I'm trying to understand the distributive law of Boolean algebra by relating it to what I know from ordinary algebra, and it seems to only work for one form of the law.
Ordinary algebra:
$x(y+z) = xy + xz$
This holds true for the first part of the distributive law.
Ordinary algebra:
$x + (yz)$
It's just that. I don't see any way of simplifying this more.
But in Boolean Algebra:
$x + (yz) = x\vee (y \wedge z) = (x\vee y) \wedge (x\vee z) = (x+y)(x+z)$
And in ordinary algebra, if you expand this, it's:
$xx+xz+xy+yz$
Could someone please help me understand this? Am I wrong in trying to relate ordinary algebra to Boolean algebra? Subquestion: if I am wrong in trying to make this connection, why?
Best Answer
A simple answer to your question is that Boolean algebra is not the same thing as 'ordinary' algebra (arithmetic). The intuition that connects $+$ with $\vee$ and $\times$ with $\wedge$ really is just intuition: the laws of addition and multiplication of numbers do not exactly coincide with the laws of the join and meet operations, and there's no reason to expect them to do so.
But there is a connection that can be made, which is if work with the integers modulo $2$, i.e. we only look at whether an integer is even or odd. In this case, it is true that $x+yz = (x+y)(x+z)$, since the quantity on the left is even (resp. odd) if and only if the quantity on the right is even (resp. odd).
A more mathematical answer for why $x \vee (y \wedge z) = (x \vee y) \wedge (x \vee z)$ has to do with duality. Every boolean algebra $\mathbb{B}$ has a dual $\mathbb{B}^{\mathrm{op}}$, defined so that the meet of $\mathbb{B}^{\mathrm{op}}$ is the join of $\mathbb{B}$, the join of $\mathbb{B}^{\mathrm{op}}$ is the meet of $\mathbb{B}$, and the top and bottom elements are swapped. This means that any equation that holds in an arbitrary Boolean algebra must also be true in an arbitrary Boolean algebra if we swap $\wedge$ and $\vee$ (and $\top$ and $\bot$).