DeMorgan's Law and Differences of Sets I'm currently trying to prove the following identity:
$(A \cup B)\triangle C \equiv (A \triangle C)\triangle(B \setminus A)$
I can easily figure that the left side reduces down to $(x \in A \land x \notin C) \lor (x \in B \land x \notin C) \lor (x \in C \land x \notin A \land x \notin B)$
But when I start working on the right side, I end up with something a bit confusing on my hands:
Let $x \in(A \triangle C)\triangle(B \setminus A) \rightarrow (x \in(A\triangle C)\setminus(B\setminus A)) \lor (x \in (B\setminus A)\setminus(A\triangle C))$
The first of those two possibilities is where it starts to break down for me, particularly when treating the second possibility in $(A \triangle C)$, which is $(C \setminus A)$. In that case, I end up with the following:
$x \in (C \setminus A)\setminus(B\setminus A)$
The logical equivalent of the difference of sets is 'and not,' so it could be rewritten
$(C \land \lnot A)\land \lnot (B \land \lnot A)$
But according to DeMorgan's Law, the negation of an end statement would be an or statement with both elements negated, meaning it becomes
$(C \land \lnot A) \land (\lnot B \lor A)$
If you distribute this, you'd end up with
$(C \land \lnot A \land \lnot B) \lor (C \land \lnot A \land A)$
That first one is fine, but if you let the contradiction cancel out, you're still now left with $x \in C$ in a way that does not exclude elements in A or B, which would mean it doesn't match up to the left side. What am I missing here?
 A: I am not really familiar with all the logical statements, but I was able to solve it using the following properties


*

*$X\cup Y=(X\Delta Y)\Delta(X\cap Y)$

*$X\Delta Y=(X\cup Y)\backslash(X\cap Y)$


and some elementary set operations. I hope that this route will also be of use to you :)
To prove: $(A\cup B)\Delta C\equiv(A\Delta C)\Delta(B\backslash A)$
Proof:
By 1. we have 
$$(A\cup B)\Delta C=(A\Delta B)\Delta(A\cap B)\Delta C=(A\Delta B)\Delta C\Delta(A\cap B)=(A\Delta C)\Delta B\Delta(A\cap B)$$
Note that we want this to equal $(A\Delta C)\Delta(B\backslash A)$, so we want to show $B\Delta(A\cap B)=B\backslash A$.
By 2. we have
$$B\Delta(A\cap B)=\Big(B\cup (A\cap B)\Big)\backslash\Big(B\cap (A\cap B)\Big)$$
Now $B\cup (A\cap B)=B$ and $B\cap (A\cap B)=B\cap A$, such that
$$B\Delta(A\cap B)=B\backslash(B\cap A)=B\backslash A$$
and we are done.
A: $
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\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
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$First, regarding your "What am I missing here?" question, as far as I can see the answer is simply this: $\;
(x \in C \land x \not\in A \land x \not\in B) \lor (x \in C \land x \not\in A \land x \in A)
\;$ is (as you already noted) equivalent to $\;
x \in C \land x \not\in A \land x \not\in B
\;$, which is one of the disjuncts of your $$
(x \in A \land x \notin C) \lor (x \in B \land x \notin C) \lor (x \in C \land x \notin A \land x \notin B)
$$

Second, as an alternative proof of $$
\tag 0
(A \cup B) \symdiff C \;=\; (A \symdiff C) \symdiff (B \setminus A)
$$ we could first use the fact that $\;\symdiff\;$ is both symmetrical and associative:
$$\calc
    (A \cup B) \symdiff C \;=\; (A \symdiff C) \symdiff (B \setminus A)
\op=\hint{$\;\symdiff\;$ is symmetric}
    (A \cup B) \symdiff C \;=\; (B \setminus A) \symdiff (A \symdiff C)
\op=\hints{$\;\symdiff\;$ is associative}
    \hint{-- this gives LHS and RHS a similar shape}
    (A \cup B) \symdiff C \;=\; ((B \setminus A) \symdiff A) \symdiff C
\op\when\hint{logic: Leibniz}
    \tag{1}
    A \cup B \;=\; (B \setminus A) \symdiff A
\endcalc$$
Now we are left with proving $\ref 1$, and the symmetry and associativity of $\;\symdiff\;$.

Third, another alternative proof of $\ref 0$ would be to use the definition $$
x \in X \symdiff Y \;\equiv\; x \in X \not\equiv x \in Y
$$ together with the fact that both $\;\equiv\;$ and $\;\not\equiv\;$ are symmetric and associative, and also mutually associative: that will quickly reduce $\ref 0$ to something with a shape like $$
P \lor Q \;\equiv\; \lnot Q \;\equiv\; P \land \lnot Q
$$ which is easy to prove using the fact that $\;R \then S\;$ is equivalent to both $\;\lnot R \lor S\;$ and $\;R \;\equiv\; R \land S\;$.
