Simplifying boolean algebra expression that contains XOR How can I simplify followed boolean algebra expression; Normally I express as simplify without XOR also this expression contains both XOR and multiple variables.
(((A + B)' * C') xor ((A' + B') * C')' ) * A
 A: Lets say X = (A + B)' * C') and Y = ((A' + B') * C')'
We know that XOR can be represented as
$X ⊕ Y = X\overline Y + \overline X Y$
So, first simplify the inners and once you have the simplified version, you can deal with the outer A.
A: I've never been a fan of the boolean algebra notation, so this is not a direct answer to your question, but more how I would do this simplification using propositional logic notation.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
Translated into that notation, you're asked to simplify $$
\big(\lnot (A \lor B) \land \lnot C \;\not\equiv\; \lnot ((\lnot A \lor \lnot B) \land C)\big) \;\land\; A
$$
We can do that as in the following calculation:
$$\calc
    \big(\lnot (A \lor B) \land \lnot C \;\not\equiv\; \lnot ((\lnot A \lor \lnot B) \land C)\big) \;\land\; A
\op=\hint{use $\;A\;$ on other side of the rightmost $\;\land\;$}
    \big(\lnot (\true \lor B) \land \lnot C \;\not\equiv\; \lnot ((\false \lor \lnot B) \land C)\big) \;\land\; A
\op=\hint{simplify}
    \big(\false \;\not\equiv\; \lnot (\lnot B \land C)\big) \;\land\; A
\op=\hint{simplify $\;\false \not\equiv P\;$ to $\;P\;$; DeMorgan}
    (B \lor \lnot C) \;\land\; A
\endcalc$$
And now it is trivial to translate that back to boolean algebra notation...
