# Boolean Algebra: Explain why (M AND (NOT N)) OR (X AND M AND N) = (M AND NOT N) OR (X AND M)?

I have no idea how this is true, by what theorem, and I literally have been thinking about this for 3 hours now. I know it's really simple, but I just must not be in the right mindset to discover this now.

Here is (one of) the exact places where this is occurring:

= K'L'MN + MN'
= K'L'M + MN'


So you can see that N from the first term is getting dropped, due to some logical constraint I can't seem to fathom. If you replace K'L' with X, and then use a 3-circle Venn Diagram, you'll see it's true there also. But I was hoping someone could explain it in words, or at least Boolean algebra lemmas/rules.

Can someone explain why this is true?

• You only have three variables, $X, M, N$, so it shouldn't be too hard to check via a truth table. – Sp3000 Sep 13 '13 at 4:47
• Oh I did, and it's true. But I can't fathom why, in my head. It's really bothering me, especially because it is so common/simple. – user2055216 Sep 13 '13 at 4:47

Let's take a look at $MN' + XMN$. We can write this as
\begin{align} MN' + XMN &= M(N'+XN) \\ &= M((N'+X)(N'+N)) \\ &=M((N'+X)\cdot1)\\ &=M(N'+X)\\ &=MN'+XM \end{align}
That's how you could do it with Boolean algebra. The only real trick was the $N'+XN=(N'+X)(N'+N)$ part, which uses the distributivity of AND and OR.
• How did you replace $N' + N$ with $1$? – goblin Sep 13 '13 at 5:39