Science and Technology news... :: View topic - My talk about Godel to the post-grads.

Virgil · Guest

In article
<87041b9c-5a16-4f8c-854d-18129e961933@k36g2000pri.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]It is not appropriate to put on the blinders of formal logic here.
They might look like a jester>s cap.
[/quote]
Formal logic is more like a magnifying glass, magnifying all logical
errors until they become clearly visible.

The mathematical custom of using that tool WM honors only in the breach
rather than in the observance.

Herbert Newman · Guest

On Fri, 10 Oct 2008 09:55:46 -0700 (PDT) MoeBlee wrote:

[quote]
Also you must be a CRANK if you don>t understand the
issue.

(1) He does understand the main issue at that juncture of our
conversation.

And even if he didn>t (which is not the case)

(2) [...] that would not in itself make him a crank.

Obviously georgie has no correct (reasonable) notion of a /crank/.[/quote]

He might check:
http://en.wikipedia.org/wiki/Crank_(person)

Herb

Virgil · Guest

In article
<18d9723f-3016-4ed5-a9a1-8ef22008bd5b@g61g2000hsf.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 01:00, Virgil <Vir...@gmale.com> wrote:
In article
6f40d13d-0e17-4fe0-998a-86a0abfc7...@v16g2000prc.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
There is no final stage. At every stage that exists, my assertion is
correct. The meta-diagonals are never in one list.

Any set of "meta-diagonals" or anything else whose members cannot be put
into one list is uncountable by definition.

Any set of items that have been constructed or at least defined as
individuals is coutable by the fact that the set finite strings over a
finite alphabet is countable.

The set of all antidiagonals of lists that can be defined or
constructed is countable but not countable.
[/quote]
Since the set of ALL lists (not merely those "constructible" lists) is
uncountable, so is the set of their anti-diagonals.

[quote]The logical expression "A and not A" is called contradiction.
[/quote]
And it frequently is assumed in WM>s arguments.

But is not in more standard mathematics.

Virgil · Guest

In article
<f0c369b2-873e-4ecf-a323-39c76bcff2be@p31g2000prf.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 01:07, Virgil <Vir...@gmale.com> wrote:

You cannot take more than a countable number of single reals (even if
they were there), because there is no chance to do so (finite
alphabet). And you cannot construct uncountably many lists.

Mathematics is not constrained by the self-mutilation of constructionism

But all constructive sets belong to your mathematics and can be
treated by your mathematics? Or are there exceptions?
[/quote]
Our mathematics is not artificially limited to only constructive sets
the way yours seems to be.

Mathematics takes place in the imagination, and it seems that our
imaginations can contain sets that your imagination is too limited to
contain.

Herbert Newman · Guest

On Fri, 10 Oct 2008 09:17:10 -0700 (PDT) georgie wrote:

[quote]
If you thought I objected, you should be able to see where exactly
it was that I objected.

This is an argument, but -obviously- an unsound one.[/quote]

Why SHOULD he be able to do that? Moreover: what relevance does that have
concerning the (true) claim that "the objections of those doubting the
validity of the diagonal argument we meet in the news are inane twaddle"?

Btw. Do YOU have any objections concerning the validity of the diagonal
argument YOURSELF? If so which ones?

Herb

Virgil · Guest

In article
<7882e8a7-a2a5-4d7c-ba45-5916b372e62e@a19g2000pra.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 01:19, Virgil <Vir...@gmale.com> wrote:
In article
6119fc84-34e3-488f-9b2f-025570d7f...@a70g2000hsh.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:

No, the definition is not in error. The definition defines something
that
does not exist.

That means it is in error, because the definition includes the
statement of existence.

Then when someone defines a Kris Kringle, or a leprechaun or a unicorn,
one must be simultaneously stating one exists?

He should be aware that the blinders of his logic might resemble a
fools cap.

A definition starts: X is ...

This "is" includes the existence of X: "X is". And the definition is
wrong, in error, mistaken, nonsense, foolish (according to the degree
of erring) if X is not.
[/quote]
Perhaps German mathematics and logic choose to handicap themselves in
this way, but outside of Germany, we are less hidebound.

We are allowed to create definitions which need not be instantiated.

We can, for instAnce, speak of a largest natural number, even though we
know that no such thing exists. According to WM' own rules, he should
not even be allowed to use the phrase "largest natural number", as it is
incapable, in his world, of even being defined, though WM has, in fact,
often done so.

Virgil · Guest

In article
<76d8dd6d-bd8e-4f36-9506-c562fbb9c628@e38g2000prn.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 01:27, Virgil <Vir...@gmale.com> wrote:
In article
c2b0f6b0-8769-4fc4-a706-7c69282f3...@2g2000hsn.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 20:30, William Hughes <wpihug...@hotmail.com> wrote:

This claim is pure nonsense, and is never uttered. William gave a
construction with the claim that there is a list that contains B.
You
say "no" to that with an argument that is irrelevant.

William claimed so. Why not claim that all reals are in a list? I
showed that it is impossible to have all antidiagonals of
constructable lists in one list. That argument cannot be refuted.

So what? No one said it could.

This set is countable.

Can YOU *construct* a list of all anti-diagonals of all constructable
lists?

I just said that that is impossible.
[/quote]
Then you have proved the set of them to be uncountable.
[quote]
Or construct a proof showing that such a construction is possible?

Such a proof would contradict my proof. I usually try to avoid such
proofs.
[/quote]
With little success.
[quote]
Absent those, you have made yet a n o t h e r unproven claim.

Every constructable item belongs to a countable set. Therefore I need
not construct all constructable lists to see that they and their
antidiagonals belong to a countable set.
[/quote]
In our mathematical world, a set is listable if and only if it is
countable, so that WM>s claiming a set which is one but not the other is
nonsense, at least in mathematics.

Perhaps if WM would give mathematically valid definitions of what he
means by each (one of these would , of course, have to differ from their
mathematical definitions) he would be in a better position to make his
case.

WM>s last definition, but improper, of "countable" was, IIRC:
A set is countable if it injects into a countable set".

Since it, as stated, is a circular definition, one can never prove a
"first" or ur-set countable from which to prove other sets countable.

Typical of WM>s mathematical ineptitude.

Scott · Guest

I still don>t see how you are explaining the problem away. Perhaps a
change in notation would be helpful. Let F and G be any wff.

I think we all agree we have:

|- (F&~F) -> G

and for all intepretations the above is true so we have:

|= (F&~F) -> G

Suppose we have (F&~F) then we get:

{ F&~F, (F&~F)->G } |- G

then for some interpretations M1 and M2 we can have:

M1: { F&~F, (F&~F)->G } |= G
M2: { F&~F, (F&~F)->G } |/= G

Thus, across all intepretations (ie, limited to the pure predicate
calculus) we have:

{ F&~F, (F&~F)->G } |- G
{ F&~F, (F&~F)->G } |/= G

My question is: We say from a contradiction any wff can be proved, but
I don>t see that. I see we can derive (symbolic manipluation to
generate the string of symbols forming the wff) any wff G, but not
prove (wff is true) every wff G since G>s truth value is dependent
upon the intepretation.

Furthermore, it seems to me that for all interpretations you will
have:

Mx: { F&~F, (F&~F)->G } |= G
Mx: { F&~F, (F&~F)->~G } |/= ~G

or

Mx: { F&~F, (F&~F)->G } |/= G
Mx: { F&~F, (F&~F)->~G } |= ~G

so across all intepretations (ie, limited to the pure predicate
calculus) we can only conclude:

{ F&~F, (F&~F)->G } |- G
{ F&~F, (F&~F)->G } |= Gv~G

Which is very different from a contradiction proves any wff.

So, can you explain why you believe

{ F&~F, (F&~F)->G } |= G

is correct and a contradiction proves any wff?

Thanks

MoeBlee · Guest

On Oct 10, 3:32 pm, Herbert Newman <nomail@invalid> wrote:
[quote]On Fri, 10 Oct 2008 14:50:35 -0700 (PDT) MoeBlee wrote:

Yet another of lwal>s nutty inferences.

Another strange thing is his unbroken believe that some cranks, say WM, are
proponents of some coherent theories. And that>s REALLY nutty (in the face
of all the evidence to the contrary).
[/quote]
If I>m not mistaken, I think his point is that he>s interested in
seeing how one might work on crank (or "crank" as he uses scare
quotes) ideas to formulate them as actual theories; I don>t have the
impression that he thinks that the crank stuff come already formulated
as theories.

MoeBlee

Virgil · Guest

In article
<6c505f72-a008-4783-8781-ff61d01ecc25@v53g2000hsa.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]So it is. Nevertheless the set of all anti-diagonals emerging from
that construction is countable.
[/quote]
The set of all constructable infinite binary strings, from whatever
sources, is countable, but is a proper subset of the set of all infinite
binary strings, which is not.

Virgil · Guest

In article
<d36fb691-e64c-404a-bbc7-8488ac92bed4@v15g2000hsa.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]
I mean that that list contains new meta-dagonals, that are not in that
list. It is impossible to all meta-diagonals in one list, unless you
make a cut and say the meta-diagonals of my final list do not count.
[/quote]
In an infinite sequence of lists, which is the "final list"?

Jesse F. Hughes · Guest

georgie <geo_cant@yahoo.com> writes:

[quote]On Oct 10, 11:55 am, MoeBlee <jazzm...@hotmail.com> wrote:

(2) Even if he did misunderstand (which he didn>t), that would not in
itself make him a crank.

The de facto standard is that the people who don>t understand the
arguments, even if they have been clearly laid out in front of them,
are the CRANKS. Since I>ve had to explain this multiple times,
we can only conclude that you don>t understand it.
[/quote]
We?

First of all, that is *not* the "de facto standard" for crankdom,
though it>s a pretty good candidate if we modify it a bit. Here>s an
off-the-cuff attempt: We might say that one is a mathematical crank
if the following are true.

(1) He alleges that perfectly standard and correct mathematical
arguments are invalid.

(2) He persists in this despite the fact that his criticisms have been
clearly refuted.

(3) He thus concludes that the overwhelming majority of mathematicians
are either ignorant or lying when they fail to alter mathematical
texts to reflect his own claims.

Simple failure to understand arguments is not enough -- especially
since, as it happens, there really *are* lots of bad arguments out
there. I>ve taught Kant>s ethics repeatedly and yet I still don>t
quite see how he goes from the claim that actions have moral worth
only if they are done for the sake of duty to his statement of the
categorical imperative (in its first form). The fact that I don>t
quite get this argument does not mean I am a crank -- not even if the
argument is correct.

But, in any case, I certainly don>t see that you>ve offered any
correct argument that Moe doesn>t understand.

Not sure why I>m treating such a silly troll with a serious reply, to
be honest. Too much time on my hands, evidently.

--
Jesse F. Hughes
"This notion that the United States is getting ready to attack Iran is
simply ridiculous. Having said that, all options are on the table,"
-- George W. Bush
** Posted from http://www.teranews.com **

Jesse F. Hughes · Guest

Herbert Newman <nomail@invalid> writes:

[quote]Obviously georgie has no correct (reasonable) notion of a /crank/.

He might check:
http://en.wikipedia.org/wiki/Crank_(person)
[/quote]
He might even try http://en.wikipedia.org/wiki/Crank_(mechanism).

It>s not a very good definition of mathematical crank, but it>s no
worse than his.

--
"By initially making it virtually impossible to maintain a heterogenous
environment of Word 95 and Word 97 systems, Microsoft offered its customers
that most eloquent of arguments for upgrading: the delicate sound of a
revolver being cocked somewhere just out of sight." --Dan Martinez
** Posted from http://www.teranews.com **

MoeBlee · Guest

On Oct 10, 4:18 pm, Scott <ToaTe...@gmail.com> wrote:
[quote]I still don>t see how you are explaining the problem away.
[/quote]
I don>t know of an actual particular problem you have found.

[quote]Perhaps a
change in notation would be helpful. Let F and G be any wff.

I think we all agree we have:

|- (F&~F) -> G

and for all intepretations the above is true so we have:

|= (F&~F) -> G
[/quote]
Right.

[quote]Suppose we have (F&~F)
[/quote]
What do you mean "we have"? We have in what context? As a line in a
proof?

[quote]then we get:

{ F&~F, (F&~F)->G } |- G
[/quote]
We have that in any case. It>s just an instance of modus ponens. (My
'we have' here means 'it is a meta-theorem'.)

[quote]then for some interpretations M1 and M2 we can have:

M1: { F&~F, (F&~F)->G } |= G
[/quote]
That notation doesn>t make sense.

For a model M, sentence P and a set of sentences G, (I>m stating for
sentences; thouth we can restate for formulas in general), we have
these notations:

|=_M P
to assert that P is true in M.

|= P
to assert that P is true in all models.

G |= P

to assert that any model in which all the members of G are true is a
model in which P is true.

We do NOT have a notation:

G |=_M P

What would it even mean?

Or, if you want to have such a notation, then please define it.

[quote]M2: { F&~F, (F&~F)->G } |/= G
[/quote]
Same remarks as above.

[quote]Thus, across all intepretations (ie, limited to the pure predicate
calculus) we have:

{ F&~F, (F&~F)->G } |- G
[/quote]
We don>t need to mention interpretations for the above.

[quote]{ F&~F, (F&~F)->G } |/= G
[/quote]
No, that is INCORRECT.

{ F&~F, (F&~F)->G } |= G
if and only if
every interpretation that satisifes {F&~F (F&~F)->G} is an
interpretation that satisfies G.

And it IS the case that every interpretation that satisifes {F&~F
(F&~F)->G} is an interpretation that satisfies G, since there is no
interpretation that satisifies {F&~F (F&~F)->G}.

[quote]My question is: We say from a contradiction any wff can be proved, but
I don>t see that. I see we can derive (symbolic manipluation to
generate the string of symbols forming the wff) any wff G,
[/quote]
What we mean by 'G is proven from a contradiction', is that we get G
by the rules of our system, using only logical axioms and the
contradiction as a premise.

And it should be clear to you by now (as you have studied enough
sentential logic) how to prove G from a contradiction, in whatever
ordinary proof system you>re using.

[quote]but not
prove (wff is true) every wff G since G>s truth value is dependent
upon the intepretation.
[/quote]
(1) I just addressed the notion of proof.

(2) The other notion is semantic. That is the notion of entailment, as
formulated in terms of interpretations. In that regard, it is the case
that a contradiction entails G. I.e., for any formula F and G we have:

F&~F |= G.

That is, any interpretation that satisfies F&~F is an interpretation
that satisifes G. Or more exactly:

For all formulas F, sets of formulas G, and interpretations M, if M
satisfies F&~F, then M satisfies G.

Or, if we>re dealing with sentences:

For all sentences F, sets of sentences G, and models M, if F&~F is
true in M, then G is true in M.

MoeBlee

Herbert Newman · Guest

On Fri, 10 Oct 2008 12:02:49 -0600 Virgil wrote:

[quote]
Perhaps German mathematics and logic choose to handicap themselves in
this way, but outside of Germany, we are less hidebound.

Don>t be silly. Maybe you know that FREGE was German, and that _he_[/quote]
provided a first accurate analysis of the _different_ meanings of "is".
(*sigh*)

"One of the basic assumptions of the first-order logic of Gottlob Frege,
reproduced in the notation of Bertrand Russell, is that the ¡be¢
conjugation, usually represented in the form of the third-person singular
'is,' is semantically ambiguous between:

1. The 'is' of predication, as in ¡Socrates is wise.¢
2. The 'is' of identity, as in ¡Mark Twain is Samuel Clemens.¢
3. The 'is' of class-inclusion, as in ¡Man is an animal.¢
4. The 'is' of existence, which can include statements such as
'Socrates is,' and the more common type of ¡there is¢ locutions."

Source
http://evans-experientialism.freewebspace.com/frege_russel_is.htm

PLEASE don>t be led astray by the fact that WM is a *complete* moron.

Herb


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