Question

Prove that: A(exclusive or)B = (A\B)U(B\A)

Answer 1

I know that A(exclusive or)B is true when the elements in A aren't in B, and that (A\B) is saying to subtract anything that A and B share, and that (B\A) is the same, but reversed, and therefore that (A\B)U(B\A) is really just saying to include all of A without the B intersect, or to use all of B without the A intersect, which is essentially saying to cut out all similarities, which is A(exclusive or)B.

My problem is how to "prove" this.

Answer 2

\[
\forall x \in (A\otimes B) \iff \forall x(x\in A\land x\notin B) \lor (x\notin A\land x\in B)
\]

Answer 3

Just so I can make sure I understand this, what this is saying is:

For all x that exists in (a XOR b), it is equal to, for all x where x is in A and x is not in B, or where x is not in A and x exists in B)

Answer 4

it's the definition.

Answer 5

Ok, thank you!

Answer 6

\[
\forall x(x\in A\land x\notin B) \lor (x\notin A\land x\in B)\iff \forall x(x\in A\setminus B) \lor ( x\in B\setminus A)
\]

Answer 7

\[
\forall x(x\in A\setminus B) \lor ( x\in B\setminus A) \iff \forall x\in (A\setminus B)\cup (B\setminus A)
\]

Answer 8

When all steps used are bi conditional, then you can go just one way. If one-way conditional steps are used, then you have to prove both ways.

Answer 9

Maybe you can help with this one real quick too,

Let T = {a, b, c}
Let S = {1, 2}

How many 1-1 functions are possible with: T->S

Is this even possible? I thought 1-1 needed to have all of the T values mapped to at most 1 S value?

Answer 10

The answer is 0.

Answer 11

Ok, that's what I thought. Thanks for clarification!