Ordering Cycle implies Equality

Theorem
Let $\left({S,\le}\right)$ be an ordered set.

Let $x_1, x_2, x_3$ be elements of $S$.

Suppose that

Then $x_1 = x_2 = x_3$.

Proof
Since $\le$ is an ordering, it is transitive and antisymmetric.

By transitivity, $x_1 \le x_3$.

Since $x_1 \le x_3$ and $x_3 \le x_1$, antisymmetry allows us to conclude that $x_1 = x_3$.

Since $x_1 = x_3$ and $x_1 \le x_2$, we must have $x_3 \le x_2$.

Since $x_3 \le x_2$ and $x_2 \le x_3$, antisymmetry allows us to conclude that $x_2 = x_3$.

Thus $x_1 = x_2 = x_3$.