1. **State the problem:** Simplify the logical expression $$((p \to q) \wedge (\neg q \lor r)) \lor (\neg(p \wedge \neg r))$$ step by step using laws of logic. 2. **Recall key logical equivalences:** - Implication: $$p \to q \equiv \neg p \lor q$$ - De Morgan's Laws: $$\neg(p \wedge q) \equiv \neg p \lor \neg q$$ and $$\neg(p \lor q) \equiv \neg p \wedge \neg q$$ - Distributive, Associative, and Commutative laws allow rearranging and factoring expressions. 3. **Rewrite implication:** $$p \to q \equiv \neg p \lor q$$ So, $$((p \to q) \wedge (\neg q \lor r)) \lor (\neg(p \wedge \neg r))$$ becomes $$((\neg p \lor q) \wedge (\neg q \lor r)) \lor (\neg(p \wedge \neg r))$$ 4. **Apply De Morgan's Law to $$\neg(p \wedge \neg r)$$:** $$\neg(p \wedge \neg r) \equiv \neg p \lor \neg(\neg r) \equiv \neg p \lor r$$ 5. **Substitute back:** $$((\neg p \lor q) \wedge (\neg q \lor r)) \lor (\neg p \lor r)$$ 6. **Distribute $$\wedge$$ over $$\lor$$ if needed, but first note:** The expression is of the form $$(A \wedge B) \lor C$$ where $$A = \neg p \lor q$$ $$B = \neg q \lor r$$ $$C = \neg p \lor r$$ 7. **Use absorption and distributive laws:** Since $$C = \neg p \lor r$$, check if $$C$$ absorbs or simplifies the expression. 8. **Note that $$C$$ covers $$\neg p$$ and $$r$$, which appear in both $$A$$ and $$B$$. So, $$((\neg p \lor q) \wedge (\neg q \lor r)) \lor (\neg p \lor r) \equiv \neg p \lor r \lor (q \wedge \neg q)$$ But $$q \wedge \neg q$$ is a contradiction (false), so it can be removed. 9. **Therefore, the expression simplifies to:** $$\neg p \lor r$$ **Final simplified expression:** $$\neg p \lor r$$