In article <37a98f77-6bc9-4cae-88e5-fcbadd1d0a75@q2g2000vbr.googlegroups.com> WM <mueckenh@rz.fh-augsburg.de> writes: > On 27 Mai, 15:50, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: ... > > > > > 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 think so, but you have to prove that it is valid for infinite complete linear sets. Note that "classical logic is obtained from finite sets". Nowhere in that quote the word linear is mentioned.
> > > 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.
The reverse of E n A m P(m, n) -> A m E n P(m, n) is E n A m P(m, n) <- A m E n P(m, n) which is [***], neither [*] nor [**].
> > 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.
It is not true for the infinite set of naturals. (1) define FISON(n) be the set of naturals from 1 to n, that is: {1, ..., n}. (2) A{m in N} E(n in N} such that FISON(m) subset FISON(n), trivial, take n = m + 1. (3) E{n in N} A{m in N} such that FISON(m) subset FISON(n), trivially false, take m = n + 1. Which part of this proof is wrong?
It clearly shows that E n A m P(m, n) <- A m En P(m, n) is false. Here with: E n meaning E{n in N} A n meaning A{n in N} P(m, n) meaning FISON(m) subset FISON(n). The first part of the implication is false while the second part of the implication is true, and so the implication is false (all by classical logic).
> > 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.
But as Weyl did not include "linear" in his words, how can that quote support your claim?
> > > > What is contested is that: > > > > En Am: m =< n <== Am En m =< n [***] > > > > implies [*]. And *that* is the form you do use. ... > > 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.
You just state without proof. Where in my proof above that it is false did I go wrong?
> And you > should recognize that actually complete linear sets obey [*].
Strange, I give above a proof that it does not hold. I did not use "actual infinity" nor "potentially infinity", only logic and the axioms of ZF.
> 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 [*].
Why is that a necessary consequence? Because you want it to be so? Can you give a *mathematical* reason? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
