Ordering Cycle implies Equality

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

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

Suppose that

Then $x_1 = x_2 = x_3$.

Proof
Because $\preceq$ is an ordering, it is transitive and antisymmetric.

By transitivity, $x_1 \preceq x_3$.

Because $x_1 \preceq x_3$ and $x_3 \preceq x_1$, antisymmetry allows us to conclude that $x_1 = x_3$.

Because $x_1 = x_3$ and $x_1 \preceq x_2$, we must have $x_3 \preceq x_2$.

Because $x_3 \preceq x_2$ and $x_2 \preceq x_3$, antisymmetry allows us to conclude that $x_2 = x_3$.

Thus $x_1 = x_2 = x_3$.