Why does $AB + BC + CA = (A\oplus B)C + AB$ not imply $BC + CA = (A\oplus B)C$ in boolean algebra? I am new to Logical Inequalities. Please bear with me if I am inexplicably stupid.
The following is a Proven Equality:
$$AB + BC + CA = (A\oplus B)C + AB$$
I noticed that I cannot "cancel out" $AB$ from both sides of the identity, i.e.
$$BC + CA \neq (A\oplus B)C$$
In mathematical equations, we cancel out stuff all the time. For example,
$$X + 3 = Y + 3 \quad\implies\quad X = Y$$
Can this not be done with Logical Statements? Is there a reason why this is such a way?
 A: In general, in a Boolean algebra it is not true that if equal terms appear on both sides of an equivalence, then we may "cancel" them (cancellation law). In other words,
\begin{align}
A+C = B + C \,\not\!\!\!\implies A = B
\end{align}
Indeed, consider the case where $A = 1 = C$ and $B = 0$. Then, $A + C = 1 + 1 = 1 = 0 + 1 = 1 = B$ but $A = 1 \neq 0 = B$. 
You have the phenomenon in set theory, and essentially for the same reason. To see the analogy, consider the boolean $+$ (the logical "or") as the set union $\cup$. We have have that:
\begin{align}
A \cup C = B \cup C \, \not\!\!\!\implies A = B
\end{align}
Indeed, take $A = \{a\}$, $B = \{b\}$ and $C = \{a,b\}$, with $a \neq b$. Then, $A \cup C = \{a,b\} = B \cup C$ but $A = \{a\} \neq \{b\} = B$. 

The deep reason is that boolean $+$, as well as set union $\cup$, is idempotent, differently than arithmetical $+$ in natural numbers. Idempotence means that $a + a = a$ for any $a$ in your domain. 
This is what happens with any $a \in \{0,1\}$ and boolean $+$, as well as with any set and set union $\cup$.
Idempotence makes impossible to define a "minus operation" that is the inverse of the idempotent $+$.
A: This is why I don't write "$+$" for or, but rather for $\operatorname{Xor}$. The $\operatorname{Xor}$ (eXclusive or) operator is better behaved. 
If you set $+$ to be $\operatorname{Xor}$, you get the equivalence you wanted : 
$$ \varphi + \psi = \varphi + \psi' \Longleftrightarrow \psi = \psi'$$
This is because every $\varphi$ has an inverse for $\operatorname{Xor}$, namely itself since $\varphi \operatorname{Xor} \varphi = 0$.
A: Boolean algebra has no minus operation since the equation $X + A = B$ has no unique solution. Consider the case $A=B=1.$ Then both $X=0$ and $X=1$ are solutions.
A: It is worth clarifying here: there is only one single rule at all times for equations: If $x=y$, then $f(x) = f(y)$, where $f$ is some mapping. Nothing more, nothing less.
What about cancellation? Isn't it true that, if $f(x) = x+ 3$, then $f(x)=f(y)$ implies $x = y$? Does this make a rule?
No. Forget it. What really happens behind this is that we define $g(x)=x-3$, so
$g(f(x)) = x$. And we apply the one single rule and we get $g(f(x)) = g(f(y))$, which simplifies to $x=y$. Is that clear?
Now go ahead and find a function $g$ such that when composed with $f(x) = x \oplus A$ recovers $g(f(x))=x$. There is no such function. And this is why cancellation doesn't hold.
