Formal Generalization Problem - Solve it

[Victor Porton]

[I repost message which you not all received. Also I claimed that I
(partially) solved the problem myself. No my solution was with errors. Can
anybody solve?]

During my math logic research (draft "21 Century Math Method" online:
http://ex-code.com/~porton/math/method.html)

I found a simply formulated problem. I formulated it just today and have not
yet solved it. Whether it is a hard open problem or not, it is interesting.

It is about formalization of the philosophical concept of "generalization".

Please help to solve it. I will reward you with a weblink if you solve
before myself.

See here:
http://ex-code.com/~porton/math/news/2005-05-07-generalization-problem.html

--
Victor Porton (por...@ex-code.com; http://ex-code.com/~porton/)
* Mathematics research, Christian revelations, software *

[Jim Spriggs]

Victor Porton wrote:

> During my math logic research (draft "21 Century Math Method" online:
> http://ex-code.com/~porton/math/method.html)

What a load of tosh.

[Victor Porton]

I also mistaken in the conditions of the problem (namely in definition of
the operation -top->, which I call top-substitution).

Corrected version of the problem:
http://ex-code.com/~porton/math/news/2005-05-10-generalization-problem.html

Victor Porton <por...@ex-code.com> wrote:
> Oh, a mistake, below is a corrected conjecture which is more likely to be
> true:
>
> Conjecture. The enough and necessary answer to the problem is the
> following predicate: "If any subformula X of A is specialization of any
> subformula Y of A then X=Y." (That is it is required and necessary that no
> subformula of A is a specialization of a different subformula of A.)

This problem was raised during my math logic research ("21 Century Math
Method"):
http://ex-code.com/~porton/math/method.html)

>> I think this can be proved by mapping formulas to multigraphs (maybe with
>> "colorized" edges) and then specialization functions would become
>> isomorphisms of one graph to a part of another graph. This would make the
>> proof obvious, I think.

--
Victor Porton (http://ex-code.com/~porton/)