Question

$(S1)(p\Rightarrow q)\vee (p\wedge (\sim q))$ is a tautology $(S2)((\sim p)\Rightarrow (\sim q))\wedge ((\sim p)\vee q)$ is a contradiction. Then

• A. both (S1) and (S2) are correct
• B. only (S2) is correct
• C. only ( S 1) is correct
• D. both ( S 1) and (S2) are wrong

Answer 1

Answer: (C)

$p$	$q$	$p\Rightarrow q$	$\sim q$	$p\wedge \sim q$	$(p\Rightarrow q)\vee (p\wedge \sim q)$
T	T	T	F	F	T
T	F	F	T	T	T
F	T	T	F	F	T
F	F	T	T	F	T

$\sim p$	$\sim q$	$\sim p\Rightarrow \sim q$	$\sim p\vee q$	$((\sim p)\Rightarrow (\sim q))\wedge (\sim p)\vee q)$
F	F	T	T	T
F	T	T	F	F
T	F	F	T	F
T	T	T	T	T