Symmetric set difference and odd number of sets containing x Let $A_1,A_2,...,A_n$ be unempty sets, $n\in \mathbb N$.
$$x\in A_1\triangle A_2\triangle...\triangle A_n\;\iff x\in odd\; number\;of\;sets$$
Our assistant proved this by mathematical induction as many others, but I have another question. Link to the related question:Prove through induction that $a \in A_1 \triangle A_2 \triangle \ldots \triangle A_n $ $\iff$ $|{\{i|a \in A_i}\}| $ is odd
$1)$ base $n=1, \;x\in A_1 \iff \; x \in odd\;number\; of\; sets $
assumption: the statement$\;x\in A_1\triangle A_2\triangle...\triangle A_n\;\iff x\in odd\; number\;of\;sets$ is true for some number $n\in \mathbb N.$
$2)$ step of the induction:
$x\in A_1\triangle A_2\triangle...\triangle A_n\triangle A_{n+1}\iff$ 
$\Big(x\in A_1\triangle A_2\triangle...\triangle A_n\;\wedge x\notin A_{n+1}\Big)\lor\Big(x\notin A_1\triangle A_2\triangle...\triangle A_n\;\wedge x\in A_{n+1}\Big)$
$\Big(x \in odd\;number\;of\;sets\;A_1,A_2,...A_n\; \wedge x\notin A_{n+1}\Big)\lor\Big(x \in even\;number\;of\;sets\;A_1,A_2,...A_n\; \wedge x\in A_{n+1}\Big)$
Now, when I have this in mind, how can I prove it without induction? Is there any other way?
 A: Here's a proof that I feel is more intuitive.  First of all, define the XOR ("exclusive or") operator $\oplus$ as follows:
$$
p \oplus q\overset{\text{ def}}{\iff} (\lnot p \land q) \lor (p \land \lnot q).
$$
Thus, $\triangle$ is defined by
$$
x \in A_1 \triangle A_2 \iff (x\in A_1) \oplus (x\in A_2).
$$
With that, the key is to see that $\oplus$ is really addition modulo $2$.  In particular, here is a table for addition modulo 2 and a truth table for $\oplus$:
$$
\begin{array}{c|cc}
+ & 0 & 1\\
\hline 
0 & 0 & 1\\
1 & 1 & 0
\end{array} \qquad 
\begin{array}{c|cc}
\oplus & F & T\\
\hline 
F & F & T\\
T & T & F
\end{array}
$$
In other words, the identification $F \mapsto 0$, $T \mapsto 1$ is an isomorphism of abeliean groups.  Thus, evaluating the statement
$$
x \in A_1 \triangle A_2 \triangle \cdots \triangle A_n
$$
amounts to evaluating
$$
[[(x \in A_1) \oplus (x_1 \in A_2)] \oplus \cdots \oplus (x \in A_n)],
$$
which is equivalent to evaluating the sum
$$
[[x_1 + x_2] + \cdots + x_n]
$$
modulo $2$ where $x_i$ is $0$ if $x \notin A_i$ and $1$ if $x \in A_i$.  In other words, the result is true iff $x_1 + \cdots + x_n$ is equal to $1$ modulo $2$, which is to say that $x_1 + \cdots + x_n$ is odd.
A: Let $f_x: (\mathcal{P}(E), \Delta) \to (\Bbb{Z}/2\Bbb{Z}, +)$ be defined by
$$
f_x(A) = \begin{cases}
1 &\text{if $x \in A$} \\
0 &\text{otherwise}
\end{cases}
$$
It is easy to see that $f_x(A \Delta B) = f_x(A) + f_x(B)$. In other words, $f_x$ is a semigroup morphism. It follows that $f_x(A_1 \Delta \dotsm \Delta A_n) = 1$ if and only if $x$ belongs to an odd number of $A_i$'s.
