prove that for sets
if and only if
There is also a different way to solve this that sometimes works very well with symmetric difference.
Define the function by if and if . Here is any set and is any object.
Also, define the function as follows:
(If we associate 1 with True and 0 with False, then . Also, is addition modulo 2.) One can check that is commutative and associative.
Show that for all , , and .
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 for all . We can add (using ) an expression to both sides where is , , , or . E.g., if we add to both sides, we get = . From here it is easy to show the desired equation.
Hello santiagos11I think this is easiest to prove by re-writing it as the equivalent statement using logical propositions. In other words:Prove that if are logical propositions, then if and only if , where 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 . It is a simple matter to show that the table for produces the same result.
Grandad