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The problem asks to obtain the **Principal Conjunctive Normal Form (PCNF)** of the logical expression:

\[
(\neg p \rightarrow r) \land (q \leftrightarrow p)
\]

without using a truth table. Let's proceed step by step:

### Step 1: Express each connective in terms of basic operators
We can rewrite implications and biconditionals in terms of logical conjunction (∧), disjunction (∨), and negation (¬).

1. **Implication**: 
   \[
   \neg p \rightarrow r \equiv p \lor r
   \]
   This is because \( A \rightarrow B \) is logically equivalent to \( \neg A \lor B \).

2. **Biconditional**:
   \[
   q \leftrightarrow p \equiv (q \land p) \lor (\neg q \land \neg p)
   \]
   This is the logical equivalence definition for biconditionals.

Thus, the given expression becomes:
\[
(p \lor r) \land ((q \land p) \lor (\neg q \land \neg p))
\]

### Step 2: Distribute the conjunction
We need to convert the expression into a conjunctive normal form (CNF), which is a conjunction of disjunctions.

Using the distributive property, distribute the first part \((p \lor r)\) over the second part:
\[
(p \lor r) \land ((q \land p) \lor (\neg q \land \neg p))
\]

We distribute as follows:

1. Distribute \(p \lor r\) over \((q \land p)\):
   \[
   (p \lor r) \lor (q \land p) \equiv (p \lor r \lor q) \land (p \lor r \lor p)
   \]
   Since \(p \lor p \equiv p\), this simplifies to:
   \[
   (p \lor r \lor q) \land (p \lor r)
   \]

2. Distribute \(p \lor r\) over \((\neg q \land \neg p)\):
   \[
   (p \lor r) \lor (\neg q \land \neg p) \equiv (p \lor r \lor \neg q) \land (p \lor r \lor \neg p)
   \]

Now we combine these results:
\[
(p \lor r \lor q) \land (p \lor r) \land (p \lor r \lor \neg q) \land (p \lor r \lor \neg p)
\]

### Step 3: Simplify
There are no further simplifications needed. Thus, the **Principal Conjunctive Normal Form (PCNF)** is:

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

This is the final PCNF.

---

Would you like further clarification on any step? Here are some related questions:

1. How do we convert biconditional operators to basic logical operators?
2. What are the steps involved in converting to CNF?
3. What are some common mistakes when converting logical expressions to PCNF?
4. Can we verify this result using a truth table?
5. How does the distributive property apply in logic?

**Tip**: When converting to CNF, remember to first express all implications and biconditionals in terms of conjunctions, disjunctions, and negations.

Obtain the PCNF of (¬p → r) ∧ (q ↔ p) without using a truth table.

Learn how to convert the logical expression (¬p → r) ∧ (q ↔ p) to its Principal Conjunctive Normal Form (PCNF) step-by-step without using a truth table. This problem covers key concepts in propositional logic, including implications and biconditionals.

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