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If statement $\mathrm{p}$ and $\mathrm{q}$ are true and $\mathrm{r}$ and $\mathrm{s}$ are false, then truth values of $\sim(\mathrm{p} \rightarrow \mathrm{q}) \leftrightarrow(\mathrm{p} \wedge \mathrm{s})$ and $(\sim \mathrm{p} \rightarrow \mathrm{q}) \wedge(\mathrm{r} \leftrightarrow \mathrm{s})$ are respectively.
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The correct answer is:
T, T
$\begin{aligned} & \sim(\mathrm{p} \rightarrow \mathrm{q}) \leftrightarrow(\mathrm{r} \wedge \mathrm{s}) \\ & \equiv \sim(\mathrm{T} \rightarrow \mathrm{T}) \leftrightarrow(\mathrm{F} \wedge \mathrm{F}) \\ & =\sim(\mathrm{T}) \leftrightarrow \mathrm{F} \equiv \mathrm{F} \leftrightarrow \mathrm{F} \equiv \mathrm{T} \\ & \text { Also }(\sim \mathrm{P} \rightarrow \mathrm{q}) \wedge(\mathrm{r} \leftrightarrow \mathrm{s}) \\ & \equiv(\sim \mathrm{T} \rightarrow \mathrm{T}) \wedge(\mathrm{F} \leftrightarrow \mathrm{F}) \\ & \equiv(\mathrm{F} \rightarrow \mathrm{T}) \wedge(\mathrm{T}) \equiv \mathrm{T} \wedge \mathrm{T} \equiv \mathrm{T}\end{aligned}$
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