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.


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 $\cup$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.