Boolean Algebra - Xor simplification I have a boolean equation: $e(g \oplus (g + b))$ and it is simplified to $e(\lnot g)b$.
I do not see how this simplification is done. What i did was the following:
$e(g \oplus (g + b)) --> e(g(\lnot(b+g)) + (\lnot g)(b + g))$
$--> e(g(\lnot b) + g(\lnot g) + (\lnot g)b + (\lnot g)g)$
$--> e(g(\lnot b) + (\lnot g)b)$
But this is not correct. What am i doing wrong and what is the correct way to reduce the boolean equation?
 A: Your mistake was with De Morgan's Law.  $(b+g)'=b'g'$.  So you wind up with a $gb'g'$ term which goes away since g and not g can't both be true.
A: $\begin{align}e\cdot(g \oplus (g + b)) = & e\cdot(g\cdot\lnot(b+g) + \lnot g\cdot(b + g))
\\= & e\cdot(g\cdot\lnot b\color{red}{\cdot}\lnot g + \lnot g\cdot b + \lnot g\cdot g)
\\= & e\cdot\lnot g\cdot b
\end{align}$
A: To expand upon Graham's answer, 
$e \cdot(g \oplus (g + b)) = e \cdot((g \cdot \neg(g + b)) + (\neg g \cdot (g + b))$ [Definition of $\oplus$]
$\quad \quad \quad \quad \quad \quad \;\;\; = e \cdot((g \cdot (\neg g \cdot \neg b)) + (\neg g \cdot (g + b)))$ [De Morgan's]
$\quad \quad \quad \quad \quad \quad \;\;\; = e \cdot (((g \cdot \neg g) \cdot b) + ((\neg g \cdot g) + (\neg g \cdot b)))$ [Associativity, distribution]
$\quad \quad \quad \quad \quad \quad \;\;\; = e \cdot (((\bot) \cdot b) + ((\bot) + (-g \cdot b)))$ [Complementation]
$\quad \quad \quad \quad \quad \quad \;\;\; = e \cdot ((\bot) + (\neg g \cdot b))$ [Identity - conjunction, identity - disjunction]
$\quad \quad \quad \quad \quad \quad \;\;\; = e \cdot (\neg g \cdot b)$ [Identity - disjunction]
This link on the laws of propositional logic lays out all of the equivalences and rules allowed in the system :) Hopefully this is useful!
A: Using $x+y\iff x \oplus y\oplus xy$ gives an expression in a Boolean ring where all "normal" laws of addition and multiplication holds:
$e(g\oplus (g+b))=e(g\oplus g\oplus b\oplus gb) =e(b\oplus bg)=eb\oplus ebg=eb(1\oplus g)=eb(\neg g)\;$, since $g\oplus g=0$
