How to prove that $(A \bigtriangleup B) - C = (A - C) \bigtriangleup (B - C)$? 
Let $A,B,C$ be sets. Prove that $(A \bigtriangleup B) - C = (A - C) \bigtriangleup (B - C)$.

I've already checked the Venn diagrams and it seems like said identity is true but I have no idea how to prove it using known properties of sets and axioms from set theory. It'd be awesome if you could help me, thank you!
 A: Go for two-way containment.
From left to right: if $x\in(A\triangle B) - C$ then $x$ is in exactly one of $A,B$ and not in $C$. Say, it's in $A$, then $x\in(A-C)$ but $x\notin B-C$, and thus $x\in(A-C)\triangle(B-C)$.
Can you do the other direction? 
A: $A-B=A\cap B' $
$(A\cup B)-C\ =\ (A\cup B)\cap C'\ =\ (A\cap C')\cup (B\cap C')$ 
(De morgan's law)      
$=(A-C)\cup(B-C)$
A: Since a point is either in a set or not, you only need to compare two truth tables with 3 variables or use some Boolean manipulation. This is entirely equivalent to using $\ ^c, \cup, \cap$, but I find it easier:
Fix some point $x$ and let $a$ be true iff $x \in A$ and similarly for $b,c$.
Then you want to show that $(a \bar{b} + \bar{a}b) \bar{c} $ is the same as 
$a \bar{c} \overline{b \bar{c}} + \overline{a \bar{c}}b \bar{c}$.
We have $(a \bar{b} + \bar{a}b) \bar{c} = a \bar{b}\bar{c} + \bar{a}b\bar{c}$ and
$a \bar{c} \overline{b \bar{c}} + \overline{a \bar{c}}b \bar{c} = a \bar{c} (\bar{b}+c) + (\bar{a}+c)b \bar{c} =a \bar{b}\bar{c} + \bar{a}b\bar{c}$.
Since this holds for all $x$ we have the desired result.
A: If $x \in (A \bigtriangleup B) - C$
Then we have that $x \in A\bigtriangleup B$ and that $x \not \in C$
If $x \in A\bigtriangleup B$ there are two cases.  Either 
1) $x \in A$ but $x \not \in B$ or
2) $x \in B$ but $x \not \in A$.
If 1) Then $x \in A$ and $x \not \in C$ so $x \in A-C$.  
But $x \not \in B$.  And if $x \not \in B$ it is not true that both $x \in B$ and $x \not \in C$ (although it is true $x \not \in C$, it isn't true that $x \in B$).
So $x \in A-C$ but $x \not \in B-C$ so $x \in (A-C)\bigtriangleup (B-C)$.
If 2) Then $x \in B$ and $x \not \in C$ so $x \in B-C$.  
But $x \not \in A$.  And if $x \not \in A$ it is not true that both $x \in A$ and $x \not \in C$ (although it is true $x \not \in C$, it isn't true that $x \in A$).
So $x \in B-C$ but $x \not \in A-C$ so $x \in (A-C)\bigtriangleup (B-C)$
So either way $x \in (A-C)\bigtriangleup (B-C)$.
So we have $(A\bigtriangleup B) - C \subset (A-C)\bigtriangleup (B-C)$.
....
Now if $y \in (A-C)\bigtriangleup (B-C)$
we have two possible cases:
Either 1) $y \in A-C$ but $y\not \in B-C$ or
2) $y \in B-C$ but $y \not \in A-C$.
If 1) Then $y \in A-C$ so $y\in A$ and $y \not \in C$.  But $y \not \in B-C$ which means if $y \in B$ than $y$ can't not be $C$ so $y \in C$.  By $y$ isn't in $C$ so $y$ can't be in $B$ at all.
So we have $y \in A$ and $y \not \in B$.  So $y \in A\bigtriangleup B$.  And $y \not \in C$ so $y \in (A\bigtriangleup B)-C$.
If 2) then $y \in B-C$ so $y\in B$ but $y\not \in C$.  But $y \not \in A-C$ so there is no way $y\in A$ unless $y \in C$. but $y\not\in C$ so $y\not \in A$.
So we have $y \in B$ and $y\not \in A$ so $y \in A\bigtriangleup B$.  And $y\not \in C$ so $y \in (A\bigtriangleup B)-C$.
So either way 
$y \in (A\bigtriangleup B)-C$
So $(A-C)\bigtriangleup (B-C)\subset (A\bigtriangleup B)-C$.
So $(A-C)\bigtriangleup (B-C)$ and $(A\bigtriangleup B)-C$ are subsets of each other and they are therefore equal to each other.
