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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

1. True
2. False
3. Same as truth table of $Q$
4. Same as truth table of $S$

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.

### 1 comment

What if R is false in your ist point ?

1 vote