Prove if $A\Delta B\subseteq A\Delta C$ then $A\cap C\subseteq B\subseteq A\cup C$ I'm a bit stuck on the following problem:

Prove if $A\Delta B\subseteq A\Delta C$ then $A\cap C\subseteq B\subseteq A\cup C$.

At first, I write the statement like that:
$$(A\Delta B\subseteq A\Delta C)\implies(A\cap C\subseteq B)\land(B\subseteq A\cup C)$$
And then handle the two parts separated by $\land$ independently. On each part I use proof by contradiction, so I will try to prove the two following statements:
I. $A\cap C\not\subseteq B$
II. $B\not\subseteq A\cup C$
In case of $A\cap C\not\subseteq B$:
It means that every $x$ will meet the requirement of $(x\in A)\land(x\in C)\land(x\not\in B)$.
Now there is the problem that I can't understand how to proceed from here. I know that I should find a statement which conflicts with the if requirement, but it looks tricky.
I would appreciate your help.
 A: $A\cap C\not\subseteq B$ means that there exists such an $x$, and not for all $x$, that doesn't make sense.
So, assume $x\in A,\ x\in C,\ x\notin B$. Then we have $x\in A\triangle B$ while $x\notin A\triangle C$.
Can you similarly do the other case? 
A: If I. is true, then there is some element $x$ such that $x \in A \cap C$ and $x \notin B$. Then $x \in A \triangle B$ but $x \notin A \triangle C$.

If II. is true, then there exists some $x$ such that $x \in B$ and $x \notin A$ and $x \notin C$. Then $x \in A \triangle B$ but $x \notin A \triangle C$.
A: If $x\in A\cap C$, then $x\in A \land A\in C$.
Therefore $x\not\in A\ominus C$, therefore (by $A\ominus B\subseteq A\ominus C$), $x\not\in A\ominus B$.
Therefore $x\in B$, because if $x\not\in B$,then  $x\in A\ominus B$, which is a contradiction.

If $x\in B$, and $x\not\in A \land x\not\in C$, then $x\in A\ominus B \land x\not\in A\ominus C$, which is a contradiction.
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I came up with this proof later on.
We construct the truth table for the Boolean variables $A,B,C$, and calculate $A\otimes B, A\otimes C$, where $X\otimes Y$ is $X$ XOR $Y$.
This is the same as sets in that $x\in A, x\in B, x\not\in C$ is the same as saying $A=1, B=1, C=0$. All possibilities are covered.
The truth table looks like:
\begin{array} {|c|c|c|c|c|}
\hline
A&B&C&A\otimes B&A\otimes C\\
\hline
0&0&0&0&0\\
0&0&1&0&1\\
0&1&0&1&0\\
0&1&1&1&1\\
1&0&0&1&1\\
1&0&1&1&0\\
1&1&0&0&1\\
1&1&1&0&0\\
\hline
\end{array}
The definition of subset is that:
$X\subset Y \implies (x\in X \implies x\in Y)$
and corresponds to the truth table:
\begin{array} {|c|c|c|}
\hline
X&Y&X \subset Y\\
\hline
0&0&1\\
0&1&1\\
1&0&0\\
1&1&1\\
\hline
\end{array}
which is to say that if $X\le Y$ then the SUBSET relation holds, otherwise it doesn't.
The only two lines for $A\ominus B\not\subset A\ominus C$ are:
\begin{array} {|c|c|c|c|c|}
\hline
A&B&C&A\otimes B&A\otimes C\\
\hline
0&1&0&1&0\\
1&0&1&1&0\\
\hline
\end{array}
So, as we are given the SUBSET relation, if $x\in B$, then $x\in A \lor A\in C$ (because if in neither, the SUBSET relation fails). Therefore $B\subseteq A\cup C$.
And also, if $x\in A\land C$, then it must be in $B$, otherwise the SUBSET relation fails. Therefore $A\cap C\subset B$.
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And another:
If $X$ and $Y$ are Boolean variables, then the value of $X$ XOR $Y$ is given by this formula.
$X\otimes Y=X+Y-2XY$
We have been told:
$A\otimes B\subseteq A\otimes C$
$A\otimes B\le A\otimes C$
which translates to:
$A+B-2AB\le A+C-2AC$
$B(1-2A)\le C(1-2A)$
This is simplified by considering either $A=0$ or $A=1$, to give:
$A=0: B\le C$
$A=1: C\le B$.
For the rest, we need formula for AND and OR, which are:
$X\land Y=XY$
$X\lor Y=X+Y-XY$
So we need to prove:


*

*$AC\le B$
and


*$B\le A+C-AC$
These follow easily by using the two cases for $A$.


*

*$A=0$ gives $0\le B$, which is always true. $A=1$ gives $C\le B$, which is confirmed by the formula.

*$A=0$ reduces to $B\le C$, which is true by the formula for $A=0$. $A=1$ reduces to $B\le A(=1)$, which is always true.
A: $A-B$ = $A\cap B^c$ (property1)

So we are given that (using property 1)
($A\cap B^c$) $\cup$ ($B\cap A^c$) $\subseteq$ ($A\cap C^c$) $\cup$ ($C\cap A^c$)
Obviously $A\cap B^c$ and $C\cap A^c$ have no elements common , similarly $B\cap A^c$ and $A\cap C^c$ also have no element common ($A$ and $A^c$ cannot have a common element )
It can be concluded $A\cap B^c$ $\subseteq$ $A\cap C^c$ ....(i) and also $B\cap A^c $ $\subseteq$ $C\cap A^c$.....(ii)

Now if $L \ \subseteq M$ implies $M^c \ \subseteq \ L^c$ (property 2) 
We will use this property on (i) and (ii) 

Hence we have 
($A\cap C^c$)$^c \ \subseteq$  ($A\cap B^c$)$^c$ and do the same with the second one 

Using de morgan's rule we have 
$ A^c \cup C$ $\subseteq$ $A^c \cup B$ , now obviously 
$C-A^c$ $\subseteq \ B$ , because these elements cannot be found in $A^c$ 
Using property 1 we have  $C\cap A$ $\subseteq$ $B$ 

If you repeat similar prodecure to second , you will end with 
$C^c-A$ $\subseteq$ $B^c$ .
So we have $C^c\cap A^c$ $\subseteq$ $B^c$ 
Using property 2 
$B$ $\subseteq$ ($A^c\cap C^c$)$^c$
Again using de morgan's rule 
We have $B$ $\subseteq$ ($A\cup C$)
