Let's work through simplifying these logical expressions using the rules of logic step by step.

(a) $p \lor (p \land (p \lor q)) \equiv p$

1. Apply Absorption Law: $p \lor (p \land q) \equiv p$ This simplifies the inner expression: $p \lor (p \land (p \lor q)) \equiv p \lor p \equiv p$ Therefore, the simplified form is: $p \equiv p$

(b) $\neg((p \lor q) \land r) \lor \neg q \equiv q \land r$

1. Apply De Morgan's Law on the negation: $\neg((p \lor q) \land r) \equiv \neg(p \lor q) \lor \neg r \equiv (\neg p \land \neg q) \lor \neg r$ The full expression becomes: $(\neg p \land \neg q) \lor \neg r \lor \neg q \equiv \neg p \lor \neg r \lor \neg q$

2. Now compare both sides, but we can’t simplify it directly to $q \land r$ by basic logic laws alone. We’ll need additional steps (such as truth tables) to verify equivalence for this expression.

(c) $p \lor q \lor (\neg p \land \neg q \land r) \equiv p \lor q \lor r$

1. Distribute the first part using Distributive Law: $(p \lor q) \lor (\neg p \land \neg q \land r)$
2. Apply Absorption Law to simplify: $p \lor q \lor r$ Therefore, the simplified form is: $p \lor q \lor r$

(d) $(\neg(p \lor q) \Rightarrow (p \land q \land r)) \equiv p \land q$

1. Express the implication using the Implication Law: \equiv (p \lor q) \lor (p \land q \land r)$$
2. Apply Absorption Law: $(p \lor q) \equiv p \land q$ Therefore, the simplified form is: $p \land q$

These are the simplified results using logical laws. Would you like further clarifications or details on any step?

Here are 5 related questions:

1. Can you list the basic logical equivalence laws?
2. How can De Morgan's laws simplify complex expressions?
3. Why does absorption simplify $p \lor (p \land q)$ to $p$?
4. What is the difference between logical implication and biconditional?
5. How can truth tables verify logical equivalences?

Tip: When simplifying logical expressions, always start by identifying which laws (like De Morgan's or distribution) directly apply to inner expressions.