how to prove: $A=B$ iff $A\bigtriangleup B \subseteq C$ I am given this: $A=B$ iff $A\bigtriangleup B \subseteq C$. And $A\bigtriangleup B :=(A\setminus B)\cup(B\setminus A)$. 
I dont know how to prove this and I dont know where to start. 
please give me guidance
 A: Hint: For an arbitrary set $C$, what is the one and only set that is the subset of every set?
So given $\,A\triangle\,B \subseteq C$, where $C$ is any arbitrary set, what does this tell you about the set $A\triangle B$?   
And what does that tell you about the relationship between $A$ and $B$?
A: Here is a another way to do this, where the empty set almost automatically falls out of the calculation.
You have to prove that $A = B \;\equiv\; \langle \forall C :: A \Delta B \subseteq C \rangle$.  (Note that I write $\equiv$ where many others write $\Leftrightarrow$.)
One heuristic that often works when proving statements about sets, is to use set extensionality ("two sets are equal iff they have the same elements") and something similar for $\subseteq$ etc.  That translation brings us from the level of sets to the logic level, where simpler laws usually apply.  For example, we have the following law (which can be used as the definition) for $\Delta$: $$
(0) \;\;\;\;\; x \in A \Delta B \equiv x \in A \not\equiv x \in B
$$ Now looking at the problem in these terms, you have to prove that $$
\langle \forall x :: x \in A \equiv x \in B \rangle
$$ is equivalent to $$
\langle \forall C :: \langle \forall x :: x \in A \Delta B \Rightarrow x \in C \rangle \rangle
$$
Another heuristic is that if you need to prove a relationship (here: equivalence), you start with the most complex side (here: the last expression), and transform until you've reached the other side.  So we can calculate as follows:
$$
\begin{align}
& \langle \forall C :: \langle \forall x :: x \in A \Delta B \Rightarrow x \in C \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\Delta$ -- really the only thing we can do"} \\
& \langle \forall C :: \langle \forall x :: (x \in A \not\equiv x \in B) \Rightarrow x \in C \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: expand $\Rightarrow$ -- work towards our goal by introducing $x \in A \equiv x \in B$"} \\
& \langle \forall C :: \langle \forall x :: (x \in A \equiv x \in B) \lor x \in C \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: exchange quantifications -- to bring $\forall C$ nearer the only place it is used"} \\
& \langle \forall x :: \langle \forall C :: (x \in A \equiv x \in B) \lor x \in C \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: move $x \in A \equiv x \in B$ outside of $\forall C$ -- again to move $\forall C$ nearer its use"} \\
& \langle \forall x :: (x \in A \equiv x \in B) \lor \langle \forall C :: x \in C \rangle \rangle \\
\end{align}
$$
At this point we are almost where we want to be, except for that pesky $\langle \forall C :: x \in C \rangle$.  To make progress, we have to stop and think: given any $x$, when is $\langle \forall C :: x \in C \rangle$ true? I.e., which $x$ is an element of  every set?  The answer, obviously, is that there is no such $x$, since the empty set has no elements:
$$
\begin{align}
& \langle \forall C :: x \in C \rangle \\
\Rightarrow & \;\;\;\;\;\text{"choose $C := \emptyset$"} \\
& x \in \emptyset \\
\equiv & \;\;\;\;\;\text{"definition of $\emptyset$"} \\
& \textrm{false} \\
\end{align}
$$
(Note that this is a proof by contradiction to show that $\langle \forall C :: x \in C \rangle \equiv \textrm{false}$.) Therefore we can complete our first calculation: 
$$
\begin{align}
& \langle \forall x :: (x \in A \equiv x \in B) \lor \langle \forall C :: x \in C \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"by our second calculation"} \\
& \langle \forall x :: (x \in A \equiv x \in B) \lor \textrm{false} \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& \langle \forall x :: x \in A \equiv x \in B \rangle \\
\end{align}
$$
Putting all of this together, we've proved that $\langle \forall C :: A \Delta B \subseteq C \rangle \equiv A = B$, which is what you set out to prove.
