# Duality Principle (Boolean Algebras)

This proof is about Duality Principle in the context of Boolean Algebra. For other uses, see Duality Principle.

## Theorem

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

Then any theorem in $\struct {S, \vee, \wedge, \neg}$ 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 $\struct {S, \wedge, \vee, \neg}$:

 $(\text {BA}_1 0)$ $:$ $S$ is closed under $\vee$, $\wedge$ and $\neg$ $(\text {BA}_1 1)$ $:$ Both $\vee$ and $\wedge$ are commutative $(\text {BA}_1 2)$ $:$ Both $\vee$ and $\wedge$ distribute over the other $(\text {BA}_1 3)$ $:$ Both $\vee$ and $\wedge$ have identities $\bot$ and $\top$ respectively $(\text {BA}_1 4)$ $:$ $\forall a \in S: a \vee \neg a = \top, a \wedge \neg a = \bot$

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.

$\blacksquare$