Boolean Algebra Simplification of $(A + B + C') \cdot(A'\cdot B' + C)$

Question

Good Afternoon all,

I am having trouble simplifying the expression $(A + B + C') \cdot(A'\cdot B' + C)$, I've tried converting $A'\cdot B'+ C$ to $( A' + B' + C)$ and factoring it with the other term but I don't think this is right.

Could anyone solve this step by step, maybe with which rule applied, so I can see how to do it?

Thanks!

Welcome to MSE. Could you maybe clarify what $A,B,C$ are? Since you are tagging your question with boolean-algebra but you used unusual signs for this branch such as $+$ it is not clear what exactly you are talking about. Just add these details :) — mrtaurho, Sep 22 '18 at 21:02

score 1 · Accepted Answer · answered Sep 23 '18 at 10:11

Using your notation we have: $$\begin{align} (A+B+C')\cdot(A'\cdot B'+C) & = A\cdot (A'\cdot B'+C) +B\cdot (A'\cdot B'+C)+C'\cdot (A'\cdot B'+C) \\ & =(A A' B'+ A C) + (B A' B'+ B C)+(C' A' B'+C' C) \\ & =0 +A\cdot C+0+B\cdot C+A'\cdot B'\cdot C'+0 \\ & =(A+B)\cdot C+(A+B)'\cdot C' \\ & =(A+B)\odot C \end{align}$$ where $\odot$ is the Exclusive NOR.

The laws used in each line are respectively distributive, distributive, complement, distributive and de Morgan, and finally definition of Exclusive NOR.

Thanks! I appreciate you posting the rules this really helped! — T S, Sep 23 '18 at 14:10

Boolean Algebra Simplification of $(A + B + C') \cdot(A'\cdot B' + C)$

1 Answers1