Math Help - A∆(b∆c) = (a∆b)∆c

1. A∆(b∆c) = (a∆b)∆c

How to prove these equalities:
A∆(B∆C) = (A∆B)∆C
A∩(B∆C) = (A∩B)∆(A∩C)
A∩B = (A∆B)∆(A∪B)

2. Originally Posted by Garas
How to prove these equalities:
A∆(B∆C) = (A∆B)∆C
A∩(B∆C) = (A∩B)∆(A∩C)
A∩B = (A∆B)∆(A∪B)

Write down what is each thing:

$x\in A\triangle(B\triangle C)\Longrightarrow x\in A\,\,or\,\,x\in B\triangle C$ but not in both , so $x\in A\,\,or\,\, x\in B\,\,or\,\,x\in C$ but

not in both $A,B$ and not both in $A, C$ , and etc.

Tonio

3. I'm not sure that i understand what you are saying.

4. Originally Posted by Garas
I'm not sure that i understand what you are saying.
Tonio is proposing you to prove:

$A\Delta(B\Delta C)\subset (A\Delta B)\Delta C$ and $(A\Delta B)\Delta C\subset A\Delta(B\Delta C)$

using the definition of $M\Delta N$:

$M-N=(M\cup N)-(M\cap N)$

An alternative is to use the property of the characteristic funcion:

$\chi _{U\Delta V}=\chi _U\cdot (1-\chi_V)+\chi _V\cdot (1-\chi_U)$

and prove:

$\chi_{A\Delta(B\Delta C)}=\chi_{(A\Delta B)\Delta C}$

But I don't know if you have studied the $\chi$ function.

Regards.

5. I haven't studied that kind of function

6. Originally Posted by Garas
I haven't studied that kind of function

It never minds: you must know what the symmetric difference of two set is in order to cope with this problem,

and that's all you need (and to know basic set theory, of course).

Tonio

7. I know some basics of theory and definition of symmetric difference, but i think that i don't know some transformations which are needed in order to prove those equations, because when i'm trying to solve this it becomes to complicated and i dont know how to move father from that point.

8. Originally Posted by Garas
I know some basics of theory and definition of symmetric difference, but i think that i don't know some transformations which are needed in order to prove those equations, because when i'm trying to solve this it becomes to complicated and i dont know how to move father from that point.
Traditionally this is one of the most dreaded proofs on all of basic set theory.
There is one way around the messy proof by cases, many cases.
That is the use of the characteristic function.
But you say you have not seen that function.
So in that case, you have to do it by cases.
It is very easy to get lost in all the cases.

9. Originally Posted by Garas
How to prove these equalities:
A∆(B∆C) = (A∆B)∆C
A∩(B∆C) = (A∩B)∆(A∩C)
A∩B = (A∆B)∆(A∪B)
Just in case a picture helps reduce disorientation and dread...

Imagine two sets B and C (as shaded circles) floating towards each other to form a combined venn diagram. But as the two float over each other, any shading in one that meets shading in the other fizzles momentarily then disappears. Then we get shading in either set but not both - as per symmetric difference.

Now if we imagine the same with set A and our picture of B∆C...

... we notice the result is symmetrical, and might just as well have been produced from either of the other two combinations, e.g...

Now suppose the momentary fizzling succeeds in validating the shading, so that it drops out otherwise... then the shading in the result shows the intersection...

This isn't so symmetrical, but we notice we can reach the same picture by getting the symmetric difference of two simpler intersections, as per your second equality...

And for the third one...

10. It's is easy with diagrams but i don't think that that is a preferred method. This was one of the home work problems and i don't think that many people will solve that. Professor maybe thought that this is a easy problem without trying to solve it.
Can someone try to solve this using characteristic function?

11. Originally Posted by Garas
It's is easy with diagrams but i don't think that that is a preferred method. This was one of the home work problems and i don't think that many people will solve that. Professor maybe thought that this is a easy problem without trying to solve it.
Can someone try to solve this using characteristic function?
It's not that hard, just messy.
Spoiler:

Let $U$ be some fixed universe and $B\subseteq U$ fixed. The, define $\varphi:2^U\times2^U\to 2^U:\left(A,C\right)\mapsto \left(A\Delta B\right)\Delta C$. By commutativity of $\Delta$ it suffices to show that $\varphi\left(A,C\right)=\varphi\left(C,A\right)$. Note then that

\displaystyle \begin{aligned}\varphi\left(A,C\right) &= \left(\left(A\Delta B\right)-C\right)\cup\left(C-\left(A\Delta B\right)\right)\\ &= \left(\left(\left(A-B\right)\cup\left(B-A\right)\right)\cap C'\right)\cup\left(C-\left(\left(A\cup B\right)-\left(A\cap B\right)\right)\right)\\ &= \left(\left(\left(A\cap B'\right)\cup\left(B\cap A'\right)\right)\cap C'\right)\cup\left(\left(C\cap A\cap B\right)\cup\left(C-\left(A\cup B\right)\right)\\ &= \left(\left(A\cap B'\cap C'\right)\cup\left(B\cap A'\cap C'\right)\right)\cup\left(\left(C\cap A\cap B\right)\cup\left(C\cap A'\cap B'\right)\right)\\ &= \left(B\cap A'\cap C'\right)\cup\left(B\cap A\cap C\right)\cup\left(B'\cap A\cap C'\right)\cup\left(B'\cap A'\cap C\right)\end{aligned}

From where it's clear that $\varphi\left(A,C\right)=\varphi\left(C,A\right)$. Since $B$ was arbitrary it follows that $\left(A\Delta B\right)\Delta C=A\Delta\left(B\Delta C\right)$ for all $A,B,C\in 2^U$.

You can also treat $\varphi$ in the above as a boolean function and check that it's true that $\varphi(A,C)=\varphi(C,A)$ for all combinations of $A,C=0,1$ (the empty set and the unvierse).