# Boolean algebra and closure axiom

A Boolean algebra is an algebraic system (B,$$∨$$,$$∧$$,$$¬$$), where $$∨$$ and $$∧$$ are binary, and $$¬$$ is a unary operation.

One of the Boolean algebra axiom is: If $$a$$ and $$b$$ are elements of $$B$$, then $$(a ∨ b)$$ and $$(a ∧ b)$$ are in $$B$$.

i.e. the set $$B = (1111,0011,0110,1010,0000,1100,1001,0101)$$ I can't use as carrier for Boolean algebra, because the result of operation $$0011 ∨ 0110 = 0111$$ is't in set $$B$$.

Is it correct? Do I correctly think about closure for $$∨$$ and $$∧$$ operators?

• Yes, that's right. Dec 20, 2018 at 12:54
• @MauroALLEGRANZA because I need binary vectors. Dec 20, 2018 at 12:56
• @RobertIsrael may be you can send me link to book which explains how can I prove, that some set with operations, defined on it, is Boolean algebra? Dec 20, 2018 at 12:58

The problem is how to define the $$B$$. The $$B$$ is not a concrete collection that has some concrete elements. For help get around this, you should think the $$B$$ is a collection that be formalizing defined, such as $$\mathbb{N}$$, $$\mathbb{R}$$, etc. You can say $$x \in \mathbb{B}$$ but you cannot list all the elements of $$\mathbb{B}$$.
So the closure axiom can be denote: For all $$a$$, $$b$$ if $$a, b \in$$ $$\mathbb{B}$$, then $$(a \wedge b )\in \mathbb{B}$$