Modular Arithmetic Is Consistent
RussellE
Modular Arithmetic (MA) must be
omega inconsistent.
These are the axioms of MA:
Ex (S(x)=0)
Ax ((S(x)=S(y)) -> (x=y))
Ax (x+0=x)
AxAy (x+S(y)=S(x+y))
Ax (x*0=0)
AxAy (x*S(y)=x*y+x)
First Order Induction
Ay (P(0,y) and Ax (P(x,y) -> P(S(x),y))) -> Ax P(x)
where y is a finite number of free variables
These are the standard axioms of
first order Peano Arithmetic (PA)
except the first axiom has been negated.
http://en.wikipedia.org/wiki/Peano_axioms#First-order_theory_of_arithmetic
I am also including an omega rule.
Omega Rule
P(0)
P(0) and P(S(0))
P(0) and P(S(0)) and P(S(S(0)))
,,,
If all such sentences are true then Ax P(x).
The omega rule is an informal form of induction.
http://en.wikipedia.org/wiki/%CE%A9-consistent_theory
Unlike PA, it is easy to prove MA is consistent.
MA has finite models based on modular arithmetic.
http://en.wikipedia.org/wiki/Modular_arithmetic
This is a model of MA with one individual:
U: {0}
S(0) = 0
0+0 = 0
0*0 = 0
In a finite model, both induction and the omega rule
can be replaced with a single axiom of the form:
P(0) and P(S(0)) and ... and P(Sn(0))
-> Ax P(x)
where Sn(0) is some closed term in MA.
We can prove all finite models of MA are
indeed finite. Consider the following
theorems of MA and PA:
( (0=0) and Ax(x=0 -> S(x)=0) ) -> Ax(x=0)
(0=0 or 0=S(0)) and
Ax(x=0 or x=S(0) -> S(x)=0 or S(x)=S(0))
-> Ax(x=0 or x=S(0))
(0=0 or 0=S(0) or S(S(0))) and
Ax(x=0 or x=S(0) or x=S(S(0))
-> S(x)=0 or S(x)=S(0)) or S(x)=S(S(0)) )
-> Ax(x=0 or x=S(0) or x=S(S(0)))
...
We can create one of the theorems above
for any closed term in the language of MA.
The first theorem is for the term 0, the
second theorem is for S(0), etc.
These theorems are true in any model
of MA or PA. They show if S(z)=0, and z
can be replaced with a closed term, then
the model is finite.
Assume MA has an infinite model.
Assume there is an e, and S(e)=0.
If e is a closed term, the model
is finite. So, e must be is a
non-standard natural number.
Because e is not a closed term in
the language of MA, we can prove
Ax(S(x) != 0) using the omega rule.
This contradicts Ex (S(x)=0).
This proves any infinite model of MA
must be omega inconsistent.
Most definitions of induction schema
use substitution. We can substitute
terms in the language for P(x) in our
induction axioms.
http://plato.stanford.edu/entries/schema/#2
If we only allow x to be substituted
by closed terms in the language of MA
then we can prove Ax(S(x) != 0) using
first order induction (assuming we
have an infinite model).
If we allow x to be substituted by
individuals in the model then we
can prove a "counter-example" to
Ax(S(x) != 0), namely S(e) = 0.
Notice that a counter-example in the
model to Ax(S(x)=0) assumes the
"model" is consistent.
Also, the "witness" to the counter-example,
ie, S(e)=0, can not be written as a wff in
the language of MA. e is not a closed term.
Not only is any infinite model of MA
omega inconsistent, any infinite "model"
is simply inconsistent if we restrict
substitution in our induction axioms
to closed terms in the language.
MA is consistent because it has finite
models and no infinite models.
Russell
- Integers are an illusion
Virgil
<f4c51683-1e9d-4e0f...@h12g2000pro.googlegroups.com>,
RussellE <reas...@gmail.com> wrote:
> I show why any infinite model of
> Modular Arithmetic (MA) must be
> omega inconsistent.
But non-modular arithmetic of the integers being consistent is what
validates all modular arithmetic.
--
Frederick Williams
>
> I show why any infinite model of
> Modular Arithmetic (MA) must be
> omega inconsistent.
What does it mean to say of a model that it is omega inconsistent?
> We can prove all finite models of MA are
> indeed finite.
No doubt we can.
> [...]
>
> This proves any infinite model of MA
> must be omega inconsistent.
>
> [...]
--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting
Frederick Williams
>
> [...]
>
> I am also including an omega rule.
So you're MA is no longer a first order theory.
>
> where Sn(0) is some closed term in MA.
I suspect that you mean that it's S(S(...S(0)...)) with n S's.
Frederick Williams
>
> RussellE wrote:
> >
> > [...]
> >
> > I am also including an omega rule.
>
> So you're MA is no longer a first order theory.
The reason I say that is that what you know about first order theories
may no longer be true when the omega rule is used.
RussellE
<freddywilli...@btinternet.com> wrote:
> RussellE wrote:
>
> > I am also including an omega rule.
>
> So you're MA is no longer a first order theory.
I am sure I see why this is true.
Normally, the omega rule is given as:
P(0) and P(1) and P(2) ... -> Ax P(x)
This would be an infinitely long statement
and would not be allowed in a first order theory.
My omega rule is a schema for wff in the
language of MA. All of these sentences
are allowed in a first order theory.
P(0)
P(0) and P(S(0))
...
If we can prove Ax P(x) using first order
induction then we can prove Ax P(x) with
the omega rule. The converse in not true.
I have shown how certain statements
proven with the omega rule can't be
proven with first order induction.
Why do we assume the omega rule
is not first order and the induction
schema is first order?
> > where Sn(0) is some closed term in MA.
>
> I suspect that you mean that it's S(S(...S(0)...)) with n S's.
Yes.
S(0) + S(0) would also be a closed term.
I think I can show any closed term can be replaced
with S(S(...S(0)...)) with n S's.
It seems to me first order induction allows us
to include infinitely long sentences into the language.
Assume we have a model of PA and e is a
non-standard natural number.
We can't rule out the possibility our model
has a non-standard natural number.
If we prove Ax (S(x) != 0) using first order induction
we are saying S(e) != 0. My problem is e can't
be expressed with a finite string in the language
of PA. S(e) != 0 would be an infinitely long
sentence if we substitute e for x.
Any model of MA should be able to
substitute some closed term for x
in the axiom Ex (S(x)=0).
The model I descibed with U = {0}
would satisfy this axiom with:
S(0) = 0
Virgil
<97c6470b-e510-4947...@j14g2000prn.googlegroups.com>,
RussellE <reas...@gmail.com> wrote:
The consistency of modular arithmetic follows from the consistency
integer arithmetic.
--
Jesse F. Hughes
> On Jun 29, 10:18 am, Frederick Williams
> <freddywilli...@btinternet.com> wrote:
>> RussellE wrote:
>>
>> > I am also including an omega rule.
>>
>> So you're MA is no longer a first order theory.
>
> I am sure I see why this is true.
> Normally, the omega rule is given as:
>
> P(0) and P(1) and P(2) ... -> Ax P(x)
>
> This would be an infinitely long statement
> and would not be allowed in a first order theory.
>
> My omega rule is a schema for wff in the
> language of MA. All of these sentences
> are allowed in a first order theory.
>
> P(0)
> P(0) and P(S(0))
> ...
>
> If we can prove Ax P(x) using first order
> induction then we can prove Ax P(x) with
> the omega rule. The converse in not true.
> I have shown how certain statements
> proven with the omega rule can't be
> proven with first order induction.
No, this amounts to an inference rule with infinitely many premises.
That's no less dubious than an infinitely long statement. (For one
thing, to apply this rule in a proof, the proof must have infinitely
many statements! You cannot apply your "rule" in a finite proof.)
Maybe there are logics where such an inference rule is sensible, but it
sure ain't FOL[1].
Since the theorem you're working hard to deny (Löwenheim-Skolem) is a
theorem about FOL, this approach isn't doing you any good.
> Why do we assume the omega rule
> is not first order and the induction
> schema is first order?
The induction scheme is an infinite set of axioms. It's not at all the
same as an inference rule which takes an infinite set of premises.
Footnotes:
[1] For the record, I don't know about such logics, but perhaps someone
has studied such oddball inference rules.
--
"Reality has a fascinating ability to check us when we get a little too
big for our britches... Make no mistake. There isn't a mathematician alive
today that I can't now touch, and not a mathematical career on the planet
that I can't now affect." --James Harris, render of worlds
RussellE
OK. The omega rule is a proof with an infinite
number of steps.
> Maybe there are logics where such an inference rule is sensible, but it
> sure ain't FOL[1].
>
> Since the theorem you're working hard to deny (Löwenheim-Skolem) is a
> theorem about FOL, this approach isn't doing you any good.
>
> > Why do we assume the omega rule
> > is not first order and the induction
> > schema is first order?
>
> The induction scheme is an infinite set of axioms. It's not at all the
> same as an inference rule which takes an infinite set of premises.
How is induction different from the omega rule?
When we say Ax P(x) don't we mean all these
sentences are true?
P(0)
P(S(0))
P(S(S(0)))
...
Most of the descriptions I have seen
of induction schemas talk about substitution.
If we can substitute x in P(x) with a
non-standard natural number then
we are allowing infinitely long sentences
to be part of the language.
How is this different from allowing an
infinitely long proof?
I know one method is to assign a
constant to each individual in the model.
This allows the sentence P(x) to be finite
for any x, but we have added an
infinite number of constants to the
language.
Russell
- 2 many 2 count
Frederick Williams
>
> Why do we assume the omega rule
> is not first order and the induction
> schema is first order?
Because of the way first order theory is defined.
> It seems to me first order induction allows us
> to include infinitely long sentences into the language.
That's just nonsense. There are languages with infinitely long
sentences, but they are not first order languages. The presence of
induction has got nothing to do with it. How can it? Languages are
defined before axioms are mentioned, and that obviously must be the
case.
Frederick Williams
>
> OK. The omega rule is a proof with an infinite
> number of steps.
No it isn't. It's a rule with an infinite number of premisses
> How is induction different from the omega rule?
Just look up the definitions.
> Most of the descriptions I have seen
> of induction schemas talk about substitution.
> If we can substitute x in P(x) with a
> non-standard natural number
Which we can't. It is terms (such as numerals) that get substituted for
variables (with provisos to stop variable clashes).
> then
> we are allowing infinitely long sentences
> to be part of the language.
How does that follow?
Jesse F. Hughes
No, not "steps".
An infinite number of premises.
Any proof which applies the rule is thus necessarily infinitely long.
>> Maybe there are logics where such an inference rule is sensible, but it
>> sure ain't FOL[1].
>>
>> Since the theorem you're working hard to deny (Löwenheim-Skolem) is a
>> theorem about FOL, this approach isn't doing you any good.
>>
>> > Why do we assume the omega rule
>> > is not first order and the induction
>> > schema is first order?
>>
>> The induction scheme is an infinite set of axioms. It's not at all the
>> same as an inference rule which takes an infinite set of premises.
>
> How is induction different from the omega rule?
We do not require an infinite number of steps to prove a universal
statement.
Haven't you seen such a proof?
> When we say Ax P(x) don't we mean all these
> sentences are true?
>
> P(0)
> P(S(0))
> P(S(S(0)))
> ...
It entails that, yes, and more. It entails that *every* object in the
domain satisfies P, not just that the closed terms do.
But so what? The simple sentence P also entails infinitely many
propositions, include P & P, P & P & P and so on.
The fact that (Ax)Px entails infinitely many other propositions is
irrelevvant.
> Most of the descriptions I have seen
> of induction schemas talk about substitution.
> If we can substitute x in P(x) with a
> non-standard natural number then
> we are allowing infinitely long sentences
> to be part of the language.
This is, of course, nonsense. You don't substitute semantic entities
into syntactic formula.
Could you give me a reference for some of these very strange
presentations?
> How is this different from allowing an
> infinitely long proof?
You have simply misunderstood the meaning of the universal statement.
You think that it involves substituting infinitary objects into
formulas, but it doesn't.
> I know one method is to assign a
> constant to each individual in the model.
> This allows the sentence P(x) to be finite
> for any x, but we have added an
> infinite number of constants to the
> language.
No one said that the set of constants for a language must be finite, you
know.
--
Jesse F. Hughes
"So far as this negative attitude toward life is concerned, Buddhism
is merely Taoism a little touched in its wits."
-- Lin Yutang, /My Country and My People/
RussellE
wrote:
> RussellE wrote:
>
> > Most of the descriptions I have seen
> > of induction schemas talk about substitution.
> > If we can substitute x in P(x) with a
> > non-standard natural number
>
> Which we can't.
We already know if MA has an infinite model,
Ex (S(x)=0) must be satisfied by a non-standard
natural number. Call this non-standard natural e.
If we can't substitute e into S(x)=0 then how
can we say e satisfies the axiom?
> It is terms (such as numerals) that get substituted for
> variables (with provisos to stop variable clashes).
S(0) is a term in the language of MA. 42 is not.
RussellE
> RussellE <reaste...@gmail.com> writes:
>
> > Most of the descriptions I have seen
> > of induction schemas talk about substitution.
> > If we can substitute x in P(x) with a
> > non-standard natural number then
> > we are allowing infinitely long sentences
> > to be part of the language.
>
> This is, of course, nonsense. You don't substitute semantic entities
> into syntactic formula.
>
> Could you give me a reference for some of these very strange
> presentations?
http://plato.stanford.edu/entries/schema/#2
Notice the article talks about Num(x) when
describing first order induction.
> > How is this different from allowing an
> > infinitely long proof?
>
> You have simply misunderstood the meaning of the universal statement.
> You think that it involves substituting infinitary objects into
> formulas, but it doesn't.
It does if the object can not be represented by
a closed term in the language.
> > I know one method is to assign a
> > constant to each individual in the model.
> > This allows the sentence P(x) to be finite
> > for any x, but we have added an
> > infinite number of constants to the
> > language.
>
> No one said that the set of constants for a language must be finite, you
> know.
Nothing says a theory has to have an
infinite number of constants, either.
I defined MA to have one constant: 0.
The language of MA is 0, S, +, and *.
Every model of MA has a closed
term that satisfies Ex (S(x)=0).
If U={0}, there is a model where S(0)=0.
if U={0,1}, there is a model where S(S(0))=0.
etc
1treePetrifiedForestLane
RussellE
wrote:
> > The consistency of modular arithmetic follows from the consistency
> > integer arithmetic (base-one .-)
Obviously.
Any model proves the theory is consistent.
We are just quibbly over how many models it has.
Frederick Williams
> We already know if MA has an infinite model,
> Ex (S(x)=0) must be satisfied by a non-standard
> natural number. Call this non-standard natural e.
> If we can't substitute e into S(x)=0 then how
> can we say e satisfies the axiom?
You're confusing numbers and numerals.
Frederick Williams
> Every model of MA has a closed
> term
Models don't have terms. Languages have terms.
> that satisfies Ex (S(x)=0).
You might mean "satisfies S(x)=0".
Jesse F. Hughes
> On Jul 2, 5:08 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>>
>> > Most of the descriptions I have seen
>> > of induction schemas talk about substitution.
>> > If we can substitute x in P(x) with a
>> > non-standard natural number then
>> > we are allowing infinitely long sentences
>> > to be part of the language.
>>
>> This is, of course, nonsense. You don't substitute semantic entities
>> into syntactic formula.
>>
>> Could you give me a reference for some of these very strange
>> presentations?
>
> http://plato.stanford.edu/entries/schema/#2
> Notice the article talks about Num(x) when
> describing first order induction.
Yes. So? That's a fairly common formulation of the principle of
induction (though not one used in ZFC typically). I'm afraid I don't
see your point.
It does not discuss substituting "non-standard natural numbers" for x,
nor does it say that this formula is true iff it is true for each
substitution of a closed term for x.
I don't see that this passage explains your claims at all. Perhaps you
could be more explicit.
>> > How is this different from allowing an
>> > infinitely long proof?
>>
>> You have simply misunderstood the meaning of the universal statement.
>> You think that it involves substituting infinitary objects into
>> formulas, but it doesn't.
>
> It does if the object can not be represented by
> a closed term in the language.
No, that's not so.
Here's one way to explain what it means to say (Ax)Px is true: As you
know, a predicate P is interpreted as a subset [[P]] of the carrier U of
the model. (Ax)Px is true iff [[P]] = U.
In other words, just in case every element of the carrier satisfies P.
There is nothing there about substituting objects into formulas. It's
about the interpretation of the predicate in the carrier of the model.
>> > I know one method is to assign a
>> > constant to each individual in the model.
>> > This allows the sentence P(x) to be finite
>> > for any x, but we have added an
>> > infinite number of constants to the
>> > language.
>>
>> No one said that the set of constants for a language must be finite, you
>> know.
>
> Nothing says a theory has to have an
> infinite number of constants, either.
> I defined MA to have one constant: 0.
> The language of MA is 0, S, +, and *.
> Every model of MA has a closed
> term that satisfies Ex (S(x)=0).
No. You know that for every model, there is an element u of the carrier
such that [[S]](u) = [[0]], that is, that the interpretation of S maps u
to the interpretation of 0.
You cannot claim that every model has a closed term satisfying S(x) =
0. That is not what your axiom says.
>
> If U={0}, there is a model where S(0)=0.
> if U={0,1}, there is a model where S(S(0))=0.
> etc
Oh, I agree in the case of all the finite models. So what? So then
magically also for any infinite models?
That's no argument.
--
But in our enthusiasm, we could not resist a radical overhaul of the
system, in which all of its major weaknesses have been exposed,
analyzed, and replaced with new weaknesses.
-- Bruce Leverett (presumably with apologies to Ambrose Bierce)
Marshall
>
> Oh, I agree in the case of all the finite models. So what? So then
> magically also for any infinite models?
>
> That's no argument.
Heh. It's the same as a bunch of WM's "arguments" go, isn't it?
"Here's all the finite cases, so magically also for any infinite
case."
Actually, it could potentially be a crank axiom.
Marshall
RussellE
wrote:
> RussellE wrote:
> > Every model of MA has a closed
> > term
>
> Models don't have terms. Languages have terms.
For every model of MA there is a closed term in
the language that satisfies S(x)=0.
> > that satisfies Ex (S(x)=0).
>
> You might mean "satisfies S(x)=0".
Yes
RussellE
wrote:
> RussellE wrote:
> > We already know if MA has an infinite model,
> > Ex (S(x)=0) must be satisfied by a non-standard
> > natural number. Call this non-standard natural e.
> > If we can't substitute e into S(x)=0 then how
> > can we say e satisfies the axiom?
>
> You're confusing numbers and numerals.
Probably. Please explain the difference.
I assume we can substitute numerals for x.
Do non-standard natural numbers have numerals?
Transfer Principle
And this thread has now become an insult thread.
Of course, this does show one key difference between the posters
who get these insults and those who give these insults. The
insulted posters tend to find it desirable for the properties of
finite objects to transfer to infinite objects, but the insulters
don't find such a transfer desirable.
Thus, in the case of the aforementioned poster WM and the OP of
this thread RE, both state that since they can show that objects
like the actual infinite set N of all natural numbers or an
infinite model of MA must have some property not inherited from
FISON's or the finite models of MA, they conclude that the
respective infinite objects (N in the case of WM, the infinite
model of MA in the case of RE) don't exist. And then Hughes and
Spight start insulting such posters.
Those who hold the view that infinite sets should inherit most of
the properties of finite sets (including both those who become
finitist if they can't find an infinite set with such properties,
and those who seek to invent new infinite sets which do inherit
such properties) should be able to do so without receiving the
insults that they regularly do. Likewise, those who hold the
standard view that properties of finite sets need not transfer to
infinite sets should be able to do so without insults. But in
this thread, the usual insults come from the usual posters.
Yes, Google -- at least for now -- appears to be working again,
and so here I am to break up these one-sided insult threads.
Transfer Principle
wrote:
> RussellE wrote:
> > We already know if MA has an infinite model,
> > Ex (S(x)=0) must be satisfied by a non-standard
> > natural number. Call this non-standard natural e.
> > If we can't substitute e into S(x)=0 then how
> > can we say e satisfies the axiom?
> You're confusing numbers and numerals.
Let's see what Ed Nelson has to say re: numbers vs. numerals.
http://www.math.princeton.edu/~nelson/papers/faith.pdf
"A numeral is an expression of the
form 0, S0, SS0, and so forth, generically denoted by n. The mark 0
denotes the number zero, and if the numeral n denotes a certain number
then Sn denotes its successor, the next number after it. Thus the
numerals serve as names for the numbers."
OK. But what about RE's axiom "Ex (S(x)=0)"? Nelson writes:
"AvF is true in case for all numerals n, the formula F_v[n] is true."
Here Nelson uses the universal quantifier, but we can easily write
the negation of RE's axiom:
Ax (~Sx=0)
So F is the formula "~Sx=0" and 'v' is the variable 'x'. So now we
ask, is the case that for all _numerals_ n, the formula F_x[n]
must be true?
This appears to be the case. The fact that F_x[e] -- i.e., the
result of substituting 'e' for 'x' in the formula F -- happens to
be false is immaterial, since 'e' is _not_ a _numeral_ (i.e., of
the form S...S0), and Nelson only requires F_v[n] to be true for
all _numerals_ n, and e is not a _numeral_.
So the negation of RE's axiom is false, so the alleged infinite
model of MA is _not_ really a model of MA. And so, at least based
on Nelson's definition of "number" and "numeral," RE is correct.
In this thread, Virgil writes that the consistency of MA follows
from that of PA. But of course, we already know that Nelson is
skeptical about the consistency of PA.
Marshall
> On Jul 2, 12:20 pm, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > That's no argument.
> > Heh. It's the same as a bunch of WM's "arguments" go, isn't it?
> > "Here's all the finite cases, so magically also for any infinite
> > case."
> > Actually, it could potentially be a crank axiom.
>
> And this thread has now become an insult thread.
> [...]
> Yes, Google -- at least for now -- appears to be working again,
> and so here I am to break up these one-sided insult threads.
You won't be breaking up anything, you pompous buffoon.
Unless by "breaking up" the thread you mean interspersing
your posts in and among the real ones. Now run off and
count the letters in "pompous buffoon," there's a good boy.
Marshall
Jesse F. Hughes
> On Jul 2, 12:20 pm, Marshall <marshall.spi...@gmail.com> wrote:
>> On Jul 2, 11:40 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> > Oh, I agree in the case of all the finite models. So what? So then
>> > magically also for any infinite models?
>> > That's no argument.
>> Heh. It's the same as a bunch of WM's "arguments" go, isn't it?
>> "Here's all the finite cases, so magically also for any infinite
>> case."
>> Actually, it could potentially be a crank axiom.
>
> And this thread has now become an insult thread.
It has?
Great!
You're the greatest waste of oxygen I know, and this includes tire
fires, you tiresome pain in the ass.
Boy, it's great that it finally became an insult thread!
[...]
> Yes, Google -- at least for now -- appears to be working again,
> and so here I am to break up these one-sided insult threads.
Oh, happy day! Imagine everyone's relief.
--
"I have charlieboo.com registered and lots of ideas. We will have to
wait until people stop throwing lots of money at me for saving
people's lives and my Chinese Harvard researcher gals stop knocking."
-- Charlie-Boo Volkstorf: ain't that a man?
Jesse F. Hughes
> On Jul 2, 11:15 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
>> RussellE wrote:
>> > Every model of MA has a closed
>> > term
>>
>> Models don't have terms. Languages have terms.
>
> For every model of MA there is a closed term in
> the language that satisfies S(x)=0.
No, that's not true.
Why do you think it is true?
>
>> > that satisfies Ex (S(x)=0).
>>
>> You might mean "satisfies S(x)=0".
>
> Yes
--
Jesse F. Hughes
"It's not a very good combination: adolescence and existentialism."
-- Dragnet TV show (1970)
Transfer Principle
You won't be able to make me run off that easily!
By "breaking up," I mean that your four-against-one insult
parade is now four-against-two.
Joshua Cranmer
> You're the greatest waste of oxygen I know, and this includes tire
> fires, you tiresome pain in the ass.
Does that include oxidation of ferrous materials?
--
Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth
Transfer Principle
> RussellE <reaste...@gmail.com> writes:
> > the language that satisfies S(x)=0.
> No, that's not true.
> Why do you think it is true?
Because of what Ed Nelson says about this issue:
http://www.math.princeton.edu/~nelson/papers/faith.pdf
"AvF is true in case for all numerals n, the formula F_v[n] is true."
As the existential quantifier Ev is equivalent to ~Av~, Nelson's
statement is equivalent to:
EvF is true in case there exists a numeral n making F_v[n] true.
Now as Ex(Sx=0) is an axiom of MA, it must be true in every model of
MA (including any infinite models of MA). Let F be the formula "Sx=0"
and v be the variable x to obtain:
Ex(Sx=0) is true in case there exists a numeral n making Sx=0 true.
Now let's see what Nelson means by "numeral":
"A numeral is an expression of the
form 0, S0, SS0, and so forth, generically denoted by n."
So there must be an expression of the form S...S0 making Sx=0 true,
so based on Nelson's definitions, RE is right. QED
RussellE
> RussellE <reaste...@gmail.com> writes:
> > On Jul 2, 11:15 am, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> >> RussellE wrote:
> >> > Every model of MA has a closed
> >> > term
>
> >> Models don't have terms. Languages have terms.
>
> > For every model of MA there is a closed term in
> > the language that satisfies S(x)=0.
>
> No, that's not true.
>
> Why do you think it is true?
Because it is true.
Show me a model where it is false.
The integers are not a model of MA.
Jesse F. Hughes
> On 7/2/2011 6:07 PM, Jesse F. Hughes wrote:
>> You're the greatest waste of oxygen I know, and this includes tire
>> fires, you tiresome pain in the ass.
>
> Does that include oxidation of ferrous materials?
Certainly. In fact, most rusting heaps of garbage are rather better
conversationalists than TP. (Which, of course, only draws attention to
my ineptitude in choosing who to converse with.)
--
"I've noticed [...] I routinely have been putting up flawed equations
with my surrogate factoring work. My take on it is that I have some
deep fear that the work is too dangerous and am sabotaging myself."
-- James S. Harris
Jesse F. Hughes
> On Jul 2, 6:08 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > For every model of MA there is a closed term in
>> > the language that satisfies S(x)=0.
>> No, that's not true.
>> Why do you think it is true?
>
> Because of what Ed Nelson says about this issue:
>
> http://www.math.princeton.edu/~nelson/papers/faith.pdf
>
> "AvF is true in case for all numerals n, the formula F_v[n] is true."
>
> As the existential quantifier Ev is equivalent to ~Av~, Nelson's
> statement is equivalent to:
>
> EvF is true in case there exists a numeral n making F_v[n] true.
I can't speak for Nelson, but in general, it is not the case that (Ex)Px
is true only if there is a closed term t such that Pt.
This is a simple fact. I imagine Nelson is aware of it and that the
context makes his meaning clear (though I did not check). But
regardless of what Nelson says, the fact holds.
To take a simple example, consider any FOL= theory with no constants.
Then this theory proves (Ex)(x = x) and indeed (Ex)(x = x) is *true* in
every model of the theory, even though there are *no* closed terms at
all.
>
> Now as Ex(Sx=0) is an axiom of MA, it must be true in every model of
> MA (including any infinite models of MA). Let F be the formula "Sx=0"
> and v be the variable x to obtain:
>
> Ex(Sx=0) is true in case there exists a numeral n making Sx=0 true.
>
> Now let's see what Nelson means by "numeral":
>
> "A numeral is an expression of the
> form 0, S0, SS0, and so forth, generically denoted by n."
>
> So there must be an expression of the form S...S0 making Sx=0 true,
> so based on Nelson's definitions, RE is right. QED
No, he's not, regardless of what Nelson says.
That's simply not how logic works.
--
Jesse F. Hughes
"Maybe I screwed up on one of my assumptions [...]. Otherwise, um,
it's very easy to factor, and things are about to get really, really
weird." -- James S. Harris
Frederick Williams
>
> On Jul 2, 11:11 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > RussellE wrote:
> > > We already know if MA has an infinite model,
> > > Ex (S(x)=0) must be satisfied by a non-standard
> > > natural number. Call this non-standard natural e.
> > > If we can't substitute e into S(x)=0 then how
> > > can we say e satisfies the axiom?
> >
> > You're confusing numbers and numerals.
>
> Probably. Please explain the difference.
Numbers are part of the model, numerals are part of the language.
> I assume we can substitute numerals for x.
You may substitute terms for variable (avoiding clashes). Numerals are
terms.
> Do non-standard natural numbers have numerals?
You may name them if you wish.
Jesse F. Hughes
I agree that the integers are not a model.
I haven't done the work to describe an infinite model of MA, but David
Libert has. Look at his post if you want to see what a model looks
like.
In any case, your "argument" that it is true is nothing more than appeal
to ignorance.
Aatu Koskensilta
> This is a simple fact. I imagine Nelson is aware of it and that the
> context makes his meaning clear (though I did not check).
Nelson's meaning is perfectly clear. He's saying that an arithmetical
assertion (Ex)Px is true just in case there is a numeral n such that Pn
is true. It's completely unclear why Transfer Principle thinks this of
any relevance here.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, dar�ber muss man schweigen."
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
RussellE
wrote:
> RussellE wrote:
>
> > On Jul 2, 11:11 am, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> > > RussellE wrote:
> > > > We already know if MA has an infinite model,
> > > > Ex (S(x)=0) must be satisfied by a non-standard
> > > > natural number. Call this non-standard natural e.
> > > > If we can't substitute e into S(x)=0 then how
> > > > can we say e satisfies the axiom?
>
> > > You're confusing numbers and numerals.
>
> > Probably. Please explain the difference.
>
> Numbers are part of the model, numerals are part of the language.
A numeral is short hand for a term in the language.
> > I assume we can substitute numerals for x.
>
> You may substitute terms for variable (avoiding clashes). Numerals are
> terms.
>
> > Do non-standard natural numbers have numerals?
>
> You may name them if you wish.
Only by adding constants to the language.
Can a non-standard natural number be assigned
a numeral in the language 0,S,+, and *?
RussellE
> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>
> > This is a simple fact. I imagine Nelson is aware of it and that the
> > context makes his meaning clear (though I did not check).
>
> Nelson's meaning is perfectly clear. He's saying that an arithmetical
> assertion (Ex)Px is true just in case there is a numeral n such that Pn
> is true. It's completely unclear why Transfer Principle thinks this of
> any relevance here.
Ex (S(x)=0) is an axiom of MA.
Which numeral can replace x in an infinite model?
I am pretty sure MA is decidable which means
we can eliminate quantifiers:
http://en.wikipedia.org/wiki/Quantifier_elimination
Nelson is perfectly clear about the induction schema too:
"AvF is true in case for all numerals n, the formula F_v[n] is true."
Ax (S(x) != 0) is true of all numerals in any infinite
"model" of MA.
Nelson is not the only one to define induction this way.
http://plato.stanford.edu/entries/schema/#2
Stanford has a predicate Num(x) which means
x has a numeral. Every description of first order
induction I have read talks about substituting
terms in the language. (Wikipedia just says
"for every definition of a subset of the naturals")
RussellE
>
> To take a simple example, consider any FOL= theory with no constants.
I assume you mean a theory with identity.
> Then this theory proves (Ex)(x = x)
Only if you have axioms of identity.
> and indeed (Ex)(x = x) is *true* in
> every model of the theory,
Assuming the theory has a model,
which it doesn't because the
theory has no constants.
> even though there are *no* closed terms at
> all.
A theory with no models is inconsistent.
Frederick Williams
I don't see how it can.
Frederick Williams
>
> On Jul 2, 10:19 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >
> > To take a simple example, consider any FOL= theory with no constants.
>
> I assume you mean a theory with identity.
>
> > Then this theory proves (Ex)(x = x)
>
> Only if you have axioms of identity.
A FOL= theory _does_ have axioms of identity. If it didn't, it would be
a FOL theory with a binary predicate which is written as "=" but doesn't
behave like identity unless the extra-logical axioms make it.
> > and indeed (Ex)(x = x) is *true* in
> > every model of the theory,
>
> Assuming the theory has a model,
> which it doesn't because the
> theory has no constants.
Eh? Why does the existence of a model require the language to have
constants?
MoeBlee
> And this thread has now become an insult thread.
>
> Of course, this does show one key difference between the posters
> who get these insults and those who give these insults. The
> insulted posters tend to find it desirable for the properties of
> finite objects to transfer to infinite objects, but the insulters
> don't find such a transfer desirable.
Do you mean in this particular thread, or in threads in general?
Because there are insults from all kinds of people.
> Thus, in the case of the aforementioned poster WM
WM throws insults. Just recently he claimed that Daryl, of all people,
is stupid ("extremely limited brain capacity" or whatever the exact
phrase), also that I and a few other people are stupid (in whatever
particular way WM couched it then). Also WM's post are insults to
intelligence and reason to start with.
MoeBlee
MoeBlee
> I show why any infinite model of
> Modular Arithmetic (MA) must be
> omega inconsistent.
Models are not things we check for omega consistency.
You STILL don't know what a model is, and won't bother just to find
out in an ordinary textbook.
MoeBlee
MoeBlee
> A numeral is short hand for a term in the language.
Every numeral IS a term of the language.
MoeBlee
Jesse F. Hughes
> On Jul 3, 7:33 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>>
>> > This is a simple fact. I imagine Nelson is aware of it and that the
>> > context makes his meaning clear (though I did not check).
>>
>> Nelson's meaning is perfectly clear. He's saying that an arithmetical
>> assertion (Ex)Px is true just in case there is a numeral n such that Pn
>> is true. It's completely unclear why Transfer Principle thinks this of
>> any relevance here.
>
> Ex (S(x)=0) is an axiom of MA.
> Which numeral can replace x in an infinite model?
None, surely.
> I am pretty sure MA is decidable which means
> we can eliminate quantifiers:
> http://en.wikipedia.org/wiki/Quantifier_elimination
>
> Nelson is perfectly clear about the induction schema too:
>
> "AvF is true in case for all numerals n, the formula F_v[n] is true."
This is so, since every natural number is named by a numeral and when he
says "is true" he means "is true in N".
But this has nothing to do with your claim. Not every element of your
infinite model is named by a term.
> Ax (S(x) != 0) is true of all numerals in any infinite
> "model" of MA.
>
> Nelson is not the only one to define induction this way.
> http://plato.stanford.edu/entries/schema/#2
>
> Stanford has a predicate Num(x) which means
> x has a numeral. Every description of first order
> induction I have read talks about substituting
> terms in the language. (Wikipedia just says
> "for every definition of a subset of the naturals")
>
>
> Russell
> - Integers are an illusion
--
Jesse F. Hughes
"The three principal virtues of a programmer are Laziness, Impatience, and
Hubris."-- Larry Wall in the Perl5 Manpages
Jesse F. Hughes
> On Jul 2, 10:19 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>
>> To take a simple example, consider any FOL= theory with no constants.
>
> I assume you mean a theory with identity.
Yes.
>> Then this theory proves (Ex)(x = x)
>
> Only if you have axioms of identity.
Surely, but by FOL=, I intended to include such axioms.
>
>> and indeed (Ex)(x = x) is *true* in
>> every model of the theory,
>
> Assuming the theory has a model,
> which it doesn't because the
> theory has no constants.
Wrong. *EVERY* set is the carrier of a model for this theory.
You really are mightily confused.
>> even though there are *no* closed terms at
>> all.
>
> A theory with no models is inconsistent.
Are you claiming that every FOL= theory with no constants is
inconsistent? This is fascinating! This must apply to the theory of
FOL= with no non-logical (or non-identity) axioms, right?
Could you be so kind as to demonstrate how you prove a contradiction in
this theory? And how, once we add a constant, the proof becomes
invalid?
Thanks much.
--
Jesse F. Hughes.
Me: It's very sad when one's husband or wife dies.
Quincy (Age 4 1/2): Yeah. You might want to tell them something and
you just can't. [Long pause] Like "Take out the trash."
RussellE
wrote:
> RussellE wrote:
>
> > On Jul 2, 10:19 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > > To take a simple example, consider any FOL= theory with no constants.
>
> > I assume you mean a theory with identity.
>
> > > Then this theory proves (Ex)(x = x)
>
> > Only if you have axioms of identity.
>
> A FOL= theory _does_ have axioms of identity. If it didn't, it would be
> a FOL theory with a binary predicate which is written as "=" but doesn't
> behave like identity unless the extra-logical axioms make it.
>
> > > and indeed (Ex)(x = x) is *true* in
> > > every model of the theory,
>
> > Assuming the theory has a model,
> > which it doesn't because the
> > theory has no constants.
>
> Eh? Why does the existence of a model require the language to have
> constants?
I don't know for sure.
Can a theory without constants have a language?
What is the language of the theory Ex (x=x)?
Can we have a model without a language?
MoeBlee
> Can a theory without constants have a language?
I guess you mean, can a language not have any individual constants
(i.e. no 0-place function symbols)?
Yes.
(By 'language' here, in this context, we mean an ordinary first order
language.)
A language has a quantifier, denumerabaly many variables, (adequately
many) sentential connectives, and, for some n >= 0, (at least one) n-
place predicate symbol (if for no n, such that n>0, do we have an n-
place predicate symbol, then the language can only do sentential
formulas), and, optionally, whatever other n-place function symbols (a
constant is a 0-place function symbol).
The predicate symbol '=' would suffice for the clause about having at
least one predicate symbol.
> What is the language of the theory Ex (x=x)?
ANY language. The theory with axiom Ex x=x is itself the theory of the
pure predicate calculus in WHATEVER language you want to specify the
theory for.
Every (first order, which is the context in the rest of this) theory
is axiomatized by an adequate set of logical axioms, and then is the
set of formulas (in some GIVEN language, whatever language you
specify) provable from the logical axioms along with a some set,
possibly empty, of non-logical axioms, by the first order logic rules
of inference. (If the rules of inference are strong enough (such as
with a natural deduction system) then the set of logical axioms could
be empty.)
And adding Ex x=x as an axiom has no extra power over identity theory
itself (in WHATEVER language), which is the theory (in WHATEVER
language) of the predicate calculus with identity, since Ex x=x is
provable in identity theory.
> Can we have a model without a language?
No. And that's a foolish question. Foolish because if you knew what a
model IS then there's not even sense asking the question.
Your great foolishness in refusing STILL to get a book and read up on
this subject continues to amaze me.
MoeBlee
RussellE
> RussellE <reaste...@gmail.com> writes:
>
>
> > "AvF is true in case for all numerals n, the formula F_v[n] is true."
>
> This is so, since every natural number is named by a numeral and when he
> says "is true" he means "is true in N".
>
> But this has nothing to do with your claim. Not every element of your
> infinite model is named by a term.
Its not my infinite model.
You are the one claiming MA has an infinite model.
And it seems you agree any infinite model of MA
must have an individual not namable in the language.
Using the standard definition of first order
induction I can prove Ax (S(x) != 0) in any
structure you claim is an infinite model.
Russell
- Never never means never is set theory
Jesse F. Hughes
> On Jul 3, 2:03 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
>> RussellE wrote:
>>
>> > On Jul 2, 10:19 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>
>> > > To take a simple example, consider any FOL= theory with no constants.
>>
>> > I assume you mean a theory with identity.
>>
>> > > Then this theory proves (Ex)(x = x)
>>
>> > Only if you have axioms of identity.
>>
>> A FOL= theory _does_ have axioms of identity. If it didn't, it would be
>> a FOL theory with a binary predicate which is written as "=" but doesn't
>> behave like identity unless the extra-logical axioms make it.
>>
>> > > and indeed (Ex)(x = x) is *true* in
>> > > every model of the theory,
>>
>> > Assuming the theory has a model,
>> > which it doesn't because the
>> > theory has no constants.
>>
>> Eh? Why does the existence of a model require the language to have
>> constants?
>
> I don't know for sure.
> Can a theory without constants have a language?
Of course! Learn just a modicum of logic!
> What is the language of the theory Ex (x=x)?
This is an ill-formed question. There are many different theories whose
only axiom is that. These theories differ by the language specified.
Among those languages is one with no function symbols or constants.
> Can we have a model without a language?
Again, the question makes no sense. Part of any specification of a
theory is a specification of the language. One is free, of course, to
specify that the language has no constants or function symbols (so that
the only terms of the language are variables).
--
Jesse F. Hughes
"The Hammer has arrived."
-- James S. Harris, Feb. 14 2006
Jesse F. Hughes
> On Jul 3, 2:26 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>>
>> > Nelson is perfectly clear about the induction schema too:
>>
>> > "AvF is true in case for all numerals n, the formula F_v[n] is true."
>>
>> This is so, since every natural number is named by a numeral and when he
>> says "is true" he means "is true in N".
>>
>> But this has nothing to do with your claim. Not every element of your
>> infinite model is named by a term.
>
> Its not my infinite model.
> You are the one claiming MA has an infinite model.
Firstly, it is not me that claims it, but it is a consequence of the
compactness theorem -- assuming that you drop the oddball infinite
premise "inference rule".
> And it seems you agree any infinite model of MA
> must have an individual not namable in the language.
Not named in the given language, yes.
> Using the standard definition of first order
> induction I can prove Ax (S(x) != 0) in any
> structure you claim is an infinite model.
Then give the proof.
--
Jesse F. Hughes
"So, either I've found the world's first perfect, non-quantum, random
number generator, or there's a way to make this [factorization] method
work." -- James S. Harris: It's Win-Win, Baby!
David Libert
> On Jul 3, 2:26=A0pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>>
>> > Nelson is perfectly clear about the induction schema too:
>>
>> > "AvF is true in case for all numerals n, the formula F_v[n] is true."
>>
>> This is so, since every natural number is named by a numeral and when he
>> says "is true" he means "is true in N".
>>
>> infinite model is named by a term.
>
> Its not my infinite model.
> You are the one claiming MA has an infinite model.
> And it seems you agree any infinite model of MA
> must have an individual not namable in the language.
> Using the standard definition of first order
> structure you claim is an infinite model.
>
>
> Russell
> - Never never means never is set theory
It was my infinite model, from:
[1] David Libert "Re: A Contradiction in Terms"
sci.logic, sci.math June 16, 2011
http://groups.google.com/group/sci.logic/msg/9bd08626f435b6a1
[2] David Libert "Re: A Contradiction in Terms"
sci.logic June 16, 2011
http://groups.google.com/group/sci.logic/msg/1eaa9cd91bbe5d6e
I forgot to crosspost [2] to sci.math.
[1]-[2] was before you added the omega rule to MA .
For the omega rule, see
[3] http://en.wikipedia.org/wiki/Omega-rule#.CF.89-logic
I have heard and read discussions of the omega rule before, but
I had never heard much discussion of semantics for it.
If you want to talk about models for MA, and are including the
omega rule in MA, it is good to set conventions about semanatics
for the omega rule.
Here is one possible version of semantics for the omega rule.
Given a theory T, expressed in logic over the omage rule, consider
the the of theorems of T, using the T axioms and also logic over those
with the omega rule.
Make a new theory T1 in conventional first order logic, defined to
have axioms all those theorems from T over the omega rule.
One sematnics could be to define the models of T with the omega rule
to be exactly the models in the sense of FOL of T1.
Here is a second reasonable semaantics. Given a theory T over the omega
rule, define semantics for it to consider those models M such that the
true sentences of M are closed under the omega rule.
Given a model M of T in this sense, suppose an existential sentence
Exist X phi(x) is true in M.
If for every numeral n M satisfied ~ phi(n), then by closure of
M truth undert the omega rule M would satisfy All x ~phi(x).
Therefore, if M satisfies Exist x phi(x), then there is a nuneral
n so M satfies phi(n).
This last is enought to get the submodel of M with universe denotations
of all numerals is an elementary submodel of M.
That elementary submodel of M therefore satisfies all T's non-logical
axioms.
Being a structure of all denotations of numerals, it is also closed
under the omega rule.
So given any model M in the 2nd sense of T, we also have the a model
in the same sense of its denoations of numerals.
This suggests a third semantics for the omega rule. Namely models
where every individual is the denotation of some numeral.
[3] does mention this third semantics, and calls those omega-models.
In
[4] David Libert "Re: A Contradiction in Terms"
sci.logic June 17, 2011
http://groups.google.com/group/sci.logic/msg/0c3288241ea7a3b5
I proposed a similar semantics for your MA. That was back before you
had introduced theomega rule.
[4] semamtics was to consider all models in which each individual
is the denotation of a closed term.
Not every closed term is a numeral. Numerals are built from
constant 0 and S.
Closed terms allow that 0 and S, but also + and *.
MA even without the omega rule proves a schematum of theorems,
for each closed term t there is a numeral n so
MA without the omega rule proves t = n.
So the third semantics of omega-models, and the term model
sematnics of [4], are equivalent for theory MA, with or
without the omega-rule.
I had already posted in [4] that your claims of no infinite
models for MA are supported for that [4] semantics.
So if you want to take your claim about the MA with omega rule
as about the 2nd or 3rd semantics for the omega rule,
then your claim is supported.
And if you want to phrase things as formal proofs, there would be
a corresponding formal proof in ZF meta-language about these
semantics.
But when you write:
>Using the standard definition of first order
>induction I can prove Ax (S(x) != 0) in any
>structure you claim is an infinite model.
if you mean that to be a formal proof in MA, even usiing the
omega rule, then the first sematnics above can be applied
to such a purported proof.
That semantics still supports the compactness theorem
so the [1]-[2] methods can still produce an infinite model.
So there is no formal MA proof, even using the omega rule
to support your claim above about infinite models.
It all comes down to what we mean by model. If you use semantics
versions 2-4, then it is supported.
But the 1st semantics seems reasonable to me to analuse purely
formal proofs, even with the omga rule.
By the way, when I pseak of proofs here, it is as approrpiate for
the omega rule, so no longer FOL. Such a proof notion can be
formalized in meta-theoery ZF, ZF just as a FOL theory.
--
David Libert ah...@FreeNet.Carleton.CA
Marshall
Ordinarily on usenet I tend to avoid making any particular
claims for my abilities, but perhaps I may do so in this case.
It strikes me that I have a facility, a talent even, for contempt.
While such folks as Jesse and Moe make entirely competent
forays into this area, I say frankly and without prejudice
that they just don't achieve my level of withering disdain nor
venomous ridicule. Perhaps their viewpoints are diluted by
actual knowledge of the material; whatever.
In any event, the reason I mention this is only to single out
the other poster on sci.math who I believe is competitive
with my gifts at scorn, derision, insult, what-have-you, and
that would be WM. He simply oozes contempt for others;
poisonous condescension rolls off him effortlessly.
Marshall
Frederick Williams
>
> I don't know for sure.
> Can a theory without constants have a language?
> What is the language of the theory Ex (x=x)?
> Can we have a model without a language?
In the thread "Axioms for Integers", on 9 June, I wrote
"It would benefit you enormously to read a book on logic just to find
out
how first order languages are specified, how models are specified, and
how they relate to one another. You remarked elsewhere that you prefer
working things out for yourself. Have you wondered what would happen if
the people who reply to your posts respected your preference?"
I recently came upon Jane Bridge's "Beginning model theory" and I
wondered if Amazon had a second hand one for a few pennies that you
might like to read (no you wouldn't: you prefer
working things out for yourself), but the cheapest one is �130.95
(http://www.amazon.co.uk/gp/offer-listing/0198531575/ref=dp_olp_used?ie=UTF8&qid=1309757893&sr=1-1&condition=used).
Crumbs. I paid �4.50 for mine.
Frederick Williams
>
> I recently came upon Jane Bridge's "Beginning model theory" and I
> wondered if Amazon had a second hand one for a few pennies that you
> might like to read (no you wouldn't: you prefer
> working things out for yourself), but the cheapest one is �130.95
> (http://www.amazon.co.uk/gp/offer-listing/0198531575/ref=dp_olp_used?ie=UTF8&qid=1309757893&sr=1-1&condition=used).
> Crumbs. I paid �4.50 for mine.
This led me to look at some other things
(http://www.amazon.co.uk/s/ref=nb_sb_noss?url=search-alias%3Dstripbooks&field-keywords=chang+and+keisler#):
Chang & Keisler, 1973 from �158.97, 1990 from �481.41. Mine, 1973,
�19.75.
If we sold our books we could bail out Greece.
Transfer Principle
> > This is a simple fact. I imagine Nelson is aware of it and that the
> > context makes his meaning clear (though I did not check).
> Nelson's meaning is perfectly clear. He's saying that an arithmetical
> assertion (Ex)Px is true just in case there is a numeral n such that Pn
> is true. It's completely unclear why Transfer Principle thinks this of
> any relevance here.
Here's why Nelson's statement is relevant: let's go one step at a time
from the direct quote from Nelson to an equivalent statement that's
directly relevant to RE.
1. "AvF is true in case for all numerals n, the formula F_v[n] is
true."
Let F be the formula "~Sx=0" and v be the variable 'x' to obtain:
2. "Ax(~Sx=0) is true in case for all numerals n, ~Sx=0 is true."
Now what does "in case" mean? I'm not sure whether it's a conditional
"if" or biconditional "iff." I'll give your side the benefit of the
doubt and take it to mean the mere conditional "if."
3. "Ax(~Sx=0) is true if for all numerals n, ~Sx=0 is true."
Rewriting so that the hypothesis is first:
4. "If ~Sx=0 is true for all numerals n, then Ax(~Sx=0) is true."
This statement is equivalent to its contrapositive:
5. "If ~Ax(~Sx=0) is true, then ~Sx=0 isn't true for all numerals n."
Converting ~Ax~ to Ex gives us:
6. "If Ex(Sx=0) is true, then Sx=0 is true for some numeral n."
Now we relate this instance of Nelson's statement to RE: We
have that Ex(Sx=0) is an axiom of the theory MA. Therefore it's
true in all models of MA. So from 6. above, we conclude that Sx=0
is true for some numeral n in each model of MA.
Now the numeral n consists of exactly n (i.e., finitely many) 'S'
symbols, followed by '0'. (Members of both sides of the debate
have agreed to this definition of "numeral.") So our conclusion
tells us that in each model of MA, the numeral n+1, with exactly
n+1 (i.e., finitely many) 'S' symbols, followed by '0', is equal
to the numeral '0'.
Thus each model of MA is finite, and each nonisomorphic model of
MA has n+1 elements, with a different value of n for each noniso
model of MA. And this is exactly RE's claim.
Therefore Nelson's statement is very relevant to RE's claim as we
just used Nelson to prove RE, step by step. QED.
Transfer Principle
> > Of course, this does show one key difference between the posters
> > who get these insults and those who give these insults. The
> > insulted posters tend to find it desirable for the properties of
> > finite objects to transfer to infinite objects, but the insulters
> > don't find such a transfer desirable.
> Do you mean in this particular thread, or in threads in general?
> Because there are insults from all kinds of people.
This thread in particular, as well as certain other threads.
Look, suppose a poster were to state that most properties of finite
sets ought to transfer to infinite sets. Now this idea doesn't hold
in standard theory, so how do those on the standard side usually
respond to such a poster? With insults, of course.
Now in this thread, RE claims that certain properties of finite
objects (here finite models of the theory MA) ought to transfer to
the infinite models. And how did Spight respond?
"'Here's all the finite cases, so magically also for any infinite
case.' Actually, it could potentially be a crank axiom."
Here Spight insults those who believe that properties of the
finite cases should transfer to the infinite cases. But I don't
single out Spight here -- his response is typical. Those on the
standard side regularly give this type of insult towards those
who want properties of finite sets to overspill to infinite sets.
> > Thus, in the case of the aforementioned poster WM
> WM throws insults. Just recently he claimed that Daryl, of all people,
> is stupid ("extremely limited brain capacity" or whatever the exact
> phrase), also that I and a few other people are stupid (in whatever
> particular way WM couched it then).
I assume that these insults were in the long thread "Yet another
supertask" -- especially sice McCullough is mentioned right in
the title of that thread. I've not read that thread in its
entirety, but I will give your side the benefit of the doubt that
WM really made those insults (since I have seen insults in posts
from WM and those on his side before).
Yes, there are insulters on both sides of the debates. I'd prefer
that no one on _either_ side of the debate gave insults. It's
only human nature to respond to insults with more insults.
That's why I jumped into this thread. I've seen what insults
like Spight's usually lead to -- more and more insults being
thrown on both sides. I hoped that my entry into this thread
prevented more insults directed at RE -- even if it meant that
insults would be directed at _me_ instead of RE. I'm trying to
make it so that posters can espouse alternate views without
being insulted for them.
Frederick Williams
>
> Now the numeral n consists of exactly n (i.e., finitely many) 'S'
> symbols, followed by '0'. (Members of both sides of the debate
> have agreed to this definition of "numeral.")
Has Russell agreed that the numerals are just finitely many 'S's,
followed by '0'? (Actually, he uses brackets for function application,
but that's an insignificant detail.) In
news:bc4c80b1-4735-4f3c...@e17g2000prj.googlegroups.com
he writes
A numeral is short hand for a term in the language.
What does he mean? S(0+0)*S(S(0)) is a term, is it a numeral?
Frederick Williams
>
> A numeral is short hand for a term in the language.
May we be told just what the numerals are? Are they
0, S(0), S(S(0)), S(S(S(0))), ...
and nothing else?
Transfer Principle
> RussellE <reaste...@gmail.com> writes:
> > A theory with no models is inconsistent.
> Are you claiming that every FOL= theory with no constants is
> inconsistent? This is fascinating! This must apply to the theory of
> FOL= with no non-logical (or non-identity) axioms, right?
> Could you be so kind as to demonstrate how you prove a contradiction in
> this theory? And how, once we add a constant, the proof becomes
> invalid?
Hmmm. Let's try to apply Nelson to FOL= without nonlogical axioms:
"AvF is true in case for all numerals n, the formula F_v[n] is true."
In pure FOL= there are no nonlogical axioms, no constants, and thus
no numerals. Thus, the hypothesis "for all numerals n, the formula
F_v[n] is true" must hold _vacuously_.
But then we conclude that AvF must be true for _any_ formula F --
and so Av(~F) would be true as well. Hmmm.
Well, I admit that you (Hughes) have me stumped here. You say that
you haven't read Nelson, so perhaps someone familiar with Nelson
could please inform me the correct conclusion when one tries to
apply Nelson's statements to pure FOL= without constants.
RussellE
wrote:
> Frederick Williams wrote:
>
> > I recently came upon Jane Bridge's "Beginning model theory" and I
> > wondered if Amazon had a second hand one for a few pennies that you
> > might like to read (no you wouldn't: you prefer
> > working things out for yourself), but the cheapest one is 130.95
> > Crumbs. I paid 4.50 for mine.
>
> This led me to look at some other things
>
> Chang & Keisler, 1973 from 158.97, 1990 from 481.41. Mine, 1973,
> 19.75.
>
> If we sold our books we could bail out Greece.
Thanks. I will try to read some of these.
Russell
- Never never means never in set theory
RussellE
> RussellE <reaste...@gmail.com> writes:
>
> Firstly, it is not me that claims it, but it is a consequence of the
> compactness theorem -- assuming that you drop the oddball infinite
> premise "inference rule".
>
> > And it seems you agree any infinite model of MA
> > must have an individual not namable in the language.
>
> Not named in the given language, yes.
Now you are doomed.
> > Using the standard definition of first order
> > induction I can prove Ax (S(x) != 0) in any
> > structure you claim is an infinite model.
>
> Then give the proof.
>
I will use the definition for
first order induction given at:
http://plato.stanford.edu/entries/schema/#2
[F(0) & Ax((Num(x) & F(x)) -> F(sx)] -> Ax(Num(x) -> F(x))
where Num(x) means has a numeral.
Somewhere it says Num(x) can be dropped
in PA because it is always true.
We know the integers are not a model of MA.
Assume we have a model like the one
Dave Limbert described.
Positive integers are mapped to even
naturals and negative integers are
mapped to odd naturals. -1 becomes 1.
At first, it seems like everything
is fine. We have S(1) = 0, so we
have a numeral that satisfies our axiom.
The problem is that "1" is not
a numeral in this model.
Because successor is linked to
addition and mutiplication, our
successor function looks like:
0, 2, 4, ..., 5, 3, 1
Notice the "..." in the middle.
This means there is no numeral
of the form Sn for any odd
natural number.
Using first order induction,
I can prove Ax (Num(x) -> S(x) != 0 )
in this structure.
Notice Num(x) is always true in MA
just like it is in PA.
RussellE
wrote:
> RussellE wrote:
>
> > A numeral is short hand for a term in the language.
>
> May we be told just what the numerals are? Are they
>
> 0, S(0), S(S(0)), S(S(S(0))), ...
>
> and nothing else?
Yes. Numerals are of the form 0, S(0), ... ,
and nothing else.
I think any closed term can be put into this form.
For example, S(0+0)*S(S(0)) would be
S(0)*S(S(0)) which is S(S(0)).
Marshall
> On Jul 3, 2:44 pm, MoeBlee <modem...@gmail.com> wrote:
>
> > On Jul 2, 5:00 pm, Transfer Principle <dwalke...@cp.csudh.edu> wrote:
> > > Of course, this does show one key difference between the posters
> > > who get these insults and those who give these insults. The
> > > insulted posters tend to find it desirable for the properties of
> > > finite objects to transfer to infinite objects, but the insulters
> > > don't find such a transfer desirable.
> > Do you mean in this particular thread, or in threads in general?
> > Because there are insults from all kinds of people.
>
> This thread in particular, as well as certain other threads.
>
> Look, suppose a poster were to state that most properties of finite
> sets ought to transfer to infinite sets. Now this idea doesn't hold
> in standard theory, so how do those on the standard side usually
> respond to such a poster? With insults, of course.
I'm not sure that I've insulted any specific person in this thread,
you stupid fuck.
> Now in this thread, RE claims that certain properties of finite
> objects (here finite models of the theory MA) ought to transfer to
> the infinite models. And how did Spight respond?
>
> "'Here's all the finite cases, so magically also for any infinite
> case.' Actually, it could potentially be a crank axiom."
>
> Here Spight insults those who believe that properties of the
> finite cases should transfer to the infinite cases.
Wrong as usual, asswipe. What I'm insulting is those who
believe a thing *without justification*. The key word for
the insult is "magically" not "also." But whatever learning
disability it is that you have (and here I'm not being insulting;
you obviously have some sort of ASD or social disability
and that is not your fault) doesn't excuse the arrogance
that leads you to think you can stand as judge for people's
motivations. In this you're like an excessively proud blind
person who insists he can drive. Being blind is no disgrace;
being that fucking arrogant is. A man's got to know his
limitations.
> That's why I jumped into this thread. I've seen what insults
> like Spight's usually lead to -- more and more insults being
> thrown on both sides. I hoped that my entry into this thread
> prevented more insults directed at RE -- even if it meant that
> insults would be directed at _me_ instead of RE. I'm trying to
> make it so that posters can espouse alternate views without
> being insulted for them.
And it's been pointed out many times that you're a dumb
fuck for saying that.
Marshall
Frederick Williams
>
> I will use the definition for
> first order induction given at:
> http://plato.stanford.edu/entries/schema/#2
>
> [F(0) & Ax((Num(x) & F(x)) -> F(sx)] -> Ax(Num(x) -> F(x))
>
> where Num(x) means has a numeral.
> Somewhere it says Num(x) can be dropped
> in PA because it is always true.
I've looked at http://plato.stanford.edu/entries/schema/#2, and searched
the whole of http://plato.stanford.edu/entries/schema, nowhere does it
say that Num(x) means has a numeral (you mean "Num(x) means x has a
numeral" don't you?). It seems to mean "x is a number".
We don't know what a numeral is, but
http://plato.stanford.edu/entries/schema/#2 speaks of first-order number
theory, so they could just be 0, s0, ss0, sss0, ..., but Num is true of
(e.g.) ss0 * sss0 also.
> The problem is that "1" is not
> a numeral in this model.
It can hardly be a problem. Numerals are part of the language they are
not in the model. Will you ever get it?
Frederick Williams
> Yes. Numerals are of the form 0, S(0), ... ,
> and nothing else.
>
> I think any closed term can be put into this form.
No doubt, but it needs proving (by induction in the meta language after
term has been defined recursively, I would think).
> For example, S(0+0)*S(S(0)) would be
> S(0)*S(S(0)) which is S(S(0)).
--
David Libert
> On Jul 3, 5:29=A0pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>>
>> Firstly, it is not me that claims it, but it is a consequence of the
>> compactness theorem -- assuming that you drop the oddball infinite
>> premise "inference rule".
>>
>> > And it seems you agree any infinite model of MA
>> > must have an individual not namable in the language.
>>
>> Not named in the given language, yes.
>
> Now you are doomed.
>
>> > Using the standard definition of first order
>> > structure you claim is an infinite model.
>>
>> Then give the proof.
>>
>
> I will use the definition for
> first order induction given at:
> http://plato.stanford.edu/entries/schema/#2
>
> [F(0) & Ax((Num(x) & F(x)) -> F(sx)] -> Ax(Num(x) -> F(x))
>
> where Num(x) means has a numeral.
> Somewhere it says Num(x) can be dropped
> in PA because it is always true.
I looked up that reference and I didn't find any definition
there of Num(x).
Tarski defined truth in a model:
http://en.wikipedia.org/wiki/Truth_definition
http://plato.stanford.edu/entries/tarski-truth/
I read a comment by one of the big names i model theory
(I have forgotten who), that model theory was
essenatially invented by Tarski.
Tarski had many Ph.D students in Berkeley, that became
may of the major model theorists.
In particular, I think the most definative reference on
model and theory is the textbook by Chang and Kiesler, and
both of them were Tarski's Ph.D students at Berkeley:
http://en.wikipedia.org/wiki/Chen_Chung_Chang
http://en.wikipedia.org/wiki/Howard_Jerome_Keisler
The standard definition in this trext and all others I
am aware of and every course I have ever taken is Tarski's
definition of truth.
Sometimes people can present alternate semantics but if so they
will usually define them.
Tarski's definition of truth rests on his definition of satisfaction,
which is an inductive defintion, of what it means for a model to
satisfy a formula with free variables, where the free variables
are interpreted by elements of the domain of the model.
The inductive clauses of the ssatisfaction definition for
quantifiers themselves quatntify in the meta-language over
all such interpretations, meaning over all elements of the domain.
This is how the ordinary semantics of FOL get wnatification
over all the elements of the domain of its models.
That is the idea of a model. The domain of the model tells you
what the quantifiers range over.
It is an exercise in beginning model theory, the sort of thing
from standard text books, that for theories with arbibrarily
karge finite models or infinite models, there are models where this
standard semantics is disntinct from the case you raise above, of
only quatnifiying over denotations of closed terms.
Your reference didn't spell out that Num(x) is numerals only.
Standard references in model theory, as those above, or texts
such as Chang and Kiesler, give the usual semantics as I have been
saying.
> We know the integers are not a model of MA.
> Assume we have a model like the one
> Dave Limbert described.
> Positive integers are mapped to even
> naturals and negative integers are
> mapped to odd naturals. -1 becomes 1.
Libert, btw :) . Anyway, from your description abpve it sounds
like you are descriving a model isomorphic to the standard integers
Z. That was not my model, and my article xplicitly noted that.
The rest of your writing below is talking about quantification
ranging over numeral denotations. So it is simply not about the
usual semantics as I wrote above.
> At first, it seems like everything
> is fine. We have S(1) =3D 0, so we
> have a numeral that satisfies our axiom.
>
> The problem is that "1" is not
> a numeral in this model.
> Because successor is linked to
> addition and mutiplication, our
> successor function looks like:
>
> 0, 2, 4, ..., 5, 3, 1
>
> Notice the "..." in the middle.
> This means there is no numeral
> of the form Sn for any odd
> natural number.
>
> Using first order induction,
> I can prove Ax (Num(x) -> S(x) !=3D 0 )
> in this structure.
> Notice Num(x) is always true in MA
> just like it is in PA.
You can prove there is every denoation of a numeral
x has S(x) != 0.
If that is what you meant by your formula, then yes it can
be proven. And that is probably what you meant, because it is
interpreting Num(x) as you wrote above.
But that formual you wrote is not in the language of MA.
MA cannot define this Num().
Your proof would be in meta-theory outside MA.
Your added after in MA Num(x) is always true. If that instead is
the intpretation to take, then the result can't be proven in MA.
My model is a counterexample. Even for proofs with omega rule
using the 1st semantics, as I posted recently.
> Russell
> - Integers are an illusion
--
David Libert ah...@FreeNet.Carleton.CA
David Libert
> On Jul 3, 2:29=A0pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > A theory with no models is inconsistent.
>> inconsistent? =A0This is fascinating! =A0This must apply to the theory of
>> FOL=3D with no non-logical (or non-identity) axioms, right?
>> Could you be so kind as to demonstrate how you prove a contradiction in
>> invalid?
>
> Hmmm. Let's try to apply Nelson to FOL=3D without nonlogical axioms:
>
> "AvF is true in case for all numerals n, the formula F_v[n] is true."
>
> no numerals. Thus, the hypothesis "for all numerals n, the formula
> F_v[n] is true" must hold _vacuously_.
>
> But then we conclude that AvF must be true for _any_ formula F --
> and so Av(~F) would be true as well. Hmmm.
>
> Well, I admit that you (Hughes) have me stumped here. You say that
> you haven't read Nelson, so perhaps someone familiar with Nelson
> could please inform me the correct conclusion when one tries to
I had not previously heard of this Nelson paper, but I looked it
up after reading your article. By the way, interesting web site of
Nelson, he has other papers and even some of his books online there.
His paper is talking about the platonist conception of truth for
arithmetic.
So, though he doesn't phrase it that way, he is basically talking
about truth for the language of arithmetic in the standard model
of arithmetic.
I had just posted about standard FOL semantics:
[1] David Libert "Re: Modular Arithmetic Is Consistent"
sci.logic, sci.math July 4, 2011
http://groups.google.com/group/sci.math/msg/67b7bd310c020f3f
The Tarski truth definition works properly for the problem case
you mention above, not making the anomaly you note above about
everything being true.
Nelson makes no claim about his defintion applying to many theories
or being general. he just gives it on one case.
In that one case he gives, his simpler definition is equivalent
to the general Tarski definition.
And his definition is simpler. He can talk about sentences and not
open formulas, and gets to avoid mentioning interpretations of free
variables in his inductive definition.
So his definition is appropriate and useful in his context.
That doesn't mean it generalizes to other contexts, as you note
above with the anomaly.
I had previously discussed a general semantics similar to what
you write: just quantifying over the denotations of closed terms:
[2] David Libert "Re: A Contradiction in Terms"
sci.logic June 17, 2011
http://groups.google.com/group/sci.logic/msg/0c3288241ea7a3b5
In [2] I noted this semantics has no recursively presented proof system
because the simple axiom set PA has complete Pi^1_1 consequences.
So you could try for a semantically presented logic like that,
but it doesn't have the nice corresponding proof sytstem as does
Tarski's semantics.
I also later posted about similar semantics to this for the omega
rule:
[3] David Libert "Re: Modular Arithmetic Is Consistent"
sci.logic, sci.math July 4, 2011
http://groups.google.com/group/sci.math/msg/69a9b4f8fefc3ea9
So we can talk about semantics as you are saying. It is just
nonstandard, and has pitfalls beyond the standard (ie sometimes
everything is true, as you noted, or those Pi^1_1 complete
consqeuences of simple axioms).
Also the fact that Nelson did something similar in a special
case is not strong evidence that we must treat other cases this
way.
--
David Libert ah...@FreeNet.Carleton.CA
Frederick Williams
>
> [...]
>
> The standard definition in this [Chang and Kiesler] trext and all others I
> am aware of and every course I have ever taken is Tarski's
> definition of truth.
One might also make reference to J.C.C. McKinsey's "new definition" in
Synthese. It is used, for example, in Sacks's "Saturated model theory".
Tim Golden BandTech.com
Go TP. Thanks for standing for our freedom to construct. Let the
mimics be labelled as such and their biblical approach will wither.
I do take interest in this thread, though some of the logical ASCII
notation is foreign to me, the treatment of the modulo number as
fundamental is not. Isn't it true that unlike the natural numbers that
we now have a family of number systems? What about modulo one numbers?
This is where MoeBlee's(as I recall) 1=0 stance can hold. Yet the most
primitive means of counting seems to be these modulo one numbers:
three = 1 1 1 ,
five = 1 1 1 1 1 .
The radix system is a level out from the modulo construction, but is
clearly dependent upon the modular assumption. When we wish to
represent numbers larger than the modular form we already have
mechanisms to do so. This is the farce of the natural number when they
state that for any n some operation holds. As soon as you pick a
truely large n such as
568997
in base ten then you are now relying upon modular principles to engage
the natural numbers, and so I am in support of RussellE's awareness
and point out that the modulo number is already more fundamental than
the natural number. And of course this is the gentleman who challenged
sign of the integers as well, which is a modulo two construction
tacked on to the natural number. Yes it is a bit more than that when
we get into geometry, but the mod-two description is accurate.
Jesse F. Hughes
> On Jul 3, 7:33 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>> > This is a simple fact. I imagine Nelson is aware of it and that the
>> > context makes his meaning clear (though I did not check).
>> Nelson's meaning is perfectly clear. He's saying that an arithmetical
>> assertion (Ex)Px is true just in case there is a numeral n such that Pn
>> is true. It's completely unclear why Transfer Principle thinks this of
>> any relevance here.
>
> Here's why Nelson's statement is relevant: let's go one step at a time
> from the direct quote from Nelson to an equivalent statement that's
> directly relevant to RE.
>
> 1. "AvF is true in case for all numerals n, the formula F_v[n] is
> true."
It seems to me that the context for Nelson's statement is something like
this:
Let L be the usual language of arithmetic and F a formula. Then (Av)F
is true (in N) iff every element of N satisfies F. But, as we know,
every element of N is named by a closed term (i.e., a numeral). Thus
(Av)F is true iff every substitution F(S(S...S(0)...)) is true.
This is a fact about the particular structure N and the usual
interpretation of L in N. (I'm probably butchering terminology here,
but hopefully my meaning is clear.)
Nelson never claimed that, in general, for every language L and
interpretation of L in some structure S, S |= (Av)F iff every term t
satisfies F(t).
You are applying Nelson's specific claim about arithmetical sentences
out of its context.
--
Jesse F. Hughes
"I think the Iraqi people owe the American people a huge debt of
gratitude." -- G.W. Bush in January, 2007
Tim Golden BandTech.com
> On Jul 2, 9:53 am, 1treePetrifiedForestLane <Space...@hotmail.com>
> wrote:
>
> > > The consistency of modular arithmetic follows from the consistency
> > > integer arithmetic (base-one .-)
>
> Obviously.
> Any model proves the theory is consistent.
> We are just quibbly over how many models it has.
Wouldn't it have been fine to write
S(X) = 0 (existence of the wrap value X)
P(0) = P(1) = P(2) ... = P(X) = True -> Ax P(x) ?
where x and X are two different but related members. I don't honestly
understand the significance of the omega consistency, but see this way
out of any worry about the ellipsis usage. If anything instantiation
is the bigger bear since we will need to fabricate symbols for the
high systems. This is perhaps a high interpretation of the natural
numbers; their uniqueness is embodied in the progression of the MA
systems, these systems requiring unique symbols whose arbitrary nature
can only be upheld by instantiation, unless we engage a true graphical
representation. Our reliance upon a modulo ten representation to
engage a modulo twelve representation cannot be clean, though it is
unique. Here perhaps is a conclusive statement on the modulo one
number, but I'll leave it cryptic.
To resolve the ellipses by instantiating X as the finite wrap, upper
bound, or the rotator. I'm for a rotational context because when this
comes to be applied as sign the geometry is rotational and
multidimensional. This then would leave the old sloppy renditions of
the natural number as the basis of the modulo number (A clear
ambiguity for we should not construct a simple concept from a more
complicated one). This would bifurcate mathematics momentarily I
suspect, since the new rendition will be much cleaner, but this would
then allow for a cleanup to occur, whereas the confusions of the
existent system will only lay up even more confusion. Anyway this
awareness is important whichever side you land upon.
The natural number can be taken as constructed from the MA math that
RussellE poses. The full family chain can be layed down with MA as a
more fundamental math which does eventually compose the natural number
MA(0),
MA(1),
MA(2),
MA(3),
...,
MA(Natural Numbers) = Natural Numbers.
There is still an ambiguity at the outset, but he has chosen to engage
the zero first, and it does appear to be consistent. Now, when we call
MA(1) the modulo two numbers shouldn't we admit that modern
mathematics contains some ambiguity here in the simplest of places?
Such ambiguity is unacceptable, and it results from a historical
accumulation, which returns to the earlier awareness.
- Tim
RussellE
> Transfer Principle <dwalke...@cp.csudh.edu> writes:
> > On Jul 3, 7:33 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> >> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>
> > 1. "AvF is true in case for all numerals n, the formula F_v[n] is
> > true."
>
> It seems to me that the context for Nelson's statement is something like
> this:
>
> Let L be the usual language of arithmetic and F a formula. Then (Av)F
> is true (in N) iff every element of N satisfies F. But, as we know,
> every element of N is named by a closed term (i.e., a numeral). Thus
> (Av)F is true iff every substitution F(S(S...S(0)...)) is true.
PA and MA don't allow us to define N.
Numerals are all we can talk about.
Induction in PA and MA is over natural numbers.
If P(x) is true for every numeral then Ax P(x).
Nelson's definition is applicable to MA.
Assume you prove Ax P(x) using FO induction in PA.
Can I claim your proof fails because P(w) is false
where w is omega? No doubt you would object
and say omega is not a natural number.
I am making the same objection.
I insist your model of MA have a
natural number, x, such that S(x)=0.
And a bijection with the natural numbers
is not the same as the natural numbers.
I will insist S(x)=0 be satisfiable with
a numeral.
Jesse F. Hughes
> On Jul 4, 6:37 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Transfer Principle <dwalke...@cp.csudh.edu> writes:
>> > On Jul 3, 7:33 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>> >> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>>
>> > 1. "AvF is true in case for all numerals n, the formula F_v[n] is
>> > true."
>>
>> It seems to me that the context for Nelson's statement is something like
>> this:
>>
>> Let L be the usual language of arithmetic and F a formula. Then (Av)F
>> is true (in N) iff every element of N satisfies F. But, as we know,
>> every element of N is named by a closed term (i.e., a numeral). Thus
>> (Av)F is true iff every substitution F(S(S...S(0)...)) is true.
>
> PA and MA don't allow us to define N.
> Numerals are all we can talk about.
We already know what the structure N is. We're discussing how to
determine whether a particular proposition is true in N or not. In so
doing, Nelson makes use of the fact that every element of N is named by
a term.
This fact is evidently not true in the infinite model of MA[1]. Wishing
won't make it so.
> Induction in PA and MA is over natural numbers.
> If P(x) is true for every numeral then Ax P(x).
> Nelson's definition is applicable to MA.
No, tisn't. There's a big difference between MA and PA. In PA, we have
an interesting fact that every model of PA contains N as a submodel.
There is no similar fact about MA.
> Assume you prove Ax P(x) using FO induction in PA.
> Can I claim your proof fails because P(w) is false
> where w is omega? No doubt you would object
> and say omega is not a natural number.
I would say that w is not a term in the language and so your claim is
nonsensical.
> I am making the same objection.
> I insist your model of MA have a
> natural number, x, such that S(x)=0.
Too friggin' bad. You can't insist that (trivial shuffling of the
carrier of the model notwithstanding).
> And a bijection with the natural numbers
> is not the same as the natural numbers.
> I will insist S(x)=0 be satisfiable with
> a numeral.
Footnotes:
[1] I'm speaking of MA without the so-called omega rule.
--
Jesse F. Hughes
"The way that she did what she did
when she did what she did to me
made me think of you." --- Delbert McClinton
RussellE
> RussellE <reaste...@gmail.com> writes:
> > On Jul 4, 6:37 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Transfer Principle <dwalke...@cp.csudh.edu> writes:
> >> > On Jul 3, 7:33 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> >> >> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>
> >> > 1. "AvF is true in case for all numerals n, the formula F_v[n] is
> >> > true."
>
> >> It seems to me that the context for Nelson's statement is something like
> >> this:
>
> >> Let L be the usual language of arithmetic and F a formula. Then (Av)F
> >> is true (in N) iff every element of N satisfies F. But, as we know,
> >> every element of N is named by a closed term (i.e., a numeral). Thus
> >> (Av)F is true iff every substitution F(S(S...S(0)...)) is true.
>
> > PA and MA don't allow us to define N.
> > Numerals are all we can talk about.
>
> We already know what the structure N is.
We don't know what the structure of N is in PA.
PA doesn't have an axiom of infinity.
N is some finite set in MA.
> We're discussing how to
> determine whether a particular proposition is true in N or not. In so
> doing, Nelson makes use of the fact that every element of N is named by
> a term.
Every element of N has a name in PA but not in MA?
> This fact is evidently not true in the infinite model of MA[1]. Wishing
> won't make it so.
No, it won't.
> > Induction in PA and MA is over natural numbers.
> > If P(x) is true for every numeral then Ax P(x).
> > Nelson's definition is applicable to MA.
>
> No, tisn't. There's a big difference between MA and PA. In PA, we have
> an interesting fact that every model of PA contains N as a submodel.
> There is no similar fact about MA.
N = {0} is a sub-model of every model of MA.
What is the significance of having a sub-model?
> > Assume you prove Ax P(x) using FO induction in PA.
> > Can I claim your proof fails because P(w) is false
> > where w is omega? No doubt you would object
> > and say omega is not a natural number.
>
> I would say that w is not a term in the language and so your claim is
> nonsensical.
>
> > I am making the same objection.
> > I insist your model of MA have a
> > natural number, x, such that S(x)=0.
>
> Too friggin' bad. You can't insist that (trivial shuffling of the
> carrier of the model notwithstanding).
I would say that w is not a term in the language and so your claim is
nonsensical.
Frederick Williams
>
> RussellE wrote:
>
> > Yes. Numerals are of the form 0, S(0), ... ,
> > and nothing else.
> >
> > I think any closed term can be put into this form.
Btw, I took "can be put into this form" to mean:
If t is any closed term then there is a numeral s such that
t = s is provable in MA.
> No doubt, but it needs proving (by induction in the meta language after
> term has been defined recursively, I would think).
Marshall
wrote:
> RussellE wrote:
>
> > I will use the definition for
> > first order induction given at:
> >http://plato.stanford.edu/entries/schema/#2
>
> > [F(0) & Ax((Num(x) & F(x)) -> F(sx)] -> Ax(Num(x) -> F(x))
>
> > where Num(x) means has a numeral.
> > Somewhere it says Num(x) can be dropped
> > in PA because it is always true.
>
> the whole ofhttp://plato.stanford.edu/entries/schema, nowhere does it
> say that Num(x) means has a numeral (you mean "Num(x) means x has a
> numeral" don't you?). It seems to mean "x is a number".
>
> theory, so they could just be 0, s0, ss0, sss0, ..., but Num is true of
> (e.g.) ss0 * sss0 also.
>
> > The problem is that "1" is not
> > a numeral in this model.
>
> It can hardly be a problem. Numerals are part of the language they are
> not in the model. Will you ever get it?
All signs point to "no."
Marshall
Jesse F. Hughes
> On Jul 4, 1:10 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > On Jul 4, 6:37 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> Transfer Principle <dwalke...@cp.csudh.edu> writes:
>> >> > On Jul 3, 7:33 am, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
>> >> >> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>>
>> >> > 1. "AvF is true in case for all numerals n, the formula F_v[n] is
>> >> > true."
>>
>> >> It seems to me that the context for Nelson's statement is something like
>> >> this:
>>
>> >> Let L be the usual language of arithmetic and F a formula. Then (Av)F
>> >> is true (in N) iff every element of N satisfies F. But, as we know,
>> >> every element of N is named by a closed term (i.e., a numeral). Thus
>> >> (Av)F is true iff every substitution F(S(S...S(0)...)) is true.
>>
>> > PA and MA don't allow us to define N.
>> > Numerals are all we can talk about.
>>
>> We already know what the structure N is.
>
> We don't know what the structure of N is in PA.
To be honest, it is apparent that you are utterly ignorant of logic and model
theory and there's little point in continuing. Sorry, Russell, but this
conversation is not particularly interesting to me any more.
Your response really is just plain ignorant. I leave the remainder
below so that others can judge the appropriateness of my reaction.
In the meantime, good luck overthrowing first-order logic without
bothering to understand it.
--
Jesse F. Hughes
"[I]t's the damndest thing. There's something wrong with every last
one of you, and I *never* thought that was a possibility. But now I
feel it's the only reasonable conclusion." --JSH sees some sorta light
Frederick Williams
> To be honest, it is apparent that you are utterly ignorant of logic and model
> theory and there's little point in continuing. Sorry, Russell, but this
> conversation is not particularly interesting to me any more.
That's a pity, but never mind for I see that Tim Golden Balls has joined
the discussion. No doubt his lambent intellect will help Russell see
the light.
RussellE
wrote:
> Frederick Williams wrote:
>
> > RussellE wrote:
>
> > > Yes. Numerals are of the form 0, S(0), ... ,
> > > and nothing else.
>
> > > I think any closed term can be put into this form.
>
> Btw, I took "can be put into this form" to mean:
>
> If t is any closed term then there is a numeral s such that
> t = s is provable in MA.
Yes. That is what I meant.
> > No doubt, but it needs proving (by induction in the meta language after
> > term has been defined recursively, I would think).
Why would I need to prove this in the meta theory?
It looks simple to prove this for addition.
Just count the number of S's in the closed term.
S(0)+S(0+S(0)) = S(S(S(0))).
Mutiplication might be tougher.
Frederick Williams
>
> On Jul 4, 2:58 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > Frederick Williams wrote:
> >
> > > RussellE wrote:
> >
> > > > Yes. Numerals are of the form 0, S(0), ... ,
> > > > and nothing else.
> >
> > > > I think any closed term can be put into this form.
> >
> > Btw, I took "can be put into this form" to mean:
> >
> > If t is any closed term then there is a numeral s such that
> > t = s is provable in MA.
>
> Yes. That is what I meant.
>
> > > No doubt, but it needs proving (by induction in the meta language after
> > > term has been defined recursively, I would think).
>
> Why would I need to prove this in the meta theory?
I take that to mean that you think that
If t is any closed term then there is a numeral s such that
t = s is provable in MA.
is provable in MA. Well, go on then, prove it. Note that you will need
to express the predicate
x is a closed term in MA
in the language of MA. Since MA is "about" numbers, not closed terms,
how will you do that? The usual way is to use G\"odel numbers, so you
need to express the predicate
x is the G\"odel number of a closed term
in the language of MA. And, similarly with
x is the G\"odel number of a numeral,
x is the G\"odel number of an equation between terms,
x is the G\"odel number of a theorem.
Then you'll need to do all the arithmetic to get the "G\"odelized"
version of the claim to be proved.
> It looks simple to prove this for addition.
I take that to mean that you think that
If t is any closed term in which '*' does not occur, then
there is a numeral s such that t = s is provable in MA.
Well, go on then, prove it. Note that you will need to express the
predicate
x is a closed term in which '*' does not occur
in the language of MA. Etc, etc, etc.
> Just count the number of S's in the closed term.
How do you even _express_ 'the number of S's in the closed term' in the
language of MA?
> S(0)+S(0+S(0)) = S(S(S(0))).
If you literally count the S's in the above, and reason about the count,
then you are reasoning in the meta language. But I bet you haven't
considered all the possible expressions that can occur on the LHS.
Hence my 'after term has been defined recursively' in the previous
post.
> Mutiplication might be tougher.
If you would find tasks T and S impossible, what does it mean to say
that S might be tougher than T?
MoeBlee
> On Jul 3, 2:44 pm, MoeBlee <modem...@gmail.com> wrote:
>
> > On Jul 2, 5:00 pm, Transfer Principle <dwalke...@cp.csudh.edu> wrote:
> > > Of course, this does show one key difference between the posters
> > > who get these insults and those who give these insults. The
> > > insulted posters tend to find it desirable for the properties of
> > > finite objects to transfer to infinite objects, but the insulters
> > > don't find such a transfer desirable.
> > Do you mean in this particular thread, or in threads in general?
> > Because there are insults from all kinds of people.
>
> This thread in particular, as well as certain other threads.
Yes, in certain threads the only people making insults are the people
making insults against cranks. And in certain other threads, there are
cranks making insults too.
> Look, suppose a poster were to state that most properties of finite
> sets ought to transfer to infinite sets. Now this idea doesn't hold
> in standard theory, so how do those on the standard side usually
> respond to such a poster? With insults, of course.
I've pointed out to you too many millions of times that people get
called 'crank' and worse not just for proposing some alternative
mathematics but rather for the crank WAY they argue for their
alternaive and against set theory.
Please stop ignoring that point.
> Now in this thread, RE claims that certain properties of finite
> objects (here finite models of the theory MA) ought to transfer to
> the infinite models. And how did Spight respond?
>
> "'Here's all the finite cases, so magically also for any infinite
> case.' Actually, it could potentially be a crank axiom."
>
> Here Spight insults those who believe that properties of the
> finite cases should transfer to the infinite cases.
My read is that Marshall is making fun of RussellE in particular,
since RussellE has shown himself over thousands of posts to be a
hopeless crank. Not a crank for having some alternative to propose,
but rather a crank for having the properties of being a crank.
> > > Thus, in the case of the aforementioned poster WM
> > WM throws insults. Just recently he claimed that Daryl, of all people,
> > is stupid ("extremely limited brain capacity" or whatever the exact
> > phrase), also that I and a few other people are stupid (in whatever
> > particular way WM couched it then).
>
> I assume that these insults were in the long thread "Yet another
> supertask" -- especially sice McCullough is mentioned right in
> the title of that thread. I've not read that thread in its
> entirety, but I will give your side
I'm not on a "side".
> the benefit of the doubt that
> WM really made those insults (since I have seen insults in posts
> from WM and those on his side before).
>
> Yes, there are insulters on both sides of the debates.
Good that you recognize that, and it belies your first line of
argument in this thread, which was to make it seem that only anti-
cranks do the insulting.
> I hoped that my entry into this thread
> prevented more insults directed at RE
RussellE DESERVES the insults. Actually, he deserves even more insult
than he gets.
MoeBlee
MoeBlee
> My read is that Marshall is making fun of RussellE in particular,
> since RussellE has shown himself over thousands of posts to be a
> hopeless crank. Not a crank for having some alternative to propose,
> but rather a crank for having the properties of being a crank.
To add though: As to properties of fintie sets carrying over to
infinite sets, it's okay with me that one specifies certain that one
wishes to carry over from finite to infinite and then sets up a theory
in which that holds. But there is a reasonable limit to that. After
all, where the definition of 'infinite' is 'not finite' then at least
one property is not carried over, viz. finiteness. Then it is
reasonable that other properties are not going to carry over too.
Anyway, as far as I'm concerned, set up whatever theory you want, but
if someone argues for certain notions and against certain other
notions in the way of a crank, then that person deserves to be called
a crank.
MoeBlee
Transfer Principle
> On Jul 4, 12:09 am, Transfer Principle <dwalke...@cp.csudh.edu> wrote:
> > Here Spight insults those who believe that properties of the
> > finite cases should transfer to the infinite cases.
> believe a thing *without justification*. The key word for
> the insult is "magically" not "also." But whatever learning
> disability it is that you have (and here I'm not being insulting;
> you obviously have some sort of ASD or social disability
> and that is not your fault) doesn't excuse the arrogance
> that leads you to think you can stand as judge for people's
> motivations.
Notice that my number one goal isn't to stand as judge for
people's motivations. My number one goal is actually to
_eliminate_ (or at least reduce) the _insults_. It's just my
hope that by attempting to judge others' motivations for
insulting, I can eliminate those motivations and thus
eliminate the insults themselves, which is my true number
one goal. I know better than to think that I can eliminate
the insulters merely by asking the insulters to stop. The
best way to get the insulters to stop is by making them not
_want_ to insult them anymore.
Now attempting to find the motivations in order to stop the
insults has been largely unsuccessful -- especially if, as
you say, I lack the ability to find the motivations to any
degree of accuracy.
> A man's got to know his limitations.
I'd be fully willing to admit that I lack the ability to
judge motivations accurately -- as soon as I find something
that _is_ within my ability that _will_ stop the insults. I
couldn't care less whether I can judge others' motivations
correctly -- what I _do_ care about is _eliminating_ or
reducing the _insults_.
So I want to skip the middleman. I want to skip finding the
motivations behind the insults, and proceed directly to
eliminating them.
Of course, I don't expect the _insulters_ to tell me how I
can, with my limited ability, reduce their insults. This is
something that I have to figure out myself.
Transfer Principle
> [2] David Libert "Re: A Contradiction in Terms"
> sci.logic June 17, 2011
> http://groups.google.com/group/sci.logic/msg/0c3288241ea7a3b5
> In [2] I noted this semantics has no recursively presented proof system
> because the simple axiom set PA has complete Pi^1_1 consequences.
Thanks for the link.
So based on Tarski's definition, RE is wrong, and Tarski's
definition, not Nelson's, generalizes beyond PA/standard N.
So I wonder what I can do to make RE's ideas work. As usual,
this won't be easy, and I expect to receive no help
whatsoever in doing so.
Transfer Principle
wrote:
> On Jul 4, 3:09 am, Transfer Principle <dwalke...@cp.csudh.edu> wrote:
> > I'm trying to
> > make it so that posters can espouse alternate views without
> > being insulted for them.
> Go TP. Thanks for standing for our freedom to construct. Let the
> mimics be labelled as such and their biblical approach will wither.
Welcome to the thread!
> I am in support of RussellE's awareness
> and point out that the modulo number is already more fundamental than
> the natural number. And of course this is the gentleman who challenged
> sign of the integers as well, which is a modulo two construction
> tacked on to the natural number. Yes it is a bit more than that when
> we get into geometry, but the mod-two description is accurate.
Ah yes, your polysigned numbers.
The problem in this thread is that those on the standard side
are trying to make RE accept _infinite_ models of MA. This would
be like having polysigned numbers with _infinitely_ many signs!
I assume that any sort of "P_infinity" would be impossible. In
"P_infinity," repeatedly multiplying by -1 generates new signs:
1(-1) = -1
(-1)(-1) = +1
(+1)(-1) = *1
(*1)(-1) = #1
(#1)(-1) = %1 (or whatever the 5th sign is)
(%1)(-1) = (6th sign)1
etc.
with none of these taking you back to 1. So there would be
infinitely many distinct signs. Yet somehow, -1 still has a
multiplicative inverse, some strange sign ?1 such that:
(?1)(-1) = 1
Which sign would "?1" be? It can't be the 1st sign, the 2nd
sign, the 3rd sign, etc.
I assume the above is impossible in polysigned numbers. Yet
this is exactly what the standard side is trying to impose on
RE -- some sort of "mod infinity" model of MA.
So you can see what we're up against. "Mod infinity" and
"sign infinity" appear to be impossible concepts, yet I can't
see how to rule them out completely.
MoeBlee
> Notice that my number one goal
Notice that my number one goal concerning you is to get you to stop
lying as often as you do.
MoeBlee
Transfer Principle
But your own line:
> I'm not on a "side".
is belied by the following:
> RussellE DESERVES the insults. Actually, he deserves even more insult
> than he gets.
That puts you squarely on a side. Which side? The side opposite that
of RE, since you say that RE deserves more insults. It puts you on the
same side as those who have already insulted RE, like Spight.
I'm on a side as well. Which side? I'd like to oppose the insulters,
but you say that there are often insulters on both sides. But more
often than not, I side with the minority, since I believe that the
minority, due to their lack of numbers, are more victimized by the
insults of the majority than vice versa.
> I've pointed out to you too many millions of times that people get
> called 'crank' and worse not just for proposing some alternative
> mathematics but rather for the crank WAY they argue for their
> alternaive and against set theory.
> Please stop ignoring that point.
Then let me search for another _way_ to make the argument in favor
of RE's alternative. If the _way_ the argument is made is the true
reason behind the insults, then I should be able to find an
argument that's not insultable and the insults should go away. If
rigorous axioms are needed, then let me search for axioms.
But I have a feeling that the insults will remain no matter what
axioms I try to give.
Remember, I don't care what the true reason behind the insults is,
but only that the insults themselves disappear.
Jesse F. Hughes
> The problem in this thread is that those on the standard side
> are trying to make RE accept _infinite_ models of MA.
Those bastards!
Of course RE should be free to reject infinite models! That's what
freedom and liberty are all about!
--
"So why talk [about my factoring method] out on Usenet? Because it's a
highly public place so I'm unlikely to disappear[...] You people are
my protection. [...] You may be what's keeping me free and walking out
in the open air." -- James S. Harris, theory guy on the edge.
Jesse F. Hughes
Well, precisely what ideas are you going to try to make work?
His primary claim, as I understand it, is that a particular first-order
theory has models of every finite size, but no infinite models. Is this
what you're going to try to make "work"?
--
Jesse F. Hughes
"Well, I guess that's what a teacher from Oklahoma State University
considers proper as Ullrich has said it, and he is, in fact, a teacher
at Oklahoma State University." -- James S. Harris presents a syllogism
RussellE
I have noticed the people most likely to call me a crank
are the ones least likely to give a valid argument.
I guess ad hominem attacks are all one has
left when there is no reasonable rebuttal.
http://en.wikipedia.org/wiki/Ad_hominem
If you really think I am such a crank why do
you keep reading my posts?
Another favorite tactic of the "anti-cranks"
is to slip in an assumption and later use
this assumption to "prove" the crank is wrong.
The Compactness theorem is an example
in this thread. Several people have claimed
Compactness proves MA has infinite models.
http://en.wikipedia.org/wiki/Compactness_theorem
"a set of first-order sentences has a model
if and only if every finite subset of it has a model"
Nothing in this description says anything about
MA having infinite models. There is no question
about MA having models (unlike PA).
Wikipedia goes on to say how compactness
implies Robinson's principle. A simple proof
is given that uses the following sequence:
1+1 != 0
1+1+1 != 0
1+1+1+1 != 0
.........
All of these sentences are assumed to be
true and if one of them is false this proves
a contradiction. Of course, while this would
prove a contradiction in some theory based
on PA, they don't prove a contradiction in MA.
S(0)+S(0) = 0 in the one element model of MA.
So, this proof of Robinson's principle would
not work in MA.
The article does go on to say how compactness
and the Löwenheim–Skolem theorem prove
a theory with arbitrarily large finite models
has an infinite model.
http://en.wikipedia.org/wiki/Upward_L%C3%B6wenheim%E2%80%93Skolem_theorem
Löwenheim–Skolem
"states that if a countable first-order theory has an infinite model,
then for every infinite cardinal number k it has a model of size k".
Löwenheim–Skolem starts out by assuming a FO theory
has an infinite model.
I am pretty sure Compactness and Löwenheim–Skolem
require the assumption that the set of all natural numbers exists.
Neither PA nor MA have such an assumption.
(Prove me wrong by providing a proof of
Löwenheim–Skolem using only the axioms of PA.)
Now for some ad hominem arguments of my own.
Most people seem to agree it is reasonable to
assume every element in a model of PA can be
represented by a numeral. MA has the same
axioms as PA with one axiom negated
(ignoring the omega rule).
Why is it reasonable to assume every element
of a model of PA has a numeral but not reasonable
to assume this in MA? I suspect the only reason is
then I can prove MA has no infinte models.
Russell
- The universe is one dimensional
RussellE
> Transfer Principle <dwalke...@cp.csudh.edu> writes:
> > On Jul 4, 2:20 am, ah...@FreeNet.Carleton.CA (David Libert) wrote:
> >> [2] David Libert "Re: A Contradiction in Terms"
> >> sci.logic June 17, 2011
> >> http://groups.google.com/group/sci.logic/msg/0c3288241ea7a3b5
> >> In [2] I noted this semantics has no recursively presented proof system
> >> because the simple axiom set PA has complete Pi^1_1 consequences.
>
> > Thanks for the link.
>
> > So based on Tarski's definition, RE is wrong, and Tarski's
> > definition, not Nelson's, generalizes beyond PA/standard N.
>
> > So I wonder what I can do to make RE's ideas work. As usual,
> > this won't be easy, and I expect to receive no help
> > whatsoever in doing so.
>
> Well, precisely what ideas are you going to try to make work?
>
> His primary claim, as I understand it, is that a particular first-order
> theory has models of every finite size, but no infinite models. Is this
> what you're going to try to make "work"?
>
That will be tough in a system that "generalizes beyond PA/standard
N".
David Libert
RussellE (reas...@gmail.com) writes:
[Deletion]
> I am pretty sure Compactness and Lowenheim Skolem
> require the assumption that the set of all natural numbers exists.
> Neither PA nor MA have such an assumption.
> (Prove me wrong by providing a proof of
> Lowenheim Skolem using only the axioms of PA.)
> Russell
> - The universe is one dimensional
I was using compactness to prove the existence of an infinite model
for MA in the sense of usual FOL semantics.
Yes, as you say, I take the background for the compactness theorem to
include the set of all natural numbers exists, as an assumption behind
my claimed proof of the existentce of a certain model.
And so you make the point above.
Using your comment, I can prove many further amazing theorems.
Thm : there are no sets.
Proof:
A set has no structure, just elements which can be = or != .
So the thoery corresponding to a simple set is the pure theory of
=.
But in the the thoery of =, we can't arithmentize syntax, so inside
the theory of = we can't prove a model of the theory of = exists.
QED
Thm: there are no groups.
Proof:
The theory for a group is the theory of groups. In the theory of groups
we cannot arithmetize syntax, and do meta-matheamtics.
So in the theory of groups we cannot formalize a proof that models of the
theory of groups exist.
QED
Thm: there are no linear orders
Proof:
In the theory of linear orders we cannont formalize meta-mathematics and
prove in that theory there is model of that theory.
QED
Similarly I can prove there are no vector spaces. Also no topological
spaces.
Such power in your methods.
--
David Libert ah...@FreeNet.Carleton.CA
David Libert
RussellE (reas...@gmail.com) writes:
[Deletion]
> Another favorite tactic of the "anti-cranks"
> is to slip in an assumption and later use
> this assumption to "prove" the crank is wrong.
>
> The Compactness theorem is an example
> in this thread. Several people have claimed
> Compactness proves MA has infinite models.
> http://en.wikipedia.org/wiki/Compactness_theorem
Right an assumption: the Compactness Theorem.
Compactness "Theorem" ?
Compactness Theorem, or Compactness Assumption ?
If anyone ever refutes me using a theorem, I can just answer
that theorem was an assumption they "slipped in" .
Anyway, beyond Theorem versus Assumtion, the subject matter at
hand shows an underhanded tactic is indeed being used here.
The subject under discussion was whether MA has inifinite models.
So the topic at hand was about the existence of models.
And what underhanded tactic was employed?
To begin talking about the Compactness Theorem, which is a theorem
about the existence of models.
What relevance could a theorem about the existence of models have
to a topic about the existence of models?
Obviously at this point, without making an effort at trying to understand
what the other people are saying, we can just dismiss such comments out
of hand.
What else coukd they be but a tactic?
> Russell
> - The universe is one dimensional
--
David Libert ah...@FreeNet.Carleton.CA
Frederick Williams
> Most people seem to agree it is reasonable to
> assume every element in a model of PA can be
> represented by a numeral. MA has the same
> axioms as PA with one axiom negated
> (ignoring the omega rule).
>
> Why is it reasonable to assume every element
> of a model of PA has a numeral but not reasonable
> to assume this in MA?
You can extend the language of any first order theory T by adding to it
a constant c_x for every x in the domain(*) of a model of TA. If the
domain is a set of numbers then the c_x may very reasonably be called
numerals.
(* Different words are used: domain = underlying set = universe =
carrier.)
David Libert
RussellE (reas...@gmail.com) writes:
[Deletion]
> Most people seem to agree it is reasonable to
> assume every element in a model of PA can be
> represented by a numeral. MA has the same
> axioms as PA with one axiom negated
> (ignoring the omega rule).
>
> Why is it reasonable to assume every element
> of a model of PA has a numeral but not reasonable
> to assume this in MA? I suspect the only reason is
> then I can prove MA has no infinte models.
You say an assumption above about what goes on in "a model"
of PA.
Which model? There are many.
If you mention an assumption about "a model" without
qualifying which model it is, does that mean you are saying
most people agree it is reasonable to assume this about every
model of PA ?
This assumption, that every element in a model of PA can be
represented by a numeral, is true of the standard model of
PA.
And any model of PA for which is true is isomorphic to the
standard model.
There are 2^Aleph_0 pairwise non-isomorphic countable
models of PA.
Every one of those models except for 1 (the standard model)
does not have the property of every element being represented
by some numeral.
So among countable PA models, the vote up to isomorphism
on the property you state, is 2^Aleph_0 no, 1 yes.
If we look beyond countable PA models, to all PA models,
there are a proper class of isomorphism types of all PA
models, allowing all possible cardinalities.
So among all models, the vote on this property up to
isomorphism is a proper class versus 1.
If this property were true, that every model of PA
has every element represented by a numeral, then up to
isomorphism there would only be one PA model the standard
one.
Normal set theory proves the standard PA model satisfies
Con(PA).
So if every PA model has all elements represented by a
numeral, then Godel's Incompleteness Theorem would be refuted.
Also this claim would immediately contradict the Lowenheim
Skolem Theorem, as you had been writing about.
So are you saying most people seem to agree it is reasonable to
assume a vote of proper class vs 1 against is a 100% vote for,
and Godel's Theorem is refuted and the Lowenheim Skolem
Theorem is refuted?
Or maybe when you write "a model" you only mean the standard
model.
So one model of PA (up to isomorphism) has this property, and
others do not.
Maybe that is the assumption you are saying, which is true.
But if that is the assumption, it corresponds to what we have
been saying about MA.
We agree with you some models of MA have each element represented
by a numeral.
Namely the finite models, as all of us have agreed exist.
But in analogy with the 2nd reading of the PA assumption,
we would be saying for MA that other MA models don't have this
property.
Which is all the rest of us have been saying all along.
> Russell
> - The universe is one dimensional
--
David Libert ah...@FreeNet.Carleton.CA
Tim Golden BandTech.com
I suspect that this can be had by using the finite model. For any two
large values we can always select an MA(X) such that their sum will
not wrap. Any system whose X is greater than A and B (the two values)
will suffice.
I suppose the struggle on infinity is the same old struggle, but the
point to me is that these MA systems can be claimed to finish out as
the natural numbers.
This puts MA theory beneath the natural number. We do not need the
number three to count to two.
It's good to get a response from you TP. I've been getting little to
no feedback here lately. The fact that RussellE has challenged the
sign of the integer here is very meaningful to me, and the fact that
that sign has modulo two behavior is very important. In effect one of
RussellE's next steps could be the generalization of sign. I look
forward to reading more. The ambiguities that I keep mentioning still
do not rise to the surface in the discussion here. For instance the
"modulo-two" number can be denoted by S(S(0)))=0 but the value two of
tradition is nonexistent within RE's construction. This is essentially
the graphical representation that I argue for, where we are forced to
the primitive form of calling this modulo-two system
modulo 1 1 ,
or whatever arbitrary symbol one wishes to use. Symbollism is so close
here. There is trouble when we reuse the same symbols for the mod-
three system as for the mod-two system, but it can work. The option to
claim unique symbols for each of these systems is a classification
that can be done via structure, and this must be so under the
awareness of those wrapping sums that you are discussing.
Frederick Williams
> a constant c_x for every x in the domain(*) of a model of TA. If the
I don't know where the A after the T came from, they're not even near
one another on the keyboard.
MoeBlee
> On Jul 5, 9:14 am, MoeBlee <modem...@gmail.com> wrote:
>
> > On Jul 4, 2:09 am, Transfer Principle <dwalke...@cp.csudh.edu> wrote:
> > > Yes, there are insulters on both sides of the debates.
> > Good that you recognize that, and it belies your first line of
> > argument in this thread, which was to make it seem that only anti-
> > cranks do the insulting.
>
> But your own line:
>
> > I'm not on a "side".
>
> is belied by the following:
>
> > RussellE DESERVES the insults. Actually, he deserves even more insult
> > than he gets.
That's my opinion. Other people might agree with that opinion. But I
don't state that opinion in the sense of being "on a side".
> Then let me search for another _way_ to make the argument in favor
> of RE's alternative. If the _way_ the argument is made is the true
> reason behind the insults, then I should be able to find an
> argument that's not insultable and the insults should go away. If
> rigorous axioms are needed, then let me search for axioms.
>
> But I have a feeling that the insults will remain no matter what
> axioms I try to give.
You have lots of feelings.
> Remember, I don't care what the true reason behind the insults is,
> but only that the insults themselves disappear.
Remember, I don't give a damn about your prissy little agenda.
MoeBlee
MoeBlee
> On Jul 5, 9:14 am, MoeBlee <modem...@gmail.com> wrote:
>
> > On Jul 4, 2:09 am, Transfer Principle <dwalke...@cp.csudh.edu> wrote:
>
> > RussellE DESERVES the insults. Actually, he deserves even more insult
> > than he gets.
>
> I have noticed the people most likely to call me a crank
> are the ones least likely to give a valid argument.
You are TOO MUCH! I have given you post after post after post of
technical explanation about a subject you have so many opinions about
yet you can't even be bothered to learn it's first basics.
I have given you exact, explicit, detailed technical corrections to
your confusions and your hopelessly INvalid arguments in hundreds and
hundreds of posts over about eight years.
And you have the chutzpah to say I'm "least likely to give a valid
argument".
You're a real case.
> If you really think I am such a crank why do
> you keep reading my posts?
I'm self-punishing.
> Another favorite tactic of the "anti-cranks"
> is to slip in an assumption and later use
> this assumption to "prove" the crank is wrong.
Name one single instance where I have "slipped in an assumption".
> The Compactness theorem is an example
> in this thread.
The compactness theorem is provable, at least, from the ordinary set
theoretic axioms. There's no "slipping in" of assumptions.
Also, you mentioned ad hominem arguments. I say you're a crank, but
that is NOT my ARGUMENT about the mathematics itself. I've given you
explicit technical explanations as to the mathematics, yet you
continue to make a jackass out of yourself by going on and on using
terminology you don't even understand and won't even learn. My
comments that you are a crank regard you as a poster; I don't
represent that those comments themselves are mathematical arguments.
MoeBlee