Given the following two statements:(S_{1}): ( | vedclass

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(C) For $(S_{1}): (q \vee p) \rightarrow (p \leftrightarrow \sim q)$If $p = T$ and $q = T$,then $(T \vee T)$ $\rightarrow (T \leftrightarrow F)$ $\Rig

Vedclass · Accepted Answer

both $(S_{1})$ and $(S_{2})$ are not correct.

Vedclass · Answer

(C) For $(S_{1}): (q \vee p) \rightarrow (p \leftrightarrow \sim q)$If $p = T$ and $q = T$,then $(T \vee T)$ $\rightarrow (T \leftrightarrow F)$ $\Rightarrow T$ $\rightarrow F = F$. Since it is not true for all truth values,$(S_{1})$ is not a tautology.For $(S_{2}): \sim q \wedge (\sim p \leftrightarrow q)$If $p = F$ and $q = F$,then $\sim F \wedge (\sim F \leftrightarrow F)$ $\Rightarrow T \wedge (T \leftrightarrow F)$ $\Rightarrow T \wedge F = F$.If $p = T$ and $q = F$,then $\sim F \wedge (\sim T \leftrightarrow F)$ $\Rightarrow T \wedge (F \leftrightarrow F)$ $\Rightarrow T \wedge T = T$.Since there exists a case where the truth value is $T$,$(S_{2})$ is not a fallacy (contradiction).Thus,both $(S_{1})$ and $(S_{2})$ are incorrect.

Given the following two statements:
$(S_{1}): (q \vee p) \rightarrow (p \leftrightarrow \sim q)$ is a tautology.
$(S_{2}): \sim q \wedge (\sim p \leftrightarrow q)$ is a fallacy.
Then:

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Given the following two statements:$(S_{1}): (q \vee p) \rightarrow (p \leftrightarrow \sim q)$ is a tautology.$(S_{2}): \sim q \wedge (\sim p \leftrightarrow q)$ is a fallacy.Then: