[(S 1)~(p q) (p ( q))] is a tautology [(S 2)~(( p) ( q)) (( p) q)] is a contradiction.

Question

$$(\mathrm{S} 1)~(p \Rightarrow q) \vee(p \wedge(\sim q))$$ is a tautology

$$(\mathrm{S} 2)~((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$$ is a contradiction.

Then

Accepted Answer

$	herefore \mathrm{S} 1$ is correct.

$	herefore \mathrm{S} 2$ is incorrect.

Solutions · 1