# GATE2005-40

2.2k views

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

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

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
0
Good Explanations sir
6
When in confusion, better to use truth table, surely to give correct answer
0
how to find out if reverse implication is not true? nevermind I understood
0

A→B is implication and B→ A is the reverse implication, in the same way, u can check that...

0
why option A is wrong ?
1

$X\equiv(P∨Q)→R$

$X\equiv\lnot(P∨Q)\vee R$

$X\equiv (\lnot P \wedge \lnot Q)\vee R$

$X\equiv \bar{P}.\bar{Q}+R$

$Y\equiv(P→R)∨(Q→R)$

$Y\equiv(\lnot P\vee R)∨(\lnot Q\vee R)$

$Y\equiv\bar{P}+R+\bar{Q}+R$

$Y\equiv\bar{P}+\bar{Q}+R$

How $X \equiv Y$ gives always true?

It is not possible to $X \equiv Y$ gives always true(Tautology).

A good explanation is already given by arjun sir. I'm going to share my approach- (using rules of boolean algebra, checking options one by one)

$X=(P\vee Q)\rightarrow r$

$=\overline{p+q} +R$

$=\bar{P}• \bar{Q} +R$

$Y=(P\rightarrow R) \vee (Q\rightarrow R)$

$=(\bar{P} + R) + (\bar Q +R)$

$=\bar P +\bar Q + R$

A. Clearly $X\;\text {not}\equiv Y$

B. X$\rightarrow$Y $=\bar X+Y$

$=(\overline{\bar P \bar Q+R}) + (\bar P +\bar Q + R)$

$=(P+Q) \color{green}{\bar R }\; +(\bar P +\bar Q)+ \color{green}R$

$=(P+Q+R)(\bar R+R)+(\bar P+\bar Q)$

$=(P+Q+R)+(\bar P+\bar Q)$

$=1\;(TRUE)$  So option B is correct..

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