Prove $A \bigtriangleup B = B \bigtriangleup A$

Question

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!

If you want to prove it in a formal way, what definition of $A \Delta B$ have you? Both of $(A \cup B) \setminus (A \cap B)$ and $(A\setminus B) \cup (B\setminus A)$ are common definitions. — Daniel Fischer, Jul 15 '13 at 19:24
No sorry, I read it few times. any way I will show it to my lecturer and ask him to explain that. thanks! — Ofir Attia, Jul 15 '13 at 20:08
@Ofir Attia:anyway characteristic function method is easy way to show these kind of equality — M.H, Jul 15 '13 at 20:12

score 8 · Accepted Answer · answered Jul 15 '13 at 19:24

8

Well if you're allowed to use the fact that the union and intersection operations are commutative, then we have: $$ A \bigtriangleup B = (A \cup B )\setminus (A \cap B )=(B \cup A )\setminus (B \cap A )= B \bigtriangleup A $$

answered Jul 15 '13 at 19:24

Adriano

41,576

well, there is another option without the fact that union and intersection operations are commutative?, its ok by me, but I just want to know more. – Ofir Attia Jul 15 '13 at 19:30
You only need the commutativity of the union if you use the second definition, $(A\setminus B)\cup(B\setminus A)$. – celtschk Jul 15 '13 at 19:39

score 4 · Answer 2 · answered Jul 15 '13 at 19:39

4

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 $B\Delta A$.

answered Jul 15 '13 at 19:39

M.H · Answer 3 · 2013-07-15T19:40:12.967

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 $$(1_A-1_B)^2=(1_B-1_A)^2 \to 1_{A \Delta B}=1_{B \Delta A}\iff A \bigtriangleup B = B \bigtriangleup A$$

score 0 · Answer 4 · answered Jul 18 '13 at 05:48

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.

Prove $A \bigtriangleup B = B \bigtriangleup A$

4 Answers4