how does Indirect NonExistence compare with Constructivism school of math? #197; 2nd ed; Correcting Math
Archimedes Plutonium
Wikipedia. I want to give a sense of the state of
knowledge on this topic of truth in logic and math.
In my search to see if anyone covered Indirect
NonExistence, about the closest that it seems anyone reached this
topic is the Constructivist
School of Logic.
--- quoting from Wikipedia on Constructivist School ---
http://en.wikipedia.org/wiki/Nonconstructive_proof
In mathematics, a constructive proof is a method of proof that
demonstrates the existence of a mathematical object with certain
properties by creating or providing a method for creating such an
object. This is in contrast to a nonconstructive proof (also known as
an existence proof or pure existence theorem) which proves the
existence of a mathematical object with certain properties, but does
not provide a means of constructing an example.
(snip)
Constructivism is the philosophy that rejects all but constructive
proofs in mathematics. Typically, supporters of this view deny that
pure existence can be usefully characterized as "existence" at all:
accordingly, a non-constructive proof is instead seen as "refuting the
impossibility" of a mathematical object's existence, a strictly weaker
statement.
(snip)
This proof is non-constructive because it relies on the statement
"Either q is rational or it is irrational" — an instance of the law of
excluded middle, which is not valid within a constructive proof. The
non-constructive proof does not construct an example a and b; it
merely gives a number of possibilities (in this case, two mutually
exclusive possibilities) and shows that one of them — but does not
show which one — must yield the desired example.
(snip)
A more substantial example is the graph minor theorem. A consequence
of this theorem is that a graph can be drawn on the torus if, and only
if, none of its minors belong to a certain finite set of "forbidden
minors". However, the proof of the existence of this finite set is not
constructive, and the forbidden minors are not actually specified.
They are still unknown.
--- end quoting Wikipedia ---
I was surprized to learn that law of excluded middle
is not valid in a constructive proof.
What I aim to show is that Indirect NonExistence
is an invalid proof argument, no matter what school
of mathematics one pledges allegiance.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped in places)
>
> --- quoting from Wikipedia on Constructivist School ---
> http://en.wikipedia.org/wiki/Nonconstructive_proof
>
>
> Constructivism is the philosophy that rejects all but constructive
> proofs in mathematics. Typically, supporters of this view deny that
> pure existence can be usefully characterized as "existence" at all:
> accordingly, a non-constructive proof is instead seen as "refuting the
> impossibility" of a mathematical object's existence, a strictly weaker
> statement.
>
I find this somewhat amusing that the school of Constructivist
mathematicians,
(and I am in favor of them) is picking apart the concept of "existence
and
impossibility" and rejecting law of excluded middle and axiom of
choice etc etc.
Amusing because I am focused on a more narrow issue of combining
Reductio
Ad Absurdum with an outcome of either the existence of something or
the
nonexistence of something.
So it looks as though Constructivists and I are on parallel paths.
But what I want to do is push the envelope outwards of making Truth
Tables
that says whether a mathematical proof is valid or invalid if it uses
the
Indirect Method.
So where Constructivists place their views as a "philosophical
school." I want
to be more aggressive and to show where Nonconstructivists are doing
invalid and fake arguments of proofs. So where Constructivists are
happy
to separate themselves away from NonConstructivists, I want to
actually
open up the fight and to show that NonConstructivists are doing fake
math.
So I think what I have to do now is to construct Truth Tables showing
that
Indirect NonExistence is a probabilistic conclusion, and not a valid
conclusion.
I suspect the Constructivists did not go that far, but stopped short
of demonstrating
that NonConstructivists have filled math with fake-proofs.
Archimedes Plutonium
Mathematics were
seen in the past as mostly that of a alternative school of thought
from the mainstream
school that allowed almost anything and any form of logic. Call it
logic-unfettered
but I would call it logic gone berserk.
I also want to thank Wikipedia for the first time in which I have
discovered and
developed some important new math and which I had to rely on no other
sources
other than Wikipedia for Truth Tables. I did not have to buy another
book or
hunt down another book for sourcing Truth Tables and Constructivist
School
but relying only on Wikipedia.
I have this thing pretty well pinpointed. And let me survey how it
works, and the
flaw in modern day mathematics. The Constructivist School is written
in Wikipedia
for anyone to lookup, and this School focuses on "existence proof" and
they
distinguish between variations of "existence" and they are not
accepting of
law of excluded middle nor the Axiom of Choice.
What I bring to the mix is Indirect NonExistence versus Direct
Existence. This
facet was overlooked by Constructivists. This facet actually allows
Constructivists
to compare two types of existence proofs and to declare one as valid
and the
other as invalid reasoning.
So for the first time in mathematics history the Constructivist School
of Math
will emerge as the Mainstream and the true school of math whereas the
Old
laissez faire School where anything goes is seen as corrupt-logic.
The Old School probably had a name for themselves in allowing almost
any
contraption to be permitted as valid logic and so a name such as
laissez faire
captures their attitude towards mathematics just as constructivism
properly
captures the attitude of an opposing school of thought.
--- quoting Wikipedia on truth tables and constructivism ---
p and q
p q p and q
T T T
T F F
F T F
F F F
If p Then q
p q p --> q
T T T
T F F
F T T
F F T
Not p
p not p
T F
F T
http://en.wikipedia.org/wiki/Modus_ponens
Justification via truth table
The validity of modus ponens in classical two-valued logic can be
clearly demonstrated by use of a truth table.
If p Then q
p q p --> q
T T T
T F F
F T T
F F T
In instances of modus ponens we assume as premises that p → q is true
and p is true. Only one line of the truth table - the first -
satisfies these two conditions. On this line, q is also true.
Therefore, whenever p → q is true and p is true, q must also be true.
--- end quoting Wikipedia ---
Notice that in the method of Modus Ponens, only one truth table line
makes for a valid proof.
Well, notice that in all Indirect Proofs there is the existence of
something. So we end up in
Indirect Proofs with always the Existence of Something, and the
NonExistence of something
would then be of a different truth table line for validity. This means
that in all Indirect Proofs,
the Indirect Existence proof would be valid but the Indirect
NonExistence would be at best
probabilistic, or as the Constructivists would say invalid.
Archimedes Plutonium
--- quoting from a website that discusses RAA ---
http://www.iep.utm.edu/reductio/
The mathematical school of so-called intuitionism has taken a definite
line regarding the limitation of reductio argumentation for the
purposes of existence proofs. The only valid way to establish
existence, so they maintain, is by providing a concrete instance or
example: general-principle argumentation is not acceptable here. This
means, in specific, that one cannot establish (∃x)Fx by deducing an
absurdity from (∀x)~Fx. Accordingly, intuitionists would not let us
infer the existence of invertebrate ancestors of homo sapiens from the
patent absurdity of the supposition that humans are vertebrates all
the way back. They would maintain that in such cases where we are
totally in the dark as to the individuals involved we are not in a
position to maintain their existence.
--- end quoting ---
Also in that website they mention that Russell and Whitehead in their
PM assessed the
Reductio Ad Absurdum argument in propositional logic as this formula:
(~p →p) →p
Now I do not know whether "intuitionists" and "constructivists" are
one and the same group
of believers only with a different name tag? And I do not remember
what the name tag
given for those that were not a member of the intuition or
constructivism schools of thought,
and I am just going to call them the laissez-faire-mainstream.
So I have come to a position in this book where I intersect with the
Constructivist School
and their concern over "existence proofs." And my concern allows us to
actually tell whether
the Constructivists were correct and whether the Mainstream was wrong.
So far the Constructive School has been more of a philosophy or
attitude about what is acceptable
and unacceptable. But my position with Indirect Existence and Indirect
NonExistence is
that we can actually test the Constructivism School against the
Mainstream and one will
win and the other will lose. And so far it looks as though the
Constructivism School is going
to win and thus take the position as the rightful and true mathematics
school.
My claim is that in all Indirect Proofs there is a step or statements
either the "existence of
something" or the "nonexistence of something." And that a valid
Indirect Method has only
the existence of something, and the probabilistic Indirect Method has
the "nonexistence
of something."
How do I prove this? Well as Whitehead and Russell indicate that the
general form of the
Indirect Method is as written above and that allows for the Truth
Tables to be applied.
And within the Indirect Argument is a statement of the Existence of
something and that
would be a valid argument with only one supporting Truth Table line as
unique as the
modus ponens truth table line. So if the Indirect Method then yields a
NonExistence of
Something could not possibly have the same truth-table unique valid
line but must be
a probabilistic conclusion.
Comment: I was rather pleased to see another example using
paleontology in the above
reference to invertebrates. And I already gave the example that the
science of
paleontology with its fossil record is a superb example of Indirect
Existence Method of
proof, but that the Indirect NonExistence simply because we have no
fossils of a believed
adaption does not mean it never existed.
So what I am doing is proving or making clear that the Indirect Method
used in mathematics
has very much limitations and that the validity of a proof in
mathematics using Indirect Method must be Indirect Existence and that
no math proof can tolerate Indirect NonExistence
of some entity. So that these alleged math proofs of Wiles's FLT,
Appel & Haken's 4 Color
Mapping, Hales's Kepler Packing; transcendental alleged proofs; recent
Poincare C. alleged,
Green-Tao alleged theorem, all of these alleged proofs that use
Indirect NonExistence are
not valid math proofs but are fakes.
Now some of them are really true ideas that require a math proof that
is true such as 4Color
Mapping requires only the Jordan Curve Theorem. But most of the above
fakes are never
going to have a valid proof because the conjecture was false from the
start. FLT and Poincare
Conjecture were false ideas just as Green-Tao's idea that you can have
any length (and thus
infinite) of a progression of primes is a false idea.
So it is a fortuitious time that someone (me) intercedes and stops
this nonsense of having
fake proofs in mathematics. For example, if anyone believes that the
Green-Tao is a true
piece of mathematics would be only a short step away from a proof of
the Infinitude of Twin
Primes since we have an infinitude of a N-tuple class of primes. But
from all indications of
mathematics, the Twin Primes are not infinite but are finite since the
set of all primes thins
out the further out one goes in counting, indicates that as the Twin
Primes dies out, that some
higher order N-tuple primes covers up the death of the Twin Primes,
then the same for the
Quad primes, then the Hex primes. So Logic says that Twin Primes are
more than likely a
finite set which dies out while some N-tuple class takes over, for
only that sort of scenerio
can explain why primes become rare and thin out the higher we go yet
still be infinite.
So it is good that I intercede and put a brake on this nonsense in
mathematics of the
Indirect NonExistence proof method for it was never a valid method of
mathematics, and
that mathematics had become a warehouse of fake proofs for almost a
century now.
Jesse F. Hughes
> So I have come to a position in this book where I intersect with the
> Constructivist School and their concern over "existence proofs." And
> my concern allows us to actually tell whether the Constructivists
> were correct and whether the Mainstream was wrong. So far the
> Constructive School has been more of a philosophy or attitude about
> what is acceptable and unacceptable. But my position with Indirect
> Existence and Indirect NonExistence is that we can actually test the
> Constructivism School against the Mainstream and one will win and
> the other will lose. And so far it looks as though the
> Constructivism School is going to win and thus take the position as
> the rightful and true mathematics school.
Wonderful, but constructivists agree that the following form
Assume (Ex)P
Derive Q & ~Q
Conclude ~(Ex)P
is valid, and it is this form (not your silly attempts at a
Fitch-style form) that you've been calling "indirect non-existence".
--
Jesse F. Hughes
"Time and again, history has shown that people who think their beliefs
trump reality lose, and lose badly. Luckily, I don't have to listen
to you." -- James Harris on reality avoidance
Marshall
> --- quoting from a website that discusses RAA ---http://www.iep.utm.edu/reductio/
>
> The mathematical school of so-called intuitionism has taken a definite
> line regarding the limitation of reductio argumentation for the
> purposes of existence proofs. The only valid way to establish
> existence, so they maintain, is by providing a concrete instance or
> example: general-principle argumentation is not acceptable here. This
> means, in specific, that one cannot establish (∃x)Fx by deducing an
> absurdity from (∀x)~Fx.
Can anyone tell me if it is possible to have a model in which
(∀x)~Fx is false and (∃x)Fx is also false? I don't see how
it could be possible; what am I missing? And if it's not
possible, I don't see how the above intuitionist objection
could be considered to have any value. Is their assertion
a mathematical one, or is it dogma?
Marshall
Aatu Koskensilta
> Can anyone tell me if it is possible to have a model in which
> (∀x)~Fx is false and (∃x)Fx is also false? I don't see how it could
> be possible; what am I missing?
To assert
~(x)~Fx --> (Ex)Fx
is, in constructive mathematics, to assert we have a method for
extracting from a proof of an absurdity from (x)~Fx an object x such
that Fx. In general, there is no apparent way of doing so, and hence
this classically valid principle is not justified on the
constructivist reading. (If you're classically minded, look up Kripke
models, which provide a (classical) semantics for intuitionistic
logic.)
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Marshall
> Marshall <marshall.spi...@gmail.com> writes:
> > Can anyone tell me if it is possible to have a model in which
> > (∀x)~Fx is false and (∃x)Fx is also false? I don't see how it could
> > be possible; what am I missing?
>
> To assert
>
> ~(x)~Fx --> (Ex)Fx
>
> is, in constructive mathematics, to assert we have a method for
> extracting from a proof of an absurdity from (x)~Fx an object x such
> that Fx.
OK. I understood that. This is a redefinition of the symbols
from their more usual meaning, though, isn't it? Or perhaps
an overlay on top of and in addition to the usual meaning.
It seems a confusing terminological approach, but every
human endeavor is full of such.
Perhaps I could re-ask my question in a slightly different
way? Is it possible for a mathematical structure
http://en.wikipedia.org/wiki/Structure_(mathematical_logic)
to exist in which (∀x)~Fx is false and (∃x)Fx is also false?
It seems to me that it is clearly not possible.
> In general, there is no apparent way of doing so, and hence
> this classically valid principle is not justified on the
> constructivist reading.
Sure, but does it then follow that there is no useful
distinction to be had between knowing how to find
a value x such that Fx and knowing that such a value
exists? It seems to me that such a distinction is a
useful one, in which case the constructivist stricture
seems dubious, unreasonable, anti-pragmatic.
Or maybe it's an arbitrary restriction placed
into a specific context in order to derive some
specific benefit within that context? For example,
one might specify a restricted computational
model, and intentionally reduce the power of the
primitives in order to guarantee termination of
all programs?
I just don't understand the motivation, nor does
the reasoning seem justified.
> (If you're classically minded, look up Kripke
> models, which provide a (classical) semantics for intuitionistic
> logic.)
Alas, that sort of thing makes my eyes glaze over.
Marshall
Archimedes Plutonium
Marshall wrote:
>
> Can anyone tell me if it is possible to have a model in which
> (∀x)~Fx is false and (∃x)Fx is also false? I don't see how
> it could be possible; what am I missing? And if it's not
> possible, I don't see how the above intuitionist objection
> could be considered to have any value. Is their assertion
> a mathematical one, or is it dogma?
>
>
> Marshall
You make it sound as if it is terribly difficult to fetch such a model
when it is quite easy. That is why I lead into this discussion with
plenty of examples:
The entire science subject of Paleontology is a case of Indirect
Existence
versus Indirect NonExistence.
So, for (∀x)~Fx is false and (∃x)Fx is false.
Model 1:
Every case of fossil bones has no record indication of rock throwing
Homo sapiens
species different from any hominid species that is not rock throwing.
Thus Human evolution never evolved stonethrowing bone anatomy.
In short, this example says that because there is no fossil to prove
Stonethrowing
then humans never evolved via Stonethrowing.
Model 2:
Every case of fossil bones has no record of warm blooded dinosaurs,
and all cases
of fossil bones show cold blooded dinosaurs. Thus, there existed a
warm blooded
dinosaur is false.
Model 3:
Using the Pagels's book Cosmic Code pages 254-261:
Every case of of turning off the EM force by a demon ends up with the
disappearance
of atoms so that no atom exists. However, because there is the Strong
Nuclear
force and the Weak Nuclear force, that their prescence causes atoms to
come
into existence.
Model 4:
The Hensel P-adics are a representation of the Reals. For every Hensel
P-adic there
are no transcendentals nor irrationals, yet the idempotents and sqrt
(-1) in 5-adics exist.
Model 5:
For every physics case study of a nearby star, there never was a
gravity wobble of its
orbit so the existence of a exoplanet is false. But just because we
cannot measure a
wobble does not mean no exoplanet exists in that nearby star!
Indirect NonExistence is what the Constructivists object to, and this
is the first time
someone has put a finger on the source of the problem. Within the RAA
argument or
proof lies a statement either for the existence of something or
against the existence
of something. The proper care whether it is existence or nonexistence
belies a
path of validity or fakery.
Archimedes Plutonium
Aatu Koskensilta wrote:
> Marshall <marshal...@gmail.com> writes:
>
> > Can anyone tell me if it is possible to have a model in which
> > (∀x)~Fx is false and (∃x)Fx is also false? I don't see how it could
> > be possible; what am I missing?
>
> To assert
>
> ~(x)~Fx --> (Ex)Fx
>
> is, in constructive mathematics, to assert we have a method for
> extracting from a proof of an absurdity from (x)~Fx an object x such
> that Fx. In general, there is no apparent way of doing so, and hence
> this classically valid principle is not justified on the
> constructivist reading. (If you're classically minded, look up Kripke
> models, which provide a (classical) semantics for intuitionistic
> logic.)
>
> --
> Aatu Koskensilta (aatu.kos...@uta.fi)
This is a day in the sun for constructivists Aatu, and not one to get
buried
in remote abstracts.
Show models that are convincing, not that are remote cryptic.
The problem in Kripke's days was that Constructivism did not have
the use of Truth-Tables to further their argument of their school of
thought.
Theirs was just an "alternative" school to the mainstream school.
What I provide with the Indirect Existence versus Indirect
NonExistence
is an actual testing between the Constructivists and the Mainstream.
This
reminds me of the EPR versus Bohr before John Bell came along to
provide
a test between EPR and Bohr with the Bell Inequality, so that either
EPR
was true or it was false and we find out that Bohr was correct.
So in much the same way as Bell Inquality provided a test between two
opposing ideas in physics. That AP's Indirect Existence versus
Indirect NonExistence provides a test as to whether Constructivists
were
all along correct in their suspicions about math existence proofs or
whether
the mainstream was correct all along.
It looks as though the Constructivists were correct all along.
For if anyone examines Indirect Method Proofs, that the inevitable
existence or
nonexistence of something occurs within the "proper steps of the
proof." For example
the Euclid Infinitude of Primes proof has to have the existence of the
Euclid Number
in the Indirect Method and so it is a Indirect Existence proof. But in
the case
of Wiles's FLT or Hales's KPP or Appel& Haken's 4Color or Green-Tao's
prime progression
there is not an arguement for the Indirect Existence of something but
an Indirect
NonExistence of something.
So the day in the Sun for Constructivists has arrived. Where they now
can take
over the mainstream of mathematics and depose the old-mainstream as
fraudsters.
So let us show respect for the Constructivists who have been quietly
and patiently
set forth their case for over a century and who are no poised to
become the
mainstream of mathematics, not the outsiders anymore.
So, Aatu, provide crisp and clear models, not some remote abstract
examples.
Provide examples as to why not have the Moebius theorem that says 4
mutual adjacencies
is the upper limit of mutual adjacencies, why not this theorem as the
heart of the proof
of 4 Color Mapping rather than the ridiculous Indirect NonExistence
offered by Appel & Haken.
(Sorry, that I wrote Jordan Curve theorem rather than Moebius theorem
for 4 Color Mapping
a few posts back.)
Now is the day in the Sun, Aatu, for the Constructivists, so provide
crisp and clear examples.
Jan Burse
> Marshall <marshal...@gmail.com> writes:
>
>> Can anyone tell me if it is possible to have a model in which
>> (∀x)~Fx is false and (∃x)Fx is also false? I don't see how it could
>> be possible; what am I missing?
>
> To assert
>
> ~(x)~Fx --> (Ex)Fx
>
> is, in constructive mathematics, to assert we have a method for
> extracting from a proof of an absurdity from (x)~Fx an object x such
> that Fx. In general, there is no apparent way of doing so, and hence
> this classically valid principle is not justified on the
> constructivist reading. (If you're classically minded, look up Kripke
> models, which provide a (classical) semantics for intuitionistic
> logic.)
>
Yes, because otherwise double negation elimination holds,
which is not intuitionistically valid in general. The
double negation is the propositional image of the above,
when you take an F where x does not occure in it.
So the above is not really an insight into intuitionistcal
quantifiers, its more a property of the intuitionistic
implication.
Actually it is not so easy to find examples where intuitionism
vs classical matters for quantifiers. One such instance comes
from failure of Glivenko's theorem in intuitionistic first order
logic. And it is:
(x)~~Fx --> ~~(x)Fx
Which is not intuitionistically valid. I am not so sure whether
all given first order kripke models for intuitionistic first
order logic respect this correctly.
Best Regards
Jan Burse
> I am not so sure whether all given first order kripke models
> for intuitionistic first order logic respect this
> correctly.
At least when I was playing around with it, I made some
errors. Lets make a brief claim Aatus example and may claim
that it less shows intuitionistic Quantifiers is related
to Barcans formula (which does hold intuitionistically),
and my example is related to the converse Barcan formula
(which does not hold intuitionistically).
Best Regards
Ramsay
|OK. I understood that. This is a redefinition of the symbols
|from their more usual meaning, though, isn't it? Or perhaps
|an overlay on top of and in addition to the usual meaning.
|It seems a confusing terminological approach, but every
|human endeavor is full of such.
Well, "more usual" seems to mean essentially "most popular".
Do we want to make it a popularity contest? The intuitionist
interpretation of the logical connectives and quantifiers is
much more natural than it seems at first glance if you're
only familiar with classical logic (the "usual").
|Perhaps I could re-ask my question in a slightly different
|way? Is it possible for a mathematical structure
|
|http://en.wikipedia.org/wiki/Structure_(mathematical_logic)
|
|to exist in which (∀x)~Fx is false and (∃x)Fx is also false?
|It seems to me that it is clearly not possible.
It's intuitionistically valid to reason that if (Ex)Fx is
false then (Ax)~Fx is true. So indeed, they cannot both
be false.
Intuitionism distinguishes between "they cannot both be
false" and "one of them is true". If the two statements
are A and B, then "they cannot both be false" is
~(~A & ~B) while "one of them is true is A v B. The
former is equivalent to ~~(A v B). The latter implies
that there's a way to find one of them that is true.
|Sure, but does it then follow that there is no useful
|distinction to be had between knowing how to find
|a value x such that Fx and knowing that such a value
|exists? It seems to me that such a distinction is a
|useful one, in which case the constructivist stricture
|seems dubious, unreasonable, anti-pragmatic.
The distinction is still there. You could for example
be told by a reliable person that something exists
without being told how to find it.
Intuitionism and constructive mathematics generally
is more systematic about making distinctions, not
less. When you "know that a value exists" classically,
you are making some kind of discovery, which you have
arranged to describe using "existential" language.
If (as is typically the case) what has happened is that
you've found a contradiction arises from assuming that
no such value exists, then might it not be more
illuminating to say ~(Ax) ~Fx rather than Ex Fx?
That strikes me as being a very direct way of stating
what you know, as opposed to "one exists but I don't
know how to find it", which could be the case if
someone else has found out how to find it but only
told you that it is possible. The apparent advantage
of using classical reasoning is just that you get to
"simplify" ~(Ax)~Fx down to (Ex)Fx, but then have this
side explanation for why (Ex)Fx has a weaker meaning
for you than constructive mathematics gives it. And
then you have this additional way of boosting up the
meaning of the quantifier by saying, "and I know how
to find it". I don't find it obviously easier to go
through all those maneuvers just to say either that
~(Ax)~Fx or (Ex)Fx.
Keith Ramsay
Marshall
Thanks for the detailed reply.
I guess what I'm fixating on is what kind of values the
logic is manipulating. Classically, we have two values,
which certainly makes it the simplest system possible.
Clearly we have a more complicated semantics intuitionistically,
and I'm unclear what that might mean. It seems the
usual perspective doesn't care about that so much
however; maybe I'm just stuck on the wrong question.
Marshall
Nam Nguyen
> On Oct 11, 12:07 pm, Marshall <marshall.spi...@gmail.com> wrote:
>
> |Sure, but does it then follow that there is no useful
> |distinction to be had between knowing how to find
> |a value x such that Fx and knowing that such a value
> |exists? It seems to me that such a distinction is a
> |useful one, in which case the constructivist stricture
> |seems dubious, unreasonable, anti-pragmatic.
>
> The distinction is still there. You could for example
> be told by a reliable person that something exists
> without being told how to find it.
> When you "know that a value exists" classically,
> you are making some kind of discovery, which you have
> arranged to describe using "existential" language.
There's a question (an issue) related to this conversation
that I've thought of for a while but don't know for certain
if the answer would be correct.
Basically is it possible that, e.g. arithmetically, ExPx is
true without a concrete example? For instance, is it conceivable
that ~GC is true without the need of citing one single example?
To understand the question, let's take a look at the 2 related
formulas: GC and cGC (cGC is the counter-Goldbach formula defined
as cGC df= "There are infinite counter examples against GC).
If GC has proof in, say Q, the proof is finite of course.
But it's conceivable if GC isn't provable it's so because
its "proof" is infinite! We can see it that way by defining
GCn as a reduction of GC over the first n even numbers >= 4.
If by proof length we mean the total number of symbols used
in the proof then as n increases, the proof length of each GCn
would also increase. Which means the overall "proof" length
of GC is infinite and hence GC would have no proof (though it
could be true!).
By the very same analogy, it's conceivable that cGC isn't provable
(though true) because its proof is infinite. Now, with GC we could
prove ~cGC and with cGC we could prove ~GC. So, if Q be consistent
- or equivalently our claimed "the naturals" be indeed a model
of Q - either GC or cGC but not both is true in Q.
Suppose then cGC be true in the naturals (as a model of Q)
then it's conceivable that there's no proof for cGC. And
since it's also conceivable that the shortest proof of ~GC(x)
is:
cGC -> ~GC(x)
[where of course ~GC is of the form ExPx.]
---
In summary, it's conceivable that ExPx is true without us ever be
able to find one _concrete_ example x where Px is true!
Strange indeed!
Nam Nguyen
[I meant to say further:]
it's also conceivable that ~GC is true but we don't _know_ why!
Nam Nguyen
>
> In summary, it's conceivable that ExPx is true without us ever be
> able to find one _concrete_ example x where Px is true!
>
> Strange indeed!
>
Imho, that it's entirely possible there exists a counter example to
Goldbach Conjecture while it's impossible to know a specific value
for the counter example should be a wake-up call to those who happen
to believe our reasoning based on the naturals is free of any
philosophical consideration.
Two foundational building blocks of modern FOL reasoning are:
(a) The certainty of symbol manipulation-game known as proofs.
It's here that it's either yes or no if there is a proof
and there's simply nothing else.
(b)The uncertainty of model interpretation. It's here about certain
arithmetical/mathematical complexity of infinity that the truth
of some statements is not reachable via our cognitive capability.
Hence our model interpretations can never be as certain as being
absolute and anything we _assert_ in making interpretations about
these statements are subjective and philosophical.
If we adhere to the limitations in finite reasoning implied by (a),
we have to accept, e.g., the philosophical position that it's impossible
to know every mathematical facts, such as the syntactical (in)consistency
of Q.
If we don't adhere to the above "philosophy" about reasoning limitation,
and adopt another philosophy that we know what the naturals be, then it'd
be an accepted philosophy with consequences about our unknownability:
such as a why GC is true, or ~GC is true without a concrete example.
But in either stands, there's still limitation in our reasoning: in our
knowledge in syntactical provability, and in our making assumption - and
not facts - about what the naturals be.
Archimedes Plutonium
Nam Nguyen wrote:
> Nam Nguyen wrote:
>
> >
> > In summary, it's conceivable that ExPx is true without us ever be
> > able to find one _concrete_ example x where Px is true!
> >
> > Strange indeed!
> >
>
> Imho, that it's entirely possible there exists a counter example to
> Goldbach Conjecture while it's impossible to know a specific value
Here are six counterexamples to the Goldbach Conjecture:
60000....00000
80000....00000
100000....00000
99999.....999998
99999.....999996
99999.....999994
Counterexamples for Fermat's Last Theorem:
Here are an infinite number of solutions to FLT:
The expression a^n+b^n=c^n is true for all n,
given the following values.
a= ...9977392256259918212890625
b= ...0022607743740081787109376
c= ...0000000000000000000000001
When your view of Natural Numbers in mathematics is only from the
Backview
with never an acknowledgment of the FrontView, well, you live in
closet attic space.
Open up the windows, open up the door, go outside and take in both the
FrontView
along with your solo view of the rearend.
Jesse F. Hughes
> Counterexamples for Fermat's Last Theorem:
>
> Here are an infinite number of solutions to FLT:
> The expression a^n+b^n=c^n is true for all n,
> given the following values.
> a= ...9977392256259918212890625
> b= ...0022607743740081787109376
> c= ...0000000000000000000000001
>
> When your view of Natural Numbers in mathematics is only from the
> Backview with never an acknowledgment of the FrontView, well, you
> live in closet attic space.
Frontview? What are the front views of a and b?
--
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
A
Did you ever offer up a definition of "arrangement" that makes precise
your attempt at definition of the real numbers as "all possible digit
arrangements"?
Archimedes Plutonium
> Archimedes Plutonium <plutonium.archime...@gmail.com> writes:
> > Counterexamples for Fermat's Last Theorem:
>
> > Here are an infinite number of solutions to FLT:
> > The expression a^n+b^n=c^n is true for all n,
> > given the following values.
> > a= ...9977392256259918212890625
> > b= ...0022607743740081787109376
> > c= ...0000000000000000000000001
>
> > When your view of Natural Numbers in mathematics is only from the
> > Backview with never an acknowledgment of the FrontView, well, you
> > live in closet attic space.
>
> Frontview? What are the front views of a and b?
>
a = 26098212819......9977392256259918212890625
b = 73901787180......0022607743740081787109376
I suppose a man of your wit needs the frontview of c also.
Jesse F. Hughes
> On Oct 13, 6:41 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Archimedes Plutonium <plutonium.archime...@gmail.com> writes:
>> > Counterexamples for Fermat's Last Theorem:
>>
>> > Here are an infinite number of solutions to FLT:
>> > The expression a^n+b^n=c^n is true for all n,
>> > given the following values.
>> > a= ...9977392256259918212890625
>> > b= ...0022607743740081787109376
>> > c= ...0000000000000000000000001
>>
>> > When your view of Natural Numbers in mathematics is only from the
>> > Backview with never an acknowledgment of the FrontView, well, you
>> > live in closet attic space.
>>
>> Frontview? What are the front views of a and b?
>>
>
> a = 26098212819......9977392256259918212890625
> b = 73901787180......0022607743740081787109376
>
> I suppose a man of your wit needs the frontview of c also.
>
No, but I'd sure be pleased if you could state n and confirm that
raising these positive integers to the power n and adding them yields
a much smaller positive integer. I'm not too good at calculating with
these buggers.
--
"And I'll reinforce the point that you are an enemy of humanity, that
my predecessors are people like Gauss, Euler, Newton, Archimedes and
others who you are spitting upon as you do it to me by trying to keep
their discipline trashed as it is now." -- James S. Harris
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped)
> > > Counterexamples for Fermat's Last Theorem:
> >
> > > Here are an infinite number of solutions to FLT:
> > > The expression a^n+b^n=c^n is true for all n,
> > > given the following values.
> > > a= ...9977392256259918212890625
> > > b= ...0022607743740081787109376
> > > c= ...0000000000000000000000001
>
> a = 26098212819......9977392256259918212890625
> b = 73901787180......0022607743740081787109376
>
There was a genuine use for Hensel p-adics afterall, as a guidance.
To me the Hensel p-adics are simply another way of writing the Reals,
just as the Reals in decimal form or the Reals as Fractions are just
different forms of writing the Reals.
But I want to say that the above when added together is not, and
then again, can be the
number 10000....00001 or the number 0000....00001. So this is
confusing
since the idea of FrontView with BackView to all numbers of
mathematics.
So that in p-adics one sees the addition of a + b as a smaller number
than either a or b.
So what is really happening? I do not have all the answers. I do know
we are
past 266! which is safe realm of Finite and beyond the Incognitum
which is
far beyond 266! and then far beyond the Incognitum with such as the
above
a,b and numbers like 9999....99999.
One way of clarifying the above is to consider that the infinity of
the Counting
Numbers goes from 0 to 1 to 2 to 3 all the way up to 9999...9998 then
9999...9999
and then add 1 more and we have what I called the South Pole number of
10000....00000 where 0 is the North Pole number. Geometry is tied to
numbers
and that as you go to infinity, numbers bend around and come back to
where
they started from. That was my other book, and I do not have the time
to discuss
it here.
Only to say that some will wonder how a + b in the above can be a
smaller number
than c?
But if you look at it in terms of geometry that the adding of a + b
above is one unit
beyond 9999....99999 And let us suppose we put a decimal point in
front of a and b
and we have thus:
a' = 0.26098212819......9977392256259918212890625
b' = 0.73901787180......0022607743740081787109376
Then what we end up with in a' + b' is 1.00.... or the number 1.
So we can begin to see that all the Counting Numbers from 0 to 1 to
999...9999
are all represented by a Real Number between 0 and 1.00.....
So that there exists two Real Numbers a' and b' above that obey the
Fermat's
Last Theorem. But that is no news to anyone since FLT has all
solutions in
Reals.
What I am showing in the above counterexamples is that when
mathematics refuses
to define Finite versus Infinite, then the Counting Numbers are just
the Reals.
If you define Finite, as any proper well behaved mathematician should
do, since
math is after all the "science of precision" then if you define Finite
as that of
266! and all math proofs have to be proven only in the sphere of
Finiteness,
since Physics ceases to exist beyond Finite, that FLT is true to 266!
and Goldbach
is probably also true to 266!
When math ceases to be the science of precision, and ignores defining
what it
is to be Finite, then math has these fake proofs and a subject gone
berserk.
So back to the Indirect NonExistence and anyone can read my other book
AP-adics.
Archimedes Plutonium
Archimedes Plutonium wrote:
> Archimedes Plutonium wrote:
> (snipped)
>
> > > > Counterexamples for Fermat's Last Theorem:
> > >
> > > > Here are an infinite number of solutions to FLT:
> > > > The expression a^n+b^n=c^n is true for all n,
> > > > given the following values.
> > > > a= ...9977392256259918212890625
> > > > b= ...0022607743740081787109376
> > > > c= ...0000000000000000000000001
>
> >
> > a = 26098212819......9977392256259918212890625
> > b = 73901787180......0022607743740081787109376
>
> a' = 0.26098212819......9977392256259918212890625
> b' = 0.73901787180......0022607743740081787109376
>
> Then what we end up with in a' + b' is 1.00.... or the number 1.
I really should proofread before I hit the "send key".
The above Hensel p-adic would equal 1000....00001 but the Hensel
p-adics have no FrontView so to the Hensel p-adics a + b = 000...0001
In the case of a' + b' of Reals between 0 and 1 that should read
1.000...0001
Archimedes Plutonium
this
moment and not terribly advancing on the Indirect NonExistence
research.
But in mathematics, it so often happens that we run a few examples
through
our minds and come to a statement such as "hard to imagine Goldbach
Conjecture
not being true". Or we run through some examples of FLT and similarly
saying hard to imagine it being false. But this is the problem of
mathematics
that has only a BackView of the numbers it wants to talk about or in
the case
of the Reals where it has only the FrontView of each Real and never
the BackView.
In the case of Goldbach Conjecture, no honest and sincere
mathematician ever
raised these doubts. That if the primes become thinner and thinner and
rare and more
rare the higher you go in the numbers. Here I am not talking about
10^10^10^10 or
such staircases of exponents for that is a tiny number compared to say
the number
2000....000000 or 4444....444444, not to mention numbers like
99999....99999.
So in the history of mathematics where all those minds focused on
Goldbach Conjecture,
that none seemed to have had these thoughts:
1) if the Counting Numbers go from 0 to 1 to 2 to all the way up to
999...9999
2) if the Primes thin out as the Prime Number Theorem x/Ln(x)
Then given those two facts that we can accept, then it is very hard
and extremely
difficult to see how every even number is the sum of two primes.
If the Prime Number Theorem is true and the Counting Numbers go from 0
to
9999....99999 then there maybe a infinite-substring of all composites
say from
about 97......97 to 9999....99999 there were no more primes. Under
such a
rarefied thinning out where there is an infinite sequence of nothing
but composites
it is extremely difficult to see how a Goldbach Conjecture could ever
hold up under
that sort of stress and strain.
To be sure, there are an infinitude of primes but given an interval
that is infinitely
long of composites, and since the primes thin out, then the Goldbach
Conjecture,
on an honest appraisal should be seen as false.
Archimedes Plutonium
Jesse F. Hughes wrote:
> Archimedes Plutonium <plutonium....@gmail.com> writes:
>
> > On Oct 13, 6:41 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Archimedes Plutonium <plutonium.archime...@gmail.com> writes:
> >> > Counterexamples for Fermat's Last Theorem:
> >>
> >> > Here are an infinite number of solutions to FLT:
> >> > The expression a^n+b^n=c^n is true for all n,
> >> > given the following values.
> >> > a= ...9977392256259918212890625
> >> > b= ...0022607743740081787109376
> >> > c= ...0000000000000000000000001
> >>
> >> > When your view of Natural Numbers in mathematics is only from the
> >> > Backview with never an acknowledgment of the FrontView, well, you
> >> > live in closet attic space.
> >>
> >> Frontview? What are the front views of a and b?
> >>
> >
> > a = 26098212819......9977392256259918212890625
> > b = 73901787180......0022607743740081787109376
> >
> > I suppose a man of your wit needs the frontview of c also.
> >
>
> No, but I'd sure be pleased if you could state n and confirm that
> raising these positive integers to the power n and adding them yields
> a much smaller positive integer. I'm not too good at calculating with
> these buggers.
>
Admittedly those are difficult for anyone who never worked with Hensel
P-adics and maybe I should abandon them in future discourse as too
remote for newcomers.
So let me start doing counterexamples of the easier type for FLT. So
that
even a High School student of math can derive his/her own
counterexample to FLT.
So let me start the ball a rolling with
a = 0.1
b = 0.2
And I use numbers between 0 and 1 so I can easily convert to Counting
Numbers
and I use the fact that the Reals are algebraically complete so I can
chose any
a, and b and n and find out what the c is going to be.
and I chose the exponent n = 3
So I have
0.1^3 + 0.2^3 = c^3
and the cube root(0.009) = 0.208008382
So what counting number or positive integer is c to satisfy that
equation?
So, now converting those to Counting Numbers I have
a = 1000....00000
b= 20000....00000
n = 000...00003
c = 208008382......
And I do not have the BackView of c
Trop
<plutonium.archime...@gmail.com> wrote:
> Sorry if it seems like I cut and paste too much of
> Wikipedia. I want to give a sense of the state of
> knowledge on this topic of truth in logic and math.
>
> In my search to see if anyone covered Indirect
> NonExistence, about the closest that it seems anyone reached this
> topic is the Constructivist
> School of Logic.
>
> --- quoting from Wikipedia on Constructivist School ---http://en.wikipedia.org/wiki/Nonconstructive_proof
>
> demonstrates the existence of a mathematical object with certain
> properties by creating or providing a method for creating such an
> object. This is in contrast to a nonconstructive proof (also known as
> an existence proof or pure existence theorem) which proves the
> existence of a mathematical object with certain properties, but does
> not provide a means of constructing an example.
>
> (snip)
>
> Constructivism is the philosophy that rejects all but constructive
> proofs in mathematics. Typically, supporters of this view deny that
> pure existence can be usefully characterized as "existence" at all:
> accordingly, a non-constructive proof is instead seen as "refuting the
> impossibility" of a mathematical object's existence, a strictly weaker
> statement.
>
>
> This proof is non-constructive because it relies on the statement
> "Either q is rational or it is irrational" — an instance of the law of
> excluded middle, which is not valid within a constructive proof. The
> non-constructive proof does not construct an example a and b; it
> merely gives a number of possibilities (in this case, two mutually
> exclusive possibilities) and shows that one of them — but does not
> show which one — must yield the desired example.
>
> (snip)
>
> A more substantial example is the graph minor theorem. A consequence
> of this theorem is that a graph can be drawn on the torus if, and only
> if, none of its minors belong to a certain finite set of "forbidden
> minors". However, the proof of the existence of this finite set is not
> constructive, and the forbidden minors are not actually specified.
> They are still unknown.
> --- end quoting Wikipedia ---
>
> is not valid in a constructive proof.
>
> What I aim to show is that Indirect NonExistence
> is an invalid proof argument, no matter what school
> of mathematics one pledges allegiance.
>
> Archimedes Plutoniumwww.iw.net/~a_plutonium
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
What I agree with is
> So the day in the Sun for Constructivists > has arrived. Where they now
> can take
> over the mainstream of mathematics and > depose the old-mainstream,
and that is, particularly, because of computer science - main child of
intuitionism and constructivism. Hilbert's foundational program is
reduction of all of intuitive (non-constructive) math to something
obvious for all mathematicians (including intuitionists). Namely, to
constructible or calculable entities. And when you would try to define
what is the constructible or calculable entities you inevitably will
come up with notion of algorithm in some form. As a result we have a
Turing machine, programming languages, formal languages etc. Only
today the official science has recognized the particular role of
Brouwer's work on intuitionism.
But as for your position, it is very obscure to me. I can't see any
precise definition of your notion of indirect non-existence proof. Is
that what Jesse F. Hughes wrote:
Assume (Ex)P
Derive Q & ~Q
Conclude ~(Ex)P
?
If it isn't, can you define it and show it in more concrete
mathematical proof example?
> FLT and Poincare
> Conjecture were false ideas just as
> Green-Tao's idea that you can have
> any length (and thus
> infinite) of a progression of primes is a
> false idea.
Are those things ideas? I think they are statements.
Trop
Keith Ramsay
|I guess what I'm fixating on is what kind of values the
|logic is manipulating.
It depends on what you mean.
Propositional or predicate calculus is always manipulating
propositions or predicates, whether you are thinking of it
as classical or intuitionistic. You get away with supposing
in the classical case that "ultimately" the propositions
all have "truth value" either "true" or "false" because
you have assumed (pretty much by fiat) that they do, but
it's not clear what the really gets you. Constructivists
sometimes think differently about what propositions are,
but it's somewhat difficult to convey. The kind of
verificationist philosophy described in Dummett's Elements
of Intuitionism might give you an idea.
For some purposes you can think of classical logic as
having "values" in Boolean algebras and intuitionistic
logic as having "values" in Heyting algebras. (That's
essentially just a restatement of certain logical laws
in algebraic form, though.)
Kripke models provide a way to think about intuitionistic
logic. Instead of thinking about what is true now, think
about what might or might not be discovered at future
stages in the game. Then without taking off your classical
thinking cap you can get a prosaic way of thinking about
intuitionistic logic.
Keith Ramsay
Keith Ramsay
|Basically is it possible that, e.g. arithmetically, ExPx is
|true without a concrete example? For instance, is it conceivable
|that ~GC is true without the need of citing one single example?
Making this into a precise question might take some
work. Thinking constructively, an arithmetic claim
of the form Ex P(x) just says that there exists an
example, and being an integer it is concrete.
Thinking classically, you could well prove the
existence of an integer satisfying P without being
able to prove that any specific integer satisfies P.
There's a metatheorem ensuring that if you can prove
a statement of the form Ex P(x) where P(x) has all
of its quantifiers bounded (so that whether P(x) is
true is computable) (in a system S taken from some
wide assortment of classical systems) then there is
also a constructive proof (in a corresponding
constructive system S+). Even without a metatheorem
to help, though, one would usually reason (classically)
that the existence of such an x means that we can
find it by checking integers one by one.
If you have a more complicated predicate, there's no
general guarantee. It's conceivable for example that
one could prove (classically) that there are finitely
many Fermat primes without being able to prove
specifically how many there are.
BTW let's please set follow-ups so that sci.physics
doesn't have to put up with this discussion.
Keith Ramsay
Marshall
Again, thanks; this is very helpful.
Marshall
spudnik
to Goldbach's conjecture," without some precipitating idea, of How;
most of us'd just assume, that such a hypothesis might be used
to prove that the cojecture is true, by a direct contradiction.
as for such a proof of the fiveness of the Fermat-Gauss primes,
they have already been "mapped" to the regular polyhedra, but
that is only a possible part of an "i.a.o.i" or iff proof,
of necessity & sufficiency. (note that one can use the words
in many ways, to assign one of other quality to either
of the twp parts of the proof, per the article in *MM*;
it is largely a matter of linguacy, subsumed
under the hare-brained study of the *trivium* in grade-school --
blah !-)
> If you have a more complicated predicate, there's no
> general guarantee. It's conceivable for example that
> one could prove (classically) that there are finitely
> many Fermat primes without being able to prove
> specifically how many there are.
thus:
Hales' proof is unfinished, as covertly acknowledged
by the existence of his "Flyspeck" program, although
it is actually admitted, here & there (in Szpiro's book e.g.)
I'd already stated this in an item that JSH started, and
abandoned.
"mathematical proof" was defined to most satisfaction
by Leibniz, as proving "necessity & sufficiency;" at least,
if you can't do that, you'll still have a problem (although
proving only one or the other is also good:
"the tetrahedron" as a neccesity for 4 colors
on a map e.g.).
thus:
Michelson-Morley was not a null;
there were regular, very small anomalies over a year.
several researchers improved upon M-M's results, at least one
of whom had an accompanying theory.
[this was covered in the Larouchiac science magazine,
ne Fusion, but may not be online:
http://www.21stcenturysciencetech.com/ .-]
--Cap'n'trade is an arbitrageur's delight;
an actual tax on imported oil is a tariff e.g..
".... beyond that level (today, it is over 300 ppm),
the effectiveness of carbon dioxide as a greenhouse gas decreases
exponentially (Figure 1). This exponential
decline in the effectiveness of carbon dioxide as a
greenhouse gas is not contested by the believers in
global warming; they simply ignore it.
There is an experiment anyone can do to understand
this principle: Take a sheet of paper, and place it
over a window with sunlight coming through. You ...."
http://larouchepub.com/lar/Articles_2009/Articles_2009/Cap_and_Trade.pdf
http://larouchepub.com/lar/Articles_2009/Where_Punt_sp09.pdf
Nam Nguyen
> On Oct 12, 3:35 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
> |Basically is it possible that, e.g. arithmetically, ExPx is
> |true without a concrete example? For instance, is it conceivable
> |that ~GC is true without the need of citing one single example?
>
> Making this into a precise question might take some
> work.
In retrospect, I shouldn't have phrased the question via model
truth; intuitions and philosophical schools of thought seems
more of a hindrance and than assistance in exploring this
kind of questions, imho. Let me rephrase the issue in different
way (with different question), using finite syntactical proof
consideration.
Would it be possible that a formula F be undecidable in _any_
extension of the Robinson Arithmetic Q formal system?
Marshall
> Let me rephrase the issue in different
> way (with different question), using finite syntactical proof
> consideration.
>
> Would it be possible that a formula F be undecidable in _any_
> extension of the Robinson Arithmetic Q formal system?
How about that extension of Q that one obtains from
adding F?
Marshall
Nam Nguyen
There's no guarantee that Q + F isn't inconsistent, hence F
wouldn't be undecidable in that case.
Marshall
You don't actually know what any of these words mean,
do you? This is like trying to have a technical conversation
with John Jones.
Marshall
Nam Nguyen
> On Oct 17, 11:05 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>> Marshall wrote:
>>> On Oct 17, 4:28 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>>>> Let me rephrase the issue in different
>>>> way (with different question), using finite syntactical proof
>>>> consideration.
>>>> Would it be possible that a formula F be undecidable in _any_
>>>> extension of the Robinson Arithmetic Q formal system?
>>> How about that extension of Q that one obtains from
>>> adding F?
>> There's no guarantee that Q + F isn't inconsistent, hence F
>> wouldn't be undecidable in that case.
>
> You don't actually know what any of these words mean,
> do you?
Why do you think that?
> This is like trying to have a technical conversation
> with John Jones.
My original question is a yes-or-no question. You didn't answer
it; then asked me another question, which I technically replied to
you?
Why didn't you address technical issues, instead of acting like
a crank or a low-class Inquisitor?
Nam Nguyen
> Keith Ramsay wrote:
>> On Oct 12, 3:35 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>> |Basically is it possible that, e.g. arithmetically, ExPx is
>> |true without a concrete example? For instance, is it conceivable
>> |that ~GC is true without the need of citing one single example?
>>
>> Making this into a precise question might take some
>> work.
>
> In retrospect, I shouldn't have phrased the question via model
> truth; intuitions and philosophical schools of thought seems
> more of a hindrance and than assistance in exploring this
> kind of questions, imho. Let me rephrase the issue in different
> way (with different question), using finite syntactical proof
> consideration.
>
> Would it be possible that a formula F be undecidable in _any_
> extension of the Robinson Arithmetic Q formal system?
I'd like to take it back this question. That's not what I meant to
ask and of course there are inconsistent extensions of Q in which
F can't be undecidable.
Let me think it for a bit more before rephrasing the issue again.
Jan Burse
> Jan Burse schrieb:
>> I am not so sure whether all given first order kripke models for
>> intuitionistic first order logic respect this
> > correctly.
>
> At least when I was playing around with it, I made some
> errors. Lets make a brief claim Aatus example and may claim
> that it less shows intuitionistic Quantifiers is related
> to Barcans formula (which does hold intuitionistically),
> and my example is related to the converse Barcan formula
> (which does not hold intuitionistically).
>
> Best Regards
Ok, I tried to prove (x)~~Ax -> ~~(x)Ax in minimal logic
to see what is missing. And ended up that double negation
elimination will solve the problem.
Which was a little bit disappointing since I thought the
problem will show a dependency in between. So my suspicion
is now, that on the semantic side the way validity can be
defined and on the syntactic side some rules, although a
quantifier does not explicitly occur, statements about
quantifiers are already made.
Nevertheless I found an interesting paper, which I am
eager to read:
"M. J. Cresswell (1995). Incompleteness and the
Barcan Formula. Journal of Philosophical Logic 24 (4).
A (normal) system of prepositional modal logic is said
to be complete iff it is characterized by a class of
(Kripke) frames. When we move to modal predicate logic
the question of completeness can again be raised. It is
not hard to prove that if a predicate modal logic is
complete then it is characterized by the class of all
frames for the propositional logic on which it is based.
Nor is it hard to prove that if a propositional modal
logic is incomplete then so is the predicate logic based
on it. But the interesting question is whether a complete
propositional modal logic can have an incomplete extension.
In 1967 Kripke announced the incompleteness of a predicate
extension of S4. The purpose of the present article is to
present several such systems. In the first group it is the
systems with the Barcan Formula which are incomplete, while
those without are complete. In the second group it is those
without the Barcan formula which are incomplete, while those
with the Barcan Formula are complete. But all these are
based on propositional systems which are characterized by
frames satisfying in each case a single first-order sentence."
Aatu Koskensilta
> My original question is a yes-or-no question. You didn't answer it;
> then asked me another question, which I technically replied to you?
Surely you recall your yes-or-no question has been answered already?
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Aatu Koskensilta
> Nam Nguyen wrote:
>
>> Would it be possible that a formula F be undecidable in _any_
>> extension of the Robinson Arithmetic Q formal system?
>
> I'd like to take it back this question. That's not what I meant to ask
> and of course there are inconsistent extensions of Q in which F can't
> be undecidable.
No need to bring in inconsistent extensions. There is no formula F
undecidable in all consistent extensions of Robinson arithmetic.
Nam Nguyen
> Nam Nguyen <namduc...@shaw.ca> writes:
>
>> Nam Nguyen wrote:
>>
>>> Would it be possible that a formula F be undecidable in _any_
>>> extension of the Robinson Arithmetic Q formal system?
>> I'd like to take it back this question. That's not what I meant to ask
>> and of course there are inconsistent extensions of Q in which F can't
>> be undecidable.
>
> No need to bring in inconsistent extensions. There is no formula F
> undecidable in all consistent extensions of Robinson arithmetic.
>
So, which _syntactically_ consistent extension of Q that GC isn't undecidable?
Aatu Koskensilta
> Aatu Koskensilta wrote:
>
>> No need to bring in inconsistent extensions. There is no formula F
>> undecidable in all consistent extensions of Robinson arithmetic.
>
> So, which _syntactically_ consistent extension of Q that GC isn't
> undecidable?
Either Q + GC or Q + ~GC.
Nam Nguyen
> Nam Nguyen <namduc...@shaw.ca> writes:
>
>> My original question is a yes-or-no question. You didn't answer it;
>> then asked me another question, which I technically replied to you?
>
> Surely you recall your yes-or-no question has been answered already?
>
Really? I must have forgotten that in all the cheap-shot Inquisition
thrown at me.:-)
Nam Nguyen
> Nam Nguyen <namduc...@shaw.ca> writes:
>
>> Aatu Koskensilta wrote:
>>
>>> No need to bring in inconsistent extensions. There is no formula F
>>> undecidable in all consistent extensions of Robinson arithmetic.
>> So, which _syntactically_ consistent extension of Q that GC isn't
>> undecidable?
>
> Either Q + GC or Q + ~GC.
>
How do you know, e.g., Q + GC isn't inconsistent?
Aatu Koskensilta
> Aatu Koskensilta wrote:
>> Nam Nguyen <namduc...@shaw.ca> writes:
>>
>>> So, which _syntactically_ consistent extension of Q that GC isn't
>>> undecidable?
>>
>> Either Q + GC or Q + ~GC.
>
> How do you know, e.g., Q + GC isn't inconsistent?
I don't. How is this relevant?
Nam Nguyen
> Nam Nguyen <namduc...@shaw.ca> writes:
>
>> Aatu Koskensilta wrote:
>>> Nam Nguyen <namduc...@shaw.ca> writes:
>>>
>>>> So, which _syntactically_ consistent extension of Q that GC isn't
>>>> undecidable?
>>> Either Q + GC or Q + ~GC.
>> How do you know, e.g., Q + GC isn't inconsistent?
>
> I don't. How is this relevant?
>
Doesn't my my question above have "which _syntactically_ consistent
extension of Q"? I you don't know the extension Q + GC isn't syntactically
inconsistent then it could be inconsistent and in which case how would
that answer my the question?
Nam Nguyen
I hope the following would be a correct re-phrasing of the issue.
Syntactically speaking, would it be possible that a formula F be undecidable
in all consistent extensions of the Robinson Arithmetic Q formal system?
Aatu Koskensilta
> Doesn't my my question above have "which _syntactically_ consistent
> extension of Q"?
If GC is true then Q + GC is a consistent extension of Q in which GC is
decidable, and if it is false, then Q is a consistent extension of Q in
which GC is decidable. Our knowledge of the truth or falsity of GC is
irrelevant.
Nam Nguyen
> Nam Nguyen <namduc...@shaw.ca> writes:
>
>> Doesn't my my question above have "which _syntactically_ consistent
>> extension of Q"?
>
> If GC is true then Q + GC is a consistent extension of Q in which GC is
> decidable, and if it is false, then Q is a consistent extension of Q in
> which GC is decidable. Our knowledge of the truth or falsity of GC is
> irrelevant.
>
Be careful. Keep answering my question with *if*-this-then-and-*if*-that-then
and you might end up agreeing with me there's a meta mathematical statement
that has no absolute truth, which would signal the advent of relativity in
mathematical reasoning.
Note that I emphasized on "_syntactically_" to signal that the question and
the answer be discussed syntactically, not model-theoretically. Let me owe
you a coming back to your "If GC is true ..." paragraph above in a later in
the post but let me share here with you et al. the motivations of my
question(s).
In my recent replies to KR, the key question I have is:
>> Syntactically speaking, would it be possible that a formula F be
>> undecidable in all consistent extensions of the Robinson Arithmetic Q
>> formal system?
---> The Objective
The objective I have in this question is to establish a meta mathematical
statement (about FOL) that would have no absolute truth though in its form
it'd appear as if there would be an absolute truth value. Lets let (I) be
the statement:
(I) F is undecidable in all consistent extensions of the Q.
If the answer to my question above is "Yes, it's possible" then (I)
would be the meta statement whose truth is relativistic, as so as
one would find in an analogy in the following statement about physics:
(1) Two particular events e1 and e2 happened simultaneously.
---> The Strategy
So that no one could accuse me as doing a "philosophical" debate here,
(I) should be a genuine meta mathematical statement, which it will be
as long as F is a wff. But even after that, how could we entertain (I)
as being relativistic while we live in a binary world where the LEM
light clearly casts the dualistic exclusive shadows of "true" and
"false"? The answer and the strategy I'm proposing here is:
the layering of LEM in the hierarchy of first-order-to-meta-level
of reasoning. (And there are infinite levels in the hierarchy, in
the sense for any current level beginning with the first-order level,
there is always a meta level higher where we could make assertions
_about_ this current level.)
What this strategy means is that LEM application on a lower level
would be trumped/superseded by that of any higher level, in the sense
that it'd be _illogical_ to talk about truth value of a statement
in a level without knowing the prerequisite truth value of any higher
level.
For example, given the formula F' = Axy[x+y=0] and the formal system
T' = {F'} let's examine the the proof of F'' = (F' \/ ~(x=x)) in T',
in light of this strategy. What this strategy stipulates is that at
whatever a level we consider the question F'' be a theorem or not
(viz-a-viz LEM), in a higher level where contemplating _about_ this
question, the question _whether or not a proof is even possible at all_,
which is LEM in this higher level. In this particular instance, and
however ridiculously trivial or circular, I for one can determine
that it's _possible_ to answer the question whether or not F'' is a
theorem, and hence my answer would be logically valid.
---> In summary.
If the Moon were Blue any nonsensical conclusions could be had.
Similarly, it's equally nonsensical to assert an answer when,
in a higher meta level, it's genuinely impossible to assert an answer.
And that's a prelude to relativity in mathematical reasoning. For if
something is impossible, in the LEM binary world, we can relativize
any one of the 2 values, which is something we can't do if it's
possible. (And there are things that are possible in mathematical
reasoning, such as the proof of F'' above.)
Nam Nguyen
This paragraph missed the ending. I meant to say:
"...in a higher level where contemplating _about_ this question,
the question _whether or not a proof is even possible at all_ ...
must be pre-determined".
Marshall
>
> Syntactically speaking, would it be possible that a formula F be undecidable
> in all consistent extensions of the Robinson Arithmetic Q formal system?
Suppose F is independent of Q. Then any extension of Q that
includes F as an axiom is consistent, and decides F.
Suppose Q decides F. Then any consistent extension of Q
decides F. This is a purely syntactic analysis.
Did I get that right, Aatu? I like this game; it's fun.
Marshall
Nam Nguyen
Did you get what right? This is supposed to be a question with
a resounding "Yes", "No", or "I don't know" answer. Your answer
here is none of those!
I give you a hint. The answer to my question is piggying back on
the answer on the syntactical (in)consistency of Q. So it's a
prerequisite to answer the question on Q's (in)consistency,
before you even have a chance to _resoundingly_ answer my
question. (And even that would only give you a chance)!
> I like this game; it's fun.
If you say so. Why don't you share the fun with us all, _after_
you find out the _resounding_ answer on the (in)consistency of Q?
Keith Ramsay
|Be careful. Keep answering my question with *if*-this-then-and-*if*-
that-then
|and you might end up agreeing with me there's a meta mathematical
statement
|that has no absolute truth, which would signal the advent of
relativity in
|mathematical reasoning.
If you want to start talking about "absolute truth"
you need to have some kind of grasp of what you
mean by the term.
If I adjoin to Q a new primitive predicate A(n)
and provide no new axioms, then lots of statements
pertaining to A are then independent of Q. Is A(0)
true? Can't tell. Is A(1) true? Can't tell. As
long as I don't supply you with an interpretation
(so that A has a specific meaning) these sentences
don't constitute specific statements any more than
"x=3" does before we say what x represents. This is
a kind of relativity that always exists. It comes
from just not specifying what "A" refers to.
If we consider sentences in the language of Q,
lots of people will say that each one is either true
or false in an absolute way, even if we have no way
of determining which it is. There's no use in trying
to attack this point of view by means of twiddling
around with elements of mathematical logic. You need
to make some kind of philosophical argument about
what is required for a statement to be absolutely
true or absolutely false. (And if you do, plenty of
people won't be persuaded. That's just how
philosophy works.)
|Note that I emphasized on "_syntactically_" to signal that the
question and
|the answer be discussed syntactically, not model-theoretically.
There's no distinctive "model theoretical"
consistency distinct from syntactic consistency.
[...]
|In my recent replies to KR, the key question I have is:
|
| >> Syntactically speaking, would it be possible that a formula F be
| >> undecidable in all consistent extensions of the Robinson
Arithmetic Q
| >> formal system?
As Aatu explained, it's not.
If F were independent of all consistent extensions
of Q, then since Q is consistent, then F would be
independent of Q. Hence there would be no proof of
~F in Q, and Q+F would be consistent. By the
original assumption again, F would be independent
of Q+F. But F is provable in Q+F, a contradiction.
The original assumption, then, that F is independent
of all consistent extensions of Q is impossible.
This proof does not rely on the law of excluded
middle. It goes straight from the assumption that
F is independent of all consistent extensions of Q
to a contradiction without dividing by cases. There
is no "either-or" needed.
|---> The Objective
|
|The objective I have in this question is to establish a meta
mathematical
|statement (about FOL) that would have no absolute truth though in its
form
|it'd appear as if there would be an absolute truth value.
It's futile to try unless and until you get some
basis for claiming that a statement has no
absolute truth value.
I don't think your strategy of resorting to meta-
levels is promising at all. To show that you need
to go to a "meta" level you need already to show
that there's some issue with the original language
that calls for it.
There are cases where we would say that a sentence
in a language doesn't have a fixed meaning. We
determine that not by considering it syntactically,
but by considering what it means in the given
context. If you would prefer not to get off into
some philosophical argument, then as there's no
non-philosophical way of addressing issues of
meaning or ultimate truth, you just have to set
the whole question aside.
Keith Ramsay
Marshall
> Marshall wrote:
> > On Oct 18, 8:29 am, Nam Nguyen <namducngu...@shaw.ca> wrote:
> >> Syntactically speaking, would it be possible that a formula F be undecidable
> >> in all consistent extensions of the Robinson Arithmetic Q formal system?
>
> > Suppose F is independent of Q. Then any extension of Q that
> > includes F as an axiom is consistent, and decides F.
>
> > Suppose Q decides F. Then any consistent extension of Q
> > decides F. This is a purely syntactic analysis.
>
> > Did I get that right, Aatu?
>
> Did you get what right?
What part of "Aatu" didn't you understand?
> This is supposed to be a question with
> a resounding "Yes", "No", or "I don't know" answer. Your answer
> here is none of those!
Only because you failed to understand it. It was a resounding
"no." Syntactically speaking.
> I give you a hint.
Hints are for teachers to give to their students. Despite the
fact that I am a beginner in this field, you are not qualified
to teach me anything, except perhaps as a negative example.
> The answer to my question is piggying back on
> the answer on the syntactical (in)consistency of Q. So it's a
> prerequisite to answer the question on Q's (in)consistency,
> before you even have a chance to _resoundingly_ answer my
> question. (And even that would only give you a chance)!
>
> > I like this game; it's fun.
>
> If you say so. Why don't you share the fun with us all, _after_
> you find out the _resounding_ answer on the (in)consistency of Q?
Q is consistent. Did you not know that?
Marshall
Nam Nguyen
I don't have a _syntactical_ proof that Q is consistent. I don't
think even Godel could prove Q's consistency purely on syntactical
means (such as rules of inference and Q's axioms).
You seem to be able to prove that. How?
Nam Nguyen
> On Oct 18, 4:26 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
> |Be careful. Keep answering my question with *if*-this-then-and-*if*-
> that-then
> |and you might end up agreeing with me there's a meta mathematical
> statement
> |that has no absolute truth, which would signal the advent of
> relativity in
> |mathematical reasoning.
>
> If you want to start talking about "absolute truth"
> you need to have some kind of grasp of what you
> mean by the term.
OK. I got what you meant but I don't think you got mine.
Sure, in the absolute ... absolute sense there there's no
absolute truth. For example, the statement S1 = "Axy[x+y=0] is
provable in the formal system T = {Axy[x+y=0]}" is not absolutely
true: it depends (hence is relative to) on what we mean
by "formal system", "provable", logical framework, etc...
But my point is after we've fixed (relativized) a framework
to do the reasoning, there are truths that will be absolutely
true although (I suspect) there are other truths that can't
be absolute, in the sense that for them to be "absolutely true"
at some level of reasoning, further relativization that need
to be done.
For example, _after_ we fixed a reasoning in the familiar FOL=
frame work, then the meta statement S1 is "absolutely true".
(Just like "2 events happened simultaneously" would be 'absolutely
true/false', relative to a certain frame of reference in SR).
But that doesn't seem to me we'd have the right to say other
meta statements must also be absolutely true or wrong, given what
we mean by "absolutely true/false" here.
I hope I've been clear on this point.
>
> If I adjoin to Q a new primitive predicate A(n)
> and provide no new axioms, then lots of statements
> pertaining to A are then independent of Q. Is A(0)
> true? Can't tell. Is A(1) true? Can't tell. As
> long as I don't supply you with an interpretation
> (so that A has a specific meaning) these sentences
> don't constitute specific statements any more than
> "x=3" does before we say what x represents. This is
> a kind of relativity that always exists. It comes
> from just not specifying what "A" refers to.
I actually don't know how all this is relevant to my
rephrased question which begins with "Syntactically
speaking, ...". The truth or falsehood (if any) of my
question would be that of a _meta statement_, and A(0),
A(1) being true or false here are not in the same
meta-hierarchy level the statement implied in my question.
>
> If we consider sentences in the language of Q,
> lots of people will say that each one is either true
> or false in an absolute way, even if we have no way
> of determining which it is.
Again, my meta question is about the syntactical proofs
of syntactical wff's. So, we don't seem to be arguing the
same things.
> There's no use in trying
> to attack this point of view by means of twiddling
> around with elements of mathematical logic. You need
> to make some kind of philosophical argument about
> what is required for a statement to be absolutely
> true or absolutely false. (And if you do, plenty of
> people won't be persuaded. That's just how
> philosophy works.)
>
> |Note that I emphasized on "_syntactically_" to signal that the
> question and
> |the answer be discussed syntactically, not model-theoretically.
>
> There's no distinctive "model theoretical"
> consistency distinct from syntactic consistency.
Oh. There actually is a distinction. I can define consistency without
at all depending on then notions of model truth, interpretation,
semantic, or the like in FOL. Otoh, you can *not* define consistency
without a dependency of existence of syntactical formulas. And that
makes a distinction.
>
> [...]
> |In my recent replies to KR, the key question I have is:
> |
> | >> Syntactically speaking, would it be possible that a formula F be
> | >> undecidable in all consistent extensions of the Robinson
> Arithmetic Q
> | >> formal system?
>
> As Aatu explained, it's not.
And I've explained that that's an illogical explanation. If a
required knowledge of an explanation is impossible to ascertain,
how could one say the explanation is logically valid?
How would one even know there isn't such a thing as a consistent
extension of Q (syntactically speaking) to begin with? And if so,
what does a logical explanation to an illogical question mean?
For the record, my question is *not* something like:
"If Q is consistent, would it be possible that ....?".
>
> If F were independent of all consistent extensions
> of Q, then since Q is consistent, then F would be
> independent of Q. Hence there would be no proof of
> ~F in Q, and Q+F would be consistent. By the
> original assumption again, F would be independent
> of Q+F. But F is provable in Q+F, a contradiction.
> The original assumption, then, that F is independent
> of all consistent extensions of Q is impossible.
>
> This proof does not rely on the law of excluded
> middle. It goes straight from the assumption that
> F is independent of all consistent extensions of Q
> to a contradiction without dividing by cases. There
> is no "either-or" needed.
>
> |---> The Objective
> |
> |The objective I have in this question is to establish a meta
> mathematical
> |statement (about FOL) that would have no absolute truth though in its
> form
> |it'd appear as if there would be an absolute truth value.
>
> It's futile to try unless and until you get some
> basis for claiming that a statement has no
> absolute truth value.
I did give some basis in that same post. And I've done some
more explanation above (about "absolute truth").
>
> I don't think your strategy of resorting to meta-
> levels is promising at all.
I think you know I'm not the only one who have tried to resort to
meta levels in mathematical reasoning.
> To show that you need
> to go to a "meta" level you need already to show
> that there's some issue with the original language
> that calls for it.
The meta levels are already there and there's nothing wrong
with the syntactical languages of FOL. Sorry, I'm not quite sure
we're talking about the same issues here.
>
> There are cases where we would say that a sentence
> in a language doesn't have a fixed meaning. We
> determine that not by considering it syntactically,
> but by considering what it means in the given
> context. If you would prefer not to get off into
> some philosophical argument, then as there's no
> non-philosophical way of addressing issues of
> meaning or ultimate truth, you just have to set
> the whole question aside.
Sorry, we've either talked past each other or have not been
on the same pages but my question and issue are syntactically
based and would have nothing to do with FOL semantics and model.
Nam Nguyen
> Keith Ramsay wrote:
>> On Oct 18, 4:26 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>> [...]
>> |In my recent replies to KR, the key question I have is:
>> |
>> | >> Syntactically speaking, would it be possible that a formula F be
>> | >> undecidable in all consistent extensions of the Robinson Arithmetic Q
>> | >> formal system?
>>
>> As Aatu explained, it's not.
>
> And I've explained that that's an illogical explanation. If a
> required knowledge of an explanation is impossible to ascertain,
> how could one say the explanation is logically valid?
>
> How would one even know there isn't such a thing as a consistent
> extension of Q (syntactically speaking) to begin with? And if so,
> what does a logical explanation to an illogical question mean?
>
> For the record, my question is *not* something like:
>
> "If Q is consistent, would it be possible that ....?".
Let me present a simple example. Suppose one is sitting on a train
next to a person who has lost his eye sight since birth and who
suddenly asks "Is it true that there's no life in the our Blue Moon?".
Should one go ahead and explain true or false to the question?
Or should one ask back: "What do you mean by 'our Blue Moon'?" ?
Marshall
> Marshall wrote:
> > On Oct 18, 8:01 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
> >> Marshall wrote:
> >>> I like this game; it's fun.
> >> If you say so. Why don't you share the fun with us all, _after_
> >> you find out the _resounding_ answer on the (in)consistency of Q?
>
> > Q is consistent. Did you not know that?
>
> I don't have a _syntactical_ proof that Q is consistent. I don't
> think even Godel could prove Q's consistency purely on syntactical
> means (such as rules of inference and Q's axioms).
So ... you didn't know it.
> You seem to be able to prove that. How?
The fact that it has a model is proof that it is consistent.
Marshall
Nam Nguyen
The definition of a model _requires all the theorems be true_ in the
model. There's a fact that _nobody_ knows a particular formula is a
theorem in Q, then how could *you* or anybody _claim_ anything is model
of Q?
Not to mention that you've deliberately kept avoiding the word "syntactical"
here. There's no magic to the word 'model': it could be completely replaced
by a syntactical definition. So if you don't have a syntactical proof for
Q's consistency, invoking the word 'model' would be in vain!
Marshall
> Marshall wrote:
> > On Oct 18, 8:55 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
> >> Marshall wrote:
> >>> On Oct 18, 8:01 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
> >>>> Marshall wrote:
> >>>>> I like this game; it's fun.
> >>>> If you say so. Why don't you share the fun with us all, _after_
> >>>> you find out the _resounding_ answer on the (in)consistency of Q?
> >>> Q is consistent. Did you not know that?
> >> I don't have a _syntactical_ proof that Q is consistent. I don't
> >> think even Godel could prove Q's consistency purely on syntactical
> >> means (such as rules of inference and Q's axioms).
>
> > So ... you didn't know it.
>
> >> You seem to be able to prove that. How?
>
> > The fact that it has a model is proof that it is consistent.
>
> The definition of a model _requires all the theorems be true_ in the
> model. There's a fact that _nobody_ knows a particular formula is a
> theorem in Q, then how could *you* or anybody _claim_ anything is model
> of Q?
We only have to check that the axioms are true in the model.
We don't have to check each and every theorem. All the axioms
of Q are true of the natural numbers with the usual operations.
Hence N is a model of Q.
> Not to mention that you've deliberately kept avoiding the word "syntactical"
> here. There's no magic to the word 'model'
Neither is there any magic in the word "syntactical."
> it could be completely replaced
> by a syntactical definition.
Likewise, a syntactical proof can be completely replaced
by a model-theoretic one, which is what I did. A theory
is consistent if and only if it has a model. Q has a model: N.
This proves that Q is consistent.
You may prefer syntactic approaches. It is fine for you to
have such a preference, but it is mistaken to believe that
syntactic approaches have any kind of monopoly on truth.
> So if you don't have a syntactical proof for
> Q's consistency, invoking the word 'model' would be in vain!
And *I* say, if you don't have a model-theoretic proof, then
invoking the word "syntax" is in vain! No, I don't really say
that, actually; that's not true; they are each equally as good.
Marshall
Aatu Koskensilta
> The fact that it has a model is proof that it is consistent.
There is no need to drag in any models. The consistency of Robinson
arithmetic can be proved finitistically (a finitist proof can be found
in Shoenfield). Apparently this doesn't count as "syntactic".
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
Nam Nguyen
> On Oct 18, 11:25 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>> Marshall wrote:
>>> On Oct 18, 8:55 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>>>> Marshall wrote:
>>>>> On Oct 18, 8:01 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
>>>>>> Marshall wrote:
>>>>>>> I like this game; it's fun.
>>>>>> If you say so. Why don't you share the fun with us all, _after_
>>>>>> you find out the _resounding_ answer on the (in)consistency of Q?
>>>>> Q is consistent. Did you not know that?
>>>> I don't have a _syntactical_ proof that Q is consistent. I don't
>>>> think even Godel could prove Q's consistency purely on syntactical
>>>> means (such as rules of inference and Q's axioms).
>>> So ... you didn't know it.
>>>> You seem to be able to prove that. How?
>>> The fact that it has a model is proof that it is consistent.
>> The definition of a model _requires all the theorems be true_ in the
>> model. There's a fact that _nobody_ knows a particular formula is a
>> theorem in Q, then how could *you* or anybody _claim_ anything is model
>> of Q?
>
> We only have to check that the axioms are true in the model.
*If* you could _check_. But can you? The axioms might be short and the
number of axioms might be finite but the number of n-tuples corresponding
to an n-ary symbol (e.g. '<') might be _infinite_ and *your checking*
would be in vain! For example, how do you _check_ the following Q's axiom:
(x < y \/ x =y \/ y < x)
to be true in your purported natural-number "model"?
> We don't have to check each and every theorem.
But who has suggested to to do that? My "The definition of a model
_requires all the theorems be true_ in the" only meant if you don't
know truth of one particular theorem in your purported model, then
that theorem could lead to an inconsistency: because so far you only
assume - but don't know for fact - all the axioms be true.
> All the axioms are true of the natural numbers with the usual
> operations. Hence N is a model of Q.
As pointed out above, you can't even establish in your purported
model the truth of one single axiom [x < y \/ x =y \/ y < x], let
alone the truth of "All the axioms".
>> Not to mention that you've deliberately kept avoiding the word "syntactical"
>> here. There's no magic to the word 'model'
>
> Neither is there any magic in the word "syntactical."
I've posted here on and off for about 10 years and one could never
find my posts defending or proposing there's magic to syntacticalism,
as your side would defend magic to models. In fact, occasionally I
even pointed out that there's no magic and only _objective rigidity_
in syntactical game-of-symbol reasoning, or that we'd be humble
(as we should) and see our own limitation of mathematical knowledge,
through concrete syntactical manipulation of formulas
It's your side that believes in the subjective magic of the so-called
the natural numbers or the standard model of arithmetic, even when
it might be impossible for your side to construct a truth value
of 1 single short formula in your purported model.
>> it could be completely replaced
>> by a syntactical definition.
>
> Likewise, a syntactical proof can be completely replaced
> by a model-theoretic one, which is what I did.
You're bluffing here. The very word "model" means model _of_ an
axiom-system, which is syntactical. You can never find a model
of something that's not syntactical in its very nature, in FOL
reasoning. Hence you can never "replace" a syntactical proofs
by models. On the other hand you can define a truth of a formula
or theorem _completely by syntactical means_!
I couldn't believe I heard you saying that. Are you suggesting
that we throw syntactical rules of inference away, and completely
replace them with sheer intuitions about "true sentences"?
> A theory is consistent if and only if it has a model.
You actually can't know that in general. There are formal systems
that Godel work wouldn't address; these are systems with infinite
numbers of logical symbols. And in these systems, (in)consistency
can be determined only by standard definition of (in)consistency:
the syntactical one.
> Q has a model: N. This proves that Q is consistent.
As pointed out above with the axiom [x < y \/ x =y \/ y < x],
you don't know what N exactly is as a model of anything, let
alone a model of Q.
>
> You may prefer syntactic approaches. It is fine for you to
> have such a preference, but it is mistaken to believe that
> syntactic approaches have any kind of monopoly on truth.
You have it backward. The reason why people invent syntactical
means such as _finite_ formulas, rules of inference, proofs is
because it has the sole monopoly of trust in reasoning, due
to the objective and rigid and un-bias characteristics vested in
these syntactical means.
>> So if you don't have a syntactical proof for
>> Q's consistency, invoking the word 'model' would be in vain!
>
> And *I* say, if you don't have a model-theoretic proof, then
> invoking the word "syntax" is in vain!
As pointed out above, you wouldn't know what "model-theoretic proof"
for the formal systems with infinite number of logical symbols.
But for these, the (in)consistency, proofs would still follow the
usual syntactical definitions! So what you say here simply isn't true.
> No, I don't really say
> that, actually; that's not true; they are each equally as good.
As I've mentioned above, they aren't equally good:
(a) models _must_ depend on formal systems to be meaningful;
(b) there are formal systems that nobody could possibly come up
with any models.
Trop
> Nam Nguyen <namducngu...@shaw.ca> writes:
> > Doesn't my my question above have "which _syntactically_ consistent
> > extension of Q"?
>
> If GC is true then Q + GC is a consistent extension of Q in which GC is
> decidable, and if it is false, then Q is a consistent extension of Q in
> which GC is decidable. Our knowledge of the truth or falsity of GC is
> irrelevant.
>
> --
>
> "Wovon mann nicht sprechen kann, darüber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
In this argument you presuppose that it's possible to answer the
question whether GC is true or not. But that's only HYPOTHESIS! A
third possibility may arise: we will never know whether GC is true or
not, i. e. there is no argumentation for or against GC. That's why
your argument is incomplete, you must consider that third possibility
too, i. e. you must argue without any reference to notion of truth of
GC.
Sergey Tropanets
Trop
The same remark as my previous one to Aatu's proof but with
substitution "whether GC is true or not" for "F is independent of Q or
not". You are still out of intuitionism.
Sergey Tropanets
Aatu Koskensilta
> The same remark as my previous one to Aatu's proof but with
> substitution "whether GC is true or not" for "F is independent of Q or
> not". You are still out of intuitionism.
As Keith notes, the result is intuitionistically correct. Suppose we
have an F independent of all consistent (axiomatisable) extensions of
Q. From this it follows that Q + F is a consistent (axiomatisable)
extension of Q in which F is undecidable; But Q + F |- F, which is
absurd. So there is no F independent of all consistent (axiomatisable)
extensions of Q.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
Marshall
> [snip]
I'm not certain, but I don't think you wrote a single correct
thing in that entire long post. I really don't see any
point in continuing the conversation. Sorry.
Marshall
Nam Nguyen
> Marshall <marshal...@gmail.com> writes:
>
>> The fact that it has a model is proof that it is consistent.
>
> There is no need to drag in any models. The consistency of Robinson
> arithmetic can be proved finitistically (a finitist proof can be found
> in Shoenfield). Apparently this doesn't count as "syntactic".
Really? Given syntactical rules of inference are designed to prove but
_not to disprove_ formulas in a formal system, how did Shoenfield disprove
a particular non-theorem in Q using purely syntactical means such as
inference rules and Q's axioms - without borrowing non-syntactical means
such as natural numbers, recursion, model truths or the like?
Let me repeat my question:
>> Syntactically speaking, would it be possible that a formula F be
>> undecidable in all consistent extensions of the Robinson Arithmetic Q
>> formal system?
Now in a response to Marshall, I said:
>> The answer to my question is piggying back on the answer on the
>> syntactical (in)consistency of Q. So it's a prerequisite to answer
>> the question on Q's (in)consistency, before you even have a chance
>> to _resoundingly_ answer my question.
We could think of "piggying back" as nothing more than just a form of
"encoding": I was "delegating" the question/answer of F's undecidability
in all of Q's consistent extension to that of Q itself. If you can
answer the question of Q's (in)consistency _syntactically speaking_
then you can answer my question. If it's impossible to have a knowledge
about of Q's syntactical (in)consistency then it's equally impossible
to answer my question speaking.
Please remember that it's *not* the consistency of Q and Q's extensions
per se that I'm after. It's _a valid impossibility_ I'm looking for.
I actually could come up with a slightly different question using model
theoretic approach and you'd still face a difficult task: accepting an
impossibility as a thesis or rejecting it. I just thought that doing it
syntactically, we'd have more rigidity and hence more clarity about
what's possible and what's impossible, than the muddy intuition-based
philosophies about model truths.