Brouwerian Lattice is Upper Bounded

Theorem
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a Brouwerian lattice.

Then $S$ is upper bounded.

Proof
By assumption:
 * $S \ne \varnothing$

By definition of non-empty set:
 * $\exists s: s \in S$

By definition of Brouwerian lattice:
 * $s$ has relative pseudocomplement with respect to $s$

By definition of relative pseudocomplement:
 * $\max \left\{ {x \in S: s \wedge x \preceq s}\right\}$ exists and equals $s \to s$

Let $x \in S$.

By Meet Precedes Operands:
 * $s \wedge x \preceq s$

Then
 * $x \in \left\{ {x \in S: s \wedge x \preceq s}\right\}$

Thus by definition of maximum:
 * $x \preceq s \to s$

Thus by definition:
 * $s \to s$ is upper bound for $S$

Thus by definition
 * $S$ is upper bounded.