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Topic: Matheology § 201
Replies: 32   Last Post: Jan 28, 2013 2:26 PM

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 Virgil Posts: 8,833 Registered: 1/6/11
Re: Matheology � 201
Posted: Jan 28, 2013 2:16 PM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 28 Jan., 08:44, William Hughes <wpihug...@gmail.com> wrote:
> > On Jan 27, 11:23 pm, William Hughes <wpihug...@gmail.com> wrote:
> >
> >
> >
> >
> >

> > > On Jan 27, 11:16 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > On 27 Jan., 23:02, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > > On Jan 27, 10:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > > On 27 Jan., 21:40, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > > > > On Jan 27, 6:46 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > > > > On 27 Jan., 18:32, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > > > > > > On Jan 27, 6:05 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > > > > > <snip>
> >
> > > > > > > > > >..the diagonal
> > > > > > > > > > cannot differ from all lines
> > > > > > > > > > (it differs from every line, though).

> >
> > > > > > > > > The fact that the diagonal differs from every line is
> > > > > > > > > enough to show (induction) that the diagonal is not
> > > > > > > > > equal to any line in the list.

> >
> > > > > > > > No.
> >
> > > > > > > Let the antidiagonal be d and the nth line be l(n)
> >
> > > > > > > We know that for each n in |N, d is not equal to l(n)
> >
> > > > > > > You have agreed that this implies
> >
> > > > > > > There is no m in |N such that d equals l(m)
> >
> > > > > > No.
> >
> > > > > You contradict yourself.  You have agreed
> > > > > that if P(n) is true for every n then
> > > > > the is no n such that P(n) is false.

> >
> > > > Don't turn the words in my mouth.
> >
> > > You have apparently forgotten the
> > > thread in which you agreed to this.

>
> I did not agree to that,

> >
> > A simple proof.
> >
> > The statements
> >
> >     i.  for every natural number n, P(n) is true
> >     ii. there exists a natural number m such that P(m)
> >         is false
> >
> > cannot both be true at the same time.
> > If you prove that i. is true then it follows
> > that ii. is false.-

>
> You have not yet understood.

He appears to understand a great deal better than WM does.

> For every natural n, P(n) is true does
> not imply anything about the existence of P(m).

"i.  for every natural number n, P(n) is true" certainly implies the
existence of a lot of P(m)'s as well as their truth.

> You have not yet
> grasped that there is no "for all".

There is outside of Wolkenmuekenheim.
>
> Of course for every n it is true that there are infinitely many
> naturals beyond.
> Of course there is no m such that there is no natural beyond.
> But if you assume that all natural exist, then there is no natural
> beyond all.
>
> You must unederstand that, therefore, there is no "all naturals".

If there is "no natural beyond all", then we already have "all naturals".
At least outside Wolkenmuekenheim

> All
> we can do is state P(n) for every n. Nevertheless it is not
> necessarily true for all n.

So WM claims he can find an instance in which
For every n in |N P(n) and for some n in |N not P(n)

Note that outside of WN's Wolkenmuekenheim
[NOT for all in |N P(n)]
if and only if
[Exists n in N such that not P(n)]
--