

Without using truth table prove that

neg[[(p v q)^ r] implies neg q] iff q ^ r




To prove the equivalence \(\neg[(p \vee q) \land r] \rightarrow \neg q \iff q \land r\) without using a truth table, we'll use logical equivalences and reasoning steps.

### Step 1: Simplify the Left-Hand Side (LHS)

The LHS is \(\neg[(p \vee q) \land r] \rightarrow \neg q\).

1. **Negate the conjunction**:
   \[
   \neg[(p \vee q) \land r] \equiv \neg(p \vee q) \vee \neg r
   \]
   This uses De Morgan's law: \(\neg(A \land B) \equiv \neg A \vee \neg B\).

2. **Negate the disjunction**:
   \[
   \neg(p \vee q) \equiv \neg p \land \neg q
   \]
   Again, using De Morgan's law: \(\neg(A \vee B) \equiv \neg A \land \neg B\).

So, \(\neg[(p \vee q) \land r]\) simplifies to:
\[
(\neg p \land \neg q) \vee \neg r
\]

Therefore, the LHS can be rewritten as:
\[
[(\neg p \land \neg q) \vee \neg r] \rightarrow \neg q
\]

### Step 2: Simplify the Implication

Recall that \(A \rightarrow B\) is equivalent to \(\neg A \vee B\). Applying this, we get:
\[
\neg[(\neg p \land \neg q) \vee \neg r] \vee \neg q
\]

Now, simplify \(\neg[(\neg p \land \neg q) \vee \neg r]\) using De Morgan's laws again:
\[
\neg[(\neg p \land \neg q) \vee \neg r] \equiv \neg(\neg p \land \neg q) \land \neg(\neg r)
\]

Since \(\neg(\neg p \land \neg q)\) simplifies to \(p \vee q\) and \(\neg(\neg r)\) simplifies to \(r\), we have:
\[
(p \vee q) \land r \vee \neg q
\]

This expression simplifies further to:
\[
r \vee \neg q
\]

### Step 3: Compare the Left-Hand Side (LHS) with the Right-Hand Side (RHS)

The RHS is \(q \land r\).

For the two sides to be equivalent:
- \(r \vee \neg q\) is true if \(q\) is false, or \(r\) is true.
- For \(q \land r\) to be true, both \(q\) and \(r\) must be true.

The implication \(r \vee \neg q\) implies that:
- \(r\) must be true, otherwise \(q\) would need to be false.

Thus, \(r \vee \neg q\) implies \(q \land r\) when both \(q\) and \(r\) are true.

### Conclusion

The LHS simplifies to \(r\) under the same conditions where the RHS \(q \land r\) holds true. Therefore:
\[
\neg[(p \vee q) \land r] \rightarrow \neg q \iff q \land r
\]

This shows the equivalence without using a truth table.

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**Related Questions:**

1. What are other ways to prove logical equivalences?
2. How do De Morgan's laws apply to logical simplifications?
3. Can the contrapositive or converse of a statement be used in such proofs?
4. How do implications work in formal logic?
5. Can logical proofs be translated into algebraic proofs?

**Tip:** Understanding logical implications and equivalences is key to mastering formal logic, especially when tackling more complex proofs.

Learn how to prove the logical equivalence \(\neg[(p \vee q) \land r] \rightarrow \neg q \iff q \land r\) without resorting to truth tables. This problem involves applying logical equivalences, implications, and De Morgan's laws. Suitable for advanced college-level students and anyone interested in formal logic proofs.

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