Ordering Cycle implies Equality/General Case

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

Let $x_0, x_1, \dots, x_n \in S$.

Suppose that for $k = 0, 1, \dots, n-1: x_k \preceq x_{k+1}$.

Suppose also that $x_n \preceq x_0$.

That is, suppose that $x_0 \preceq x_1$, $x_1 \preceq x_2, \ldots, x_{n-1} \preceq x_n$, and $x_n \preceq x_0$.

Then $x_0 = x_1 = \dots = x_n$.

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

From the first premise and Transitive Chaining, we conclude that $x_0 \preceq x_n$.

From the other premise we know that $x_n \preceq x_0$.

Since $\preceq$ is antisymmetric, we conclude that $x_0 = x_n$.

Also known as
refers to this property as anti-circularity.