Among the statements:(S1): (p Rightarrow q) v | vedclass

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(A) To determine if the statements are true,we construct truth tables for each.For $(S1): (p \Rightarrow q) \vee ((\sim p) \wedge q)$| $p$ | $q$ | $p

Vedclass · Accepted Answer

neither $(S1)$ nor $(S2)$ is True

Vedclass · Answer

(A) To determine if the statements are true,we construct truth tables for each.For $(S1): (p \Rightarrow q) \vee ((\sim p) \wedge q)$| $p$ | $q$ | $p \Rightarrow q$ | $\sim p \wedge q$ | $(p \Rightarrow q) \vee (\sim p \wedge q)$ ||---|---|---|---|---|| $T$ | $T$ | $T$ | $F$ | $T$ || $T$ | $F$ | $F$ | $F$ | $F$ || $F$ | $T$ | $T$ | $T$ | $T$ || $F$ | $F$ | $T$ | $F$ | $T$ |Since the last column contains an $F$,$(S1)$ is not a tautology. Thus,$(S1)$ is False.For $(S2): (q \Rightarrow p) \Rightarrow ((\sim p) \wedge q)$| $p$ | $q$ | $q \Rightarrow p$ | $\sim p \wedge q$ | $(q \Rightarrow p) \Rightarrow (\sim p \wedge q)$ ||---|---|---|---|---|| $T$ | $T$ | $T$ | $F$ | $F$ || $T$ | $F$ | $T$ | $F$ | $F$ || $F$ | $T$ | $F$ | $T$ | $T$ || $F$ | $F$ | $T$ | $F$ | $F$ |Since the last column contains both $T$ and $F$,$(S2)$ is neither a tautology nor a contradiction. Thus,$(S2)$ is False.Therefore,neither $(S1)$ nor $(S2)$ is True.

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

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Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a fallacy.
Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$ is a tautology.

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