quick question about writing consensus theorem

Question

In order to prove $bc + abc + bcd + a'(d+c) = abc + a'c + a'd$

I got it down to $abc + a'c + a'd + bc + bcd$ (LHS), and from there I factor out $bc$ from $bc + bcd$, which is $bc(1+d)$, simplifies into $bc$, what should I do after?

Answer 1

You're trying to prove $abc + a'c + a'd + bc=abc + a'c + a'd$. Write $bc=(a+a')bc=$ $abc+a'bc$, and note $a'bc+a'c=a'c$ by $x+xy=x$, and note also $abc+abc=abc$, so the LHS becomes $a'c+a'd+abc$ as desired.

A different approach: Suppose $a=0$, then $abc=0$, and the equation we want to prove reduces to $bc+bcd+d+c=d+c$, which is true applying $x+xy=x$.

Suppose $a=1$, then $a'(d+c)=0$ and the equation we want to prove reduces to $bc+bc+bcd=bc$, which is also true (we need also $x+x=x$).