  M, R  S, M, S
R  M ------(1)
R  S ----(2)
M -----(3)
S ----------(4)
R ……….(5)
From(2)and (4), By the rule of Disjunctive Syllogism, We have
R ………..(6)
But, R and R cannot be simultaneously true (Contradiction).
Hence, the given premises are inconsistent.

Ex: Verify that the following argument is valid by using the rules of
inference (Here, H1 , H2 , …. are premises and C is conclusion) :
H1 : If Joe is a Mathematician, then he is ambitious.
H2 : If Joe is an early riser, then he does not like oat meal.
H3 : If Joe is ambitious, then he is an early riser
C : Hence, if Joe is a Mathematician, then he does not like oat
meal.
 Solution: Let us make the following representations
p : Joe is a Mathematician.
q : Joe is ambitious
r : Joe is an early riser
s : Joe likes oat meal
The symbolic form of the given argument is

Contd.,
H1 : pq ….(1)
H2 : rs ….(2)
H3 : qr …..(3)
pr ……(4)
p  s ……(5)
i.e., if Joe is a Mathematician, then he does not like oat meal.
 The conclusion logically follows from the premises.
Hence, the argument is valid

Ex: Verify that the following argument is valid by using the rules of
inference (Here, H1 , H2 , …. are premises and C is conclusion) :
H1 : If Cliffton does not live in France, then he does not
speak French.
H2 : Cliffton does not drive a Datsun.
H3 : If Cliffton lives in France, then he rides a Bicycle.
H4 : Either Cliffton speaks French,or he drives a Datsun.
C : Hence, Cliffton drives a bicycle.
p : Cliffton lives in France.
q : Cliffton speaks French.
r : Cliffton drives a Datsun.
s : Cliffton drives a Bicycle.
The symbolic form of the given argument is

Contd.,
 H1 : p q …..(1)
H2 : r …..(2)
H3 : p  s …..(3)
H4 : q  r …….(4)
q ……..(5)
(1)  q  p …….(6)
From (5) and (6), By the rule of Modus ponens, we have
P ……(7)
From (3) and (7), By the rule of Modus ponens, s follows
 The conclusion logically follows from the premises.
Hence, the argument is valid

Ex:Using Symbolic logic, Show that the following
premises are inconsistent
1. If Jack misses many classes through illness,then he fails high
school.
2. If Jack fails high school, then he is uneducated.
3. If Jack reads a lot of books, then he is not uneducated.
4. Jack misses many classes through illness and reads a lot of
books.
p : Jack misses many classes through illness
q : Jack fails high school
r : Jack is uneducated
s : Jack reads a lot of books
Now, the given premises can be represented as

Contd.,
p  q …..(1) q  r …..(2)
s  r …..(3) p  s …..(4)
From (1) and (2), By transitivity, p r …..(5)
From(3), By Contra positive equivalence, r  s ….. (6)
From (5) and (6), By transitivity, we have
p s …..(7)
From(4), we have
p …..(8)
From (7) and (8), By the rule of modus ponens, s follows
From (4), s follows
But, s and s cannot be simultaneously true (Contradiction).
Hence, the given premises are inconsistent

Ex:Using Symbolic logic,prove the following argument
If A works hard, then either B or C will enjoy themselves.
If B enjoys himself, then A will not work hard.
If D enjoys himself, then C will not enjoy himself.
Therefore, If A works hard, then D will not enjoy himself .
 Solution: Let us use the following representations.
A : A works hard.
B : B will enjoy himself.
C : C will enjoy himself.
D : D will enjoy himself.
Now, we have to show that, A  D follows from
A (B  C) , B  A and D  C

Contd.,
A (B  C) ….(1) B  A ….(2)
D  C ….(3) A …. (4) ( Additional premise)
From, (1) and (4), By modus ponens, We have
(B  C) ……(5)
(2)  A  B …. (6)
From, (4) and (6), By modus ponens, B ….(7) follows.
From (5) and (7), By the rule of Disjunctive Syllogism, we have
C ….(8)
(3)  C D …. (9)
From (8) and (9), By modus ponens, D follows
Hence, By CP rule, A  D follows

Mathematical foundations of computer science

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Mathematical foundations of computer science