For $U$ the universe under consideration, $P(x)$ and $Q(x)$ predicates with free variable $x$, I want to find a useful denial of
$$ (\exists x \in {U})(Q(x) \land P(x)). $$
Now I'm aware of the fact that $\lnot (\exists x)(P(x)) = (\forall x)(\lnot P(x))$, but I don't know how to deal with $\lnot (\exists x\in U)$. Thank you in advance.
The notation $(\exists x\in U)$ is called a modified quantifier since it modifies the existential quantifier $(\exists x)$ and the notation $$ (\exists x\in U)(P(x)), $$
is actually an abbreviation for the statement
$$ (\exists x)(x\in U \land P(x)). \quad\quad (*) $$
Similarly, $(\forall x\in U)$ is actually an abbreviation for the statement
$$ (\forall x)(x\in U \to P(x)). \quad\quad (**) $$
With $(*)$ and (**) in mind, we have $$ (\exists x \in {U})(Q(x) \land P(x)) = (\exists x)(x\in U \land (Q(x) \land P(x))); $$
Its negation is
\begin{align*} \lnot(\exists x \in {U})(Q(x) \land P(x)) & \equiv (\forall x)(\lnot(x\in U) \lor \lnot(Q(x) \land P(x)))\\ &\equiv (\forall x) (\lnot(x\in U) \lor (\lnot Q \lor \lnot P(x)))\\ &\equiv (\forall x) (x\in U \to (\lnot Q \lor \lnot P(x)))\\ &\equiv (\forall x\in U) (\lnot Q \lor \lnot P(x))\\ &\equiv (\forall x\in U) (Q \to \lnot P(x)). \end{align*}
To arrive at the final result I have used the fact that $Q\to P\equiv \lnot Q\lor P$.