Math Problem Statement

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

Solution

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

(¬pr)(qp)(\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: ¬prpr\neg p \rightarrow r \equiv p \lor r This is because ABA \rightarrow B is logically equivalent to ¬AB\neg A \lor B.

  2. Biconditional: qp(qp)(¬q¬p)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: (pr)((qp)(¬q¬p))(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 (pr)(p \lor r) over the second part: (pr)((qp)(¬q¬p))(p \lor r) \land ((q \land p) \lor (\neg q \land \neg p))

We distribute as follows:

  1. Distribute prp \lor r over (qp)(q \land p): (pr)(qp)(prq)(prp)(p \lor r) \lor (q \land p) \equiv (p \lor r \lor q) \land (p \lor r \lor p) Since pppp \lor p \equiv p, this simplifies to: (prq)(pr)(p \lor r \lor q) \land (p \lor r)

  2. Distribute prp \lor r over (¬q¬p)(\neg q \land \neg p): (pr)(¬q¬p)(pr¬q)(pr¬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: (prq)(pr)(pr¬q)(pr¬p)(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:

(prq)(pr)(pr¬q)(pr¬p)(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.

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Math Problem Analysis

Mathematical Concepts

Propositional Logic
Conjunctive Normal Form (CNF)

Formulas

Implication: A → B ≡ ¬A ∨ B
Biconditional: A ↔ B ≡ (A ∧ B) ∨ (¬A ∧ ¬B)

Theorems

Distributive Property of Logic

Suitable Grade Level

Undergraduate/Graduate Level