UGC NET CSE | November 2017 | Part 3 | Question: 57

Question

Let $P, Q, R$ and $S$ be Propositions. Assume that the equivalences $P \Leftrightarrow (Q \vee \neg Q )$ and $Q \Leftrightarrow R$ hold. Then the truth value of the formula $(P \wedge Q) \Rightarrow ((P \wedge R) \vee S)$ is always

• A. True
• B. False
• C. Same as truth table of $Q$
• D. Same as truth table of $S$

Answer 1

P ⇔ (Q ∨ ¬ Q) "P should be true because RHS will be TRUE always "

Q ⇔ R "when Q is true R is true" and  "when Q is false R is false"

 $(P ∧ Q) ⇒ ((P ∧ R) ∨ S)$

there can be only 2 cases (value of S doesn't matter)

1) P = True, Q = True and R = True

        $(T ∧ T) ⇒ ((T ∧ T) ∨ S)$ 

         so this case is True

2) P =True, Q = R = False

        $(T ∧ F) ⇒ ((P ∧ R) ∨ S)$

        In case implication if the premises is false then whole statement is true

        this case is also true

The given expression is True in both the cases.

Answer is 1) True

Comments

What if R is false in your ist point ?