Boolean simplification - am I right? How do I simplify the boolean function $AB + C(B' + AB)$.
I implemented the following 
$AB + C(B' + AB) = AB + C(B'+A) = AB + CB' + CA$
I've been stuck on this for weeks :(
I feel like my answer is wrong after attempting to verify it online
 A: $$AB + C(AB) = AB$$
so 
$$AB + C(B^\prime + AB) = AB + CB^\prime$$
A: The Consensus Theorem states that:
$$XY+X'Z+YZ=XY+X'Z$$
See the pattern?  You have one term with $X$ (the $XY$) another one with $X'$ (the $X'Z$), and a third that combines whatever else occurs in the first two (the $YZ$). In that case, the third term can be removed. 
Why? Conceptually, the dual form is actually a little easier to understand:
$$(X+Y)(X'+Z)(Y+Z)=(X+Y)(X'+Z)$$
OK, so here trivially the left side logically implies the right side, but the right side also implies the left side: If $X$ is True, then given that $X'+Y$ must be true, it must be the case that $Z$ is true, and hence we have $Y+Z$ true as well. On the other hand, if $X$ is false, then $X'$ is True, and since  $X+Y$ must be true, it must be the case that $Y$ is true, and hence once again we have $Y+Z$ true as well. So, with the right side being true, $Y+Z$ will have to be true as well, and hence the whole left side is true.
For the original 
$$XY+X'Z+YZ=XY+X'Z$$
we can argue as follows: This time, the right side trivially implies the left side, and to see that the left side implies the right side, we'll do the contrapositive: we'll show that if the right side is false, then the left side is false. OK, so assume the right side is false. Then both $XY$ and $X'Z$ have to be false. Now, if $X$ is true, then since $XY$ cannot be true, $Y$ must be false, and hence $YZ$ is false. On the other hand, if $X$ is False, then $X'$ is false, and since $X'Z$ has to be false, $Z$ has to be false, and so once again $YZ$ is false. So, either way $YZ$ is false, and hence the whole left side is False as well.
You can also derive the Consensus Theorem algebraically from more elementary boolean algebra principles:
$$XY+X'Z+YZ=$$
$$XY+X'Z+(1)YZ=$$
$$XY+X'Z+(X+X')YZ=$$
$$XY+X'Z+XYZ+X'YZ)=$$
$$XY+XYZ+X'Z+X'YZ)=$$
$$XY(1+Z)+X'Z(1+Y)=$$
$$XY(1)+X'Z(1)=$$
$$XY+X'Z+$$
Finally, you can nicely see what happens here in a K-Map. Let's start with the K-Map for $XY+X'Z+YZ$:

Now let's break apart the $YZ$ term (note that another well-known Boolean Algebra principle called Adjacency states that $PQ+PQ'=P$ ... so in this case, we do: $YZ=XYZ+X'YZ$. Note that the two terms are indeed 'adjacent' in the K-Map ... hence the name):

Finally, we realize that we don't need the $XYZ$ and $X'YZ$ terms, since they are 'covered' by the $XY$ and $X'Z$ terms respectively (indeed, $P+PQ=P$ is thw ell-known principle of Absorption: the $PQ$ terms gets 'absorbed' by the $P$ term):

But of course, we should have realize that the $YZ$ form the very first K-Map was unnecessary, hence the Concensus theorem. That is, the $XY$ and the $X'Z$ terms together forms the concensus of $YZ$.
