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21 votes
21 votes
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.$(A \vee B) = T \vee B = T$
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4 votes
4 votes

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.

1 votes
1 votes

By Exportation Law

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

check the above compound proposition is a tautology or not by the indirect method of proof(proof by contradiction)  in which try to make all premises true and the conclusion false. Alternatively, we try to make all premises and negation of conclusion true and check for consistency.

 

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