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If $p \rightarrow(\sim p \vee q)$ is false, then the truth values of $p$ and $q$ are, respectively
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Verified Answer
The correct answer is:
T, F
$$
\mathrm{p} \rightarrow(\sim \mathrm{p} \vee \mathrm{q}) \equiv \mathrm{F}
$$
We know that $\mathrm{T} \rightarrow \mathrm{F} \equiv \mathrm{F}$
$$
\therefore \mathrm{p} \equiv \mathrm{T} \text { and }(\sim \mathrm{p} \vee 1) \equiv \mathrm{F}
$$
We know that $\mathrm{F} \vee \mathrm{F} \equiv \mathrm{F}$
$$
\therefore \sim \mathrm{p} \equiv \mathrm{F} \text { and } \mathrm{q} \equiv \mathrm{F}
$$
Thus p, q are T, F respectively.
\mathrm{p} \rightarrow(\sim \mathrm{p} \vee \mathrm{q}) \equiv \mathrm{F}
$$
We know that $\mathrm{T} \rightarrow \mathrm{F} \equiv \mathrm{F}$
$$
\therefore \mathrm{p} \equiv \mathrm{T} \text { and }(\sim \mathrm{p} \vee 1) \equiv \mathrm{F}
$$
We know that $\mathrm{F} \vee \mathrm{F} \equiv \mathrm{F}$
$$
\therefore \sim \mathrm{p} \equiv \mathrm{F} \text { and } \mathrm{q} \equiv \mathrm{F}
$$
Thus p, q are T, F respectively.
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