The logical statement $[\sim(\sim \mathrm{p} \vee \mathrm{q}) \vee(\mathrm{p} \wedge \mathrm{r})] \wedge(\sim \mathrm{q} \wedge \mathrm{r})$ is equivalent to

Question

Aicharya · Accepted Answer

$\begin{aligned} & {[\sim(\sim p \vee q) \vee(p \wedge r)] \wedge(\sim q \wedge r)} \ & \equiv[(p \wedge \sim q) \vee(p \wedge r)] \wedge(\sim q \wedge r) \ & 	ext {...[De Morgan's law] } \ & \equiv \mathrm{p} \wedge(\sim q \vee r) \wedge(\sim q \wedge r) \ldots 	ext {. [Distributive law] } \ & \equiv \mathrm{p} \wedge[(\sim \mathrm{q} \vee \mathrm{r}) \wedge \sim \mathrm{q}] \wedge \mathrm{r} \ldots 	ext {... Associative law] } \ & \equiv p \wedge(\sim q) \wedge r \quad 	ext {... [Absorption law] } \ & \equiv(p \wedge r) \wedge \sim q \quad \ldots[	ext { Commutative law }] \ & \end{aligned}$

The logical statement $[\sim(\sim \mathrm{p} \vee \mathrm{q}) \vee(\mathrm{p} \wedge \mathrm{r})]…

Solution