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### Problem 1: Proving Logical Equivalence
We have to prove the following statements:

#### **A.** 
\[
\neg (p \land \neg q) \lor (\neg p \land \neg q) \equiv \neg p
\]

Let's use logical equivalence steps:

1. Apply De Morgan's Law:
   \[
   \neg (p \land \neg q) = \neg p \lor \neg (\neg q) = \neg p \lor q
   \]
   Thus, the expression becomes:
   \[
   (\neg p \lor q) \lor (\neg p \land \neg q)
   \]
2. Apply the Absorption Law:
   \[
   \neg p \lor (\neg p \land \neg q) = \neg p
   \]
   Therefore, 
   \[
   \neg (p \land \neg q) \lor (\neg p \land \neg q) \equiv \neg p
   \]
   This verifies that statement A is correct.

#### **B.** 
\[
\neg ((\neg p \land q) \lor (\neg p \land \neg q)) \lor (P \lor Q) \equiv P
\]

Let's simplify:

1. Apply distribution:
   \[
   \neg p \land (q \lor \neg q) = \neg p \land \text{True} = \neg p
   \]
   Thus, the original expression simplifies to:
   \[
   \neg (\neg p) \lor (P \lor Q) \equiv p \lor (P \lor Q) 
   \]
   By the idempotent law and identity laws, we get:
   \[
   p \lor P \equiv P
   \]
   This verifies that statement B is also correct.

### Problem 2: Simplifying Logical Statements
#### **A.**
\[
\neg (\neg p \land q) \land (p \lor q)
\]

Let's simplify it step-by-step:

1. Apply De Morgan's Law:
   \[
   \neg (\neg p \land q) = p \lor \neg q
   \]
   The expression becomes:
   \[
   (p \lor \neg q) \land (p \lor q)
   \]
2. Apply distribution:
   \[
   p \lor (\neg q \land q) = p \lor \text{False} = p
   \]
   Hence, the entire expression simplifies to:
   \[
   p
   \]

#### **B.**
\[
(p \lor \neg q) \land \neg (p \land q)
\]

Let's simplify:

1. Apply De Morgan's Law:
   \[
   \neg (p \land q) = \neg p \lor \neg q
   \]
   The expression becomes:
   \[
   (p \lor \neg q) \land (\neg p \lor \neg q)
   \]
2. Apply distribution:
   \[
   (p \land \neg p) \lor (p \land \neg q) \lor (\neg q \land \neg q)
   \]
   The simplified expression is:
   \[
   p \land \neg q
   \]

### Final Results
1. For Problem 1:
   - **A:** \(\neg (p \land \neg q) \lor (\neg p \land \neg q) \equiv \neg p\)
   - **B:** \(\neg ((\neg p \land q) \lor (\neg p \land \neg q)) \lor (P \lor Q) \equiv P\)

2. For Problem 2:
   - **A:** \(\neg (\neg p \land q) \land (p \lor q) \equiv p\)
   - **B:** \((p \lor \neg q) \land \neg (p \land q) \equiv p \land \neg q\)

Would you like further details or explanations on any part?

### 5 Questions for Further Practice
1. What is De Morgan's law, and how is it used in logical equivalence?
2. Can you apply the absorption law to simplify other logical expressions?
3. How does the distribution law work in logical equivalence?
4. Can two different expressions be equivalent without using basic logical equivalences?
5. How does using truth tables help verify logical equivalences?

### Tip
When simplifying logical statements, always apply De Morgan's Law first to eliminate negations, making the remaining expressions easier to handle!

Prove the following logical equivalences and simplify logical statements as given:
1. Prove that the following statements are equivalent:
   A. ¬(p ∧ ¬q) ∨ (¬p ∧ ¬q) ≡ ¬p
   B. ¬((¬p ∧ q) ∨ (¬p ∧ ¬q)) ∨ (P ∨ Q) ≡ P
2. Simplify the following using logical equivalence laws:
   A. ¬(¬p ∧ q) ∧ (p ∨ q)
   B. (p ∨ ¬q) ∧ ¬(p ∧ q)

Learn how to prove logical equivalences and simplify propositional logic statements using key logical laws such as De Morgan's Law, Absorption Law, and Distribution Law. Ideal for undergraduate students studying computer science or logic.

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