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!

Answer 1

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.