Pseudocomplemented Lattice is Bounded

Theorem
Let $(L, \wedge, \vee, \preceq)$ be a pseudocomplemented lattice.

Then $(L, \wedge, \vee, \preceq)$ is a bounded lattice.

Proof
By the definition of pseudocomplemented lattice, $L$ has a smallest element $\bot$.

Let $x \in L$.

Then $x \wedge \bot = \bot$.

By the definition of pseudocomplemented lattice, there is a greatest element $\bot^*$ such that $\bot \wedge \bot^* = \bot$.

But then by the definition of greatest element, $x \preceq \bot^*$ for each $x \in L$, so $\bot^*$ is the greatest element of $L$.

Since $L$ has a smallest element and a greatest element, it is a bounded lattice.