(∃!x)A(x) is equivalent to (∃x)A(x) ∧ (∀y)(∀z) (A(y) ∧ A(z) ⇒ y = z)

Question

I have to prove this using real world examples and I don't know how. I tried to do this with nicotine, but the unique existential quantifier confuses me.

Answer 1

Ok, so let's say that $A(x)$ means $x$ is a perfect milkshake.

then $\exists ! x A(x)$ states that there is exactly one perfect milkshake.

The $\exists x A(x) \land \forall y \forall z ((A(y) \land A(z)) \rightarrow y=z)$ statement states two things:

$\exists x A(x)$ ... This means that there is at least one perfect milkshake

$\forall y \forall z ((A(y) \land A(z)) \rightarrow y=z)$ ... This means that if you ever have two perfect milkshakes, then they are really just one and the same ... meaning that you can never have two different perfect milkshakes, and thus you can have only at most one perfect milkshake.

But if you have at least one perfect milkshake, and you can't have more than one, then you must have exactly one!

So, the two statements say the same thing!