Elementary set theory problem: Prove\Disprove that $ab+a'b'+bc=ab+a'b'+a'c$. I'm trying to prove or disprove this following equation in Set theory
:
$ab+a'b'+bc=ab+a'b'+a'c$.
I draw a Venn diagram and saw that the union of groups in each side of equation are the same, but I had a difficulty to give a formal prove for this.
I tried to use the fact that $a'b'=a'b'(c+c')$  in order to get to a $a'c$ expression.
Thanks a lot
 A: $ab+a'b'+bc=ab+a'b'+(a+a')bc=ab+a'b'+abc+a'bc=ab+a'b'+a'bc$ because $ab+abc=ab$. $ab+a'b'+a'c=ab+a'b'+a'(b+b')c=ab+a'b'+a'bc+a'b'c=ab+a'b'+a'bc$ because $a'b'+a'b'c=a'b'$. 
A: From your Venn diagram you can see that the $bc$ term on the lefthand side adds to $ab+a'b'$ only what’s in $b$ and $c$ and not in $a$: any part of $bc$ that’s in $a$ is already taken care of by the $ab$ term. Similarly, the $a'c$ term on the righthand side adds only what’s in $c$, not in $a$, and in $b$: any part of $a'c$ that’s not in $b$ is already taken care of by the $a'b'$ term. This suggests trying to prove that both sides are equal to $ab+a'b'+a'bc$. 
For the lefthand side, remember that we noticed pictorially that $bc$ could be replaced by $a'bc$ because the $abc$ part of it was already covered by $ab$. This suggests that we should try splitting it into those two parts:
$$\begin{align*}
ab+a'b'+bc&=ab+a'b'+(a+a')bc\\
&=ab+a'b'+abc+a'bc\\
&=(ab+abc)+a'b'+a'bc\\
&=ab+a'b'+a'bc\;.
\end{align*}$$
For the righthand side we saw that $a'c$ could be replaced by $a'bc$ because the $a'b'c$ part was already covered by $a'b'$, so this time I split $a'c$ into $a'bc$ and $a'b'c$:
$$\begin{align*}
ab+a'b'+a'c&=ab+a'b'+a'(b+b')c\\
&=ab+a'b'+a'bc+a'b'c\\
&=ab+(a'b'+a'b'c)+a'bc\\
&=ab+a'b'+a'bc\;.
\end{align*}$$
The result now follows immediately.
