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1 vote

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

- True
- False
- Same as truth table of $Q$
- Same as truth table of $S$

5 votes

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