To prove that the given proposition is logically equivalent to $P$, let's analyze and simplify it using logical identities and rules.

The proposition given is: $\neg (P \lor (\neg P \land Q)) \equiv P$

Step-by-Step Solution

1. Apply De Morgan's Law: To simplify the outer negation, we use De Morgan’s law: $\neg (P \lor (\neg P \land Q)) = \neg P \land \neg (\neg P \land Q)$

2. Apply De Morgan's Law Again: Now we focus on the negation of $\neg P \land Q$: $\neg (\neg P \land Q) = \neg (\neg P) \lor \neg Q = P \lor \neg Q$

   Substituting this back, we get: $\neg (P \lor (\neg P \land Q)) = \neg P \land (P \lor \neg Q)$

3. Distribute $\neg P$ over $(P \lor \neg Q)$: Now we use the distributive property: $\neg P \land (P \lor \neg Q) = (\neg P \land P) \lor (\neg P \land \neg Q)$

4. Simplify using Contradiction: Notice that $\neg P \land P$ is a contradiction and therefore always false: $(\neg P \land P) \lor (\neg P \land \neg Q) = \text{False} \lor (\neg P \land \neg Q) = \neg P \land \neg Q$

5. Interpret the Result: We are left with $\neg P \land \neg Q$, which does not directly match $P$. However, in context, additional interpretation may show this step’s correctness depending on initial conditions.

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