Boolean Algebra simplification: $X=((AB)'C(A'+(B+C)'))'$

Question

I've had two statements I need to simplify, and I'm not sure about my work:

1. $X=((AB)'C(A'+(B+C)'))'.\quad $ With this one, do you apply DeMorgan's theorem to the interiors of the brackets and proceed from there? I got an answer of $\,A + B + C',\,$ but am not really sure whether I'm correct.

2. $Y=(((B+C)'+A')'+C)'. \quad $ I was pretty confused here as well, I tried to simplify it and got an answer of $\,A'C' + B'C',\,$ but again don't know whether I did the simplification correctly.

Thanks for any help offered!

Answer 1

I'll proceed with a detailed step-by-step simplification (or "unpacking") of the first problem. Try to follow with the appropriate justifications for each step. I did work from "inside out" (applying DeMorgan's first on $(B + C)'$). $$\begin{align} X& =((AB)'C(A'+(B+C)'))'\tag{1}\\ \\ & = \Big((AB)'C(A' + B'C')\Big)'\\ \\ & = \Big((AB)'(CA' + CB'C')\Big)' \\ \\ & = \Big((AB)'(CA' + F)\Big)' = \Big((AB)'(CA')\Big)' \\ \\ & = (AB) + (CA')'\\ \\ & = AB + C' + A \\ \\ & = (AB + A) + C' \\ \\ & = A(B + T) + C'\\ \\ & = A + C'\tag{1}\end{align}$$

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For the second simplification, your simplification is correct. If it will help to review your work, I'll proceed with some simplifications, and in this case, I'll start with applying DeMorgan's first from the outermost negation:

$$\begin{align} Y & = (((B+C)'+A')'+C)' \tag{2}\\ \\ & = ((B+C)' + A') C' \tag{DeMorgan's}\\ \\ & = (B'C' + A')C' \tag{DeMorgan's} \\ \\ & = B'C'C' + A'C' \tag{Distribution} \\ \\ & = B'C' + A'C' \equiv (A' + B')C'\end{align}$$

Answer 2

What do you mean by '$+$'? Do you mean addition(henceforth $+$) or upper bound (henceforth $\cup$), because I think you are confusing the two? In case you really mean addition, I have the following solution (note that in a boolean algebra $x'=1+x, x+x=0$ and $x\cup y=x+y+xy$ for all $x,y$):

1. $\begin{align} X &= ((AB)'C(A'+(B+C)'))'=1+(1+AB)C(1+A+1+B+C)\\&=1+(C+ABC)(A+B+C)=1+AC+BC+C+3ABC\\&=1+C+AC+BC+ABC=1+C(1+A+B+AB)=(C(A\cup B)')'\\&=A\cup B\cup C'\end{align}$


2.$\begin{align} Y=(((B+C)'+A')'+C)'=1+C+1+1+A+1+B+C=A+B\end{align}$