I've been drafting it for hours and got nowhere :/
(AΔC) U (BΔC) = (AUBUC) \ (A∩B∩C)
please help,
thanks
I will do this one way for you. You must post some effort on the other way.
Suppose that $\displaystyle t\in \left((A\Delta C)\cup(B\Delta C)\right)$ then it follows that $\displaystyle \left( {t \in A} \right) \vee \left( {t \in B} \right) \vee \left( {t \in C} \right).$
If $\displaystyle t\in \left((A\Delta C)\right) $ then $\displaystyle t\notin\left((A\cap C\right) $ which implies $\displaystyle t\notin\left(A\cap B\cap C\right).$
Thus $\displaystyle t\in \left(A\cup B\cup C\right)\setminus\left(A\cap B\cap C\right) $.
Your turn.
hey thanks for your help.
Personally, I don't realy like this method - I find it harder, plus I don't quite understand how you got from $\displaystyle t\notin\left(A\cap B\cap C\right)$ to $\displaystyle t\in \left(A\cup B\cup C\right)\setminus\left(A\cap B\cap C\right)$?
I did try using it though, started by translating all into logical signs and operators.
I worked only on the LHS, it got really messy and somehow I got to $\displaystyle t\notin C$? (maybe I got some stuff wrong in the process).
Before I resulted to this method I tried working the set identities, which backtracked me to the original sentence with every try.
I'd still rather get some hint on how to solve it using only the pure identities.
thnaks again.
It is quite easy really.
$\displaystyle \left( {t \in A} \right) \vee \left( {t \in B} \right) \vee \left( {t \in C} \right)$ implies that $\displaystyle t\in \left(A\cup B\cup C\right) $
So together with $\displaystyle t\notin\left((A\cap B\cap C\right).$ the result follows.
It is widely acknowledged that when working with symmetric difference involving more than two sets, trying to use set identities can lead into circular arguments.
It is true that $\displaystyle \left((A\Delta C)\cup(B\Delta C)\right)=(A\cap C^c)\cup(C\cap A^c)\cup (B\cap C^c)\cup (C\cap B^c)$.
Now what?
excuse me for not getting back to you for the last couple days, I was tied up with some other hw assignments.
about the first part you solved for me - this is the continuance (I guess) :
$\displaystyle t\in \left((A\Delta C)\cup(B\Delta C)\right)$ → $\displaystyle \left( {t \in A} \right) \vee \left( {t \in B} \right) \vee \left( {t \in C} \right)$
then if $\displaystyle t\in \left((B\Delta C)\right) $ then $\displaystyle t\notin\left((B\cap C\right) $ → $\displaystyle t\notin\left(A\cap B\cap C\right).$
Thus $\displaystyle t\in \left(A\cup B\cup C\right)\setminus\left(A\cap B\cap C\right) $ - so if i'm correct, you didn't really leave me much to do cause I just needed to do a symmetrical proof for B.
although I was looking for this (I didn't know it was possible) :
$\displaystyle \left((A\Delta C)\cup(B\Delta C)\right)=((A\cup C)\cap(A\cap c)^c)\cup ((B\cup C)\cap (B\cap C)^c)$
and then just opening it
but thanks anyways, I it wrote your way