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Show that the formula $\left[(\sim p \vee q) \Rightarrow (q \Rightarrow p)\right]$ is not a tautology.

Let $A$ be a tautology and $B$ any other formula. Prove that $(A \vee B)$ is a tautology.
asked in Mathematical Logic by Veteran (59.4k points)
edited by | 296 views

2 Answers

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Best answer
a.

$\left[(\sim p \vee q) \Rightarrow (q \Rightarrow p)\right] \\ = \sim (\sim p \vee q)) \vee (q \Rightarrow p) \\= (p \wedge \sim q) \vee (\sim q \vee p) \\= p \vee \sim q$.

Hence not tautology.

b.

$(A \vee B) = T \vee B = T$
answered by Veteran (342k points)
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To prove that it is not a tautology, we need to show this proposition should result as FALSE.

To make it as FALSE in implication (ex : x->y) , we need to prove x as TRUE and y as FALSE. Here x = (not p ) OR q and y = q->p.

To make y as FALSE , we need to definitely have q = TRUE and p = FALSE.If we give that , the resulting truth value of this proposition is FALSE.

answered by Loyal (6.8k points)


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