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MODULE I
MATHEMATICAL LOGIC
Statements or Propositions
A statement or proposition is defined to be a declarative sentence to which it is possible to assign a
true value true or false but not both simultaneously. The statement or proposition that cannot be broken
into simpler statements are called primary or primitive statements. They are usually denoted by the
alphabets A,B,C,... or by suffix capital letters P1,P2,...
Compound propositions
Statements or propositions which can further be simplified into atomic statements or propositions
are called compound propositions.
Eg. Ram and Rahim went up the hill.
Two valued logic
Since propositions have two values true or false the logic is called two- valued logic. They may also
be denoted by the symbols 1 or 0.
Connectives
The connecting words between two atomic statements is called connectives. The connectives
generally used in English are 'and' and 'or'.
Eg.1.Ram and Rahim went up the hill.
2.Today is a holiday or tomorrow is a holiday.
Negation
If 'P' denote a statement then the negation of 'P' is written as '‫ך‬P' and is read as 'not P'.
Eg.1.Ram is an intelligent boy.(P)
2.Ram is not an intelligent boy.(‫ך‬P)
Negation does not connect two statements but it modifies one statement and it is also taken as a
connective.
Truth Table of ‫ך‬P
P
‫ך‬P
T
F
F
T
Conjunction
If P and Q are two statements, their conjunction is a statement denoted by PΛQ which is read as P
and Q. The statement PΛQ has true value T only when both P and Q have trhth value T.In all other
cases it has truth value F.
Truth Table of PΛQ
P
Q
PΛQ
T
T
T
T
F
F
F
T
F
F
F
F
Eg. P: Jack went up the hill.
Q: Jill went up the hill.
PΛQ: Jack and Jill went up the hill.
Disjunction
Let P and Q be two given statements.Then the disjunction of the statements is denoted by PVQ and
is read as P or Q. The new statement has truth value F only if both P and Q have truth value F. In all
other cases it is T. In other words the statement PVQ is true if at least one of the statement P or Q is
true.
Truth Table of PVQ
P
Q
PVQ
T
T
T
T
F
T
F
T
T
F
F
F
Statement formula
The statement which contain one or more atomic statements and connective are called statement
formula.
Eg. ‫ך‬P, PVQ , PΛQ , ‫( ך‬PVQ), ‫(ך‬PΛQ) etc.
Problems
1.Construct the truth table for PΛ‫ך‬Q
P
Q
‫ך‬Q
PΛ‫ך‬Q
T
T
F
F
T
F
T
T
F
F
T
F
F
T
F
F
Construct the truth table of the following statements.
1) Pv‫ך‬Q
2) ‫ ┐ ( ך‬PΛ‫ך‬Q)
3) ‫ ┐ ( ך‬Pv‫ך‬Q)
Conditional
If P and Q are any two statements then the statement P→Q which is read as if P then Q
is called a conditional statement.
Eg. There is a flood : P
Crop will be destroyed : Q
Truth Table of p→Q
P
Q
P→Q
T
T
T
T
F
F
F
T
T
F
F
T
The statement P→Q has the truth value F when P has truth value T and Q has truth value F.
Otherwise the result is true.
Problems
Write in symbolic form.
If either Ram takes maths or Raj takes physics then Rahim will take english.
Ram takes maths. : P
Raj takes physics. : Q
Rahim takes english. : R
Symbolic form : (PVQ→R)
Write the truth table for the following.
1.(P→Q) Λ (Q→P)
P Q
P→Q
Q→P (P→Q) Λ (Q→P)
T T
T
T
T
T F
F
T
F
F T
T
F
F
F F
T
T
T
2.QΛ(P→Q)→P
3.(P→Q) → (Q→R)
Biconditional Statements.
If P and Q are any two statements then the statement P↔Q is called a biconditional
statement. It is read as P if and only if Q. ( P iff Q) The statement has the truthvalue T if either, P and
Q has truthvalue T , or both have truthvalue F. It is defined by the following truth table.
P
T
T
F
F
Q
T
F
T
F
PQ
T
F
F
T
Actually this statement P↔Q is equivalent to the conjunction of two implications P→Q and Q→P.
eg. P- Two lines are parallel.
Q- Two lines have same slope.
PQ – Two lines are parallel if they have same slope.
Write the truth table for the formulas 1)┐(PQ) ┐P┐Q
2)┐(P(QR)) PQ )PR)
[H.W]
Ans.(1)
P
Q
PQ ┐(PQ)
┐P
┐Q
┐P┐Q
T
T
F
F
T
F
T
F
T
F
F
F
F
T
T
T
F
F
T
T
F
T
F
T
F
T
T
T
┐(PQ) ┐P┐Q
T
T
T
T
Well formed formula
A statement formula is an expression which is a string consisting of statement variables, parenthesis
and connecting symbols which produces a statement when the variables are replaced by symbols. But
every string of these symbols is not a formula.
Definition
A well formed formula (wff) can be generated by the following rules.
1) A statement variable standing alone is a wff.
2) If A is a wff then ┐A is also a wff.
3) If A and B are wff, then AVB, AΛB, A→B and A↔B are wff.
4) A string of symbols containing statement variables, connectives and parenthesis is a wff iff it
can be obtained by finitely many applications of rules 1, 2 and 3.
eg. ┐PQ), ┐(PQ ), ( P→(PQ), ((P→Q) Λ (Q→R)↔(P→R))
The following are not wff.
1) (P→Q) →( Q) Since  Q is not wff.
2) ( P→Q
The wff is (P→Q).
Tautology (Logical truth)
A statement formula which is true regardless of the truthvalue of the statement which replace the
variable in it, is called tautology.
In other words if each entry in the final column of the truth table of a statement formula is true, then it
is called a tautology.
Similarly, a statement formula which is false regardless of the truthvalue of the statement which
replace the variable in it, is called contradiction.
In other words if each entry in the final column of the truth table of a statement formula is false, then
it is called a contradiction.
We can call tautology as identical true statement and contradiction as identical false statement.
Notes:
1) The negation of a tautology is a contradiction and negation of a contradiction
is a tautology.
2) The conjunction and disjunction of two tautologies is a tautology and the conjunction and
disjunction of two contradictions is a contradiction.
3) The disjunction of a tautology and any other formula is a tautology.
4) The conjunction of a contradiction and any other formula is a contradiction.
5)
Check whether the formula given is tautology or not.
1) (P┐P) ┐P
(P┐P) ┐P
P
┐P
T
F
F
T
F
T
T
T
P┐P
It is a tautology.
2) (P→Q) →P
3) (P→Q→R)) → ((P→Q)→(P→R))
4)(┐P→Q)→(Q→P)
[HW]
Equivalence formula
Two statements A and B are said to be equivalent to each other iff the formula AB is a tautology.
We write AB and it is read as A is equivalent to B or A is implies and implied in B.
Notes:
It is a symmetric relation.
It is transitive. Ie; if AB and BC then AC.
Prove the following.
1) (PQ) P→Q)(Q→P)
P→Q)(Q→P)
P
Q
PQ
P→Q
Q→P
T
T
T
T
T
T
T
F
F
F
T
F
F
T
F
T
F
F
F
F
T
T
T
T
Note
If the truthvalue of A is equal to the truthvalue of B then A and B are said to be equivalent formulas.
2) ┐(P→Q) P┐Q)
3) (PQ) (P┐Q) P
4) ┐(PQ) ┐P┐Q
5) ┐┐PP
6) PQQP
7) P QQP
8) (P Q)R  P (QR)
9) (PQ)R  P(QR)
10) ┐(P Q)┐P┐Q
11) P PP
12) PPP
Some important equivalent formulas
1) Idempotent laws
P PP
PPP
2) Associative laws
(P Q)R  P (QR)
(PQ)R  P(QR)
3) Commutative laws
PQQP
P QQP
4) Distributive laws
P (QR)  (P Q) ( PR)
P (QR)  (PQ)  ( PR)
5) Absorption laws
P(PQ)  P
P(P Q)P
6) Demorgan's laws
┐(P Q)┐P┐Q
┐(PQ) ┐P┐Q
7) Some more results
P FP
PTP
P T
PFF
PQ P→Q)(Q→P)
P→Q┐PQ
Q→P┐QP
┐(P→Q) P┐Q)
Replacement process
Let A be the formula P→(Q→R). The formula Q→R is a part of the formula A. We can replace
Q→R by ┐QR. Now A becomes P→ (┐QR). Let this formula be B. It can be easily verified that A
and B are equivalent to each other. This process of obtaining B from A by replacing a part of the
formula A by an equivalent formula is known as Replacement process. This can be used to obtain a
simpler formula equivalent to a given formula. Also it can be used to determine the given formula is a
tautology.
(I)Show the following without using truthtable.
1)(A→ B) (C→ B)  (A C)→ B
LHS = (A→ B) (C→ B)
┐AB) (┐C B)
B┐A) (B┐C)
 B┐A ┐C)
B┐AC )
 ┐AC )B
 AC )→ B = RHS
2) P→(Q→R) PQ) →R
3) A(A→ B)→ B T
4) P→ (Q→R) ┐P→ (P→Q)
[H W]
Tautological implications
A statement formula A is said to be tautologically imply a statement B iff A→ B is a tautology
and it is denoted by AB and it is read as A implies B.
1)Prove that ┐QP→Q) ┐P
┐Q
P
Q
T
T
F
F
T
F
F
T
T
F
F
T
┐P→ Q) P
PQQ
P (PQ)
Q PQ
Q P→Q
2)
3)
4)
5)
6)
P→Q
┐P [A]
T
F
T
T
F
F
T
T
┐QP→Q) [B]
F
F
F
T
A→ B
T
T
T
T
[H W]
Joint implications
Let H1, H2,.....,H m be m formulas. We say that H1, H2,.....,H m jointly implies a formula Q if
( H1 H2 .....H m) Q.
ie if H1, H2,.....,H m→ Q is a tautology .In this case we write H1, H2,.....,H mQ.
H1, H2,.....,Hm are called premises and Q is called conclusion.
Theorem
If H1, H2,.....,Hm,PQ then H1, H2,.....,Hm P→Q
Proof
By using the result, PQ) →R P→(Q→R)
If, If H1, H2,.....,H m,PQ, H1 H2 .....H m P→ Q is a tautology.
m → (P→Q) This is also a tautology.
 H1, H2,.....,H mP→Q
Law of duality
Let A be a formula containing the connectives ┐,andonly. If A contains any other connectives, it
may be replaced by an equivalent formula containing these connectives only. Then let A * be the
formula obtained by this replacing byand by Then A and A* are said to be duals of each
other. The connectives andare also calledduals of each other.
If A contains the special variables T or F then replace T with F and F with T to get its dual A*
eg,1) If A: (P  Q) R then A*: (P Q) R
2) P1: ┐PQ) P*:┐PQ)F
 Result
Let A and A* be dual formulas and P1, P2,.....,Pn be all the atomic variables occur in A and A*.
ie; A : A( P1, P2,.....,Pn) and
A*: A*( P1, P2,.....,Pn)
Then, ┐A( P1, P2,.....,Pn) A*( ┐ P1, ┐P2,.....,┐Pn)
eg. ┐(P  Q)  ┐P  ┐Q
eg. Show that ┐( ┐ P┐(QR)) P Q R)
Let A(P,Q,R) be ┐ P┐(QR)
Then, A* (P,Q,R) is ┐P┐(QR)
and A* (┐P,┐Q,┐R) is ┐┐P┐┐Q ┐R)
ie; P Q R)
But ┐A( P,Q,R ) A* (┐P,┐Q,┐R)
┐( ┐ P┐(QR)) P Q R).
3)
Result
If any two formulas are equivalent, then their duals are also equivalent to each other.
Ie; if A then A**
Normal forms
Here we use 'product' in place of conjunction and 'sum' in place of disjunction.
Elementary Product
A product of variables and their negation in a formula is called an elementary product.
eg. P Q, P┐Q
Elementary Sum
A sum of variables and their negation in a formula is called an elementary sum.
eg. PQ, P┐Q, P┐QR
Disjunctive normal form (DNF)
A formula which is equivalent to a given formula and which consists of sum of elementary
product is known as DNF.
1) Obtain the disjunctive normal form of P (P→ Q)
P (P→ Q)  P (┐PQ)
 ( P ┐P)(PQ)
Extended Distributive law
(PQ) (RS)  (PR)(PS) (QR)(Q S)
Obtain the DNF of 1) ┐(P→ QR) )
 ┐(┐P(QR))
 ┐┐P ┐QR)
P(┐Q ┐R)
P┐QP┐Rwhich is a dnf.
2) ((PQ)(P┐Q)) ((P┐Q)(┐P┐Q))
3) ( P┐(QR)( PQ)┐R)P)
Disjunctive normal form of a given formula is not unique. They are equivalent.
Conjunctive normal form (DNF)
A formula which is equivalent to a given formula and which consists of product of elementary
sum is known as CNF.
1) Obtain the conjunctive normal form of P (P→ Q)
P (┐PQ)
2) QP┐Q) ┐P┐Q))
3) ┐( PQ)  PQ
Minterms
minterm consists of conjunction in which each statement variable or its negation , not both appears
only once.
For two variables P and Q there are 22 minterms. These are known as Boolean conjunction of P and Q.
From the truth table of these minterms it is clear that no minterms are equivalent.
Each minterms have truth value T for exactly one combination of truth values of the variables P and Q.
Principal disjunctive normal form (PDNF).
For a given formula consisting of disjunction of minterms is known as PDNF.
(I)Obtain pdnf of 1) ┐PQ
┐PQ (┐P)Q)
(┐PQ┐Q))(QP┐P))
(┐PQ)(┐P┐Q) QP)(Q ┐P)
(┐PQ)(┐P┐Q) QP)
2) Find the pdnf of the following.
i) P→((P→ Q) ┐(┐Q┐P)
ii) ( PQ)(┐PR)QR)
iii) P(┐PQ) PQ
iv) P(┐P┐QR)
v) P( PQ)P
(II)Obtain pcnf of the following.
i) Q( P┐Q)
Q( P┐Q) (QF)( P┐Q)
(Q( P┐P)) ( P┐Q)
(Q P)(Q┐P) ( P┐Q)
ii) (Q→ P)(┐PQ)
iii) P(┐P→ (Q(┐Q → R))
iv) (┐P→R)(Q P)
The theory of inference for statement calculus
1)Determine whether the conclusion C follows logically from the premises H1 and H2
H1:P→Q, H2: P, C:Q
P (H2)
T
P→ Q (Q)
Q (C)
T
T
T
F
F
F
T
T
F
F
F
We observed that the first row is the only row in which both the premises have the value T. The
conclusion also have the value T in this case. Hence it is valid.
2) Check whether the statements given are is valid or not.
H1:P→Q, H2:┐P, C:Q
3) H1:P→Q, H2:┐(PQ), C:┐P
Without using truthtable
The truth value techniques become difficult when the no. of atomic variables is large. So we have to
consider some other method of derivation. For this purpose we may use the result of equivalence and
implications we have already seen. The important results of implications and equivalence are listed
below.
Implications
Simplification
I1:PQ P
I2:PQ Q
Addition
I3:PPQ
I4:QPQ
I5:┐PPQ
I6:QP→Q
I7:┐(P→Q)P
I8:┐(P→Q)┐Q
I9:P,QPQ
I10:┐P,PQQ
I11:P,P→Q Q (modus ponens)
I12:┐Q,P→Q┐P (modus tollens)
I13:P→Q ,Q→RP→R
I14:PQ,P→R Q→R R
Equivalence
E1 :┐┐P P
E2: PQQP
E3 : PQQP
E4: (PQ)RP(QR)
E5 : (PQ)RPQR)
E6:PQR) PQ) (PR)
E7 :P(QR) (PQ)( PR)
E8:┐( PQ) ┐P┐Q
E9 : ┐( PQ) ┐P ┐Q
E10: PPP
E11 :PP P
E12:P┐P T
E13 :P┐PF
E14: PTT
E15 :PTP
E16 :PF P
E17: PFF
E18 : P→Q┐PQ
E19:┐(P→Q) P┐Q
E20 :P→Q ┐Q→ ┐P
E21: P→(Q→R)PQ)→ R
E22 :┐(PQ )P┐Q
E23 : PQ┐PQ) ( ┐QP)
E24:PQ (PQ) (┐P┐Q)
Using these results by replacement process, we may derive a required implication.
But there is another way for a formal proof. It is a step by step derivation mentioning the justification
for each step. For this purpose, in addition to the above results we may use the following 3 rules of
inference.
Rule P
We may introduce a premise at any point of the derivation.
Rule T
We may introduce a formula S in a derivation if S is tautologically implied by one or more of the
preceding formulas in the derivation.
Rule CP
If we can derive S from R and a set of premises, then we can derive R→S from from the set of
premises alone.
iePR)→ S P→(R→S)
1)P┐Q, Q R, ┐SP, ┐R S. show that S is a valid statement.
Step No.
Statement
Reason
1
Q R
Rule P
2
┐R
Rule P
3
Q
Rule T
[from steps 1 and 2]
Q R┐R Q
4
P┐Q
Rule P
5
┐P
Rule T
6
┐SP
Rule P
7
S
Rule T
[from steps 3 and 4]
[from steps 5 and 6]
Hence S is a valid statement.
2) Show that ┐( P  ┐Q) ( ┐QR) ┐R  ┐P
3) Show that RS logically follows from CD,(CD) ┐H, ┐H A  ┐B ,(A ┐B)RS
4) Show that RPQ) is a valid conclusion from PQ, QR, PM, ┐M.
5) Show that SR is a valid conclusion from the premises PQ, PR, QS.
6) Derive P→(Q→S) from the premises P→(Q→R), Q→(R→S).
Indirect method of proof
The technique of indirect method of proof is as follows.
1. Introduce the negation of the desired conclusion as a new premise.
ie;assume the conclusion C is false and consider ┐C as an additional premise.
2. From the additional or new premise, together with the new premises derive a contradiction.
3. Assert a desired conclusion as a logical inference from the premises. Thus C follows
logically from the premises H1, H2, ........,Hn.
Def: A set of formulas H1, H2, ........,Hn is said to be consistent if their conjunction has the truth value
T for some assignment of the truth values to the atomic variables appearing in H1, H2, ........,Hn.
If for every assignment of the truth value to the atomic variable, at least one of the formulas
H1, H2, ........,Hn is false, so their conjunction is identically false. Therefore the formulas
H1, H2, ........,Hn are inconsistent.
In other words, a set of formulas
contradiction.
ie;
H1, H2, ........,Hn is consistent if their conjunction implies a
H1 H2 .......Hn R┐R
1) Show that ┐( PQ) follows from ┐P┐Q by indirect method.
C = ┐( PQ)
┐C = ┐( ┐P ┐Q)
= PQ (Additional premise.)
Step No.
Statement
Reason
1
PQ
Rule P
2
P
Rule T
3
┐P┐Q
Rule P
4
┐P
Rule T
5
┐PP
Rule T
The statement in step 5 is a contradiction. Hence we say that ┐( PQ) follows from ┐P┐Q.
Show that the following premises are inconsistent.
1. P→Q, P→R, Q→ ┐R, P.
Step No.
Statement
Reason
1
P→Q
Rule P
2
Q→ ┐R
Rule P
3
P→ ┐R
Rule T
4
PR
Rule P
5
┐R→ ┐P
Rule T
6
P┐P
Rule T
7
┐P  ┐P
Rule T [from step 6] P→Q┐PQ
8
┐P
Rule T
9
P
Rule P
10
┐PP
[from steps 1 and 2] P→Q ,Q→ R P→R
[from steps 3 and 4]
Rule T
Therefore the given premises are inconsistent.
2) By indirect method show that ┐Q ,P→Q, PRR
3) By indirect method show that P→Q, Q→R,PRR
Predicate calculus
Consider the statements – John is a student, Raj is a student. In the statement calculus we need
different statement symbols foe these statements even though they have a common feature. The
statement “John is a student” has two parts. “John” is the subject and “is a student” is the predicate.
Now we may represent the predicate “is a student” by B and the name by “j”. Then the above
statement can be written as B(j), B(r).
Generally we may say that B(x) represent “x is a student”. Here “B” represents the “predicate” “is
a student” and x is a place holder.
B(x) may be called as a statement function or predicate and x is the object or variable. Usually
predicates are represented by capital letters and object by small letters.
In general, the statement of the form “p is Q” is represented by Q(p). In the above eg. B(x) where x
is a student is a one-place predicate, since it is related to one object. Now consider “< : is less than”.
We may consider a two place predicate L(x,y) : x is less than y.
S(x,y,z) : x sit b/w y and z.
S : sit b/w.........3-place predicate
In general, an n-place predicate requires n names of objects to be inserted in fixed positions in order
to obtain a statement.
A single proposition or statement may be considered as a zero place predicate.
Quantifiers
Certain statement involves words that indicate quantity such as all, some, none, one etc. Since
such words indicate quantity , they are called quantifiers.
The quantifier “all” is the universal quantifier. The symbol, x represents the phrases, for all x,
for every x, for each x, everything x is such that, each thing x is such that etc.
The quantifier some is an existential quantifier. We denote this by the symbol x. This represent
the phrases for some x, some x such that, there exist an x such that, there is an x such that, there is at
least one x such that, etc.
eg.1) Consider the statement – All roses are red. We can rewrite the above statement as, for all x if x is
a rose then x is red.
R(x) : x is a rose.
C(x) : x is red.
Then the above statement can be written as, x(R(x)C(x))
eg.2) All men are good.
M(x) : x is a man.
G(x) : x is good.
Then the statement becomes, x(M(x)G(x))
eg.3) Some people who trust others are rewarded.
There exist x, x is a person, x trust others and x is rewarded.
P(x) : x is a person.
T(x) : x trust others.
R(x) : x is rewarded.
x (P(x)T(x)R(x))
Predicate calculus – Rules of inference
Rule P : Rule P implies given result.
Rule T : Implies derived result.
Rule US : (Universal specification)
x A(x) A(y)
Rule UG : (Universal generalization)
x A(y)x A(x)
Rule ES : (Existential specification)
x A(x) A(y)
Rule EG : (Existential generalization)
A(y)x A(x)
1) Show that x M(x) follows logically from the premises (x)(H(x) M(x)) and x H(x).
Step No.
Statement
Reason
1
x H(x)
Rule P
2
H(y)
Rule ES
3
(x)(H(x) M(x))
Rule P
4
H(y) M(y)
Rule US; P,PQQ
M(y)
x M (x)
6
Rule T
Rule EG
2) Prove that x(P(x)Q(x))x P(x)x Q(x)
3)Verify the validity
All men are mortal. Socrates is a man. He is mortal.
4) Verify the validity of the following argument
Tigers are dangerous animals. There are tigers. Therefore there are dangerous animals.
Let T(x) : x is a tiger
D(x): x is a dangerous animal
Then the inference pattern is,
(x) ( H(x) M(x))
x T(x)
x D(x)
1 x T(x)
Rule P
2 T(b)
Rule ES, 1
3 (x) ( T(x) D(x))
Rule P
4 T(b) D(b)
Rule US, 3
5 D(b)
Rule T, 2,3
6 x D(x)
Rule EG,5
Thus the inference is valid.
Universe of discourse
Consider the statement.- All men are giant.
Let : M(x) : x is a man.
G(x) : x is giant.
Then the above statement can be written as (x) ( M(x) G(x))
However if we restrict the variable x to the universe which is the class of men. Then the statement
becomes (x) (G(x)) . Such a restricted class is called universe of discourse.
Write the predicate : Given any positive integer, there exist a greater positive integer with and without
giving the universe of discourse.
P(x) : x is a positive integer.
G(y,x) : y is greater than x.
(x) ( P(x) G(y,x))
i) x ( P(x) (y) (G(y,x)))
or x ( P(x) (y) (P(y)G(y,x))
ii) x ((y)G(y,x))
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