Consider the two statements:(S1): (p rightarr | vedclass

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(C) For $(S1): (p$ $\rightarrow q) \vee (\sim q$ $\rightarrow p)$Using the implication law $(a \rightarrow b \equiv \sim a \vee b)$,we get:$(\sim p \v

Vedclass · Accepted Answer

Both $(S1)$ and $(S2)$ are true.

Vedclass · Answer

(C) For $(S1): (p$ $\rightarrow q) \vee (\sim q$ $\rightarrow p)$Using the implication law $(a \rightarrow b \equiv \sim a \vee b)$,we get:$(\sim p \vee q) \vee (q \vee p)$$= (q \vee \sim p) \vee (q \vee p) = q \vee (\sim p \vee p) = q \vee t = t$.Thus,$(S1)$ is a tautology.For $(S2): (p \wedge \sim q) \wedge (\sim p \vee q)$Using De Morgan's Law,$(\sim p \vee q) \equiv \sim (p \wedge \sim q)$.So,$(S2) = (p \wedge \sim q) \wedge \sim (p \wedge \sim q) = C$ (Contradiction/Fallacy).Thus,both $(S1)$ and $(S2)$ are true.

Consider the two statements:
$(S1): (p$ $\rightarrow q) \vee (\sim q$ $\rightarrow p)$ is a tautology.
$(S2): (p \wedge \sim q) \wedge (\sim p \vee q)$ is a fallacy.
Then:

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Consider the two statements:$(S1): (p$ $\rightarrow q) \vee (\sim q$ $\rightarrow p)$ is a tautology.$(S2): (p \wedge \sim q) \wedge (\sim p \vee q)$ is a fallacy.Then: