Q.Using the rules of negation, write the negation of the following, with justification: $p \wedge \sim q$

Negate using De Morgan's law and $\sim(q\to r)\equiv q\wedge\sim r$.

$\sim[p\wedge(q\to r)] \equiv \sim p \vee \sim(q\to r) \equiv \sim p\vee(q\wedge\sim r)$

Since $p\to X \equiv \sim p\vee X$, option (d) $p\to(q\wedge\sim r)$ is logically equivalent to $\sim p\vee(q\wedge\sim r)$, which matches the derived negation.

Construct the truth table for both sides and compare.

The columns for $\sim(p\vee q)$ and $\sim p \wedge \sim q$ are identical (F, F, F, T) in every row.

Hence $\sim(p\vee q) \equiv \sim p \wedge \sim q$ is verified (this is De Morgan's Law).

Apply De Morgan's law to $\sim[p\land(q\to r)]$.

We need $\sim[p\land(q\to r)]$.

By De Morgan's law, $\sim(A\land B)\equiv \sim A\lor\sim B$. Here $A=p$, $B=(q\to r)$, so

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

Now use the standard equivalence $\sim(q\to r)\equiv q\land\sim r$ (since $q\to r\equiv \sim q\lor r$, so $\sim(q\to r)\equiv q\land\sim r$).

Therefore

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

This matches option (d).