Interpreting the optimized way of expressing "At least two" using nested quantifiers. I'm precisely talk about:
$$\forall x\left(\exists \:y\left(P\left(y\right)\:\wedge \:\left(x\neq y\right)\right)\right)$$
The way I can read this is "Given any x, there exists a y such that y is P and isn't x" I have spent some time trying to make sense of why this would be saying "There are at least two Ps" to no avail, and representing it graphically wasn't so helpful. I'm not sure if I'm thinking about it in the right way, so any help is appreciated.
Thank you in advance, and have a great day.
 A: I think it would be more natural written as
$$\forall\,x:\exists\,y\ne x:P(y).$$
For any element, there is another element that fulfills $P$. In other words, there are two or more elements that verify $P$.

Technical note: for the empty set, the proposition is vacuously true.
A: If we have at least two objects $a$ and $b$ for which $P(a)$ and $P(b)$ holds. Then the statement you've written holds because, for any $x$ which is not $b$ there exists $y=b$. And for $x=b$ there exists $y=a$.
Now if we look at the statement you've written, usually it is taken that the domain of discourse of quantifiers should necessarily be nonempty, so there must exist at least some object $b$. Now from the statement you've written it follows that for $b$ there exists $c\neq b$ such that $P(c)$ holds. Now it also follows that for this $c$ there exists another $d$ such that $d\neq c$  and $P(d)$ holds. So we've found two objects such that that $P(c)$ and $P(d)$ hold and these objects are different.
So basically we've shown that these statements are equivalent.
A: $\forall x~\exists y~(P(y)\wedge x\neq y)$
This says: for any anything there is something else that satisfies $P$.
Assume the statement holds.   Then anything either satisfies $P$, or there is something else that does.   Thus there is at least something that satisfies $P$.   However, there must also be something else that satisfies $P$.   Therefore the statement entails that there are at least two things which satisfy $P$.

Assume that at least two things satisfy $P$.  We may say that for anything there is something else which satisfies $P$; that is $\forall x~\exists y~(P(y)\wedge x\neq y)$.   Therefore the assumption entails the statement.

Therefore the statement is true exactly when there are at least two things that satisfy $P$.
A: More natural would be: $~~\exists x: \exists y:[x\neq y \land P(x) \land P(y)]$
