Let's simplify the statement $(p \land (p \to r \lor q)) \land (r \to q)$ step by step, citing the appropriate logical laws at each stage.

Expression:

$(p \land (p \to r \lor q)) \land (r \to q)$

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Step 1: Replace the implication $p \to r \lor q$ with its equivalent using the implication law $p \to A \equiv \neg p \lor A$.

$(p \land (\neg p \lor (r \lor q))) \land (r \to q)$

Justification: Implication law $p \to A \equiv \neg p \lor A$.

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Step 2: Replace the second implication $r \to q$ with its equivalent form $r \to q \equiv \neg r \lor q$.

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

Justification: Implication law $r \to q \equiv \neg r \lor q$.

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Step 3: Apply the distributive property of $\land$ over $\lor$ to the entire conjunction $(p \land (\neg p \lor (r \lor q))) \land (\neg r \lor q)$.

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

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Step 4: Simplify $p \land (\neg p \lor (r \lor q))$ using distribution. First, distribute $p \land \neg p$ and $p \land (r \lor q)$:

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

Justification: Distributive law $p \land (A \lor B) \equiv (p \land A) \lor (p \land B)$.

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Step 5: Simplify $p \land \neg p$ to false (contradiction):

$(\text{false} \lor (p \land (r \lor q))) \land (\neg r \lor q)$

Justification: Contradiction law $p \land \neg p \equiv \text{false}$.

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Step 6: Simplify $\text{false} \lor (p \land (r \lor q))$ to $p \land (r \lor q)$, as $\text{false} \lor A \equiv A$.

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

Justification: Identity law $\text{false} \lor A \equiv A$.

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Step 7: Distribute $\land$ over $\lor$ for $(p \land (r \lor q)) \land (\neg r \lor q)$.

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

Justification: Distributive law $A \land (B \lor C) \equiv (A \land B) \lor (A \land C)$.

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Step 8: Apply distribution again for $((p \land r) \lor (p \land q)) \land (\neg r \lor q)$.

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

Justification: Distributive law for conjunction and disjunction.

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Step 9: Simplify $p \land r \land \neg r$ to false and $p \land q \land q$ to $p \land q$, as $A \land \neg A \equiv \text{false}$ and $A \land A \equiv A$:

$\text{false} \lor (p \land r \land q) \lor (p \land q \land \neg r) \lor (p \land q)$

Now, simplify the expression by removing the $\text{false}$ and redundant $p \land q$:

$p \land q$

Justification: Absorption law $A \lor A \equiv A$ and identity law for false.

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Final Simplified Expression:

$p \land q$

Let me know if you need further explanation on any step or want more details.

Here are five related questions:

1. How does the implication law apply to logical statements?
2. Why is $p \land \neg p$ always false?
3. What is the absorption law and how is it used in simplifications?
4. How does distribution work between conjunction and disjunction?
5. Can the identity law apply when both terms in a disjunction are true?

Tip: Always check for contradictions like $p \land \neg p$, as they simplify to false.