P → ((P → Q) ∧ ¬(¬Q ∨ ¬P )) find pdnf and pcnf

We are tasked with finding the **PDNF** (Principal Disjunctive Normal Form) and **PCNF** (Principal Conjunctive Normal Form) for the logical expression:

\[
P \to ((P \to Q) \land \neg(\neg Q \lor \neg P))
\]

### Step 1: Simplify the expression

We can break down the main components step by step.

#### First, use the implication rule:
\[
P \to A \equiv \neg P \lor A
\]
Where \(A = ((P \to Q) \land \neg(\neg Q \lor \neg P))\).

So:
\[
P \to ((P \to Q) \land \neg(\neg Q \lor \neg P)) \equiv \neg P \lor ((P \to Q) \land \neg(\neg Q \lor \neg P))
\]

Now let's simplify each part inside the conjunction.

#### Simplify \(P \to Q\):
\[
P \to Q \equiv \neg P \lor Q
\]

#### Simplify \(\neg(\neg Q \lor \neg P)\) using De Morgan's law:
\[
\neg(\neg Q \lor \neg P) \equiv Q \land P
\]

Now substitute these into the original expression:
\[
P \to ((\neg P \lor Q) \land (Q \land P))
\equiv \neg P \lor ((\neg P \lor Q) \land (Q \land P))
\]

### Step 2: Distribute and simplify

We now simplify the expression \((\neg P \lor Q) \land (Q \land P)\).

1. First, expand using distribution:
\[
(\neg P \lor Q) \land (Q \land P) = ((\neg P \lor Q) \land Q) \land ((\neg P \lor Q) \land P)
\]

2. Expand further:
\[
(\neg P \lor Q) \land Q = (\neg P \land Q) \lor (Q \land Q) = (\neg P \land Q) \lor Q = Q
\]
Thus, we have:
\[
((\neg P \lor Q) \land Q) \land ((\neg P \lor Q) \land P) = Q \land ((\neg P \lor Q) \land P)
\]
Now expand the second part:
\[
(\neg P \lor Q) \land P = (\neg P \land P) \lor (Q \land P) = Q \land P
\]
Thus:
\[
Q \land (Q \land P) = Q \land P
\]

So the expression simplifies to:
\[
P \to ((P \to Q) \land \neg(\neg Q \lor \neg P)) \equiv \neg P \lor (Q \land P)
\]

### Step 3: PDNF (Principal Disjunctive Normal Form)

To get the **PDNF**, we look for a disjunction of minterms (combinations of literals that make the expression true). The simplified form \(\neg P \lor (Q \land P)\) already is close to PDNF:

\[
\neg P \lor (Q \land P) = (\neg P) \lor (P \land Q)
\]

This is the PDNF:
\[
PDNF: \neg P \lor (P \land Q)
\]

### Step 4: PCNF (Principal Conjunctive Normal Form)

To get the **PCNF**, we look for a conjunction of maxterms (combinations of literals that make the expression false). For the expression \(\neg P \lor (Q \land P)\), the false cases occur when both disjuncts are false. This happens when \(P\) is true and \(Q\) is false.

So the PCNF will involve the maxterms that cover all the conditions where the formula is false:

\[
PCNF: (P \lor Q) \land (\neg P \lor Q)
\]

---

### Final Answer:
- **PDNF**: \(\neg P \lor (P \land Q)\)
- **PCNF**: \((P \lor Q) \land (\neg P \lor Q)\)

---

Would you like further details on any of the steps? Here are five related questions to explore:

1. How is De Morgan's law applied in simplifying logical expressions?
2. What is the relationship between PDNF and truth tables?
3. How can we use truth tables to confirm PDNF and PCNF?
4. What are some other methods for simplifying logical expressions?
5. How does distribution work in logical expressions compared to algebra?

**Tip**: Always check your logical expressions with truth tables for validation!

Learn how to find the Principal Disjunctive Normal Form (PDNF) and Principal Conjunctive Normal Form (PCNF) for the logical expression P → ((P → Q) ∧ ¬(¬Q ∨ ¬P)). This step-by-step guide simplifies the expression using the implication rule, De Morgan's law, and distribution. Suitable for undergraduate students studying propositional logic.

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