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-proble...
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/)
* Mathematics research, Christian revelations, software *
