Among the two statements:(S1): (p Rightarrow | vedclass

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(D) For $(S1): (p \Rightarrow q) \wedge (q \wedge (\sim q))$Since $(q \wedge (\sim q))$ is always false $(F)$,the entire expression $(p \Rightarrow q)

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both are true

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(D) For $(S1): (p \Rightarrow q) \wedge (q \wedge (\sim q))$Since $(q \wedge (\sim q))$ is always false $(F)$,the entire expression $(p \Rightarrow q) \wedge F$ is always false. Thus,$(S1)$ is a contradiction.For $(S2): (p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$We can simplify this using distributive laws:$= [q \wedge (p \vee (\sim p))] \vee [(\sim q) \wedge (p \vee (\sim p))]$$= [q \wedge T] \vee [(\sim q) \wedge T]$$= q \vee (\sim q) = T$Since the result is always true $(T)$,$(S2)$ is a tautology.Therefore,both statements are true.

Among the two statements:
$(S1): (p \Rightarrow q) \wedge (q \wedge (\sim q))$ is a contradiction and
$(S2): (p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$ is a tautology.

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Among the two statements:$(S1): (p \Rightarrow q) \wedge (q \wedge (\sim q))$ is a contradiction and$(S2): (p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$ is a tautology.