discrete tics unit-i

Celestine Bernard

LOGIC

Statement:

Any meaningful sentence.

Any sentence with subject, predicate

and verb

Ex:

1) Canada is a country

2)1+101=110

3) X=2 is a solution of =4.

Compound proposition:

A proposition consists of two or more

simple propositions.

Ex:

Consider any two propositions.

1) Roses are red in color.

2) Shirt is blue in color

Then the compound proposition is

Roses are red and shirts are blue in

color.

Connectors:

1. Disjunction:

Symbol: [V]

Operator name: And

Consider two propositions P and Q.

The symbolic representation will be

P V Q

Example:

P: It is raining today

Q: There are ten tables in this room.

P V Q: it is raining today or there are

Ten tables are in this room.

Truth table:

P

Q P V Q

T T T

T F T

F T T

F F F

2) Conjunction:

Symbol: [ʌ]

Operator name: OR

Explanation:

Consider any two propositions P and Q

Then the Symbolic representation will be

P Λ Q

Example:

P: It is raining today.

Q: there are ten tables in this room.

P Λ Q: It is raining today and there are ten

tables in this room.

Q Λ P: there are ten tables in this room

And it is raining today.

Truth table:

P

Q P Λ Q

T T T

T F F

F T F

F F F

3. Negation:

Symbol: ~

Operator name: negation

Explanation:

Consider a proposition P

The symbolic representation will be

~ P

Example:

P: London is a city

~P: London is not a city.

Truth table:

P ~P

T F

F T

Conditional Propositions:

Symbol: P Q (If P then Q)

Truth table:

P

Q P → Q

T T T

T F F

F T T

F F T

Ex:

P: 3=8 …………………………….F

Q: 3+5=8 ….………………………T

P Q T

If 3=8 then 3+5=8.

Biconditional propositions:

Symbol: P ↔ Q (P if and only if Q)

Truth table:

P

Q P ↔ Q

T T T

T F F

F T F

F F T

NOTE:

P → Q is direct proposition.

Q → P is transverse of P → Q.

~P → ~Q is inverse function.

~Q → ~P is contra positive.

Q → P ≡ ~P → ~Q

(Transverse ≡ Inverse)

P → Q ≡ ~Q → ~P

(Direct Proposition ≡ Contra positive)

P ↔ Q ≡ (P → Q) ʌ (Q → P)

(Biconditional ≡Direct Prop., Λ transverse)

Examples:

1) Construct truth table:

Q Λ (P Q) P

P Q PQ QΛ(PQ) QΛ(PQ)P

T T T T T

T F F F T

F T T T F

F F F F T

2) (P Q) Λ (Q P)

Truth table:

P Q PQ (QP) (PQ)Λ(QP)

T T T T T

T F T F F

F T F T F

F F T T T

3) ~( P Λ Q) ↔ (~P V ~Q)

P Q ~P ~Q (~PV~Q) PΛQ ~(PΛQ) ~(PΛQ)↔(~PV~Q)

T T F F F T F T

T F F T T F T T

F T T F T F T T

F F T T T F T T

Logical Equivalence:

Suppose that compound propositions

P and Q are made up of the propositions

P1, P2,P3,…………………………………….… Pn

Then P ≡ Q provided that given any truth

values of

P1, P2, P3 ,……………………………………….P n

either P and Q are both true or P and Q

both false.

Properties of Equivalence:

Idempotent law:

Disjunction Conjunction

P V P ≡ P P Λ P ≡ P

Associative law:

Disjunction

(P V Q) V R ≡ P V (Q V R) Conjunction

(P Λ Q) Λ R ≡ P Λ (Q Λ R)

Commutative law:

Disjunction

(P V Q) ≡ (Q V P) Conjunction (P Λ Q) ≡ (Q Λ P)

Distributive law:

Disjunction

P V (Q Λ R) ≡ (P V Q) Λ (P V R) Conjunction P Λ (Q V R) ≡ (P Λ Q) V (P Λ R)

Identity law:


T V P ≡ T T Λ P ≡ P F V P ≡ P F Λ P ≡ F

Component law:


P V ~P ≡ T P Λ ~P ≡ F ~T ≡ F ~F ≡ T

Absorption law:

Disjunction

P V( P V Q) P Conjunction P Λ( P Λ Q) P

DE Morgan’s law:

Disjunction

~(P V Q) ≡ ~P Λ ~Q Conjunction ~(P Λ Q) ≡ ~P V ~Q

Innovation law:

~(~P) ≡ P

Tautologies and Contradictions:

Tautology:

A statement is called a tautology (or) a

universally valid formula (or) logically

truth when the last column of the

statement is true.

Cheat sheet value: Last column will be

true

Ex: Find the tautology and contradiction.

1) P (PVQ)

Sol:

P Q PVQ P(PVQ)

T T T T

T F T T

F T T T

F F F T

Ans: It’s a Tautology

Contradiction:

A statement is called a contradiction if

the last column of the statement.

Cheat sheet value: Last column will be

false

Ex:

1) Find the tautology and contradiction.

(~Q Λ P) Λ Q

Sol:

P Q ~Q (~QΛP) (~QΛP)ΛQ

T T F T F

T F T T F

F T F T F

F F T F F

Ans: It’s a contradiction

Tautology (or) Contradiction:

Cheat sheet value: Last column will have

Both (True and False) values

Ex:

1) Find the tautology and contradiction

P Q PΛQ (PΛQ)↔P

T T T T

T F F F

F T F T

F F F T

Ans:

It’s neither tautology nor contradiction.

Exercise:

1) Construct truth table for the

following. And find whether it is

tautology or contradiction.

(P~P)~P

(~PQ)(QP)

[P(QR)][(PQ)(PR)]

[PΛ(PQ)](PΛQ)

[(P↔Q)]↔*(PΛQ)V(~PΛ~Q)+

Substitution instance:

A formula X is called a substitution

instance of another formula Y, If X can be

obtain from Y by substituting formula’s for

some variables of Y.

Ex:

P V Q ≡ ~(~P) V ~(~Q)

Important substitution formulae:

P ↔ Q ≡ (P→Q) V (Q P)

P Q ≡ ~P V Q

Cheat sheet value:

Any substitution instance of a

tautology is tautology

Replacement process:

It is just like substitution instance.

You can replace a formula by another

formula if it equals.

Ex:

P Λ Q (P V Q) Λ Q

Example sum:

Show that

P(QR) (P(~Q VR)) ((PΛQ) R)

Sol:

There are three parts. Consider any two

statements and prove.

P (QR) P (~Q V R)

~P V (~QVR)

Using PQ ~P V Q

(~P V~Q) V R

Using associative property

(P V Q) V R ≡ P V (Q V R)

~ (P Λ Q) V R

Using DE Morgan’s law

~P V ~Q ~ (P Λ Q)

(P Λ Q) R

Using ~P V Q P Q

2. Show that

{(PVQ) Λ~ [PΛ (~QV~R)]} V

{(~PΛ~Q) V (~PΛ~R)}

is a tautology

Sol:

This statement has two parts. So take

the first part and reduce it. Same way do it

for the second one. Finally combine those

two equations and make that as true.

(PVQ)Λ [~PΛ (~Q V~R)]

(PVQ) Λ ~ (~P Λ ~ (Q Λ R))


(PVQ) Λ ~~ (PV (QΛR))


(PVQ)Λ (PV (QVR))

Using ~~PP

(PVQ)Λ * (PVQ) Λ (PVR) +

Using distributive law

*(PVQ) Λ (PVQ)+ Λ (PVR)

Using associative law

(PVQ)Λ (PVR) ……………………….(1)

Now,

Take the second part and reduce it.

*(~PΛ~Q) V (~PΛ~R)+

~ (PVQ) V ~ (PVR)


~ *(PVQ) Λ (PVR)+…………………. (2)

Combine equations 1 & 2

[(PVQ) Λ (PVR)]V ~ *(PVQ) Λ (PVR)+

We know that

P V ~P ≡ T

PV~P

T

So it is a tautology.

For more formulae………….

Refer CHEAT SHEET

Attached with this short notes

Duality law:

Two formulae A and A* are said to be

duals of each other if either one can be

obtained from other one by replacing

V by Λ and Λ by V.

The connectives V and Λ are known as

duals of each other.

Cheat sheet value:

Replace...

V by Λ

Λ by V

T by F

F by T

Example:

Write the duals for the following.

1) (PVQ) Λ R (PΛQ) V R

2) (PΛQ) V T (PVQ) Λ F

Sol:

1) (PΛQ)V R (PVQ) Λ R

2) (PVQ)Λ F (PΛQ) V T

Cheat sheet value:

Proving methods:

Consider a question like this

Prove that A B CQ

Method I

By using property of equivalence

Ex: Associative law, DE Morgan’s law…

Method II

By using tautology

(i.e) make AB and CD as true.

Ex: AB T

CD T

Here you can write ABCD

Example for Method 2:

Prove that

P (QP) [~P (PQ)]

Sol:

Method II

P (QP) ~P V (~QVP)

(~PVP) V ~Q

T V~Q

T ………………………..(1)

[~P(PQ)]~P(~PVQ)

PV(~PVQ)

(P V~P) VQ

T VQ

T …………………………(2)

From 1 & 2, P (QP) [~P (PQ)]

Example for method I:

Prove that

P(QVR) (P Q) V(PR)

Sol:

R.H.S (PQ)V(PR)

(~PVQ) V (~PVR)

(~PV~P) V(QVR)

~P V (QVR)

P(QVR)

L.H.S

Tautological implications:

A statement M is said to be

tautologically implied to a statement N if

and only if MN is a tautology.


P Λ Q P

P Λ Q Q

P P V Q

~P P Q

Q P Q

~ (P Q) P

~ (P Q) ~Q

P Λ ( P Q) Q

~ Q Λ ( P Q) ~P

~P Λ ( P V Q) Q

( P Q) Λ (Q R) PR

( P V Q) Λ (P R) Λ (Q R) R

Cheat sheet value:

PΛQ P

Antecedent consequent

Proof:

1) PΛQ P

Sol:

P Q P V Q P V QP

T T T T

T F F T

F T F T

F F F T

It is a tautology.

Therefore PΛQP

Same way we can prove all statements.

Normal forms:

DNF :( Disjunctive Normal Form)

Cheat sheet value:

DNF is not unique

It should have only Λ and V symbols

It shouldn’t have any other symbols

Ex: and ↔

Syntax:

(… A ...) V (...B...)

Example:

Find DNF of the following

(I) PΛ(PQ)

(II) P[(PQ)Λ~(~QV~P)+

Sol:

(i) PΛ(PQ)

PΛ (~PVQ)

(PV~P) V (PVQ)……………….. (DNF)

(ii) P[(PQ)Λ(~QΛ~P)+

Sol:

~PV [(PQ) Λ (~QΛ~P)+

~PV *(~PVQ) Λ~~ (QVP)+

~PV *(~PVQ) Λ (PVQ)+

~PV *(~PVQ) Λ (PΛQ)+

~PV ,*~PΛ (PΛQ)+ V *QΛ (PΛQ)+-


~PV *(~PΛPΛQ) V (QΛPΛQ)+

~PV *(TΛQ) V (PΛQ)+

~PV *F V (QΛP)+

~PV (QΛP) …………………… (DNF)

Conjunctive normal form (CNF):

Cheat sheet value:

(… A …) Λ (… B …)

It should have only Λ and V symbols

It shouldn’t have any other symbols

Ex: and ↔

Example:

Find the CNF for the following

a) PΛ(PQ)

b) *PΛ~(PVR)+V , *(PΛQ)V~R+ Λ P-

Sol:

a) PΛ(PQ)

PΛ (~P V Q)………………………. (CNF)

b) *PΛ~(PVR)+V { *(PΛQ)V~R+ Λ P-

*PΛ~ (PVR)+V ,*(PΛQ) V~R+ Λ P-

*PΛ (~PΛ~R)+V ,*(PΛQ) ΛP] V

[~RΛP+- …using distributive law

*(PΛ~P) Λ~R+V ,*(PΛP) ΛQ+ V

*~RΛP+-

*FΛ~R+ V *(PΛQ) V (~RΛP)+

F V *(PΛQ) V (~RΛP)+

(PΛQ) V (~RΛP) ……………….. (CNF)

Principle disjunctive normal form (PDNF):

Let P and Q be two statement

variables. Let us construct all possible

formulas which consist of P, ~P, Q, ~Q.

The possible forms are

PΛQ, ~PΛQ, PΛ~Q, ~PΛ~Q

These formulas are called minterms or

Boolean conjunctions.

For a given formula, an equivalent

formula consisting of disjunctions of

minterms only is known as its PDNF.

Such a normal form is also called the

Sum of products canonical form.

Cheat sheet value(PDNF):

Syntax:

(…V…) Λ (…V…) Λ(…V…)

Obtain PDNF for the following.

1) ~P V Q

2) (PΛQ) V (~PΛR) V (QΛR)

Sol:

1) ~P V Q

Introduce a variable T

(~PVT) V (QVT)

Introduce the missing variable using

component law.

[~PV (~QVQ)] V [QV (PV~P)]

(~PΛQ) V (~PΛ~Q) V (QΛP)V(QΛ~P)


(~PΛQ) V (~PV ~Q) V (QΛP)

If a minterm repeats just negate 1

PDNF

2) (PΛQ) V (~PΛR) V (QΛR)

Introduce the new variable T

*(PΛQ)ΛT+V*(~PΛR)ΛT+ V*(QΛR) ΛT+

*(PΛQ) Λ(RV~R)+V*(~PΛR)Λ(QV~Q)+

V *(QΛR)(PV~P)+

(PΛQΛR) V (PΛQΛ~R) V(~PΛRΛ~Q)

V (~PΛRΛ~Q) V(QΛRΛP)V (QΛRΛ~P)

Using distributive law.

If a minterm repeats just negate 1.

(PΛQΛR) V (PΛQΛ~R) V

(~PΛQΛR) V (~PΛ~QΛR)

PDNF

Principle conjunctive normal form (PCNF):

Let us consider two variables P & Q

Let us write all possible maxterms.

The maxterms are,

P V Q

~P V ~Q

~P V Q

P V ~Q

For a given formula, an equivalent formula

consisting of conjunctions of maxterms

only is known as its PCNF. This form is also

called as the product of sums of canonical

form.

Cheat sheet value:

Syntax:

(…Λ…) V (…Λ…) V(…Λ…)

Example:

Obtain PCNF for the following.

P↔Q

Sol:

Step 1: construct the truth table

P Q Minterms P↔Q

T T P Λ Q T T F P Λ~Q F F T ~PΛQ F F F ~PΛ~Q T

*T=P & F=~P ⌂ *T=Q & F=~Q

Step 2:

Write down the minterms which has true

value and make that as A

A (PΛQ) V (~PΛ~Q)

Step 3:

Find ~A

~A (~PΛ ~Q) V (P ΛQ)

*P=~P *Q=~Q

Step 4:

Find ~ (~A)

~(~A)~{ (~PΛ ~Q) V (P ΛQ)-….PCNF of A

For more theory refer discrete maths I-2

Celestine Bernard

Discrete mathematics

Cheat Sheet

Disjunction:

Truth table:

P

Q P V Q

T T T

T F T

F T T

F F F

Conjunction:

Truth table:

P

Q P Λ Q

T T T

T F F

F T F

F F F

Negation:

Truth table:

P ~P

T F

F T

Conditional propositions:

Truth table:

P

Q P → Q

T T T

T F F

F T T

F F T

Biconditional propositions:

Truth table:

P

Q P ↔ Q

T T T

T F F

F T F

F F T

NOTE:

P → Q is direct proposition.

Q → P is transverse of P → Q.

~P → ~Q is inverse function.

~Q → ~P is contra positive.

Q → P ≡ ~P → ~Q

(Transverse ≡ Inverse)

P → Q ≡ ~Q → ~P

(Direct Proposition ≡ Contra positive)

P ↔ Q ≡ (P → Q) ʌ (Q → P)

(Biconditional ≡Direct Prop., Λ transverse)

Properties of Equivalence:

Idempotent law:


P V P ≡ P P Λ P ≡ P

Associative law:

Disjunction

(P V Q) V R ≡ P V (Q V R) Conjunction (P Λ Q) Λ R ≡ P Λ (Q Λ R)

Commutative law:

Disjunction

(P V Q) ≡ (Q V P) Conjunction (P Λ Q) ≡ (Q Λ P)

Distributive law:

Disjunction

P V (Q Λ R) ≡ (P V Q) Λ (P V R) Conjunction P Λ (Q V R) ≡ (P Λ Q) V (P Λ R)

Identity law:


T V P ≡ T T Λ P ≡ P F V P ≡ P F Λ P ≡ F

Component law:


P V ~P ≡ T P Λ ~P ≡ F ~T ≡ F ~F ≡ T

Absorption law:

Disjunction

P V( P V Q) P Conjunction P Λ( P Λ Q) P

DE Morgan’s law:

Disjunction

~(P V Q) ≡ ~P Λ ~Q Conjunction ~(P Λ Q) ≡ ~P V ~Q

Innovation law:

~(~P) ≡ P

Substitution instance:

P ↔ Q ≡ (P→Q) V (Q P)

P Q ≡ ~P V Q


P Λ Q P

P Λ Q Q

P P V Q

~P P Q

Q P Q

~ (P Q) P

~ (P Q) ~Q

P Λ ( P Q) Q

~ Q Λ ( P Q) ~P

~P Λ ( P V Q) Q

( P Q) Λ (Q R) PR

( P V Q) Λ (P R) Λ (Q R) R

Implication, Equivalence formula’s for

Inference theory:

Implication formulae:

P P V Q ……………………………….(I -1)

Q P V Q ……………………………….(I-2)

P Λ Q P ………………………………..(I-3)

P Λ Q Q ……………………………….(I-4)

P , PQ Q …………………………..(I-5)

~Q , PQ ~P ……………………….(I-6)

~P, P V Q Q ………………………….(I-7)

P Q, Q R PR ………………(I-8)

P , Q P Λ Q ……………………………(I-9)

Q PQ ……………………………….(I-10)

P V Q , PR , QR R ………….(I-11)

~P PQ ……………………………….(I-12)

~(P Q) P …………………………..(I-13)

~(P Q) ~P ………………………..(I-14)

Equivalence formulae:

~ ~ P P …………………………………(E-1)

P Q ~P V Q ……………………….(E-2)

P Q ~Q ~P ……………………(E-3)

P ↔ Q (P Q) Λ (QP) …….(E-4)

P (Q R) (P Λ Q) R …….(E-5)

~(P Λ Q) (~P V ~Q) ………………..(E-6)

(P ↔ Q ) (P Λ Q) V (~P V ~Q)…(E-7)

Quantifiers:

There are two types of quantifiers.

Universal quantifier

Symbol: (Ɏ) Meaning: All

Existential quantifier

Symbol :( ƭ) Meaning: some

Equivalent formulae :( Quantifiers)

(Ɏx) *P(x)+ ~(Ɏx) *~P(x)+

(Ɏx) *~P(x)+ ~(ƭx) P(x)

~(Ɏx) *P(x)+ (ƭx)*~P(x)+

~(Ɏx) *~P(x)+ (ƭx) *~P(x)+

(ƭx) *A(x) V B(x)+ *(ƭx)A(x) V (ƭx)B(x)+

(Ɏx)*A(x) Λ B(x)+ *(Ɏx)A(x)V(Ɏx)B(x)+

(Ɏx)*A Λ B(x)+ A V (Ɏx)B(x)

(ƭx)*A Λ B(x)+ A Λ (ƭx) B(x)

(Ɏx) A(x) B (ƭx) (A-1(x)B)

(ƭx) A(x) B (Ɏx)*A(x)B]

A(Ɏx)B(x) (Ɏx)*AB(x)]

A(ƭx)B(x) (ƭx)* AB(x)]

Implication formulae:

(Ɏx)A(x) V(Ɏx)B(x) (Ɏx)*A(x) VB(x)+

(ƭx) *A(x) Λ B(x)+ (ƭx) A(x) Λ (ƭx) B(x)

P(y) (ƭx)P(x)

(Ɏx)P(x) (ƭx) P(x)

(Ɏx)*P(x)Q(x)] (Ɏx)P(x)(Ɏx)Q(x)

Theory of inference:

Rules:

1. Universal specification(UF):

(Ɏx) A(x) A(y)

2. Universal generation(UG):

A(y) (Ɏx)A(x)

3. Existential specification(ES):

(ƭx)A(x) A(y)

4. Existential generation(EG):

A(y) (ƭx)A(x)

discrete tics unit-i

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