Logical Expression : Is it same or not?

Question

I have $p\rightarrow \left ( q\wedge r \right )$, If i negate it: It will become like below:

$\lnot \left ( p\rightarrow \left ( q\wedge r \right ) \right )$

$\lnot \left ( \lnot p\vee \left ( q\wedge r \right ) \right )$

$\therefore p\wedge \left ( \lnot q\vee \lnot r \right )$

My answer key said that $\lnot p\rightarrow \lnot \left ( q\wedge r \right )$ is the answer.

But I get like above $\therefore p\wedge \left (\lnot q\vee \lnot r \right )$ as the answer.

What's wrong? Or I'm misunderstanding of the concept......?

Answer 1

The negation of an implication is not an implication. ~$(p \to q)$ is not ~$p \to$ ~$q$. But ~($p \to q$) is equivalent to the statement $p \land$ ~$q$.

Applied to your situation:

When you negate the statement $p \to (q \land r)$, you get:

~($p \to (q \land r$))

$p \land$ ~$(q \land r)$

$p \land$ (~$q \lor$ ~$r$).

Answer 2

You're right and the answer key is wrong...

You are indeed correct that $$\lnot\Big(p \rightarrow (q\land r)\Big) \not\equiv \lnot p\rightarrow \lnot (q \land r)$$

Rather, as you have correctly shown, $$ \lnot\Big(p \rightarrow (q\land r)\Big)\equiv p \land \lnot(q \land r) \equiv p \land (\lnot q \lor \lnot r)$$

Answer 3

A simple argument that the answer key is wrong is just to plug in some values.

Suppose $p$, $q$ and $r$ are all $\sf true$. Then, evaluating, we get that $p\to(q\land r)$ is $\sf true$, and $(\neg p)\to \neg(q\land r)$ is also $\sf true$.

Thus, since there is a situation in which the two formulas have the same truth value, they cannot be each other's negations.