Epress $\exists! x P(x)$using universal quantifications, existential quantifications and logical operators

Question

Epress the quantification $\exists! x P(x)$, using universal quantifications, existential quantifications and logical operators.

Does anyone have an idea?

Answer 1

$$\exists xP(x) \land \forall x \forall y(P(x)\land P(y)\implies x=y)$$

The exact way to place parentheses depends on your particular logical system: add some if you need them. Note that the variable $y$ must be a variable that is not free in $P(x)$, and it would be best to use a variable that does not occur at all in $P(x)$.

Note also that my answer uses the binary relation of equality, which is not mentioned as an allowed symbol in your question. However, I cannot conceive of defining uniqueness without the equality relation, at least in a first-order theory.