# 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

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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.

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And your answer differs from mine...how? Well, it is formatted more attractively. – Gerry Myerson Nov 15 '11 at 23:36
@Gerry: Had it been only a matter of formatting, I’d not have spent the time. It also does much more to explain how I arrived at the argument, and hence how the OP might have done so. – Brian M. Scott Nov 15 '11 at 23:53
True. Then again, OP knows $a'b'=a'b'(c+c')$, which is the main thing needed; it was just a matter of applying it to the third term instead of the second. – Gerry Myerson Nov 16 '11 at 1:15

$ab+a'b'+bc=ab+a'b'+(a+a')bc=ab+a'b'+abc+a'bc=ab+a'b'+a'bc$ 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 $a'b'+a'b'c=a'b'$.

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