This is so obvious by inspection:
But how do you prove this?
Another way is to note that , and similarly for . This in turn can be proved by showing that where or depending on whether is in and is exclusive or. A similar method for proving associativity of is described in this post. If one starts with knowing that is associative and commutative then is obvious.
Suppose A B = A C & x in B. Show x in C.
Suppose A B = A C & x in C. Show x in B.
You don't have to get confused by the symbols. It's pretty simple:
Suppose A B = A C. Show B=C.
Suppose x in B.
Either x in A or x not in A.
Suppose x in A.
So x not in A B.
So x not in A C.
So, since x in A, if x were not in C then x would be in A\C so x would be in A C. So x in C.
Suppose x not in A.
So x in A B.
So x in A C.
So, since x not in A, we have x not in A\C, so x in C\A, so x in C.
So, whether x in A or x not in A, we have x in C.
So x in B implies x in C.
So B is a subset of C.
Then prove C is a subset of B in a similar manner.
In your post #6 from oct 2009 you should also prove the following:
If , is a group of all functions from to , and is power group of . Now if we define the transformation: such that for all in , .
So we need to prove that is a injective transformation.
( is characteristic function)