# Reduce following expression to one literal, boolean algebra

$$W'X(Z'+Y'Z)+X(W+W'YZ)$$

The goal is to reduce the following to one literal So after I expanded it out, i got the following: $$W'XZ'+W'XY'Z+WX+W'XYZ$$

Now from here, I got stuck and didn't know what to do next. So i went and checked the solutions and it gave me $$W'XZ'+W'XZ+WX$$ $$W'X+WX$$ $$X$$

Now I'm just wondering , how the last term in the equation suddenly dissapeared and I was guessing it was the consensus theorem which is something like: $$XY+X'Z+YZ = XY+X'Z$$

But I am confused because I don't know when and what form the equation has to be in order for me to able to use that theorem. If someone could show me the algebraic process to get rid of the last terms, it would be great (for my question, not for the theorem).

We essentially use only the distributive rule

• $AB + AC = A(B+C)$ or $BA + CA = (B+C)A$ and
• the identity that for any $A$, $(A + A') = 1$.

Starting from where you left off:

\begin{align} W'XZ'+W'XY'Z+WX+W'XYZ & = W'XZ' + W'XZ(Y' + Y) + WX \\ \\ & = W'XZ' + W'XZ(1) + WX \\ \\ & = W'XZ' + W'XZ + WX \\ \\ & = W'X(Z'+Z) + WX \\ \\ & = W'X(1) + WX \\ \\ & = W'X + WX \\ \\ & = (W' + W)X \\ \\ & = (1)X \\ \\ & = X \end{align}

• Thank you for quick answer, can you explain your very first step? I don't see how you manipulated it. Also I think you missed a W. How did the last term dissapear? – Aaron Feb 26 '13 at 4:28
• The first step uses the distributive rule: The common factor of the second and fourth term is $W'XZ$. So by the distributive rule $W'X\color{blue}{\bf Y'}Z + W'X\color{blue}{\bf Y}Z = W'XZ(\color{blue}{\bf Y' + Y})$ – amWhy Feb 26 '13 at 4:34
• Oh damn I missed that.. – Aaron Feb 26 '13 at 4:37
• I thought it was more complicated than it actually is – Aaron Feb 26 '13 at 4:37
• Not a problem! It sometimes gets confusing with all the LETTERS, their ordering, and keeping straight which letters are negated...We all have our days :-) – amWhy Feb 26 '13 at 4:39