is the binary relation "strictly higher than'' antisymmetric?

Question

A order on a set X is a binary relation ≻,defined as:x≻y if x≽y and ¬y≽x. Is that binary relation antisymmetric?

Answer 1

Yes, because $$\forall x,\forall y, (x=y\lor \neg(x\succ y\land y\succ x))$$

as it can be seen by the fact that by definition $$\neg(x\succ y\land y\succ x)\equiv \neg(x\succeq y\land \neg y\succeq x\land y\succeq x\land \neg x\succeq y)\equiv \neg\text{FALSE}\equiv \text{TRUE}$$

Answer 2

By definition, the relation will be antisymmetrical exactly when $\forall x\in X~\forall y\in X~((x\succ y\wedge y\succ x)\to x=y)$.

So if it is antisymmetrical, then you can take arbitrary $x,y$ from $X$, assume $x\succ y\wedge y\succ x$, and somehow derive $x=y$.

Can you?