1. The problem is to simplify and understand the expression $P \equiv (\sim r^{\sim q})$. 2. The expression uses the notation $\sim$ typically meaning negation, but the caret symbol $^$ is ambiguous here; it can mean either exponentiation or logical AND in some contexts. We clarify its meaning first: 3. If $\sim q$ and $\sim r$ mean the negation of $q$ and $r$ respectively, and if $^$ means logical AND, then $\sim r^{\sim q}$ is $\sim r \wedge \sim q$. 4. In this case, $P \equiv (\sim r \wedge \sim q)$. This means $P$ is true only when both $r$ is false and $q$ is false. 5. Alternatively, if $^$ means exponentiation, the expression is $P \equiv (\sim r)^{\sim q}$, which is unusual in logic and the problem more likely suggests logic operations. 6. Assuming logical AND, the simplified interpretation is $P$ equals $\sim r$ AND $\sim q$. 7. Using De Morgan's law, $\sim r \wedge \sim q$ is equivalent to $\sim (r \vee q)$. Therefore, $P$ also equals $\sim (r \vee q)$. 8. Hence the simplified and equivalent logical expression is $$P \equiv \sim (r \vee q)$$. This means $P$ is true only when both $r$ and $q$ are false. 9. In summary, $P$ represents the negation of the disjunction (OR) between $r$ and $q$.