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.

$\blacksquare$


Sources

  • 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.5$: Theorem $1.14$
  • 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Boolean algebra
  • 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): duality: 2.
  • 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Boolean algebra
  • 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): duality: 2.
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