Proving that a Boolean algebra only has 2 elements

Question

We define a Boolean algebra as a set $B$ of elements $a,b,\dots$ which satisfies the following axioms$^{[1]}$:

1. $B$ has two binary operators $\wedge$ or $\cdot$ (logical AND) and $\vee$ or $+$ (logical OR)
2. Idempotence
   • $a\wedge a = a \vee a = a$
3. Commutative law
   • $a \wedge b = b \wedge a$
   • $a \vee b = b \vee a$
4. Associative law
   • $a \wedge (b \wedge c) = (a \wedge b) \wedge c$
   • $a \vee (b \vee c) = (a \vee b) \vee c$
5. Absorption law
   • $a \wedge (a \vee b) = a \vee (a \wedge b) = a$
6. Mutual distributiveness
   • $a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c)$
   • $a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$
7. $B$ contains universal bounds $\emptyset$ (empty set) and $I$
   • $\emptyset \wedge a = \emptyset$
   • $\emptyset \vee a = a$
   • $I \wedge a = a$
   • $I \vee a = I$
8. $B$ has a unary operator $a\rightarrow a'$ such that
   • $a \wedge a' = \emptyset$
   • $a \vee a' = I$

How can I prove that $B$ is composed of only 2 truth elements $x, y\in B$? I feel like it should come from Axioms 7 and 8, but I couldn't come up with a rigorous proof.

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[1]http://mathworld.wolfram.com/BooleanAlgebra.html

Answer 1

A Boolean algebra need not have only two elements. For instance, if $X$ is any set, then you could take $B$ to be the set of all subsets of $X$, with $\wedge$ being intersection, $\vee$ being union, $\emptyset$ being the empty set, $I$ being $X$, and $a'$ being $X\setminus a$. All the axioms are then just familiar properties of basic set operations.