Another way to state antisymmetry of a total order?

Question

Adapted from Wikipedia:

"Set $S$ is totally ordered under $\le$" means:

$x,y \in S \implies ((x \le y \wedge y \le x) \implies x=y)$ (antisymmetry)

and

$x,y,z \in S \implies ((x \le y \wedge y \le z) \implies x \le z)$ (transitivity)

and

$x,y \in S \implies (x \le y \vee y \le x)$ (totality)

I'm trying to understand it independently of the $=$ relation. Is there another way to state antisymmetry? Does the meaning of "total order" remain equivalent if antisymmetry is changed to the following?

$x \in S \implies ((x \le y \wedge y \le x) \implies y \in S)$

EDIT: as comments/answers have pointed out, that removes meaning since $x \le y$ is only meaningful if $x,y \in S$. So the question becomes "Is there a way to state antisymmetry that doesn't use $=$?"

EDIT: I think I was going for this:

$x,y \in S \implies ((x \le y \wedge y \le x) \implies$ after removing either of them from $S$, the other $\notin S)$.

In other words:

$x,y \in S \implies ((x \le y \wedge y \le x) \implies (S' \subset S \implies (x \in S' \iff y \in S')))$

Answer 1

$x,y \in S \implies ((x \le y \wedge y \le x) \implies (S' \subset S \implies (x \in S' \iff y \in S')))$

Answer 2

The reason the antisymmetry is written this way is that total orders are also reflexive, i.e. $x\le x$.
If you want to break this, it's called a strict total order, and you can have $$x,y\in S \implies \neg (x< y \wedge y< x)$$

The OP's suggested alternative doesn't make sense since $\le$ is only defined for pairs of elements of $S$.

Edit: silly typo fixed.