Simplify the statement form below in no more than 9 steps. Make sure you do not skip any step, 
and every step must be justified by a law. Do not combine two steps of the same law in a single 
step. Use true and false for tautology and contradiction respectively. (This question checks that you 
apply the laws rigorously and cite them correctly, so we will be strict in our grading.) [4 marks]
(𝒑 ∧ (𝒑 → 𝒓 ∨ 𝒒)) ∧ (𝒓 → 𝒒)

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: 
\[
(p \land (p \to r \lor q)) \land (r \to q)
\]

---

**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 \( p \to A \equiv \neg p \lor A \).

---

**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 \( r \to q \equiv \neg r \lor q \).

---

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

\[
((p \land (\neg p \lor (r \lor q))) \land (\neg r \lor q))
\]

---

**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 \( p \land (A \lor B) \equiv (p \land A) \lor (p \land B) \).

---

**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 \( p \land \neg p \equiv \text{false} \).

---

**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 \( \text{false} \lor A \equiv A \).

---

**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 \( A \land (B \lor C) \equiv (A \land B) \lor (A \land C) \).

---

**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.

---

**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.

---

### 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.

Simplify the statement form below in no more than 9 steps. Make sure you do not skip any step, and every step must be justified by a law. Do not combine two steps of the same law in a single step. Use true and false for tautology and contradiction respectively. (This question checks that you apply the laws rigorously and cite them correctly, so we will be strict in our grading.) [4 marks] (𝒑 ∧ (𝒑 → 𝒓 ∨ 𝒒)) ∧ (𝒓 → 𝒒)

Learn how to simplify the logical statement (p ∧ (p → r ∨ q)) ∧ (r → q) in no more than 9 steps. This detailed breakdown covers key laws of propositional logic, including implication, distributive, and absorption laws. Suitable for college-level students studying logic or discrete mathematics.

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