# Boolean algebra

## Axioms

THEOREM

Let $$(S, \vee, \wedge)$$ be a Boolean algebra. Then any theorem in $$(S, \vee, \wedge)$$ remains valid if both $$\vee$$ and $$\wedge$$ are interchanged, and also $$\bot$$ and $$\top$$ are interchanged throughout the whole theorem.

Proof.     ProofWiki
THEOREM    $$- x \sqcap x = \bot$$
Proof 1.     \begin{align} - x \sqcap x &= x \sqcap - x \\ &= \bot \end{align}
Proof 2.
THEOREM    $$- x \sqcup x = \top$$
Proof 1.     \begin{align} - x \sqcup x &= x \sqcup - x \\ &= \top \end{align}
Proof 2.
THEOREM

Top is a right zero of join.

$$x \sqcup \top = \top$$
Proof 1.     ProofWiki
Proof 2.
THEOREM

Bottom is a right zero of meet.

$$x \sqcap \bot = \bot$$
Proof 1.     ProofWiki
Proof 2.
THEOREM

Bottom is the complement of top.

$$- \top = \bot$$
Proof 1.     \begin{align} - \top &= \top \sqcap - \top \\ &= \bot \end{align}
Proof 2.
THEOREM

Top is the complement of bottom.

$$- \bot = \top$$
Proof 1.     \begin{align} - \bot &= \bot \sqcup - \bot \\ &= \top \end{align}
Proof 2.
THEOREM

Join is idempotent.

$$x \sqcup x = x$$
Proof 1.     \begin{align} x &= x \sqcup \bot \\ &= x \sqcup (x \sqcap - x) \\ &= (x \sqcup x) \sqcap (x \sqcup - x) \\ &= (x \sqcup x) \sqcap \top \\ &= x \sqcup x \end{align}
Proof 2.     ProofWiki
Proof 3.
THEOREM

Meet is idempotent.

$$x \sqcap x = x$$
Proof 1.     \begin{align} x &= x \sqcap \top \\ &= x \sqcap (x \sqcup - x) \\ &= (x \sqcap x) \sqcup (x \sqcap - x) \\ &= (x \sqcap x) \sqcup \bot \\ &= x \sqcap x \end{align}
Proof 2.     ProofWiki
Proof 3.
THEOREM

Meet absorbs join.

$$x \sqcap (x \sqcup y)$$
Proof 1.     ProofWiki
Proof 2.     \begin{align} x \sqcap (x \sqcup y) &= (x \sqcup \bot) \sqcap (x \sqcup y) \\ &= x \sqcup \bot \sqcap y \\ &= x \sqcup \bot \\ &= x \end{align}
Proof 3.
THEOREM

Join absorbs meet.

$$x \sqcup (x \sqcap y)$$
Proof 1.     ProofWiki
Proof 2.
Proof 3.
THEOREM    $$- (x \sqcup y) = - x \sqcap - y$$
Proof.
THEOREM    $$- (x \sqcap y) = - x \sqcup - y$$
Proof.