On 27 Mai, 15:50, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > In article <351d0bd7-fe7d-4005-a10a-42b63a903...@q16g2000yqg.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 26 Mai, 04:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > In article <15af9b47-eac6-4d22-8720-0d6e7ebba...@e21g2000yqb.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > > > The problem boils down to the following: > > > > > > > > En Am: m =< n <==> Am En m =< n [*] > > > > En Am: m =< n ==> Am En m =< n [**] > > > > > > > > You know: Classical logic was obtained from finite sets ... > > > > Show me a finite set that obeys [**] but not [*]. > > > > > > The above is not contested: [**] implies > > > [*]. > > > > I said: For complete linear sets [*] is true. > > Not in the article to which I responded.
But frequently I made use of what you call quatifier exchange and what is allowed in case of complete linear sets. > > > You said [*] is not true, but [**] is true. > > That is not what I said.
You said: The only thing that can be stated is (symbolically): E n A m P(m, n) -> A m E n P(m, n) not the reverse, this is just basic logic.
> I said that for the case involved you have to > *prove* that it is true, because it is not generally true.
It is generally true for complete linear sets. You have to prove that it is not. > > > Weyl said: Classical logic was obtained from finite sets. > > Right. > > > Therefore I asked you: Show me a linear complete finite set, that > > makes your claim [**] right and my claim [*] wrong. > > But now you include the word "linear". Where did "Weyl" include the > word "linear"?
I did never claim that quantifier exchange is allowed in case of non- linear sets, like cyclic sets as, for instance, your dice. That would be nonsense. A simple example: Every country has a country that lies west of it. But there is no country that lies west of all countries. > > > This is so simple that it should be understandable even for someone > > who is not "very deep in logic". > > > > > What is contested is that: > > > En Am: m =< n <== Am En m =< n [***] > > > implies [*]. And *that* is the form you do use. > > > > No. I do not use the implication only, I use the full equivalence. Of > > course the equivalence includes the implication [***] as well as the > > implication [**] > > Because the implication [**] is always true, the only part of the equivalence > that is new is the implication [***].
For complete linear sets both are true, therefore [*] holds. And you should recognize that actually complete linear sets obey [*]. Only potentially infinite sets do not. But you mix up things. You claim the existence of a complete linear set but disregard the necessary consequence of completeness or linearity, namely the validity of [*].
