Question

$( S 1)(p \Rightarrow q) \vee(p \wedge(\sim q))$ is a tautology(S2) $((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$ is a contradictionThen

• 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)

only ( $S 1)$ is correct

Answer 2

p	q	p ⇒ q	∼q	p∧∼q	(p ⇒ q) ∨ (p∧∼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

∼p	∼q	∼p ⇒ ∼q	∼p ∨ q	((∼p) ⇒ (∼q)) ∧ (∼p)∨q)
F	F	T	T	T
F	T	T	F	F
T	F	F	T	F
T	T	T	T	T