To show that $\neg(p \vee (\neg p \land q)) \equiv (\neg p \land \neg q)$, we’ll use logical identities step-by-step to prove the equivalence. Here's the breakdown:

Step 1: Apply De Morgan’s law to the outer negation

Starting with: $\neg(p \vee (\neg p \land q))$ Using De Morgan’s law, which states $\neg(A \vee B) \equiv \neg A \land \neg B$, we get: $\neg(p) \land \neg(\neg p \land q)$

Step 2: Apply De Morgan’s law to the inner negation

Now we simplify $\neg(\neg p \land q)$ using De Morgan’s law again, which gives: $\neg(\neg p \land q) \equiv \neg(\neg p) \vee \neg(q)$ Since $\neg(\neg p) \equiv p$, this simplifies to: $p \vee \neg q$ Thus, we now have: $\neg(p) \land (p \vee \neg q)$

Step 3: Distribute $\neg p$ over $(p \vee \neg q)$ using the distributive law

Using the distributive law, $A \land (B \vee C) \equiv (A \land B) \vee (A \land C)$, we get: $(\neg p \land p) \vee (\neg p \land \neg q)$

Step 4: Simplify using the contradiction law

Notice that $\neg p \land p$ is a contradiction (it is always false), so: $(\neg p \land p) \vee (\neg p \land \neg q) \equiv \text{false} \vee (\neg p \land \neg q)$ Since $\text{false} \vee X \equiv X$, we get: $\neg p \land \neg q$

Conclusion

We have shown that: $\neg(p \vee (\neg p \land q)) \equiv \neg p \land \neg q$ This completes the proof.

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Would you like more details or have any questions?

Here are some related questions that expand on this topic:

1. What are the main laws of propositional logic used in simplifications?
2. How can De Morgan's laws help in simplifying complex logical statements?
3. How does the distributive law apply in Boolean algebra?
4. How can contradiction laws help in logical simplifications?
5. Why is understanding logical equivalences important in proofs and problem-solving?

Tip: Mastering De Morgan's laws is essential for simplifying negated statements in logic.