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.    

See also