# symmetric difference of sets

• Dec 1st 2009, 08:20 PM
santiagos11
symmetric difference of sets
prove that for sets $A,B,C,D$
$A \Delta B = C \Delta D$ if and only if $A \Delta C = B \Delta D$
• Dec 1st 2009, 10:11 PM
Drexel28
Quote:

Originally Posted by santiagos11
prove that for sets $A,B,C,D$
$A \Delta B = C \Delta D$ if and only if $A \Delta C = B \Delta D$

What have you tried? Although there are many I think that the definition of symmetric difference best suited to this case is $A\Delta B=\left(A\cup B\right)-\left(A\cap B\right)$.
• Dec 2nd 2009, 05:47 AM
emakarov
There is also a different way to solve this that sometimes works very well with symmetric difference.

Define the function $in$ by $in(x,A)=1$ if $x\in A$ and $in(x,A)=0$ if $x\notin A$. Here $A$ is any set and $x$ is any object.

Also, define the function $\oplus:\{0,1\}\times\{0,1\}\to\{0,1\}$ as follows:
$0\oplus0=0$
$0\oplus1=1$
$1\oplus0=1$
$1\oplus1=0$
(If we associate 1 with True and 0 with False, then $x\oplus y=\neg(x\leftrightarrow y)$. Also, $\oplus$ is addition modulo 2.) One can check that $\oplus$ is commutative and associative.

Show that $in(x,A\triangle B)=in(x,A)\oplus in(x,B)$ for all $A$, $B$, and $x$.

The part above was generic; it applies to any problem with symmetric difference and has to be done only once.

Now, from this problem's assumption and the fact above we have $in(x,A)\oplus in(x,B)=in(x,C)\oplus in(x,D)$ for all $x$. We can add (using $\oplus$) an expression $in(x,E)$ to both sides where $E$ is $A$, $B$, $C$, or $D$. E.g., if we add $in(x,B)$ to both sides, we get $in(x,A)$= $in(x, C)\oplus in(x,D)\oplus in(x,B)$. From here it is easy to show the desired equation.
• Dec 2nd 2009, 06:05 AM
Hello santiagos11
Quote:

Originally Posted by santiagos11
prove that for sets $A,B,C,D$
$A \Delta B = C \Delta D$ if and only if $A \Delta C = B \Delta D$

I think this is easiest to prove by re-writing it as the equivalent statement using logical propositions. In other words:
Prove that if $a, b,c, d$ are logical propositions, then $a \oplus b \equiv c \oplus d$ if and only if $a \oplus c \equiv b \oplus d$, where $\oplus$ denotes 'exclusive or'.
This is quite straightforward if you use a Truth Table for each of the compound propositions and show that each table produces an identical output.

The attached diagram shows the truth table for $a \oplus b \equiv c \oplus d$. It is a simple matter to show that the table for $a \oplus c \equiv b \oplus d$ produces the same result.