I am given this unique existential quantifier: $\displaystyle \exists !x P(x)$, which means there is only one single unique x in the domain that statisfy P(x)=true. But I don't understand how would this $\displaystyle (\exists x P(x)) \wedge (\forall y, (y\neq x) \rightarrow \lnot P(y))$ be equivalent?

Wouldn't $\displaystyle (\exists x P(x)) \wedge (\forall y, (y\neq x) \rightarrow \lnot P(y))$ mean there is a possibility that there is more than one $\displaystyle x$ in the domain, so long as $\displaystyle x\neq y$? There could have been like 2 $\displaystyle x$s that is different from all of the other $\displaystyle y$s that satisfy P(x)=true. Then in this case, this isn't unique anymore right?

I also attached a diagram of how I thought of this, with the green zone being the $\displaystyle y$ and the white circles within the green zone being the $\displaystyle x$. In the diagram, it exists more than one "x" and is not equals to "y", which makes me not understand the equivalence of $\displaystyle (\exists x P(x)) \wedge (\forall y, (y\neq x) \rightarrow \lnot P(y))$ to $\displaystyle \exists !x P(x)$.

thanks!