How to verify the identity $(A \bigtriangleup B) \cup C = (A \cup C) \bigtriangleup (B \setminus C)$? I have to verify the following identity $(A \bigtriangleup B) \cup C = (A \cup C) \bigtriangleup (B \setminus  C)$ using logic symbols. I have to say what it means to be an element of each set and then use logical equivalences.
This is what I have tried so far:
\begin{align}
x \in (A \bigtriangleup B) \cup C 
& \iff (x \in (A \bigtriangleup B)) \lor (x \in C) \\
& \iff (x \in ((A \setminus B) \cup (B \setminus A)) \lor (x \in C) \\
& \iff ((x \in A \setminus B) \lor (x \in B \setminus A)) \lor (x \in C)
\end{align}
Now I can remove external brackets to obtain
\begin{align}
(x \in A \setminus B) \lor (x \in B \setminus A) \lor (x \in C)
\end{align}
I now remove the $\setminus$ symbol to obtain
$$(x \in A \land \neg(x \in B)) \lor (x \in B \land \neg(x \in A)) \lor (x \in C)$$
Is this correct? I am not sure if I am on the right track. So, if you can try to explain why something is wrong, I could try to correct it.
Should I also perform similar manipulations for right-side of the equals sign (in the expression)?
 A: What you wrote is correct. Now if you do the same thing for the other side you will obtain: 
$ ( (x \in A \vee x \in C ) \wedge (x \not\in B \vee x \in C)) \vee ((x \not\in A \wedge x \not\in C)  \wedge (x \in B \wedge x \not\in C))$
Define: 


*

*$ \psi_1(x) \leftrightarrow x \in A \wedge x \not\in B $;

*$ \psi_2(x) \leftrightarrow x \in B \wedge x \not\in A $;

*$\psi_3(x) \leftrightarrow x \in C $;  

*$\varphi_1(x) \leftrightarrow (x \in A \vee x \in C ) \wedge (x \not\in B \vee x \in C);$

*$\varphi_2(x) \leftrightarrow (x \not\in A \wedge x \not\in C)  \wedge (x \in B \wedge x \not\in C) $.


We have 


*

*$\psi_1 \rightarrow \varphi_1; \psi_2 \rightarrow \varphi_2; \psi_3 \rightarrow \varphi_1$

*$ \varphi_1 \rightarrow \psi_1 \vee \psi_3; \varphi_2 \rightarrow \psi_2 \vee \psi_3$.  


Hence $\psi_1 \vee \psi_2 \vee \psi_3 \vee \leftrightarrow \varphi_1 \vee  \varphi_2$, this gives the equality. 
A: For comparison, here is another way to do this proof, by expanding the definitions and then using the laws of logic to simplify.
I will the use the simplest definition of $\;\triangle\;$ I know, which is
$$
x \in A \triangle B \;\equiv\; x \in A \not\equiv x \in B
$$
for all $\;x,A,B\;$.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Tag}[1]{\text{(#1)}}
$Starting at the most complex side of the equality, we calculate as follows:
$$\calc
x \in (A \cup C) \triangle (B \setminus  C)
\calcop\equiv{definitions of $\;\triangle,\cup,\setminus\;$}
x \in A \lor x \in C \;\not\equiv\; x \in B \land x \not\in C
\calcop\equiv{logic: write $\;P \not\equiv Q\;$ as $\;P \equiv \lnot Q\;$}
x \in A \lor x \in C \;\equiv\; \lnot(x \in B \land x \not\in C)
\calcop\equiv{logic: DeMorgan for right hand side}
x \in A \lor x \in C \;\equiv\; x \not\in B \lor x \in C
\calcop{\tag{*}\equiv}{logic: $\;\lor\;$ distributes over $\;\equiv\;$}
(x \in A \;\equiv\; x \not\in B) \;\lor\; x \in C
\calcop\equiv{logic: write $\;P \equiv \lnot Q\;$ as $\;P \not\equiv Q\;$}
(x \in A \;\not\equiv\; x \in B) \;\lor\; x \in C
\calcop\equiv{definitions of $\;\cup,\triangle\;$}
x \in (A \triangle B) \cup C
\endcalc$$
The key step here is $\Tag{*}$: the other steps just prepare for this one, and then work back to the set level.
