Top equals to Relative Pseudocomplement in Brouwerian Lattice

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

Let $a, b \in S$.

Then
 * $\top = a \to b$ $a \preceq b$

Sufficient Condition
Let
 * $\top = a \to b$

By definition of reflexivity:
 * $\top \preceq a \to b$

By Inequality with Meet Operation is Equvalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice:
 * $\top \wedge a \preceq b$

Thus
 * $a \preceq b$

Necessary Condition
Let
 * $a \preceq b$

Then
 * $a \wedge \top \preceq b$

By Inequality with Meet Operation is Equvalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice
 * $\top \preceq a \to b$

By definition of greatest element:
 * $a \to b \preceq \top$

By definition of ordered set:
 * $\top = a \to b$