I made a silly experiment: I tried to find polynomials in $\mathbb{Z}_2$ which match the boolean operations. I have found:
$$a\wedge b \mapsto ab \;\; (mod\; 2) \quad \quad a\vee b \mapsto a+ab+b \;\; (mod\; 2)\quad \quad ¬a \mapsto a+1 \;\; (mod\; 2) \tag{1}$$
I noticed that coefficients and powers are superfluous: If a coefficient is odd, we can replace it by $1$, if even we can erase the monomial. I remember when learning about logic that we can represent any boolean function with a certain procedure, but this procedure don't always give the shortest expression for a given boolean function: There are ways to simplify it such as Veitch-Karnaugh maps, for example: From the truth-table for the multiplexer, we arrive at the following expression:
$$(x\wedge ¬y \wedge ¬ z)\vee (x\wedge y \wedge ¬ z)\vee (¬x\wedge y \wedge z)\vee (x\wedge y \wedge z) \tag{2}$$
But there is the much simpler expression:
$$( x \wedge ¬z) \vee (y \wedge z) \tag{3}$$
I thought that perhaps, converting the boolean expressions such as $(2)$ to the polynomials I found in $(1)$ and knowing that coefficients and exponents are superfluous, it could yield a natural way to simplify such expressions. I converted $(2)$ using $(1)$ and simplified, it gave me:
$$xz+x+yz$$
From which I can simplify to:
$$x(z+1)+yz$$
Now it seems that this expression is very similar to $(3)$ if we interpret $z+1$ as $¬z$. I'm not sure if this is a general process that works every time or if I was just lucky. Supposing it works, I found this process to be very easy to simplify boolean expressions. I have no idea on how to prove this. Could you help me?
I'd say it's not always going to be simplifying, for instance $$a\vee b\vee c\vee d \quad=\quad a+b+c+d+ab+ac+bc+ad+bd+cd+abc+abd+bcd+acd+abcd$$