First order logic problem.

Question

This question is really playing tricks on my mind.

It goes like this...

There is one whom everybody is for, and that one person is for everybody.

So far I have this...

$$\exists{y}\forall{x} \operatorname{isFor}(x, y) \land \forall{x}\exists{z} \operatorname{isFor}(z, x) \implies y=z$$

However I am not sure if this is correct and I keep changing my answer, can somebody please point me in the right direction.

Answer 1

I believe that the following:

$\exists y \, \forall x [\text{isFor}(x,y) \wedge \text{isFor}(y,x)]$

is correct, as the first portion states that there exists a $y$ such that for all $x$, $x$ is for $y$ -- that is, 'There is one whom everybody is for', and the second portion states that, given that $y$, $y$ is for all $x$ -- 'that one person is for everybody'.

Answer 2

There is an individual who everybody is for. How to interpet the "that one person" in the second part is not clear. But let us assume that there is an implication of uniqueness.

There may still be other individual who are for everybody.

So we will say something like this: (i) there is exactly one person who everybody is for, and that person is for everybody. So in symbols, using $F(s,t)$ for $s$ is for $t$, we have $$\exists y \Bigl((\forall x F(x,y) \land F(y,x))\land \forall z(\forall w F(w,z)\longrightarrow z=y)\Bigr) .$$

There are other equivalent ways we could say precisely the same thing. For example we might put the uniqueness statement in the middle in order to imitate more closely the order of the English sentence. We could also use fewer variables, the $w$ can be replaced by $x$.