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!