Re: Interesting problem on symmetric difference

On 29 Jan 2007 22:31:44 +1100, Logan Lee <10464307@xxxxxxxxxx> wrote:


Let A+B denote A or B.
Let A.B denote A and B.

I'd say:

Let A+B denote A union B.
Let A.B denote A intersection B.


Symmetric difference of two sets A and B is
A Delta B = (A-B)+(B-A)

Prove that
A Delta (B Delta C) = (A Delta B) Delta C

_My Proof_

....is not very convincing (to say the least).


Let's assume that (A Delta C) Delta B = A Delta (C Delta B).

Haha! Good joke! You assume what you want to prove! :-)

Note that we are interested in the theorem:

For any sets A,B,C:
A Delta (B Delta C) = (A Delta B) Delta C

Renaming the bound variables doesn't change the "meaning" of the
theorem (i.e. the altered and the original formulation are logically
equivalent).

Hence we may formulate it the following way too:

For any sets A,B,C:
A Delta (C Delta B) = (A Delta C) Delta B.

(Just changed the variable names: B <-> C.)

Since "=" is symmetric this is equivalent to:

For any sets A,B,C:
(A Delta C) Delta B = A Delta (C Delta B).

Just omitting the explicit statement of the universal quantifiers, we
get:

(A Delta C) Delta B = A Delta (C Delta B).

So _assuming_ that associativity DOES hold for the delta-operation,
you "prove" that it does hold. Wow!

Moreover:

Let's assume also that you can apply distributive law to this.

You shouldn't just _assume_ things which have to be proved!

(It's _not obvious_ that the distributive law holds for the
delta-operation. Hence you have to prove this.)


F.

--

E-mail: info<at>simple-line<dot>de
.

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