Brouwerian Lattice is Upper Bounded

Theorem
Let $\struct {S, \vee, \wedge, \preceq}$ be a Brouwerian lattice.

Then $S$ is upper bounded.

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

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:
 * $\max \set {x \in S: s \wedge x \preceq s}$ exists and equals $s \to s$

Let $x \in S$.

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

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

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

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

Thus by definition:
 * $S$ is upper bounded.