Duality Principle (Boolean Algebras)

Theorem
Let $\struct {S, \vee, \wedge}$ be a Boolean algebra.

Then any theorem in $\struct {S, \vee, \wedge}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.

Proof
Let us take the axioms of a Boolean algebra $\struct {S, \wedge, \vee}$:

It can be seen by inspection, that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout does not change the axioms.

Thus, what you get is a Boolean algebra again.

Hence the result.