On Apr 21, 10:25 am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:

Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

(Daryl McCullough) wrote:

Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

Newberry says...

(Daryl McCullough) wrote:
Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

You would have to write down the axioms and rules.
Frankly, I don't believe you.

Newberry <newberr...@gmail.com> writes:
On Apr 21, 10:25 am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:

Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

It's amazing how much you know about what your system proves, and yet
you have never offered axioms.

On Apr 22, 5:03 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
Newberry <newberr...@gmail.com> writes:
On Apr 21, 10:25 am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:

Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

It's amazing how much you know about what your system proves, and yet
you have never offered axioms.

Did you read Diaz's book?

On Apr 22, 6:52=A0am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:
Newberry says...

(Daryl McCullough) wrote:
Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

You would have to write down the axioms and rules.
Frankly, I don't believe you.

There are a million ways to express that G is underivanle. Only one of
them leads to a contradiction. Goedel's theorem hangs there by a thin
thread.

Newberry says...







On Apr 22, 6:52=A0am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:
Newberry says...

(Daryl McCullough) wrote:
Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

You would have to write down the axioms and rules.
Frankly, I don't believe you.

There are a million ways to express that G is underivanle. Only one of
them leads to a contradiction. Goedel's theorem hangs there by a thin
thread.

No, that's just not true.

Let T be your theory, whatever it is.
I'm assuming that T has numerals for all the
natural numbers, and that T only has countably
many formulas. So pick some enumeration of
all the formulas.

Now, define the set S to be the
set of all pairs of naturals (x,y)
such that T proves the sentence Phi_x(y).
That is, T proves the sentence resulting
from substituting the numeral for y for
any free variables occurring in the formula
with index x.

S is a perfectly meaningful set of
pairs of naturals. So if S is meaningful,
then so is the set D, which is the set
of all natural numbers x such that
(x,x) is an element of S.

D is a perfectly meaningful r.e. set
of naturals. For any natural number n,
the statement "n is an element of D"
or "n is not an element of D" is perfectly
meaningful.

So either the set D is
definable in T, or it isn't. If it
isn't definable in T, then that shows
that your theory T is semantically
weak; there are meaningful r.e. sets
of naturals not definable in T. Such
a T is inadequate for doing mathematics
(at least computability theory, anyway).

So let's suppose that D is definable in
T. So there is a formula Q(x) formalizing
the statement "x is an element of D".
Similarly, we can also define the complement
of D: "x is a natural number that is not an
element of D" can be formalized as ~Q(x).

Since we've picked an enumeration of all
the formulas of T, ~Q(x) must be equal to
Phi_n(x) for some natural number n.
So let's consider the statement:
n is not an element of D.

I've just argued that D is a meaningful,
well-defined set of naturals. (Since D
is defined in terms of what is provable
from T, if D is *not* well-defined, it
has to be due to the fact that T itself
is not well-defined.) So the statement
"n is not an element of D" is a perfectly
meaningful statement. Can it be proved
by T? If so, then T is inconsistent.
If not, then T is semantically
incomplete.

Newberry says...







On Apr 22, 6:52=A0am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:
Newberry says...

(Daryl McCullough) wrote:
Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

You would have to write down the axioms and rules.
Frankly, I don't believe you.

There are a million ways to express that G is underivanle. Only one of
them leads to a contradiction. Goedel's theorem hangs there by a thin
thread.

No, that's just not true.

Let T be your theory, whatever it is.
I'm assuming that T has numerals for all the
natural numbers, and that T only has countably
many formulas. So pick some enumeration of
all the formulas.

Now, define the set S to be the
set of all pairs of naturals (x,y)
such that T proves the sentence Phi_x(y).
That is, T proves the sentence resulting
from substituting the numeral for y for
any free variables occurring in the formula
with index x.

S is a perfectly meaningful set of
pairs of naturals. So if S is meaningful,
then so is the set D, which is the set
of all natural numbers x such that
(x,x) is an element of S.

D is a perfectly meaningful r.e. set
of naturals. For any natural number n,
the statement "n is an element of D"
or "n is not an element of D" is perfectly
meaningful.

So either the set D is
definable in T, or it isn't. If it
isn't definable in T, then that shows
that your theory T is semantically
weak; there are meaningful r.e. sets
of naturals not definable in T. Such
a T is inadequate for doing mathematics
(at least computability theory, anyway).

So let's suppose that D is definable in
T. So there is a formula Q(x) formalizing
the statement "x is an element of D".
Similarly, we can also define the complement
of D: "x is a natural number that is not an
element of D" can be formalized as ~Q(x).

Since we've picked an enumeration of all
the formulas of T, ~Q(x) must be equal to
Phi_n(x) for some natural number n.
So let's consider the statement:
n is not an element of D.

I've just argued that D is a meaningful,
well-defined set of naturals. (Since D
is defined in terms of what is provable
from T, if D is *not* well-defined, it
has to be due to the fact that T itself
is not well-defined.) So the statement
"n is not an element of D" is a perfectly
meaningful statement. Can it be proved
by T? If so, then T is inconsistent.
If not, then T is semantically
incomplete.

Let's say D is not definable in T. Does it mean that not all p.r.

Let's say D is not definable in T.

Does it mean that not all p.r. functions are definable in T?

There are infinitely many ways to express each p.r.
function. If you eliminate a class of formulas does
not mean that you have eliminated any p.r. function.

Newberry says...

[In the following, T is some axiomatizable
theory of mathematics. We assume that every
natural number n is represented in T by some
constant, the numeral for n. We assume that
there is some enumeration Phi_n of all
the formulas of T. We let Phi_n(m) be the
result of substituting the numeral for m
for the variables occurring in the formula
with index n. Then D is the set of naturals
n such that Phi_n(n) is provable in T.]

Let's say D is not definable in T.

Then that shows either that T is ill-defined,
or is not axiomatizable, or that T is inadequate
for reasoning about mathematics (because there
are meaningful r.e. sets that are not definable
in T).

Does it mean that not all p.r. functions are definable in T?

Yes, it does. D is a recursively enumerable set. That
means that there is a primitive recursive function f(x)
such that D = { f(0), f(1), f(2), ... }. That is,
D is the image, or range, of the function f. So if
D is not definable, that means that f is not definable,
and so T does not define all primitive recursive functions.

I'm not sure whether you meant "p.r." to mean "primitive
recursive" or "partial recursive", but it doesn't matter.
The same conclusion holds for both.

There are infinitely many ways to express each p.r.
function. If you eliminate a class of formulas does
not mean that you have eliminated any p.r. function.

If there is *any* way to express the primitive recursive
function f(x) whose image is D, then there is a way to
define D. Namely, D = the set of all y satisfying
"exists x such that f(x) = y".

I think you are mistaken. If D is not definable in T then it is not an
image of anything.

The question is if there is any primitive recursive
function, which T cannot express. T can certainly define an equivalent
of D' as defined in PA. It just does not appear on T's diagonal. But I
am going to look into this in more detail.

On Apr 24, 7:33=A0am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:

If there is *any* way to express the primitive recursive
function f(x) whose image is D, then there is a way to
define D. Namely, D = the set of all y satisfying
"exists x such that f(x) = y".

I think you are mistaken. If D is not definable in T then it is not an
image of anything.

The question is if there is any primitive recursive
function, which T cannot express.

T can certainly define an equivalent of D' as defined in PA.

Newberry says...



On Apr 24, 7:33=A0am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:
If there is *any* way to express the primitive recursive
function f(x) whose image is D, then there is a way to
define D. Namely, D = the set of all y satisfying
"exists x such that f(x) = y".

I think you are mistaken. If D is not definable in T then it is not an
image of anything.

But we can easily see that D is the image of a primitive
recursive function. So either T is inadequate for representing
primitive recursive functions, or D is definable in T.

The question is if there is any primitive recursive
function, which T cannot express.

And the answer is this: If the set D is not definable,
then there is a primitive recursive function f that
T cannot express.

T can certainly define an equivalent of D' as defined in PA.

Then T is either semantically incomplete or T is inconsistent.

Newberry says...







On Apr 22, 6:52=A0am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:
Newberry says...

(Daryl McCullough) wrote:
Well, is G provable in your system?

No.

If so, then your system
is inconsistent. If not, then is the statement "G is not provable"
provable?

Yes.

If so, your system is inconsistent.

No.

You would have to write down the axioms and rules.
Frankly, I don't believe you.

There are a million ways to express that G is underivanle. Only one of
them leads to a contradiction. Goedel's theorem hangs there by a thin
thread.

No, that's just not true.

Let T be your theory, whatever it is.
I'm assuming that T has numerals for all the
natural numbers, and that T only has countably
many formulas. So pick some enumeration of
all the formulas.

Now, define the set S to be the
set of all pairs of naturals (x,y)
such that T proves the sentence Phi_x(y).
That is, T proves the sentence resulting
from substituting the numeral for y for
any free variables occurring in the formula
with index x.

S is a perfectly meaningful set of
pairs of naturals. So if S is meaningful,
then so is the set D, which is the set
of all natural numbers x such that
(x,x) is an element of S.

D is a perfectly meaningful r.e. set
of naturals. For any natural number n,
the statement "n is an element of D"
or "n is not an element of D" is perfectly
meaningful.

So either the set D is
definable in T, or it isn't. If it
isn't definable in T, then that shows
that your theory T is semantically
weak; there are meaningful r.e. sets
of naturals not definable in T. Such
a T is inadequate for doing mathematics
(at least computability theory, anyway).

So let's suppose that D is definable in
T. So there is a formula Q(x) formalizing
the statement "x is an element of D".
Similarly, we can also define the complement
of D: "x is a natural number that is not an
element of D" can be formalized as ~Q(x).

Since we've picked an enumeration of all
the formulas of T, ~Q(x) must be equal to
Phi_n(x) for some natural number n.
So let's consider the statement:
n is not an element of D.

I've just argued that D is a meaningful,
well-defined set of naturals. (Since D
is defined in terms of what is provable
from T, if D is *not* well-defined, it
has to be due to the fact that T itself
is not well-defined.) So the statement
"n is not an element of D" is a perfectly
meaningful statement. Can it be proved
by T? If so, then T is inconsistent.
If not, then T is semantically
incomplete.

Newberry <newberr...@gmail.com> writes:
On Apr 15, 3:40 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
Newberry <newberr...@gmail.com> writes:
You are joking, right?

You are joking when you say a sentence is a picture of a possible
state of affairs, right?

Why else would you keep ignoring my question about the sentence
"Liberty is a fundamental value of democracy"?

You make asillyclaim about the relationship between sentences and
pictures and then argue that our interpretation of sentences must be
wrong, because we can't describe a plausible picture for your sample
sentence.  Meanwhile, you can't describe a plausible picture for my
sample sentence.  What does that mean?  Is the problem in our
understanding or in your suggestion that sentences are like pictures?

We could sit here for weeks and debate if your phrase is a picture of
a possible state of affairs or not. Either way it is not going to
alter the fact "all round squares are green" is not.

So what?

If you acknowledge that "Liberty is a fundamental value of democracy"
is a meaningful sentence and you admit that you cannot find any
sensible "picture of a state of affairs" that corresponds to my
sentence, then it seems you have to give up on your silly insistence
that meaning is related to these pictures.

And if you give up on thatsillyinsistence, then it would not matter
whether "All round squares are green" has a corresponding picture or not.

So, *even if* I agreed that there is no picture for the latter
sentence, I again ask, "So what?"  

(By the way, you should learn to write clearly.  *Of course* my
sentence is not a picture.  No sentence is literally a picture.  A
sentence is a sequence of words satisfying certain conditions.  A
picture is something else.  You seem to have some sort of relation
between sentences and pictures in mind, but it surely cannot be
identity.)

--
"So now, The Hammer is here, and with it, the end of days.  The world will be
destroyed, and then remade, as foretold. You will be lost, with your
children, and then there will be others, and one day they will be tested, and
will pass, but that is another story." --James S Harris gets a bit excited..- Hide quoted text -

- Show quoted text -