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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. $¬Y → X$
asked in Mathematical Logic by Boss (18.3k points)
edited by | 1.3k views

1 Answer

+24 votes
Best answer
$X \equiv (P \vee Q) → R$
$\quad \equiv \neg(P \vee Q) \vee R$
$\quad \equiv (\neg P \wedge \neg Q) \vee R$
$\quad \equiv (\neg P \vee R) \wedge (\neg Q \vee R)$
$\quad \equiv (P →  R) \wedge (Q → R)$

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

Hence, B.
answered by Veteran (367k points)
edited by
0
Good Explanations sir
+3
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).

Answer:

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