Truth Table

1

WHAT IS TRUTH TABLE A truth table specifies the truth value of a compound proposition for all possible truth values of its constituent proposition.

A convenient method for analyzing a compound statement is to make a truth table to it

NEGATION (~)

If p=statement variable, then negation of ‘p’ “NOT p”, is denoted by “~p”

If p is true, ~p is false

If p is false ~p is true

TRUTH TABLE FOR ~P

CONJUNCTION (^)

If p and q is statement then conjunction is “p and q”

Denoted by “p ̂ q”

If p and q are true then true

If both or either false then False

Truth Table

2

P ̂ q

DISJUNCTION (v)

If P and q is statement then “p or q”

Denoted by “p v q”

If both are false then false

If both or either is true then true

P v q

Truth Table for this statement ~p^ q

Truth Table

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Truth Table for ~p^ (q v ~r)

Truth table for (p v q) ^ ~ (p ^q)

Double negation property ~ (~p) =p

So it is clear that “p” and double negation of “p” is equal.

Truth Table

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Example

English to symbolic

P= I am Ali

~p= I am not Ali

~ (~p) = I am Ali

So it is clear that double negation of “p” is also equal to “p”.

~ (p ^q) & ~p ^~q are not logically Equivalent.

So it is clear that “~ (p ̂ q) & ~p ̂ ~q” are not equal

De Morgan’s Law

1. The negation of “AND” statement is logically equivalent to the “OR” statement in which each component is negated.

Symbolically ~ (p ^q) = ~p v ~q

P q P ^q ~(p ^q) ~p ~q ~p v ~q

T T T F F F F

T F F T F T T

F T F T T F T

F F F T T T T

So it is clear that ~ (p ̂ q) = ~p v ~q are logically equivalent

Truth Table

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2. The negation of “OR” statement is logically equivalent to the “AND” statement in which component is negated.

Symbolically ~ (p v q) = ~p ^ ~q

So it is clear that ~ (p v q) = ~p ^ ~q is Equal.

Application

Negation for each of the following:

a. The fan is slow ‘OR’ it is very hot

b. Ali is fit ‘OR’ Akram is injured.

Solution:

a. The fan is not slow “AND” it is not very hot

b. Ali is not fit “AND” Akram is not injured

Truth Table

6

Inequalities and Demorganes laws

1. (p ^q) ^ r = P^(q ^ r)

p q r (p ^q) (q ^r) (p ^q)^r P^(q ^r)

T T T T T T T

T T F T F F F

T F T F F F F

T F F F F F F

F T T F T F F

F T F F F F F

F F T F F F F

F F F F F F F

So it clears that Colum 6 and Colum 7 are equal

2. (P ^q) v r = p ^ (q v r)?????

P q r (p ^q) (q v r) (p ^q) v r P ^(q v r)

T T T T T T T

T T F T T T T

T F T F T T T

T F F F F F F

F T T F T T F

F T F F T F F

F F T F T T F

F F F F F F F

So it clear that Colum 6 and Colum 7 are not equal.