Prove $ A\triangle(B\backslash C)=(A\triangle B)\backslash C $ iff $A\cap C=\emptyset$ Prove$$ A\triangle(B\backslash C)=(A\triangle B)\backslash C $$
If and only if $A\cap C=\emptyset$.
I should prove it using just or and and. Like this:
The first inclusion:
$$ x \in A\triangle(B\backslash C)\\
 [x \in A \backslash(B \backslash C)] \vee [x\in(B \backslash C) \backslash A]$$
I think the next step is wrong:
$$
 [x \in (A \backslash B) \backslash C)] \vee [x\in(B \backslash A) \backslash C]\\
  [x \in (A \backslash B) \vee x\in(B \backslash A) ] \land x\notin C\\
  (A\triangle B)\backslash C
  $$
 A: We wish to prove the following theorem:
$$
A\triangle(B\setminus C) = (A\triangle B)\setminus C \quad \Leftrightarrow \quad A\cap C=\emptyset
$$
I don't particularly like obfuscating what is going on behind the symmetric difference and set-minus operators, so let us unwind that a bit:
$$
[A\setminus(B\setminus C)]\cup[(B\setminus C)\setminus A] = ([A\setminus B] \cup [B\setminus A])\setminus C \quad \Leftrightarrow \quad A\cap C=\emptyset
$$
$$
[A\cap(B\cap C^c)^c]\cup[(B\cap C^c)\cap A^c] = ([A\cap B^c] \cup [B\cap A^c])\cap C^c \quad \Leftrightarrow \quad A\cap C=\emptyset
$$
$$
[A\cap(B^c\cup C)]\cup[(B\cap C^c)\cap A^c] = ([A\cap B^c] \cup [B\cap A^c])\cap C^c \quad \Leftrightarrow \quad A\cap C=\emptyset
$$
Now, if you squint really hard at that last line, you will see that the $\Leftarrow$ direction is incredibly easy to prove! Just distribute various intersections over various unions and utilize the hypothesis that $A\cap C=\emptyset$ (which is equivalent to $A\cap C^c=A$). You will be left with $X=X$, which is true.
For the $\Rightarrow$ direction, things are a bit harder. I would recommend approaching this via the contrapositive; assume that $A\cap C\neq\emptyset$ and try and prove that $A\triangle(B\setminus C) \neq (A\triangle B)\setminus C$. If you do this, then again we find that the proof is very straightforaward:
$$
[A\cap(B^c\cup C)]\cup[(B\cap C^c)\cap A^c] \stackrel{?}{=} ([A\cap B^c] \cup [B\cap A^c])\cap C^c 
$$
$$
[A\cap B^c]\cup[A\cap C]\cup[B\cap A^c\cap C^c] \stackrel{?}{=} [A\cap B^c\cap C^c] \cup [B\cap A^c\cap C^c] 
$$
On the RHS we have the union of two disjoint sets. One of those sets also appears on the LHS, but the other is a proper subset of the first two bracketed sets on the LHS. Thus, equality is impossible!
This answer is not completely rigorous, because I got lazy and didn't want to  name the element that belongs to the LHS and not the RHS for the inequality argument. However, it should be a good enough starting point for you to finish this completely.
A: No that step is correct because in the case which $$[x \in (A \backslash B) \backslash C)] \vee [x\in(B \backslash A) \backslash C]$$ $x$ doesn't belong to $C$ in either cases so that you can kick $C$ out of both the bracket cases. Note that this only proves that $$A\triangle(B\backslash C)\subseteq(A\triangle B)\backslash C $$you will also need a converse proof on this theorem (which is as the same proof of direct side of the theorem in reverse direction).
Alternative proof
$$A\triangle (B\backslash C){=\Big[(B-C)-A\Big]\cup\Big[A-(B-C)\Big]\\=\Big[B\cap C'\cap A']\cup\Big[A\cap(B\cap C')'\Big]\\=\Big[B\cap C'\cap A']\cup\Big[A\cap(B'\cup C)\Big]}$$and $$(A\triangle B)\backslash C{=\Big[(A-B)\cup(B-A)\Big]\cap C'\\=\Big[A\cap B'\cap C'\Big]\cup \Big[B\cap C'\cap A'\Big]\\=\Big[A\cap (B\cup C)'\Big]\cup \Big[B\cap C'\cap A'\Big]}$$then we need to show that $$A\cap (B\cup C)'=A\cap(B'\cup C)\iff A\cap C=\emptyset$$from the other side $$\Big[A\cap (B\cup C)'\Big]\cup\Big[A\cap C\Big]{=A\cap\Big[(B\cup C)'\cup C\Big]\\=A\cap\Big[(B'\cap C')\cup C\Big]\\=A\cap\Big[(B'\cup C)\cap (C'\cup C)\Big]\\=A\cap\Big[(B'\cup C)\cap \mathrm{U}\Big]\\=A\cap(B'\cup C)}$$where $\mathrm{U}$ is the Universal Set. Also $$\Big[A\cap (B\cup C)'\Big]\cap\Big[A\cap C\Big]{=\Big[A\cap B'\cap C'\Big]\cap\Big[A\cap C\Big]\\=A\cap B'\cap C'\cap C\\=\emptyset}$$therefore $$A\cap (B\cup C)'=A\cap(B'\cup C)$$if and only if $$ A\cap C=\emptyset$$and the proof is complete.
A: OK, we want to show 
$$A \Delta (B \setminus C) = (A \Delta B)\setminus C \Leftrightarrow A \cap C = \emptyset$$
Assume we have $A \cap C \neq \emptyset$ (so trying for the contrapositive, which seems easier as it gives us something concrete to work with). Hence we have some $p \in A \cap C$. Case 1: $p \in B$, then $p \notin (B \setminus C)$ (as $p \in C$) but also $p \in A$, so $p \in A \Delta (B \setminus C)$ but on the other hand $p \notin A \Delta B$ so $p \notin (A \Delta B) \setminus C$ so we have disproved the LHS identity in this case. Case 2: $p \notin B$, then $p \in A \Delta B$ (and still $p \in C$) so $p \notin (A \Delta B) \setminus C$, and $p \notin B \setminus C$ and still $p \in A$ so $p \in A \Delta (B\setminus C)$ as in case 1, we have disproved the identity again.
Now assume that $A \cap C = \emptyset$ holds. We want to show the LHS identity which is really two inclusions. So suppose $x \in A \Delta (B \setminus C)$, so case 1: $x \in A$ and $x \notin B \setminus C$. As $x \notin C$ (as otherwise $A \cap C \neq \emptyset$) we know also that $x \notin B$ (or else $x \in B \setminus C)$ after all) and so $x \in (A \Delta B)\setminus C$, so the left to right inclusion holds for case 1. Case 2: $x \notin A$ and $x \in B \setminus C$, and then $x \in B$ and $x \notin C$ and so $x \in (A \Delta B)\setminus C$ and so the left to right inclusion also holds in case 2. So we're done with that inclusion. Now suppose that $x \in (A \Delta B) \setminus C$ so $x \in A \Delta B$ and $x \notin C$. Again 2 cases: case 1: $x \in A$, $x \notin B$, but then $x \notin B \setminus C$ and so $x \in A \Delta (B \setminus C)$ and the right to left inclusion holds. Case 2: $x \notin A$ and $x \in B$, and then $x \in B \setminus C$ and so $x \in A \Delta (B \setminus C)$ and the right to left inclusion holds (without even using $A \cap C = \emptyset$, so that inclusion always holds). So we have the LHS equality when $A$ and $C$ are disjoint. 
A: 
These are the Venn diagrams for both expressions. As we can see, they differ precisely by $A\cap C$, which is why $A\cap C=\varnothing$ is required to make them equal. Also, notice that $(A\bigtriangleup B)\setminus C\subseteq A\bigtriangleup(B\setminus C)$ always; it's only the opposite inclusion that needs $A\cap C=\varnothing$.
If you're expected to prove the claim using just or and and logical connectors, then your first steps look good, but then I guess you should continue expanding all the definitions. Let's see how that looks for the inclusion $A\bigtriangleup(B\setminus C)\subseteq(A\bigtriangleup B)\setminus C)$ that you started working on.
$$\begin{split}
x\in A\bigtriangleup(B\setminus C) &\iff \left[x\in A\setminus(B\setminus C)\right]\vee\left[x\in(B\setminus C)\setminus A\right] \\
&\iff \left[x\in A\wedge x\notin(B\setminus C)\right]\vee\left[x\in(B\setminus C)\wedge x\notin A\right] \\
&\iff \left[x\in A\wedge\neg\left(x\in B\wedge x\notin C\right)\right]\vee\left[x\in B\wedge x\notin C\wedge x\notin A\right] \\
&\iff \left[x\in A\wedge\left(x\notin B\vee x\in C\right)\right]\vee\left[x\notin A\wedge x\in B\wedge x\notin C\right] \\
&\iff \left[x\in A\wedge x\notin B\right]\vee\left[x\in A\wedge x\in C\right]\vee\left[x\notin A\wedge x\in B\wedge x\notin C\right].
\end{split}$$
Each part of the last expression describes a piece in the Venn diagram for $A\bigtriangleup(B\setminus C)$. In particular, $\left[x\in A\wedge x\notin B\right]$ is $A\setminus B$, but it includes two pieces of the Venn diagram, and we can see that we'd like to separate them:
$$\begin{split}
x\in A\wedge x\notin B &\iff x\in A\wedge x\notin B\wedge\left(x\in C\vee x\notin C\right) \\
&\iff \left[x\in A\wedge x\notin B\wedge x\in C\right]\vee\left[x\in A\wedge x\notin B\wedge x\notin C\right].
\end{split}$$
If we put it back into the last step above, then we'll get that $x\in A\bigtriangleup(B\setminus C)$ iff
$$\left[x\in A\wedge x\notin B\wedge x\in C\right]\vee\left[x\in A\wedge x\notin B\wedge x\notin C\right]\vee\left[x\in A\wedge x\in C\right]\vee\left[x\notin A\wedge x\in B\wedge x\notin C\right].$$
Moreover, we should observe that $\left[x\in A\wedge x\notin B\wedge x\in C\right]$ is "included" in, i.e. describes a subset of, $\left[x\in A\wedge x\in C\right]$. We can work it out using or and and in the same fashion (I'll leave that to you), but the bottom line is that the former ends up being "consumed" by the latter, so we'll get that $x\in A\bigtriangleup(B\setminus C)$ iff
$$\color{blue}{\left[x\in A\wedge x\notin B\wedge x\notin C\right]\vee\left[x\notin A\wedge x\in B\wedge x\notin C\right]}\vee\color{red}{\left[x\in A\wedge x\in C\right]}.$$
Note that the blue parts are precisely what is included on the other diagram, while the red part describes $A\cap C$ — which is precisely what distinguishes the two diagrams.
Now, if we work out $(A\bigtriangleup B)\setminus C$ in a similar fashion, we'll end precisely with the blue expression. That will show that the two are equal iff $A\cap C=\varnothing$.
NOTE: I do realize that this solution seems lengthy and difficult. Yes, it is lengthy and tedious, but it's completely dictated by the pictures. From the visual point of view, I'd say it's pretty straightforward.
