Assume that my definition of a Boolean algebras is the following one: I have a set $B$ with two binary operations $\vee$ and $\wedge$ which both satisfy the commutative, associative and distributive laws. Both the operations have an identity element ($0$ and $1$) and for each element $b\in B$ there exists a complement (i.e. an element $b{'}\in B$ such that $b\vee b{'}=1$ and $b\wedge b{'}=0$).

Now I found the following definition of Boolean Algebra. Let $B$ be a set on which we define a binary operation $\cap$ and an unary operation ${'}$. We also require that those operations satisfy the following rules:

1) $x\cap y=y\cap x$ for each $x,\, y\in B$

2) $(x\cap y)\cap z=x\cap (y\cap z)$

3) $x\cap y{'}=z\cap z{'}$ iff $x\cap y=x$

Are those equivalent definitions? I was trying to work the thing out in the following way.

a) First we may put $0:=z\cap z'$ by property 3).

b) We may define $x\cup y:=(x{'}\cap y{'}){'}$ and put $0{'}=:1$ and we may prove the commutative law for $\cup$.

Anyway I cannot prove the associativity of $\cup$ or either the distributivity or even the fact the $0$ and $1$ are identity elements.


1 Answer 1


In rule 3, put $x=y=z=a$ to get $$ a \cap a' = a \cap a' \iff a \cap a = a $$ so $a\cap a = a$ in general. This implies $$ p \cap 0 = p \cap p \cap p' = p \cap p' = 0 $$ for all $p$.

Now put $x=b''$, $y=b$, $z=b'$ to get $$ b''\cap b' = b'\cap b'' \iff b'' \cap b = b'' $$ so $b'' \cap b = b''$ in general.

Finally put $x=c$, $y=c''$ to get $$ c \cap c''' = 0 \iff c \cap c'' = c $$ However, $$c\cap c''' = c\cap c' \cap c''' = 0 \cap c''' = 0 $$ so $c\cap c'' = c$ in general.

Thus $p = p\cap p'' = p''$ in general, so $'$ is an involution, which is what you need to transfer properties such as associativity from $\cap$ to $\cup$.

Now put $x=d$, $y=0'$ to get $$ d\cap 0'' = 0 \iff d\cap 0' = d $$ and since $0''=0$, the RHS of this is always true, and $1:=0'$ is an identity for $\cap$. By duality, $0$ is an identity for $\cup$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .