Verifying $(A \bigtriangleup B) \cup C=(A \cup C) \bigtriangleup (B\setminus C)$ I am trying to do self-study out of a set theory book. In one of the question sections, it asks to verify the following identity:
$$(A \bigtriangleup B) \cup C=(A \cup C) \bigtriangleup (B\setminus C)$$
My plan of attack for verification is to try and and show that if an element $x$ is an element of one set it is an element of the other. I have tried to run through this along both directions, but I keep getting lost in the plethora of terms that come up when I try to expand the expressions I get each time I try to apply logical rules to each step. Is there anyway that someone can show a step by step of how to get from one side to the other or at least give me some kind of bridging hint?
 A: Half of left-to-right: If $x \in (A \bigtriangleup B) \cup C$ then $x \in (A \bigtriangleup B)$ or $x \in C$. Suppose $x\in C$. then $x \in (A \cup C)$ and $x \notin (B\setminus C)$ so $x \in (A \cup C) \bigtriangleup (B\setminus C)$. Now do the other l-to-r case then do r-to-l. You just crank it out and keep track of where you are. Don't try to expand the expressions, just crank out the logic one case at a time. 
A: Let us consider only elements from $A\cup B\cup C$, and none elsewhere, since elements outside that domain will not be included by either construction.
We argue that $(A\triangle B)\cup C$ clearly excludes elements in the union of $A$ and $B$ unless they are also in $C$. All other elements in the domain are included.
Then we examine $(A\cup C)\triangle(B\setminus C)$ and observe that this equals $((A\setminus C)\cup C)\triangle(B\setminus C)$. Any element of $C$ is in this set. Of those elements not in $C$, only those which are also in both in $A$ and $B$ are excluded.
That is, of all the elements in $A$, $B$, and $C$, only those elements in the union of $A$ and $B$ that are not also in $C$ are excluded by both set constructions.  They are equivalent.
$$(A\triangle B)\cup C \equiv (A\cup B\cup C)\setminus((A\cap B)\setminus C) \equiv (A\cup C)\triangle(B\setminus C)$$
A: Here are the hints I have for you.$
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\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
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$
First, treat this as a simplification problem: start at the right hand side, which is the most complex side, and try to simplify, working towards the other side.
Second, solve this at the logic level: expand the set theory definitions to logical formulas, and use the laws of logic.
Third, use the following definition of $\;\sdiff\;$:
$$
\tag 0
x \in A \sdiff B \;\equiv\; x \in A \not\equiv x \in B
$$
for all $\;x,A,B\;$.
With this, the proof becomes a calculation that starts out as follows: for all $\;x\;$,
$$\calc
  x \in (A \cup C) \sdiff (B \setminus C)
\op\equiv\hint{definition $\ref 0$ of $\;\sdiff\;$; definitions of $\;\cup, \setminus\;$}
  x \in A \lor x \in C \;\not\equiv\; x \in B \land x \not\in C
\op\equiv\hints{logic: $\;P \not\equiv Q\;$ is equivalent to $\;P \equiv \lnot Q\;$}
\hints{-- $\;\equiv\;$ is often easier to manipulate, and}
\hint{this gives both sides the same shape}
  x \in A \lor x \in C \;\equiv\; x \not\in B \lor x \in C
\op\equiv\hint{logic: $\;\lor\;$ distributes over $\;\equiv\;$}
  \ldots
\endcalc$$
Now work towards the left hand side of your equality, and (using set extensionality) the proof is complete.
