Can someone explain consensus theorem for boolean algebra In boolean algebra, below is the consensus theorem
$$X⋅Y + X'⋅Z + Y⋅Z = X⋅Y + X'⋅Z$$
$$(X+Y)⋅(X'+Z)⋅(Y+Z) = (X+Y)⋅(X'+Z)$$
I don't really understand it? Can I simplify it to 
$$X'⋅Z + Y⋅Z = X' \cdot Z$$
I don't suppose so. Anyways, why can $Y \cdot Z$ be removed?

The proof that grep has given is fine, as is the one in Wikipedia, but they don’t give much insight into why such a result should be true. To get some feel for that, look at the most familiar kind of Boolean algebra: the Boolean algebra of subsets of some given set $S$, with $\cap$ for $\cdot$, $\cup$ for $+$, and $'$ interpreted as the relative complement in $S$ (i.e., $X' = S \setminus X$). In this algebra the theorem says that $$(X\cap Y) \cup (X' \cap Z) \cup (Y \cap Z) = (X\cap Y) \cup (X' \cap Z),$$ which amounts to saying that $$Y \cap Z \subseteq (X\cap Y) \cup (X' \cap Z).$$ This isn’t hard to prove, but doing so won’t necessarily give you any better feel for what’s going on. For that I suggest looking at the corresponding Venn diagram, with circles representing $X$, $Y$, and $Z$. Shade the region representing $(X\cap Y) \cup (X' \cap Z)$. Now look at the region representing $Y \cap Z$: it’s already shaded, because it’s a subset of $(X\cap Y) \cup (X' \cap Z)$. Throwing it in with $(X\cap Y) \cup (X' \cap Z)$ to make $(X\cap Y) \cup (X' \cap Z) \cup (Y \cap Z)$ adds nothing.
