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\