Question

# 1. Construct a truth table for: (¬p ∨ (p → ¬q)) → (¬p ∨ ¬q) 2....

1. Construct a truth table for: (¬p ∨ (p → ¬q)) → (¬p ∨ ¬q)

2. Give a proof using logical equivalences that (pq)(qr) and (pr) are not logically equivalent.

3.Show using a truth table that (pq) and (¬q¬p) are logically equivalent.

4. Use the rules of inference to prove that the premise p ∧ (p ¬q) implies the conclusion ¬q. Number each step and give the rule used and which statement numbers the rule uses.

Sol. 1)

 p q ~p ~q p-->~q ~pV(p-->~q) ~pV~q (~pV(p-->~q))-->(~pV~q) 0 0 1 1 1 1 1 1 0 1 1 0 1 1 0 0 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1

Table used in this question

Logical Disjunction (OR) (V):

 A B AVB 0 0 0 0 1 1 1 0 1 1 1 1

Negation(~)

 A B 0 1 1 0

Logical Implication/if-then (Conditional) (-->)

 A B A-->B 0 0 1 0 1 1 1 0 0 1 1 1

Logical conjunction (AND) (/\)

 A B A /\ B 0 0 0 0 1 0 1 0 0 1 1 1

Sol. 2)

(p-->q) V (q-->r) and  (p-->r) are not logically equivalent.

 p q r p-->q q-->r (p-->q) V (q-->r) (p-->r) 0 0 0 1 1 1 1 0 0 1 1 1 1 1 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 0 1 0 1 0 1 1 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1

From the above clearly appears that (p-->q) V (q-->r) and (p-->r) are not the same. So they are not logically equivalent.

Sol. 3)

(p-->q) and (~q-->~p) are logically equivalent.

 p q ~p ~q (p-->q) (~q-->~p) 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 1 1

The above table clearly shows that (p-->q) and (~q-->~p) are logically equivalent.

Sol. 4)

p /\ (p-->~q) --> ~q

using rule of modus ponens which states that if p and  p-->q is true then we can infer q is true. In tautology it is

rule: p /\ (p-->q) -->q

Therefore

p /\ (p-->~q) --> ~q

We can also prove the above with truth table.

 p q ~q p-->~q 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 0

So you can see when p is true , and (p-->q) is true ~q is also true.

Note: 0 is also represented as true(T) and 1 is also represented a false(F).