Mathematicians don't need the word true.

For "I think its true" say "I think its provable".

For "G is true" say "G is proven"

That is a fact! That is true, that is provable.
I won't be a hypocrite here, on with the proof!

True is not a ball, a cat, a dog, an object you can look at.
True is an INTERPRETATION, a perspective.
Interpretation only occurs when a subject is interpreting the object.

Without the subject there is no TRUE. Our universe could
be completely devoid of intelligent life, planets might
still orbit stars but that would not be a true fact, because
there would no one to dispute it, and no language to represent it.

***************************************
> >> >Truth is what's proven.
> >>
> >> Can you prove that?
> >
> >yes
> >truth
> >absolute truth
> >substantiated truth
> >derived truth
> >proven truth
> >proof
> >
> >which line is this sequence of equivalent statements is erronous?
>
> I don't accept any of those as equivalent. But if you claim that
> they are equivalent, and that truth is the same as provability,
> then you should be able to prove their equivalence.

Truth : absolute truth
{absolute truth} - {truth} = {}
{truth} - {absolute truth} = {partially true} = {}

absolute truth : substantiated truth
substantied_true -> absolute_true (truth has been justified}
absolute_true -> substantiate_true (what is known)
absolute_true <-> substantiated_true

substantiated truth : proven truth
sub·stan·ti·ate ( P ) Pronunciation Key (sb-stnsh-t)
To support with proof or evidence

proven truth : proof

THEREFORE Truth <=> Proof
*******************************************

Unfortunately, despite the illusion of rigor, mathematicians have
a longstanding habit of being gullible. Any clever lie to a mathematician
will be slotted into whatever paradigm is incomplete into which it
may hold true. To fool a mathematician one starts with the mere
subset relation. proven(facts) C true(facts). Then you make a
LIARS statement, factX <=> true(factX) = ~proven(factX).

The solution today? Mathematics in incomplete, we can't prove
factX but we *know its true*. It makes my blood boil and
'in this consistent (no LIAR statements allowed) domain' just adds
bamboo under the toenails.

Do it properly without inventing fantasy formal systems that don't know
what they're proving! true(facts) = proven(facts) U provable(facts).

factX <=> true(factX) = ~proven(factX)
factX <=> proven(facts) U provable(facts) = ~proven(factX)
factX <=> t U p = ~p
factX <=> p = ~p (extension to case t={})
factX <=> CONTRADICTION

Herc
--
No act is more patriotic than speaking out when your government
is doing the wrong thing in your name. This is not your right
but your sacred duty. And none are more treasonous than those
who would silence these voices.
-------------------------------------s-o-s------------------------------------

From:	Ogie Ogelthorpe
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 06:29:37 -0400

|-|erc wrote:
> Mathematicians don't need the word true.
>
> For "I think its true" say "I think its provable".
>
> For "G is true" say "G is proven"
>
< snipped the rest of the useless drivel>

The only thing true is that you are a certified nut job who should be
locked up before you hurt yourself or someone else.

Ogie

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 22:16:12 +1000

"Ogie Ogelthorpe" wrote in ...
> |-|erc wrote:
> > Mathematicians don't need the word true.
> >
> > For "I think its true" say "I think its provable".
> >
> > For "G is true" say "G is proven"
> >
> < snipped the rest of the useless drivel>
>
> The only thing true is that you are a certified nut job who should be
> locked up before you hurt yourself or someone else.
>
> Ogie

Results 1 - 10 of about 45 for author:ogie author:oglethorpe group:rec.org.mensa
Results 1 - 10 of about 10,500 for author:|-|erc group:rec.org.mensa

Can you mensa newbies learn the pecking order, if not some manners?

Herc

From:	Ogie Ogelthorpe
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 19:09:38 -0400

|-|erc wrote:
> "Ogie Ogelthorpe" wrote in ...
>
>>|-|erc wrote:
>>
>>>Mathematicians don't need the word true.
>>>
>>>For "I think its true" say "I think its provable".
>>>
>>>For "G is true" say "G is proven"
>>>
>>
>>< snipped the rest of the useless drivel>
>>
>>The only thing true is that you are a certified nut job who should be
>>locked up before you hurt yourself or someone else.
>>
>>Ogie
>
>
> Results 1 - 10 of about 45 for author:ogie author:oglethorpe group:rec.org.mensa
> Results 1 - 10 of about 10,500 for author:|-|erc group:rec.org.mensa
>
> Can you mensa newbies learn the pecking order, if not some manners?
>
> Herc
>
>
>
Hey loser I am higher on the evolutionary scale than you, thus the
pecking order makes you my bitch.

How is that for manners?

Ogie

From:	Lysander
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 09:46:20 -0800

D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:

> Ogie Ogelthorpe writes:
>
>>|-|erc wrote:
>>> Mathematicians don't need the word true.
>>>
>>> For "I think its true" say "I think its provable".
>>>
>>> For "G is true" say "G is proven"
>>>
>>< snipped the rest of the useless drivel>
>
>>The only thing true is that you are a certified nut job who should be
>>locked up before you hurt yourself or someone else.
>
> The distinction between "provable" and "true" is easy to demonstrate.
>
> A sentence is "provable" or "unprovable" for a given theory (a set of
> sentences). It is inappropriate to describe a sentence as being "true"
> or "false" for a theory.
>
> A sentence is "true" or "false" for a specific model (the truth value of
> a formula for a certain assignment of variables within a model is defined
> by recursion on the complexity of the formula, and the truth value of a
> sentence for a model is independent of the assignment of variables).
> It is inappropriate to describe a sentence as being "provable" or
> "unprovable" for a model.
>
> So a sentence is "provable" or "unprovable" for a theory, but not for a
> model. A sentence is "true" or "false" for a model, but not for a theory.
>
> A sentence which is provable in a theory is true in all models of the
> theory.
>
> A sentence which is unprovable in a theory is false in some model(s) of
> the theory (i.e. it is false in at least one model of the theory).
>
> A sentence which is true in all models of the theory is provable in the
> theory.
>
> For the sentence which is used in the proof of Godel's Incompleteness
> Theorem, the interpretation given in the proof is the interpretation in
> the STANDARD MODEL of the natural numbers. It is NOT the interpretation
> in all models (i.e. the given interpretation is MODEL dependent). The
> sentence is unprovable in the theory of formal arithmetic, and it is true
> in the standard model. This does not cause a contradiction, since there
> are models of formal arithmetic in which the sentence is false, and in
> NONE of these models is the interpretation that given in the proof of the
> Incompleteness Theorem.

I don't necessarily disagree with anything you say, I'm just trying
to figure out exactly what you mean.

First off, my understanding (which may be seriously flawed) of the
Incompleteness Theorem is that Goedel, using construction rules
which are legitimate in the Principia Mathematica (PM), constructed
a statement which was true (on meta-mathematical grounds) but
formally unprovable within the system. The construction method uses
factorization, so the logical requirement is for the Fundamental
Theorem of Arithmetic, and, therefore,the proof only applies to
logic systems sufficiently powerful to use that theorem. Is this
what you mean by the 'STANDARD MODEL of the natural numbers, that is
natural numbers which we can decompose by factoring? And
what other models of the natural numbers are there?

According to your 'a sentence which is unprovable in a theory is
false in ... at least one model of the theory.' Which 'model' for a
PM like logical system would you propose which makes the statement
used in the Incompleteness Theorem false?

Just asking, not arguing that such doesn't exist.

L.
--
Prediction is difficult, especially the future.
Niels Bohr

From:	David McAnally
Subject:	Re: True = [ proven \| provable ]
Date:	18 Jan 2005 10:52:38 GMT

Lysander writes:

>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:

>> Ogie Ogelthorpe writes:
>>
>>>|-|erc wrote:
>>>> Mathematicians don't need the word true.
>>>>
>>>> For "I think its true" say "I think its provable".
>>>>
>>>> For "G is true" say "G is proven"
>>>>
>>>< snipped the rest of the useless drivel>
>>
>>>The only thing true is that you are a certified nut job who should be
>>>locked up before you hurt yourself or someone else.
>>
>> The distinction between "provable" and "true" is easy to demonstrate.
>>
>> A sentence is "provable" or "unprovable" for a given theory (a set of
>> sentences). It is inappropriate to describe a sentence as being "true"
>> or "false" for a theory.
>>
>> A sentence is "true" or "false" for a specific model (the truth value of
>> a formula for a certain assignment of variables within a model is defined
>> by recursion on the complexity of the formula, and the truth value of a
>> sentence for a model is independent of the assignment of variables).
>> It is inappropriate to describe a sentence as being "provable" or
>> "unprovable" for a model.
>>
>> So a sentence is "provable" or "unprovable" for a theory, but not for a
>> model. A sentence is "true" or "false" for a model, but not for a theory.
>>
>> A sentence which is provable in a theory is true in all models of the
>> theory.
>>
>> A sentence which is unprovable in a theory is false in some model(s) of
>> the theory (i.e. it is false in at least one model of the theory).

The proof of this uses the Axiom of Choice.

>> A sentence which is true in all models of the theory is provable in the
>> theory.

As this statement is equivalent to the statement which precedes it, then
the proof of this statement also uses the Axiom of Choice.

>> For the sentence which is used in the proof of Godel's Incompleteness
>> Theorem, the interpretation given in the proof is the interpretation in
>> the STANDARD MODEL of the natural numbers. It is NOT the interpretation
>> in all models (i.e. the given interpretation is MODEL dependent). The
>> sentence is unprovable in the theory of formal arithmetic, and it is true
>> in the standard model. This does not cause a contradiction, since there
>> are models of formal arithmetic in which the sentence is false, and in
>> NONE of these models is the interpretation that given in the proof of the
>> Incompleteness Theorem.

> I don't necessarily disagree with anything you say, I'm just trying
> to figure out exactly what you mean.

> First off, my understanding (which may be seriously flawed) of the
> Incompleteness Theorem is that Goedel, using construction rules
> which are legitimate in the Principia Mathematica (PM), constructed
> a statement which was true (on meta-mathematical grounds) but
> formally unprovable within the system. The construction method uses
> factorization, so the logical requirement is for the Fundamental
> Theorem of Arithmetic, and, therefore,the proof only applies to
> logic systems sufficiently powerful to use that theorem. Is this
> what you mean by the 'STANDARD MODEL of the natural numbers, that is
> natural numbers which we can decompose by factoring? And
> what other models of the natural numbers are there?

No. The standard model is built on the set which is usually taken, i.e.
N = {0, 1, 2, ...}. The reason for the usage of the capital letters is
to emphasise that I am discussing the model of Formal Arithmetic which
is based on N, and not any of the other models of Formal Arithmetic.

> According to your 'a sentence which is unprovable in a theory is
> false in ... at least one model of the theory.' Which 'model' for a
> PM like logical system would you propose which makes the statement
> used in the Incompleteness Theorem false?

The Axiom of Choice is a nonconstructive axiom. We can prove the
existence of such a model, but that doesn't mean that we can construct
it.

The actual proof of existence uses well-orderings of certain sets, and
as the Well-Ordering Principle is equivalent to the Axiom of Choice,
this means that the proof uses the Axiom of Choice. Note also that the
Well-Ordering Principle is nonconstructive. The Well-Ordering Principle
asserts the existence of a well-ordering of a set, but not how to
construct the well-ordering.

David

-----

From:	Torkel Franzen
Subject:	Re: True = [ proven \| provable ]
Date:	18 Jan 2005 11:55:43 +0100

D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:

> The proof of this uses the Axiom of Choice.

Only in the general case, not for the countable theories that we use
in formalizing mathematics.

From:	David McAnally
Subject:	Re: True = [ proven \| provable ]
Date:	22 Jan 2005 14:58:33 GMT

Torkel Franzen writes:

>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:

>> The proof of this uses the Axiom of Choice.

> Only in the general case, not for the countable theories that we use
>in formalizing mathematics.

I see. In a countable theory, a similar numbering system can be
introduced for the symbols (constant symbols, function symbols, relation
symbols, variables, logical connectives, quantifiers, brackets, commas (?)),
and then an injective function mapping the formulae of the language to the
natural numbers, in the same manner as Goedel numbering in Formal
Arithmetic, can be defined. Then formulae can be well-ordered by the
ordering of the natural numbers.

Alternatively, it occurred to me that, in the general case, the
Well-Ordering Principle should only be needed to guarantee a well-ordering
of the constant symbols, function symbols and relation symbols, and that a
well-ordering of formulae can be induced from the well-ordering of the
constant symbols, function symbols, relation symbols and variables.

One means that I have seen for constructing a model for a consistent
theory is via witnesses. This is done by the introduction of additional
constant symbols into the language (the cardinality of the set of
introduced constant symbols is equal to the cardinality of the set of
constant symbols, function symbols and relation symbols in the original
language if that set is infinite, and aleph_0 if the set is finite - under
the Axiom of Choice, this gives sufficiently many additional constant
symbols as are required in the uncountable case, and in the countable
case, this gives sufficiently many additional symbols as required, even
without the Axiom of Choice). Denote this cardinality (and also its
initial ordinal) by l. The introduced constant symbols can be indexed by
ordinals less than the initial ordinal for the cardinality. The new
constant symbols will be denoted by {k_a : a < l}.

Formulae with one free variable, v, in the extended language are then
well-ordered, indexed by ordinals less than l (this uses the Axiom of
Choice in the case of uncountable languages, and Goedel style numbering in
the case of countable languages). Provided the symbols in the language
can be well-ordered, this cardinality is equal to the cardinality of the
set of introduced constant symbols. To the consistent set of sentences,
adjoin the sentences

(TE v \phi_a(v)) => \phi_a(k_{n_a}),

where TE denotes the existential quantifier (i.e. 'TE v' denoted "there
exists v"), \phi_a(v) denotes the formula of one free variable which is
indexed by the ordinal a < l, and k_{n_a} is an introduced constant
symbol, where n_a is the smallest ordinal n (necessarily less than l) such
that k_n does not occur in the sentence

(TE v \phi_b(v)) => \phi_b(k_{n_b})

for any b < a, and also does not occur in the formula \phi_a(v).

The extended set of sentences is still consistent.

The sentences in the extended language are then well-ordered (this can be
done by using the Axiom of Choice in the case of uncountable languages,
noting that it is sufficient to well-order the constant symbols, function
symbols and relation symbols in the extended language, and then induce a
well-ordering of the sentences, and by using a Goedel style numbering in
the case of countable languages). For each sentence in order, adjoin the
sentence if it is consistent with the present value of the set, and adjoin
the negation of the sentence if it is inconsistent with the present value
of the set. Specifically, let {\psi_a} be an enumeration of the sentences
by ordinals less than l. Let T_0 be the consistent set of sentences with
which to start, i.e. the original consistent set, and all the witness
sentences of the form

(TE v \phi_a(v)) => \phi_a(k_{n_a}}).

For an ordinal a, let T_{a+1} = T_a u {\psi_a} if \psi_a is consistent
with T_a, and let T_{a+1} = T_a u {not \psi_a} if \psi_a is inconsistent
with T_a, where, for sets A and B, A u B denotes their union. For limit
ordinal a, let T_a = U_{b < a} T_b, so that T_a is the union of T_b over
all ordinals b < a (here U_{b < a} denotes the union over all b < a). The
final set, T_l,is consistent. T_c is a maximal consistent set of
sentences.

The model is now determined as follows. Let {k_a : a < l} be the set of
introduced constant symbols. Take the equivalence relation "k_a is
equivalent to k_b" iff T_l |- k_a = k_b. Partition the set of introduced
constant symbols into equivalence classes relative to this equivalence
relation. The corresponding partition is the set of the model.

For a constant symbol C_i, TE v (v = C_i) is provable, and since there
exists an ordinal a such that T_l |- (TE v (v = C_i)) => k_a = C_i, then
T_l |- k_a = C_i. In the model, C_i is represented by the equivalence
class [k_a].

Let F_i be an n-ary function symbol. For constant symbols, k_{a_1}, ...,
k_{a_n}, TE v (v = F_i(k_{a_1},...,k_{a_n})) is provable, and since there
exists an ordinal b such that

T_l |- TE v (v = F_i(k_{a_1},...,k_{a_n}))

=> k_b = F_i(k_{a_1},...,k_{a_n}),

then T_l |- k_b = F_i(k_{a_1},...,k_{a_n}). For b_1, ..., b_n such that
T_l |- k_{a_1} = k_{b_1}, ..., T_l |- k_{a_n} = k_{b_n}, then

T_l |- k_b = F_i(k_{b_1},...,k_{b_n}).

It follows that whether or not T_l |- k_b = F_i(k_{a_1},...,k_{a_n})
depends only on the equivalence classes of k_{a_1}, ..., k_{a_n}.
Further, for any ordinal a, T_l |- k_a = F_i(k_{a_1},...,k_{a_n}) iff
k_a and k_b belong to the same equivalence class. This induces an n-ary
function f_i on the set of equivalence classes given by
f_i([k_{a_1}],...,[k_{a_n}]) = [k_b] iff

T_l |- k_b = F_i(k_{a_1},...,k_{a_n}).

The function symbol F_i is represented by f_i.

Let P_i be an n-ary relation relation symbol. If T_l |- k_{a_1} = k_{b_1},
...., T_l |- k_{a_n} = k_{b_n}, then T_l |- P_i(k_{a_1},...,k_{a_n}) iff
T_l |- P_i(k_{b_1},...,k_{b_n}), and whether or not
T_l |- P_i(k_{a_1},...,k_{a_n}) depends only on the equivalence classes
of k_{a_1}, ..., k_{a_n}. Let

R_i = {([k_{a_1}],...,[k_{a_n}]) : T_l |- P_i(k_{a_1},...,k_{a_n})},

then the set R_i represents P_i.

This model is a model for T_l, and therefore for the original theory.

In the case of Formal Arithmetic, Goedel numbering can be used to give a
sentence whose interpretation in the standard model is an assertion of its
own unprovability. This sentence is unprovable, and it is true in the
standard model. Due to its unprovability, there is also a model in which
the sentence in question is false. In this new model, the interpretation
of the sentence is not a statement of its own unprovability. On the other
hand, there will be another sentence whose interpretation of in the new
model is an assertion of its own unprovability (and it will be true in the
new model, and unprovable in Formal Arithmetic).

One such means of constructing such a model is by the use of witnesses, as
described above. In this case, the additional constant symbols will have
to be assigned their own numbers. This can be done in a similar manner to
Hilbert's Hotel sort of problem. Specifically, the numbers for the
individual variables can be changed (altered so that all the variables
have even Goedel numbers, for example), and the the spaces can then be
used for the Goedel numbers for the additional constant symbols. This
changes the Goedel numbers for the formulae, and therefore also the actual
formula which asserts its own unprovability. The new formula which
asserts its own unprovability may even involve introduced constant symbols.

David

-----

From:	Torkel Franzen
Subject:	Re: True = [ proven \| provable ]
Date:	22 Jan 2005 16:43:01 +0100

D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:

> One such means of constructing such a model is by the use of witnesses, as
> described above.

Right, this is Henkin's proof, now standard. The axiom of choice is
needed in the general case to show that every consistent set of
sentences has a consistent complete extension.

From:	David McAnally
Subject:	Re: True = [ proven \| provable ]
Date:	23 Jan 2005 23:49:10 GMT

Torkel Franzen writes:

>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:

>> One such means of constructing such a model is by the use of witnesses, as
>> described above.

> Right, this is Henkin's proof, now standard. The axiom of choice is
>needed in the general case to show that every consistent set of
>sentences has a consistent complete extension.

I had thought that the formula of a language (and, in particular, the
sentences) could be well-ordered, once you have a well-ordering of the
constant symbols, function symbols, relation symbols, and variables. I
had also thought that a well-ordering of the sentences would be sufficient
to prove that every consistent set of sentences has a consistent complete
extension.

As a consequence, I had thought that the axiom of choice would only be
needed to guarantee a well-ordering of constant symbols, function symbols
and relation symbols.

Also, I had thought that a well-ordering was needed for constant symbols,
functions symbols and relation symbols so that one can well-order those
formulae with one free variable, and therefore to introduce the witnesses
for those formulae. In other words, I had thought that, in the general
case, the axiom of choice was needed to introduce the witnesses (the axiom
of choice being specifically needed to well-order the symbols of the
(extended) language).

David

-----

From:	Torkel Franzen
Subject:	Re: True = [ proven \| provable ]
Date:	24 Jan 2005 08:59:03 +0100

D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:

> I had thought that the formula of a language (and, in particular, the
> sentences) could be well-ordered, once you have a well-ordering of the
> constant symbols, function symbols, relation symbols, and variables. I
> had also thought that a well-ordering of the sentences would be sufficient
> to prove that every consistent set of sentences has a consistent complete
> extension.
>
> As a consequence, I had thought that the axiom of choice would only be
> needed to guarantee a well-ordering of constant symbols, function symbols
> and relation symbols.

Right, that's one way of doing it. Whichever way we go about it, the
axiom of choice enters into showing that every consistent set of
sentences has a consistent complete extension.

> In other words, I had thought that, in the general
> case, the axiom of choice was needed to introduce the witnesses (the axiom
> of choice being specifically needed to well-order the symbols of the
> (extended) language).

We don't need choice to introduce the witnesses, just a sufficient
supply of new constants. Given our language L, first introduce a new
constant c_A for every formula A in L (for this, we don't need
choice). Repeat the procedure omega times, take the union L' of the
resulting languages and add axioms (Ex)A(x)->A(c) where c is the
constant associated with A(x).

From:	David McAnally
Subject:	Re: True = [ proven \| provable ]
Date:	17 Jan 2005 14:48:02 GMT

Ogie Ogelthorpe writes:

>|-|erc wrote:
>> Mathematicians don't need the word true.
>>
>> For "I think its true" say "I think its provable".
>>
>> For "G is true" say "G is proven"
>>
>< snipped the rest of the useless drivel>

>The only thing true is that you are a certified nut job who should be
>locked up before you hurt yourself or someone else.

The distinction between "provable" and "true" is easy to demonstrate.

A sentence is "provable" or "unprovable" for a given theory (a set of
sentences). It is inappropriate to describe a sentence as being "true"
or "false" for a theory.

A sentence is "true" or "false" for a specific model (the truth value of
a formula for a certain assignment of variables within a model is defined
by recursion on the complexity of the formula, and the truth value of a
sentence for a model is independent of the assignment of variables).
It is inappropriate to describe a sentence as being "provable" or
"unprovable" for a model.

So a sentence is "provable" or "unprovable" for a theory, but not for a
model. A sentence is "true" or "false" for a model, but not for a theory.

A sentence which is provable in a theory is true in all models of the
theory.

A sentence which is unprovable in a theory is false in some model(s) of
the theory (i.e. it is false in at least one model of the theory).

A sentence which is true in all models of the theory is provable in the
theory.

For the sentence which is used in the proof of Godel's Incompleteness
Theorem, the interpretation given in the proof is the interpretation in
the STANDARD MODEL of the natural numbers. It is NOT the interpretation
in all models (i.e. the given interpretation is MODEL dependent). The
sentence is unprovable in the theory of formal arithmetic, and it is true
in the standard model. This does not cause a contradiction, since there
are models of formal arithmetic in which the sentence is false, and in
NONE of these models is the interpretation that given in the proof of the
Incompleteness Theorem.

David

-----

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Tue, 18 Jan 2005 16:15:37 +1000

> A sentence which is provable in a theory is true in all models of the
> theory.

same thing. star(1) is bright, star(2) is bright, star(3) is bright ....
model 1 : star is bright
model 2 : star is bright
....

star is bright in all models <-> stars are bright is provable [defn]
stars are bright is provable <-> stars are bright is true
stars are bright

If there is no open clause (variable) left on a sentence, and given
all interpretations it would be true.... its TRUE.

>
> A sentence which is unprovable in a theory is false in some model(s) of
> the theory (i.e. it is false in at least one model of the theory).
>
> A sentence which is true in all models of the theory is provable in the
> theory.
>
> For the sentence which is used in the proof of Godel's Incompleteness
> Theorem, the interpretation given in the proof is the interpretation in
> the STANDARD MODEL of the natural numbers. It is NOT the interpretation
> in all models (i.e. the given interpretation is MODEL dependent). The
> sentence is unprovable in the theory of formal arithmetic, and it is true
> in the standard model. This does not cause a contradiction, since there
> are models of formal arithmetic in which the sentence is false, and in
> NONE of these models is the interpretation that given in the proof of the
> Incompleteness Theorem.
>
> David

So "this statement has no proof (in theory T or whatever)" is ABSOLUTELY_TRUE
as long as the set of formal systems together is inconsistent.

I could have told you that

MODEL
Standard .........G is true -> incompleteness theorem
X ....................G is false for some X
arithmetic1.........G is unproven
arithmetic2.........G is false

I'd just call that an incomplete theorem myself!

Herc
--
Mathematics today is building models to fit the errors of mathematics of yesterday.

From:	John Savard
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 14:17:40 GMT

On Mon, 17 Jan 2005 14:12:06 +1000, "|-|erc" wrote, in part:

>Without the subject there is no TRUE. Our universe could
>be completely devoid of intelligent life, planets might
>still orbit stars but that would not be a true fact, because
>there would no one to dispute it, and no language to represent it.

It is not useful to abandon the concept of "true" as distinct from
provable. For one thing, the concept of proof cannot be constructed
without having the prior concept of true.

Also, why should we make everything more complicated just because you
don't like Godel's proof?

John Savard
http://home.ecn.ab.ca/~jsavard/index.html

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Tue, 18 Jan 2005 12:47:36 GMT

"|-|erc" ha scritto

> Mathematicians don't need the word true.
>
> For "I think its true" say "I think its provable".
>
> For "G is true" say "G is proven"

How do you prove 1+1=2 ?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Tue, 18 Jan 2005 23:03:05 +1000

"LordBeotian" wrote in >
> "|-|erc" ha scritto
>
> > Mathematicians don't need the word true.
> >
> > For "I think its true" say "I think its provable".
> >
> > For "G is true" say "G is proven"
>
> How do you prove 1+1=2 ?
>

Usually like this:

Define the number 0
Define all other numbers as the successor of some other number.

0 = 0
1 = s(0)
2 = s(s(0))
3 = s(s(s(0)))
4 = s(s(s(s(0))))
and so on

Define addition as
0 + 0 = 0 rule 0
s(a) + 0 = s(a) rule 1
s(a) + b = a + s(b) rule 2

RTP 1+1 = 2 required to prove
RTP s(0) + s(0) = s(s(0)) put into peano form
LHS = s(0) + s(0) left hand side
LHS = 0 + s(s(0)) rule 2
LHS = s(s(0)) rule 1
RHS = s(s(0))
LHS = RHS
THEREFORE 1 + 1 = 2 since LHS = RHS

One plus one equals two is proven! Just don't ask me to prove 9 X 7 = 63.

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Tue, 18 Jan 2005 18:48:34 GMT

"|-|erc" ha scritto

> > > Mathematicians don't need the word true.
> > >
> > > For "I think its true" say "I think its provable".
> > >
> > > For "G is true" say "G is proven"
> >
> > How do you prove 1+1=2 ?
> >
>
>
> Usually like this:
>
> Define the number 0

Difined in which way?

> Define all other numbers as the successor of some other number.
>
> 0 = 0
> 1 = s(0)
> 2 = s(s(0))
> 3 = s(s(s(0)))
> 4 = s(s(s(s(0))))
> and so on
>
> Define addition as
> 0 + 0 = 0 rule 0
> s(a) + 0 = s(a) rule 1
> s(a) + b = a + s(b) rule 2

Uhm... are these true statements we do not prove?
Are they true "by definition"?
So not only is true what is proven but also is true what is stated as a
definition or as an axiom?
What if I chose my axiom/definitions in different ways so that I "prove"
1+1=3? Would 1+1=3 become true?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Wed, 19 Jan 2005 13:53:22 +1000

"LordBeotian" wrote in >
> "|-|erc" ha scritto
>
> > > > Mathematicians don't need the word true.
> > > >
> > > > For "I think its true" say "I think its provable".
> > > >
> > > > For "G is true" say "G is proven"
> > >
> > > How do you prove 1+1=2 ?
> > >
> >
> >
> > Usually like this:
> >
> > Define the number 0
>
> Difined in which way?

E0, 0 e N
zero is a number! you can deny 0 is a number, but then 1+1 may not equal 2.
what we're actually proving is "1 + 1 = 2 for the usual interpretation of the terms".

if 1 = todays breakfast and 2 = a razor and + = a mirror and "=" = air ticket
then "breakfast mirror breakfast ticket razor" may not always be true.

>
> > Define all other numbers as the successor of some other number.
> >
> > 0 = 0
> > 1 = s(0)
> > 2 = s(s(0))
> > 3 = s(s(s(0)))
> > 4 = s(s(s(s(0))))
> > and so on
> >
> > Define addition as
> > 0 + 0 = 0 rule 0
> > s(a) + 0 = s(a) rule 1
> > s(a) + b = a + s(b) rule 2
>
> Uhm... are these true statements we do not prove?

rule 0 and rule 1 are given by the *meaning* of +.
for the usual interpretation of plus, 0 + x = x
again, you can deny 0+x=x but again 1+1 wont equal 2.

rule 2 is the distributive law, it could be composed of these definitions
s(a) = a + 1. #1
s(a) + b = a + 1 + b = a + b + 1 = a + s(b) #2

#1 and #2 are a smaller grain than rule 2, but its just a "definition of the usual interpretion of plus"
rule 2 is not that hard to decipher as it stands.

> Are they true "by definition"?
> So not only is true what is proven but also is true what is stated as a
> definition or as an axiom?

bingo! there are few allowed truths as part of proof systems themselves,
like EXISTS X. but proofs are generally, A & A->B then B

A axiom (taken for granted, eg. the natural numbers appear in order)
A->B (given the axiom eg rule 1 and 2 and 0 defn, implies B! eg 1+1=2 )
B (1+2 = 2 is proven)

> What if I chose my axiom/definitions in different ways so that I "prove"
> 1+1=3? Would 1+1=3 become true?
>

Not for the usual interpretations of 1, 2, 3, +, =
The trick is your axioms must be equivalent to what the terms actually mean in real life.

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Wed, 19 Jan 2005 09:47:58 GMT

"|-|erc" ha scritto

> > > Define all other numbers as the successor of some other number.
> > >
> > > 0 = 0
> > > 1 = s(0)
> > > 2 = s(s(0))
> > > 3 = s(s(s(0)))
> > > 4 = s(s(s(s(0))))
> > > and so on
> > >
> > > Define addition as
> > > 0 + 0 = 0 rule 0
> > > s(a) + 0 = s(a) rule 1
> > > s(a) + b = a + s(b) rule 2
> >
> > Uhm... are these true statements we do not prove?
>
> rule 0 and rule 1 are given by the *meaning* of +.
> for the usual interpretation of plus, 0 + x = x
> again, you can deny 0+x=x but again 1+1 wont equal 2.
>
> rule 2 is the distributive law, it could be composed of these definitions
> s(a) = a + 1. #1
> s(a) + b = a + 1 + b = a + b + 1 = a + s(b) #2
>
> #1 and #2 are a smaller grain than rule 2, but its just a "definition of
the usual interpretion of plus"
> rule 2 is not that hard to decipher as it stands.

Perfect, so we encode the meaning of "1", "2", ... "+", "=" in a set of
axioms and rules, but what if our set of axioms and rule is not really
adequate to encode the whole meaning (i.e. is incomplete)?

> > Are they true "by definition"?
> > So not only is true what is proven but also is true what is stated as a
> > definition or as an axiom?
>
> bingo! there are few allowed truths as part of proof systems themselves,
> like EXISTS X. but proofs are generally, A & A->B then B
>
> A axiom (taken for granted, eg. the natural numbers appear in order)
> A->B (given the axiom eg rule 1 and 2 and 0 defn, implies B! eg
1+1=2 )
> B (1+2 = 2 is proven)
>
>
> > What if I chose my axiom/definitions in different ways so that I "prove"
> > 1+1=3? Would 1+1=3 become true?
> >
>
> Not for the usual interpretations of 1, 2, 3, +, =
> The trick is your axioms must be equivalent to what the terms actually
mean in real life.

Ok, so what if I discover a new "axiom" that is true according to the
"meaning of the terms in real life" but is not included in our "proof
system"? Could it happen?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Wed, 19 Jan 2005 21:19:16 +1000

"LordBeotian" wrote in >
> "|-|erc" ha scritto
>
> > > > Define all other numbers as the successor of some other number.
> > > >
> > > > 0 = 0
> > > > 1 = s(0)
> > > > 2 = s(s(0))
> > > > 3 = s(s(s(0)))
> > > > 4 = s(s(s(s(0))))
> > > > and so on
> > > >
> > > > Define addition as
> > > > 0 + 0 = 0 rule 0
> > > > s(a) + 0 = s(a) rule 1
> > > > s(a) + b = a + s(b) rule 2
> > >
> > > Uhm... are these true statements we do not prove?
> >
> > rule 0 and rule 1 are given by the *meaning* of +.
> > for the usual interpretation of plus, 0 + x = x
> > again, you can deny 0+x=x but again 1+1 wont equal 2.
> >
> > rule 2 is the distributive law, it could be composed of these definitions
> > s(a) = a + 1. #1
> > s(a) + b = a + 1 + b = a + b + 1 = a + s(b) #2
> >
> > #1 and #2 are a smaller grain than rule 2, but its just a "definition of
> the usual interpretion of plus"
> > rule 2 is not that hard to decipher as it stands.
>
> Perfect, so we encode the meaning of "1", "2", ... "+", "=" in a set of
> axioms and rules, but what if our set of axioms and rule is not really
> adequate to encode the whole meaning (i.e. is incomplete)?
>
> > > Are they true "by definition"?
> > > So not only is true what is proven but also is true what is stated as a
> > > definition or as an axiom?
> >
> > bingo! there are few allowed truths as part of proof systems themselves,
> > like EXISTS X. but proofs are generally, A & A->B then B
> >
> > A axiom (taken for granted, eg. the natural numbers appear in order)
> > A->B (given the axiom eg rule 1 and 2 and 0 defn, implies B! eg
> 1+1=2 )
> > B (1+2 = 2 is proven)
> >
> >
> > > What if I chose my axiom/definitions in different ways so that I "prove"
> > > 1+1=3? Would 1+1=3 become true?
> > >
> >
> > Not for the usual interpretations of 1, 2, 3, +, =
> > The trick is your axioms must be equivalent to what the terms actually
> mean in real life.
>
> Ok, so what if I discover a new "axiom" that is true according to the
> "meaning of the terms in real life" but is not included in our "proof
> system"? Could it happen?
>

yes, but in this particular case all further axioms of arithmetic are *consistant* with
the ones I put down. say that you did find an actually true axiom that contradicted
1+1=2. the proof still holds, proofs never break, to find an error in the proof you
have to specify which step was wrong. the conclusion logically follows the assumptions
so all you can do if fault one of the assumptions. if the conclusion is shown wrong the
proof is still correct, because the proof merely stated GIVEN this interpretation, the
conclusion is this. the interpretation of the result of the proof may have to shift to fit
with the common interpretation of the terms. instead of "this is proof 1+1=2", it would
be retitled to "this is proof of elementary functions emulating 1+1 giving 2".

http://www.cut-the-knot.org/do_you_know/mul_num.shtml
Definition 1 (Addition)
For every x define x+1 = x'.
For every x,y define x+y' = (x+y)'.
Theorem 4
The number x+y is well defined for all natural x and y.

Definition 2 (Multiplication)
For every x define x·1 = x.
For every x,y define x·y' = x·y + x.

There's multiplication, it won't hurt the proof! I don't think my addition rules were correct
looking at this but worked for 1 + 1. Peano axiom 5 is a bit difficult. Its a formal description
of proof by induction, but proof by induction is not that widespread.

Prove that all natural numbers (except 0) are bigger than 0
1 > 0
if x > 0 then x+1 > 0
therefore all natural numbers are bigger than 0.

Does that look like a proof?

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Wed, 19 Jan 2005 12:36:50 GMT

"|-|erc" ha scritto

> > > Not for the usual interpretations of 1, 2, 3, +, =
> > > The trick is your axioms must be equivalent to what the terms actually
> > mean in real life.
> >
> > Ok, so what if I discover a new "axiom" that is true according to the
> > "meaning of the terms in real life" but is not included in our "proof
> > system"? Could it happen?
> >
>
> yes, but in this particular case all further axioms of arithmetic are
*consistant* with
> the ones I put down.

Ok, that's the point!
If our system can be incomplete so there could exist a true statement that
is not provable within the system (that is the definition of incompleteness)
so how can you say that true=provable if you agree that true unprovable
statements could exist?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 11:04:45 +1000

"LordBeotian" wrote in >
> "|-|erc" ha scritto
>
> > > > Not for the usual interpretations of 1, 2, 3, +, =
> > > > The trick is your axioms must be equivalent to what the terms actually
> > > mean in real life.
> > >
> > > Ok, so what if I discover a new "axiom" that is true according to the
> > > "meaning of the terms in real life" but is not included in our "proof
> > > system"? Could it happen?
> > >
> >
> > yes, but in this particular case all further axioms of arithmetic are
> *consistant* with
> > the ones I put down.
>
> Ok, that's the point!
> If our system can be incomplete so there could exist a true statement that
> is not provable within the system (that is the definition of incompleteness)
> so how can you say that true=provable if you agree that true unprovable
> statements could exist?
>

the fact a system *can be* incomplete merely by missing some formula does not
fit the definition of incompleteness. incompleteness is that such a true statement
will NEVER be proven within the system, or any consistent extension of it. incompleteness
is actually the opposite of 'too few formula', it means the system is *powerful enough*
that it can express proof-denial-statements. the system got too big for its own boots.

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 07:56:43 GMT

"|-|erc" ha scritto

> > If our system can be incomplete so there could exist a true statement
that
> > is not provable within the system (that is the definition of
incompleteness)
> > so how can you say that true=provable if you agree that true unprovable
> > statements could exist?
> >
>
> the fact a system *can be* incomplete merely by missing some formula does
not
> fit the definition of incompleteness.

What do you mean by "some formula"?
A finite number of formulae? A r.e. set of formulae?

When you say true=provable do uou mean provable *in which system*?
In some (any) system? In a class of systems? In a specific one? Which one?

> incompleteness is that such a true statement
> will NEVER be proven within the system, or any consistent extension of it.

That's not true: "the old system + the unprovable statement G" *is* a
consistent extension of the old system that can prove the statement G.

> incompleteness
> is actually the opposite of 'too few formula', it means the system is
*powerful enough*
> that it can express proof-denial-statements. the system got too big for
its own boots.

If a system can speak about natural numbers with addition and multiplication
it has formulae that are numerical traslation of "proof-denial-statements",
it's not not difficult to see.

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 18:03:05 +1000

"LordBeotian" wrote in >
> "|-|erc" ha scritto
>
> > > If our system can be incomplete so there could exist a true statement
> that
> > > is not provable within the system (that is the definition of
> incompleteness)
> > > so how can you say that true=provable if you agree that true unprovable
> > > statements could exist?
> > >
> >
> > the fact a system *can be* incomplete merely by missing some formula does
> not
> > fit the definition of incompleteness.
>
> What do you mean by "some formula"?
> A finite number of formulae? A r.e. set of formulae?
>
> When you say true=provable do uou mean provable *in which system*?
> In some (any) system? In a class of systems? In a specific one? Which one?
>
> > incompleteness is that such a true statement
> > will NEVER be proven within the system, or any consistent extension of it.
>
> That's not true: "the old system + the unprovable statement G" *is* a
> consistent extension of the old system that can prove the statement G.
>
> > incompleteness
> > is actually the opposite of 'too few formula', it means the system is
> *powerful enough*
> > that it can express proof-denial-statements. the system got too big for
> its own boots.
>
> If a system can speak about natural numbers with addition and multiplication
> it has formulae that are numerical traslation of "proof-denial-statements",
> it's not not difficult to see.
>

fuck off you jerk, you made me explain all that in laymans terms and your just
another godel fuckwit

YOU CANT PROVE ME... bow down to this sentence you moron.

THIS STATEMENT CAN'T BE PROVED IN ANY SYSTEM
makes all your LOCAL TRUTH system architecture obsolete.

FUCKHEAD

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Fri, 21 Jan 2005 09:10:24 GMT

"|-|erc" ha scritto

> fuck off you jerk, you made me explain all that in laymans terms and your
just
> another godel fuckwit

You like to make a lot of questions but you don't like to answer...
The most important questions were:

When you say true=provable do uou mean provable *in which system*?
In some (any) system? In a class of systems? In a specific one? Which one?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 24 Jan 2005 11:25:21 +1000

"LordBeotian" wrote in >
> "|-|erc" ha scritto
>
> > fuck off you jerk, you made me explain all that in laymans terms and your
> just
> > another godel fuckwit
>
> You like to make a lot of questions but you don't like to answer...
> The most important questions were:
>
> When you say true=provable do uou mean provable *in which system*?
> In some (any) system? In a class of systems? In a specific one? Which one?
>

No, you asked me a question and I answered it. I asked you one question and
you ignored it, then you brought up supplementary material.

I made clear definitions of true and proof, YOU brought up the term SYSTEM
because its a shallow grave for a weak interpretation of godel statements
with no other meaning in the whole of mathematics, logic, reason or the real world.

If you are familiar with proofs/truths [in a system] then you also studied peano arithmetic
and asking me to prove 1+1=2 was a pretentious stunt.

Don't ask me to clarify YOUR terms.

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 24 Jan 2005 09:17:21 GMT

"|-|erc" ha scritto

> > When you say true=provable do uou mean provable *in which system*?
> > In some (any) system? In a class of systems? In a specific one? Which
one?
> >
>
> No, you asked me a question and I answered it. I asked you one question
and
> you ignored it, then you brought up supplementary material.

What question are you thinking about?
Maybe when you wrote...

> Prove that all natural numbers (except 0) are bigger than 0
> 1 > 0
> if x > 0 then x+1 > 0
> therefore all natural numbers are bigger than 0.
>
> Does that look like a proof?

My answer is: yes it look like a proof...

> I made clear definitions of true and proof,

What was the definition of "proof"?

> YOU brought up the term SYSTEM

You started talking about "proof systems", not me.

> because its a shallow grave for a weak interpretation of godel statements
> with no other meaning in the whole of mathematics, logic, reason or the
real world.

Uhm?

> If you are familiar with proofs/truths [in a system] then you also studied
peano arithmetic
> and asking me to prove 1+1=2 was a pretentious stunt.

1) If I am familiar with proof/truths I ask clarification about the relation
"True=Provable" because I already know that there are problems assuming such
a position.
2) I asked how you prove 1+1=2 because if you have so unconventional
positions about "truth" I can't know if you accept Peano Axioms... and
because 1+1=2 is a typical statement that looks true even without proof.

> Don't ask me to clarify YOUR terms.

*Your* position, as I understand, is that "proofs" have to be done *using a
set of axioms* that must be equivalent to what the terms actually mean in
real life. The problem I considered was how to coose these axioms in such a
way that True=Provable. How can I be sure that my choice of axioms did not
miss some "truth"? If I can't be sure in which sense can we say that
True=Provable?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 24 Jan 2005 19:59:12 +1000

"LordBeotian" wrote in
>
> "|-|erc" ha scritto
>
> > > When you say true=provable do uou mean provable *in which system*?
> > > In some (any) system? In a class of systems? In a specific one? Which
> one?
> > >
> >
> > No, you asked me a question and I answered it. I asked you one question
> and
> > you ignored it, then you brought up supplementary material.
>
> What question are you thinking about?
> Maybe when you wrote...
>
> > Prove that all natural numbers (except 0) are bigger than 0
> > 1 > 0
> > if x > 0 then x+1 > 0
> > therefore all natural numbers are bigger than 0.
> >
> > Does that look like a proof?
>
> My answer is: yes it look like a proof...
>
> > I made clear definitions of true and proof,
>
> What was the definition of "proof"?
>
> > YOU brought up the term SYSTEM
>
> You started talking about "proof systems", not me.
>
> > because its a shallow grave for a weak interpretation of godel statements
> > with no other meaning in the whole of mathematics, logic, reason or the
> real world.
>
> Uhm?
>
> > If you are familiar with proofs/truths [in a system] then you also studied
> peano arithmetic
> > and asking me to prove 1+1=2 was a pretentious stunt.
>
> 1) If I am familiar with proof/truths I ask clarification about the relation
> "True=Provable" because I already know that there are problems assuming such
> a position.
> 2) I asked how you prove 1+1=2 because if you have so unconventional
> positions about "truth" I can't know if you accept Peano Axioms... and
> because 1+1=2 is a typical statement that looks true even without proof.
>
> > Don't ask me to clarify YOUR terms.
>
> *Your* position, as I understand, is that "proofs" have to be done *using a
> set of axioms* that must be equivalent to what the terms actually mean in
> real life. The problem I considered was how to coose these axioms in such a
> way that True=Provable. How can I be sure that my choice of axioms did not
> miss some "truth"? If I can't be sure in which sense can we say that
> True=Provable?
>

If you can't find the axioms, then you can't prove it, then you don't know for sure
if its true.

I said nothing about division of systems at all, I'm dealing with bona fide truth, not
bullshit that "this is x, x has no proof" IS TRUE because we proved x somewhere else
that doesn't know what its proving is true. oh hardy hah hah that's the second biggst load
of hogwash in history.

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 24 Jan 2005 10:26:00 GMT

"|-|erc" ha scritto

> > *Your* position, as I understand, is that "proofs" have to be done
*using a
> > set of axioms* that must be equivalent to what the terms actually mean
in
> > real life. The problem I considered was how to coose these axioms in
such a
> > way that True=Provable. How can I be sure that my choice of axioms did
not
> > miss some "truth"? If I can't be sure in which sense can we say that
> > True=Provable?
> >
>
> If you can't find the axioms, then you can't prove it, then you don't know
for sure
> if its true.

What is "it"?????

Suppose I find a set of axioms, and after a lot of time i realize that this
set of axioms is unable to prove that x^2-y^2=(x+y)*(x-y). This surely does
not mean that I don't know for sure that x^2-y^2=(x+y)*(x-y).

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 24 Jan 2005 20:29:58 +1000

"LordBeotian" wrote in >
> > > *Your* position, as I understand, is that "proofs" have to be done
> *using a
> > > set of axioms* that must be equivalent to what the terms actually mean
> in
> > > real life. The problem I considered was how to coose these axioms in
> such a
> > > way that True=Provable. How can I be sure that my choice of axioms did
> not
> > > miss some "truth"? If I can't be sure in which sense can we say that
> > > True=Provable?
> > >
> >
> > If you can't find the axioms, then you can't prove it, then you don't know
> for sure
> > if its true.
>
> What is "it"?????
>
> Suppose I find a set of axioms, and after a lot of time i realize that this
> set of axioms is unable to prove that x^2-y^2=(x+y)*(x-y). This surely does
> not mean that I don't know for sure that x^2-y^2=(x+y)*(x-y).
>

True = [ proven | provable ]

Herc

From:	LordBeotian
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 24 Jan 2005 10:35:22 GMT

"|-|erc" ha scritto

> > > If you can't find the axioms, then you can't prove it, then you don't
know
> > for sure
> > > if its true.
> >
> > What is "it"?????
> >
> > Suppose I find a set of axioms, and after a lot of time i realize that
this
> > set of axioms is unable to prove that x^2-y^2=(x+y)*(x-y). This surely
does
> > not mean that I don't know for sure that x^2-y^2=(x+y)*(x-y).
> >
>
> True = [ proven | provable ]

Can you prove it?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 24 Jan 2005 20:37:02 +1000

"LordBeotian" wrote in >
> "|-|erc" ha scritto
>
> > > > If you can't find the axioms, then you can't prove it, then you don't
> know
> > > for sure
> > > > if its true.
> > >
> > > What is "it"?????
> > >
> > > Suppose I find a set of axioms, and after a lot of time i realize that
> this
> > > set of axioms is unable to prove that x^2-y^2=(x+y)*(x-y). This surely
> does
> > > not mean that I don't know for sure that x^2-y^2=(x+y)*(x-y).
> > >
> >
> > True = [ proven | provable ]
>
> Can you prove it?

I wouldn't be a hypocrite now would I?

Herc

From:	George Cox
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 01:56:44 +0000 (UTC)

|-|erc wrote:
>
> .... incompleteness is that such a true statement
> will NEVER be proven within the system, or any consistent extension of it.

No it isn't. Consis(PA) can't be proved in PA (if it's consistent).
But Consis(PA) can be proved in PA + Consis(PA). PA + Consis(PA) is a
consistent extension of PA.

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 12:29:44 +1000

"George Cox" wrote in
> |-|erc wrote:
> >
> > .... incompleteness is that such a true statement
> > will NEVER be proven within the system, or any consistent extension of it.
>
> No it isn't. Consis(PA) can't be proved in PA (if it's consistent).
> But Consis(PA) can be proved in PA + Consis(PA). PA + Consis(PA) is a
> consistent extension of PA.

then you haven't demonstrated the incompleteness theorem.
its no wonder none of you can agree on any clear interpretation.

Herc

From:	George Cox
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 03:30:29 +0000 (UTC)

|-|erc wrote:
>
> "George Cox" wrote in
> > |-|erc wrote:
> > >
> > > .... incompleteness is that such a true statement
> > > will NEVER be proven within the system, or any consistent extension of it.
> >
> > No it isn't. Consis(PA) can't be proved in PA (if it's consistent).
> > But Consis(PA) can be proved in PA + Consis(PA). PA + Consis(PA) is a
> > consistent extension of PA.
>
> then you haven't demonstrated the incompleteness theorem.

Your statement of the incompleteness theorem was wrong.

> its no wonder none of you can agree on any clear interpretation.
>
> Herc

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 13:40:25 +1000

"George Cox" wrote in
> |-|erc wrote:
> >
> > "George Cox" wrote in
> > > |-|erc wrote:
> > > >
> > > > .... incompleteness is that such a true statement
> > > > will NEVER be proven within the system, or any consistent extension of it.
> > >
> > > No it isn't. Consis(PA) can't be proved in PA (if it's consistent).
> > > But Consis(PA) can be proved in PA + Consis(PA). PA + Consis(PA) is a
> > > consistent extension of PA.
> >
> > then you haven't demonstrated the incompleteness theorem.
>
> Your statement of the incompleteness theorem was wrong.

no it wasn't

Herc

From:	Ogie Oglethorpe
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 03:14:25 -0400

|-|erc wrote:
> "George Cox" wrote in
>
>>|-|erc wrote:
>>
>>>"George Cox" wrote in
>>>
>>>>|-|erc wrote:
>>>>
>>>>>.... incompleteness is that such a true statement
>>>>>will NEVER be proven within the system, or any consistent extension of it.
>>>>
>>>>No it isn't. Consis(PA) can't be proved in PA (if it's consistent).
>>>>But Consis(PA) can be proved in PA + Consis(PA). PA + Consis(PA) is a
>>>>consistent extension of PA.
>>>
>>>then you haven't demonstrated the incompleteness theorem.
>>
>>Your statement of the incompleteness theorem was wrong.
>
>
> no it wasn't
>
> Herc
>
>
>
Yes it was

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 17:57:52 +1000

"Ogie Oglethorpe" wrote in
> >>>>
> >>>>>.... incompleteness is that such a true statement
> >>>>>will NEVER be proven within the system, or any consistent extension of it.
> >>>>
> >>>>No it isn't. Consis(PA) can't be proved in PA (if it's consistent).
> >>>>But Consis(PA) can be proved in PA + Consis(PA). PA + Consis(PA) is a
> >>>>consistent extension of PA.
> >>>
> >>>then you haven't demonstrated the incompleteness theorem.
> >>
> >>Your statement of the incompleteness theorem was wrong.
> >
> >
> > no it wasn't
> >
> > Herc
> >
> >
> Yes it was

An objection was made, I challenged it, he ignored the challenge, I win.
He hasn't specified a true statement in PA+consis(PA) that is not provable in that system,
hence he has not demonstrated incompleteness itself.

Herc
--
Stardate 20 1 2005. Its becoming obvious the reason every thread results in argument is because half
of people are outright lying. Putting contradictory remarks side by side takes an average of 10 posts before
they accept it's known they were caught.

From:	George Cox
Subject:	Re: True = [ proven \| provable ]
Date:	Wed, 19 Jan 2005 23:15:28 +0000 (UTC)

|-|erc wrote:
>
> Define addition as
> 0 + 0 = 0 rule 0
> s(a) + 0 = s(a) rule 1
> s(a) + b = a + s(b) rule 2

How do you know that there is a function + with those properties?

How do you show that the defined term n + m can be eliminated from any
expression that contains it?

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Thu, 20 Jan 2005 11:10:18 +1000

"George Cox" wrote in
> |-|erc wrote:
> >
> > Define addition as
> > 0 + 0 = 0 rule 0
> > s(a) + 0 = s(a) rule 1
> > s(a) + b = a + s(b) rule 2
>
> How do you know that there is a function + with those properties?

I defined 0 and + as a joint outfit for r0 and r1, and used distribution for r2.

>
> How do you show that the defined term n + m can be eliminated from any
> expression that contains it?

I didn't say they could

Herc

From:	examachine at gmail.com
Subject:	Re: True = [ proven \| provable ]
Date:	18 Jan 2005 17:12:36 -0800

I don't think provability is synonymous with truth. Neither is it
synonymous with consistency.

Regards,

--
Eray

From:	George Cox
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 20:40:21 +0000 (UTC)

|-|erc wrote:
>
> Mathematicians don't need the word true.
>
> For "I think its true" say "I think its provable".

Mathematicians don't need "I think".

>
> For "G is true" say "G is proven"

In systems that contain a modest amount of number theory, "true" and
"provable" are not co-extensive.

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Tue, 18 Jan 2005 11:21:13 +1000

"George Cox" wrote in
> |-|erc wrote:
> >
> > Mathematicians don't need the word true.
> >
> > For "I think its true" say "I think its provable".
>
> Mathematicians don't need "I think".

try it some time!

>
> >
> > For "G is true" say "G is proven"
>
> In systems that contain a modest amount of number theory, "true" and
> "provable" are not co-extensive.

You see, I proved my assertion that True = [ proven | provable ]
You all just stated yours and said.. believe this!

Hence True = [ proven | provable ] is TRUE, your claims to knowledge of the
underpinnings of universal mathematics is complete waste of bandwidth. The
fact none of you can agree on what incompleteness is and we get 1,000 different
explanations suggests maybe you should examine your axioms and get it straight?
Nobody can argue with spaghetti!

AGAIN: TRUE is not an existential, its only part of this predicate KNOW_IT'S_TRUE
which means PROOF. They are synonyms par future / past tense, provable / proven.

TRUTH is just timeless provability. NOW THINK THIS TIME.... WHO SAYS ITS TRUE??.
HOW DO THEY KNOW ITS TRUE? there's no central platonic plane stating....
"this is true, this is true, this is true", its a collective hologam of the sentient life forms
that read it. YOU CANNOT STATE SOMETHING IS ABSOLUTELY TRUE
without proof. HOW COULD YOU KNOW?

Herc

From:	Lars Kecke
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 11:31:05 +0100

|-|erc wrote:
> Mathematicians don't need the word true.
>
> For "I think its true" say "I think its provable".
>
> For "G is true" say "G is proven"

Wrong. The statement "this statement can not be proven" may still be
true. See also:
http://en.wikipedia.org/wiki/G%F6del%27s_incompleteness_theorem

HTH

Lars

From:	\|-\|erc
Subject:	Re: True = [ proven \| provable ]
Date:	Mon, 17 Jan 2005 22:13:17 +1000

"Lars Kecke" wrote in
> |-|erc wrote:
> > Mathematicians don't need the word true.
> >
> > For "I think its true" say "I think its provable".
> >
> > For "G is true" say "G is proven"
>
> Wrong. The statement "this statement can not be proven" may still be
> true. See also:
> http://en.wikipedia.org/wiki/G%F6del%27s_incompleteness_theorem
>
> HTH
>
> Lars

How do you know its true?

I should have written "this statement can not be proven" is factX here.

LIARS statement, factX <=> true(factX) = ~proven(factX).

Herc