Prove

This is so obvious by inspection:

But how do you prove this?

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- Nov 1st 2010, 01:45 PMnoviceSymmetric Difference
Prove

This is so obvious by inspection:

But how do you prove this? - Nov 1st 2010, 02:57 PMDrexel28
- Nov 1st 2010, 03:34 PMemakarov
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.

- Nov 1st 2010, 03:46 PMPlato
This a second way. Suppose that .

Then now either or .

If then because then .

But that means which is contradiction.

If because we have .

But that means that which is a contradiction.

So .

Likewise we prove that so . - Nov 2nd 2010, 08:15 AMnovice
- Nov 2nd 2010, 08:30 AMnovice
Yours is so elegant.

Mine looks naive:

Part 1. Suppose . Prove , so

Part 2. Suppose . Prove , so

. Thus

The scratch work for part 1 and 2 were long. They entire roll of toilet paper. - Nov 2nd 2010, 12:15 PMMoeBlee
Excellent for getting a real solid grasp of mathematical proof for such subjects as set theory is 'Logic: Techniques Of Formal Reasoning' by Kalish, Montague, and Mar. Then, with that background, I'd get 'Elements Of Set Theory' by Enderton and, perhaps as a supplement to Enderton, also 'Axiomatic Set Theory' by Suppes.

- Nov 2nd 2010, 12:36 PMMoeBlee
Maybe what you mean is:

Suppose A B = A C & x in B. Show x in C.

Maybe what you mean is:

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. - Nov 3rd 2010, 08:18 AMAlso sprach Zarathustra
To

**emakarov**:

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) - Nov 3rd 2010, 09:49 AMMoeBlee
If I'm not mistaken, the proofs given so far (except I didn't work through emakarov's version), including mine, use non-intuitionistic logic. My hunch is that that is unavoidable.