Prove $A \bigtriangleup B = B \bigtriangleup A$ I`m trying to prove the following statement:
$$A \bigtriangleup B = B \bigtriangleup A$$
I know that:
$$A \bigtriangleup B = (A \cup B )\setminus (A \cap B )= (A \setminus B) \cup (B \setminus A)$$
I can do that with truth table. but want to prove it by formal way.
Any suggestions?Thanks!
 A: Well if you're allowed to use the fact that the union and intersection operations are commutative, then we have:
$$
A \bigtriangleup B = (A \cup B )\setminus (A \cap B )=(B \cup A )\setminus (B \cap A )= B \bigtriangleup A
$$
A: In a formal way $A\Delta B$ is the set of elements belong to $A$ or $B$ but not both so these elements belong to $B$ or $A$ but not both so it's $B\Delta A$.
A: proofs by characteristic function :
for any $A \subset X$ define $1_A : X \to \{ 0,1\}$ by
$$1_A(x) = \left\{ 
\begin{eqnarray}
\begin{split}
1 & \mbox{if } x \in A \\
0 & \mbox{if } x \notin A \\
\end{split}
\end{eqnarray}\right.$$
then we have 


*

*$\color{green}{A= B \Leftrightarrow 1_A = 1_B}$

*$1_{A \cap B}=1_A \cdot 1_B$

*$1_{A \cup B}=1_A+1_B-1_A \cdot 1_B$

*$1_{A \backslash B}= 1_A-1_A1_B$

*$\color{green}{1_{A \Delta B}=1_A+1_B-2\cdot 1_A \cdot 1_B=(1_A-1_B)^2}$ 
$\color{red}{Proof}$

$$1_{A \Delta B}=1_A+1_B-2\cdot 1_A \cdot 1_B=(1_A-1_B)^2$$ and
$$1_{B \Delta A}=1_B+1_A-2\cdot 1_B \cdot 1_A=(1_B-1_A)^2$$ then clearly we have 
$$(1_A-1_B)^2=(1_B-1_A)^2 \to 1_{A \Delta B}=1_{B \Delta A}\iff A \bigtriangleup B = B \bigtriangleup A$$

A: The simplest proof I know requires the use of one of a family of basic logic laws which are well known to students using the Gries-Schneider book or familiar with the work of Dijkstra et al., but apparently not so well known otherwise.
One simple definition of $\;\bigtriangleup\;$ is that for any $\;x\;$,
$$
x \in A \bigtriangleup B \;\equiv\; x \in A \not\equiv x \in B
$$
And since $\;\not\equiv\;$ is symmetric, the proof is trivial: for any $\;x\;$
\begin{align}
& x \in A \bigtriangleup B \\
\equiv & \;\;\;\;\;\textrm{"definition of $\;\bigtriangleup\;$"} \\
& x \in A \not\equiv x \in B \\
\equiv & \;\;\;\;\;\textrm{"$\;\not\equiv\;$ is symmetric"} \\
& x \in B \not\equiv x \in A \\
\equiv & \;\;\;\;\;\textrm{"definition of $\;\bigtriangleup\;$"} \\
& x \in B \bigtriangleup A \\
\end{align}
Using set extensionality (two sets are equal iff they have the same elements) completes the proof.
