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If $\mathrm{q}$ is false and $\mathrm{p} \wedge \mathrm{q} \leftrightarrow \mathrm{r}$ is true, then is a tautology.
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Verified Answer
The correct answer is:
$(\mathrm{p} \wedge \mathrm{r}) \rightarrow \mathrm{p} \vee \mathrm{r}$
$\mathrm{q}$ is false and $\mathrm{p} \wedge \mathrm{q} \leftrightarrow \mathrm{r}$ is true
$\Rightarrow p \wedge q \equiv F$ and $r \equiv F$
$\Rightarrow \mathrm{p} \equiv \mathrm{T}$ or $\mathrm{F}, \mathrm{q} \equiv \mathrm{F}$ and $\mathrm{r} \equiv \mathrm{F}$
(1) $\mathrm{p} \vee \mathrm{r} \equiv \mathrm{T}$ or $\mathrm{F}$
(2) $(\mathrm{p} \wedge \mathrm{r}) \rightarrow(\mathrm{p} \vee \mathrm{r}) \equiv \mathrm{F} \rightarrow(\mathrm{T}$ or $\mathrm{F}) \equiv \mathrm{T}$
(3) $(\mathrm{p} \vee \mathrm{r}) \rightarrow(\mathrm{p} \wedge \mathrm{r}) \equiv(\mathrm{T}$ or $\mathrm{F}) \rightarrow \mathrm{F} \equiv \mathrm{T}$ or $\mathrm{F}$
(4) $\mathrm{p} \wedge \mathrm{r} \equiv \mathrm{F}$
$\Rightarrow p \wedge q \equiv F$ and $r \equiv F$
$\Rightarrow \mathrm{p} \equiv \mathrm{T}$ or $\mathrm{F}, \mathrm{q} \equiv \mathrm{F}$ and $\mathrm{r} \equiv \mathrm{F}$
(1) $\mathrm{p} \vee \mathrm{r} \equiv \mathrm{T}$ or $\mathrm{F}$
(2) $(\mathrm{p} \wedge \mathrm{r}) \rightarrow(\mathrm{p} \vee \mathrm{r}) \equiv \mathrm{F} \rightarrow(\mathrm{T}$ or $\mathrm{F}) \equiv \mathrm{T}$
(3) $(\mathrm{p} \vee \mathrm{r}) \rightarrow(\mathrm{p} \wedge \mathrm{r}) \equiv(\mathrm{T}$ or $\mathrm{F}) \rightarrow \mathrm{F} \equiv \mathrm{T}$ or $\mathrm{F}$
(4) $\mathrm{p} \wedge \mathrm{r} \equiv \mathrm{F}$
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