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The expression to prove is:

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

### Proof:

We will simplify the left-hand side of the equation step by step.

#### Step 1: Apply associativity of \( \lor \) (disjunction)
The expression on the left-hand side is:
\[
p \lor (\neg p \land \neg q) \lor (p \land q)
\]
By associativity of \( \lor \), this can be grouped as:
\[
(p \lor (p \land q)) \lor (\neg p \land \neg q)
\]

#### Step 2: Apply the absorption law
The absorption law states that \( p \lor (p \land q) = p \). Applying this to the first part:
\[
p \lor (\neg p \land \neg q)
\]

#### Step 3: Conclusion
We are left with:
\[
p \lor (\neg p \land \neg q)
\]
which matches the right-hand side of the equation.

Thus, we have proven:
\[
p \lor (\neg p \land \neg q) \lor (p \land q) = p \lor (\neg p \land \neg q)
\]

### Would you like further details or have any questions? 

Here are some related questions:
1. What is the absorption law in propositional logic?
2. Can you explain the associativity property in logic?
3. How does De Morgan's law apply to conjunctions and disjunctions?
4. What are some other useful logical equivalences for simplifying expressions?
5. How can we prove logical equivalences using truth tables?

**Tip:** When proving logical equivalences, always look for opportunities to apply properties like absorption, distribution, and associativity to simplify the expressions.

Prove that p ∨ (¬p ∧ ¬q) ∨ (p ∧ q) = p ∨ (¬p ∧ ¬q).

Explore the step-by-step proof of the logical equivalence: p ∨ (¬p ∧ ¬q) ∨ (p ∧ q) = p ∨ (¬p ∧ ¬q). Learn how to apply the absorption law and associativity in propositional logic to simplify complex expressions. Ideal for advanced high school or undergraduate students.

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