Is there a short way to write $(\exists x. f(x)) \implies (\exists x. f(x) \land g(x))$?

Question

Is there a way to simplify $(\exists x. f(x)) \implies (\exists x. f(x) \land g(x))$?

After all, neither $\exists x. f(x) \implies g(x)$ nor $\exists x. f(x) \land g(x)$ are equivalent to this.

Answer 1

It's equivalent to $$\forall x\neg f(x)\vee\exists x(f(x)\wedge g(x))$$ I don't think there's any way to prettify that.