That is, prove that A delta (B delta C) = (A delta B) delta C.
I managed a lengthy proof but only with the use of De Morgan's Laws - just checking if this is acceptable?
By definition, we know that . Therefore,
Now we make a little digression to simplify from the above expression:
, and then plug this into the equation above;
Next, if you exchange sets in the original expression ( ) in the following way: ; ; , you'll get that
, and from this, it follows that
(and, because the symmetric difference is a symmetric operation, i.e. , it follows that)= .
Here is a different way to prove this, or rather to reduce it to associativity of addition modulo 2. To remind, ; if at least one of is , and .
Definition. For a set and an object , let if and otherwise.
Lemma. For all sets and any object , .
Theorem. For all sets , .
Proof. For any , = .
hi math members my name is Paul Otuoma from Nairobi,Kenya. Am a student at the University of Nairobi.Am persuing Bachelor of Science, where am taking Double maths and Physics..Am glad to join this group its superb..Let me ask how comes (A delta B) is (A\B) U (B\A)?? our teacher taught us that, (A delta B)= ( A-B) U (B-A) ??? Please help..