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Let $P, Q$ and $R$ be three atomic propositional assertions. Let $X$ denote $( P ∨ Q ) → R$ and Y denote $(P → R) ∨ (Q → R).$ Which one of the following is a tautology?

1. $X ≡ Y$
2. $X → Y$
3. $Y → X$
4. $¬Y → X$
edited | 1.1k views

$X \equiv (P \vee Q) → R$
$\quad \equiv \neg(P \vee Q) \vee R$
$\quad \equiv (\neg P \wedge \neg Q) \vee R$
$\quad \equiv (\neg P \vee R) \wedge (\neg Q \vee R)$
$\quad \equiv (P → R) \wedge (Q → R)$

So, $X → Y$ is true as $(A \wedge B) → (A \vee B)$ is always TRUE but reverse implication is not always true.

Hence, B.
edited by
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Good Explanations sir
+2
When in confusion, better to use truth table, surely to give correct answer
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how to find out if reverse implication is not true? nevermind I understood
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A→B is implication and B→ A is the reverse implication, in the same way, u can check that...