According to my textbook, the negation of $\forall x, P(x) \to (Q(x) \land R(x))$ is $\exists x, P(x) \land (\lnot Q(x) \lor \lnot R(x))$, so $\lnot(\forall x, P(x) \to (Q(x) \land R(x))) \equiv \exists x, P(x) \land (\lnot Q(x) \lor \lnot R(x))$.
Why is $\lnot (\forall x, P(x) \to (Q(x) \land R(x)))$ not equivalent to $\exists x, P(x) \to (\lnot Q(x) \lor \lnot R(x))$, rather than $\exists x, P(x) \land (\lnot Q(x) \lor \lnot R(x))$? From what I can tell,
\begin{align} F \land T &\equiv F & F \to T &\equiv T \\ F \land F &\equiv F & F \to F &\equiv T \\ T \land F &\equiv F & T \to T &\equiv T \\ T \land T &\equiv T & T \to F &\equiv F \\ \end{align}
Beyond this, I'm not sure how to deduce the above conclusion; any help or intuition would be appreciated.
$$¬(∀x, P(x) → (Q(x) ∧ R(x))) \equiv \exists x, \lnot(P(x) \to (Q(x) \land R(x)))$$
$$\equiv \exists x, \lnot (\lnot P(x) \lor (Q(x)\land R(x)))\tag{$p\to q \equiv \lnot p \lor q$}$$
$$\equiv \exists x, (P(x) \land \lnot (Q(x) \land R(x)))\tag{DeMorgan's}$$
$$\equiv \exists x, (P(x) \land (\lnot Q(x) \lor \lnot R(x)))\tag{DeMorgan's}$$