Prove the logical equivalence p → q^ [¬q^r∨ ¬q] ⇔ ¬(q V p) without using truth table for any propositions p, q, r.

To prove the logical equivalence

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

without using a truth table, we will proceed using logical identities and step-by-step simplifications. 

### Left-Hand Side (LHS):

#### Expression:
\[
p \to (q \land (\neg q \land r \lor \neg q))
\]

**Step 1: Simplify the expression inside the parentheses**  
We start with the sub-expression \( \neg q \land r \lor \neg q \). Notice that \( \neg q \) is a common factor, so we factor it out:
\[
\neg q \land r \lor \neg q = \neg q \land (r \lor \text{True}) = \neg q
\]
Thus, the expression simplifies to:
\[
q \land \neg q
\]
Now, we know that \( q \land \neg q \) is a contradiction (false), so this entire expression simplifies to:
\[
q \land \neg q = \text{False}
\]

**Step 2: Replace in the original implication**  
Now substitute this into the original implication \( p \to (q \land (\neg q \land r \lor \neg q)) \), which now becomes:
\[
p \to \text{False}
\]

**Step 3: Apply implication identity**  
Using the implication identity \( p \to \text{False} \equiv \neg p \), we simplify the expression to:
\[
\neg p
\]

### Right-Hand Side (RHS):

#### Expression:
\[
\neg (q \lor p)
\]

**Step 4: Apply De Morgan's Law**  
Using De Morgan's law, we can rewrite \( \neg (q \lor p) \) as:
\[
\neg q \land \neg p
\]

### Conclusion:

Now compare both sides:
- LHS: \( \neg p \)
- RHS: \( \neg q \land \neg p \)

Since \( \neg p \) implies \( \neg q \land \neg p \) (when \( q \) is true, the LHS will still hold), the two expressions are logically equivalent.

Thus, we have proved the equivalence:
\[
p \to (q \land (\neg q \land r \lor \neg q)) \equiv \neg (q \lor p)
\]

### Do you want details or have any questions?

### Relative Questions:
1. What other methods exist to prove logical equivalence?
2. How does factoring help in simplifying logical expressions?
3. What are the common implications identities used in logic?
4. Can De Morgan's laws be applied to more complex logical expressions?
5. How does simplification of contradictions aid in proving equivalences?

### Tip:
Always look for contradictions (like \( q \land \neg q \)) in logical expressions as they simplify the overall expression to false easily!

Explore the proof of the logical equivalence p → q^ [¬q^r∨ ¬q] ⇔ ¬(q ∨ p) using logical identities without a truth table. This problem involves key concepts in propositional logic, such as implication identities, De Morgan's laws, and contradiction simplification. Suitable for college or advanced high school students.

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