355 views
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.
edited | 355 views

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.$(A \vee B) = T \vee B = T$
edited by

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.