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 approach the problem using logical equivalences and reasoning.

### Step 1: Analyze the left-hand side (LHS)

The expression on the left-hand side is:
\[
\neg[(p \vee q) \land r] \rightarrow \neg q
\]

Let's break it down:

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

So the expression becomes:
\[
\neg[(p \vee q) \land r] \equiv (\neg p \land \neg q) \vee \neg r
\]

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

### Step 2: Simplify the implication

Recall the logical implication \(A \rightarrow B\) is equivalent to \(\neg A \vee B\). So, the LHS becomes:
\[
\neg[(\neg p \land \neg q) \vee \neg r] \vee \neg q
\]

Let's simplify \(\neg[(\neg p \land \neg q) \vee \neg r]\):
- \(\neg[(\neg p \land \neg q) \vee \neg r]\) can be written as \(\neg(\neg p \land \neg q) \land \neg(\neg r)\).

Thus:
\[
(\neg(\neg p \land \neg q) \land \neg(\neg r)) \vee \neg q
\]

Simplifying further using De Morgan's law:
\[
(p \vee q) \land r \vee \neg q
\]

Since \(r \vee \neg q\) can be true if \(q \land r\) is false, the expression reduces to:
\[
\neg q \vee \neg r \vee \neg q \equiv \neg q \vee r
\]

Now the entire LHS implication reduces to:
\[
\neg q \vee r \equiv r
\]

### Step 3: Compare with the right-hand side (RHS)

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

To satisfy the equivalence \(\neg q \vee r \iff q \land r\):

- If \(q \land r\) is true, \(r\) must be true, and \(\neg q\

Learn how to prove the logical equivalence ~( (p ∨ q) ∧ r ) → ~q ↔ q ∧ r without using a truth table. This problem utilizes logical equivalences such as De Morgan's laws and involves advanced reasoning in propositional logic. Suitable for advanced students or enthusiasts of logic and mathematics.

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