Upper Bounds are Equivalent implies Suprema are equal

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

Let $X, Y$ be subsets of $S$.

Assume that
 * $X$ admits a supremum

and
 * $\forall x \in S: x$ is upper bound for $X \iff x$ is upper bound for $Y$

Then $\sup X = \sup Y$

Proof
We will prove that
 * $\forall b \in S: b$ is upper bound for $Y \implies \sup X \preceq b$

Let $b \in S$ such that
 * $b$ is upper bound for $Y$.

By assumption:
 * $b$ is upper bound for $X$.

Thus by definition of supremum:
 * $\sup X \preceq b$

By definition of supremum:
 * $\sup X$ is upper bound for $X$.

By assumption:
 * $\sup X$ is upper bound for $Y$.

Thus by definition of supremum:
 * $\sup X = \sup Y$