Hilbert introduced a construct $\epsilon x. P(x)$ for a predicate $P$ such that
$$\exists x. P(x) \implies P(\epsilon y.P(y))$$


Obviously, this is equivalent to the axiom of global choice. With this operator, we can form complex propositions conveniently:


$$Q(\epsilon x. P(x)) \text{ equiderivable to } \forall x. P(x)\implies Q(x)$$


assuming $\exists x. P(x)$. I'm wondering if the dual can be achieved. I.e. Can there be an operator $\delta x. P(x)$ such that


$$Q(\delta x. P(x)) \text{ equiderivable to } \exists x. P(x)\land Q(x)$$


I suspect that the difficulty encountered in defining this is because we implicitly assume universal closure, but not *existential* closure. Is there any material related to this?



A dual to Hilbert's $\epsilon$ operator
Nothing like this can happen.


Take $P(x)\equiv\top$, and let $S$ be any formula such that both $\exists x S(x)$ and $\exists x\neg\ S(x)$ are derivable. Then both $$(\*)\_1:\quad\exists x(P(x)\wedge S(x))$$ and $$(\*)\_2:\quad \exists x(P(x)\wedge\neg S(x))$$ are themselves derivable. However, we can't possibly have a single term (or term-like piece of syntax) $\delta$ such that $Q(\delta)$ is equiderivable with $\exists x(P(x)\wedge Q(x))$ for **either** choice of $Q\in\{S,\neg S\}$: each sentence of the latter type is outright derivable, while since $\delta$ is independent of $Q$ we can't have both $S(\delta)$ and $\neg S(\delta)$ be derivable.


For a concrete example, what should your $\delta$ do in the setting of the natural numbers and first-order Peano arithmetic when we take $P(x)\equiv\top$ and $Q(x)\equiv$ "$x$ is even"?


