On Apr 15, 1:03 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

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 a silly claim 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.

Here is what I have just found in "Philosophy in the Twentieth
Century" by A.J. Ayer p.112

"A genuine proposition pictures a possible state of affairs."

This is Ayer's exposition of Tractatus. So who is silly, Wittgenstein
or Ayer? As far as I am concerned, your attempt of a counterexample is
not particularly brilliant.

On Apr 15, 1:03 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
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.

Here is what I have just found in "Philosophy in the Twentieth
Century" by A.J. Ayer p.112

"A genuine proposition pictures a possible state of affairs."

This is Ayer's exposition of Tractatus. So who is silly, Wittgenstein
or Ayer? As far as I am concerned, your attempt of a counterexample is
not particularly brilliant.

OK, so I have now drawn a picture of what you have said.

y |0 1 2 3 4 5 6 7 8 9 n
-----------------------------------------
Phi_0(y)| * * * * * * * * * * * * *
Phi_1(y)| * * P * * * * * * * * * *
Phi_2(y)| * * * * * * * P * * * * *
Phi_3(y)| * * * P * * * * * * * * *
Phi_4(y)| * * * * * * * P * * * * *
Phi_5(y)| * * * * * * * * * * * * *
=2E.............................................
Phi_n(y)| * * * * * * * * * * * * ? ~Q(y) =3D y is not in D

I guess you are asking if Phi_n(n) is provable in T. The answer is no.
The question is how does it follow that T is semantically incomplete?

Would you care to substantiate the claims you are making here? I do
not see how any of them follows.

Newberry says...

Would you care to substantiate the claims you are making here? I do
not see how any of them follows.

You have to ask a more specific question. I gave you a proof
sketch. If some step was unclear, then ask for clarification.

The idea is this: By definition, D is the set of all natural
numbers x such that Phi_x(x) is provable in theory T.

To see that D is the image of a primitive recursive function,
just let f(x) be the function with the following informal
description: First, let's take the sentence
x*0 = 0. That's going to be some formula Phi_k(x).
Clearly, Phi_k(k) is provable, since it's just
the statement k*0 = 0.

f(x) > Try to decode x as a valid proof.
If that's impossible, then return k.
If x does decode as a valid proof, then let the
last statement be S. Now, go through every
natural number m less than x, and form the
statement Phi_m(m). If Phi_m(m) is equal
to statement S, then return m. If there is
no m less than x such that S = Phi_m(m), then
return k.

This is an informal description, but it should
be clear that it is implementable as a computer
program, and it should be clear that the image
of f(x) includes all n in D.

So if we assume that your theory T can represent
all primitive recursive functions, then there will
be a formula Q(x,y) such that for each n and m,

f(n) = m <-> T proves Q(n,m)
f(n) is unequal to m <-> T proves not Q(n,m)

So, we formalize "x is not an element of D"
as "not exists y, Q(y,x)". That will be
some formula of T, call it Phi_n(x). Then
we have

x is not an element of D
-> not exists y, Q(y,x)
-> Phi_n(x)

In particular, we have

n is not an element of D
-> not exists y, Q(y,n)
-> Phi_n(n)

So if n is not an element of D,
then, it follows that Phi_n(n)
holds (since Phi_n(x) formalizes
(x is an element of D)). But
by definition of D, if n is not
an element of D, then T does not
prove Phi_n(n). So Phi_n(n) is true
and unprovable by T. So T is semantically
incomplete.

Just because T cannot define the antidiagonal does not mean that it
cannot represent all the p.r. functions.

Newberry says...

Just because T cannot define the antidiagonal does not mean that it
cannot represent all the p.r. functions.

Yes, it certainly does mean that. I just went through the proof.
If every primitive recursive function is definable, then every
recursively enumerable set is definable. If every recursively
enumerable set is definable, then every co-r.e. set is definable.
(a co-r.e. set is a set of natural numbers whose complement is
an r.e. set). The diagonal is a co-r.e. set, so if it isn't
definable in T, then it can only be because T fails to define
all primitive recursive functions.

Newberry says...

Well, I have taken another look at this, and let me first ask you one
thing. In PA Phi_n(n) is not decidable. Does it mean that the set D is
not defined?

No, it doesn't mean that at all. The set D (for a theory T)
is defined to be the set of all natural numbers m such that
Phi_m(m) is provable in T. That's an r.e. set, which means
that for any natural number m, if m is an element of D,
then it is provable that m is an element of D.

[For reference,
let n be the index of the formula Phi_n(x) such that
for each m, if m is in D, then the negation of Phi_n(x)
is provable in T. In other words,
~Phi_n(m) <-> T proves Phi_m(m). It immediately follows
in the special case m=n that
~Phi_n(n) <-> T proves Phi_n(n)
or
Phi_n(n) <-> T does not prove Phi_n(n)]

And just because Phi_n(n) is not decidable it does not
follow that any p.r. function is not definable.

Of course not. It follows that PA is incomplete, because
it can neither prove Phi_n(n) nor its negation.

T is not much different in this respect.

That's my point. T cannot be semantically complete
any more than PA is.

Phi_n(n) has the truth value M.

If T is consistent, then Phi_n(n) has the truth value "true".

it follows that T is semantically complete, or at least it does not
follow by Goedel's argument that it is not.

Yes, it certainly does. You agree that T does not prove
Phi_n(n). The formalization in T of the statement
"T does not prove Phi_x(x)" is Phi_n(x). So the
formalization of the statement "T does not prove
Phi_n(n)" is Phi_n(n).

does not prove Phi_n(n), then Phi_n(n) is true,
since that is the formalization in T of that
statement.

The situation does not seem very different from PA but you
could argue that D is not well defined.

No, you could not. If T is a well-defined axiomatizable
theory, then the set of naturals n such that T proves
Phi_n(n) is well-defined. How could it be otherwise?

If the set of all sets that are not elements of themselves
does not have to exist I do not know why the set of ~(n el D) must
exist.

Because we can recursively enumerate all the elements
of D. So D is well-defined. If a set of naturals is
well-defined, then so is its complement.

And I would argue that the set D is not meaningfull
analogically to "this sentence is true."

The difference is that we have no way of enumerating
all true sentences, but we have a perfectly well-defined
way to enumerate all sentences provable in T.

I think that the disconnet in your argument is in "every recursively
enumerable set is definable." How do you know what every set is before
you defined it? D is an ill defined concept, such a set does not
exist, hence it is NOT an r.e. set.

As I said, D is as well-defined as T. I can certainly
believe that T is ill-defined, but if we substitute an
actualy theory, such as PA or ZF, then the corresponding
D is not only well-defined, but we can write a computer
program that will generate all the elements of D.

--
Daryl McCullough
Ithaca, NY

Well, I have taken another look at this, and let me first ask you one
thing. In PA Phi_n(n) is not decidable. Does it mean that the set D is
not defined?

And just because Phi_n(n) is not decidable it does not
follow that any p.r. function is not definable.

T is not much different in this respect.

Phi_n(n) has the truth value M.

it follows that T is semantically complete, or at least it does not
follow by Goedel's argument that it is not.

The situation does not seem very different from PA but you
could argue that D is not well defined.

If the set of all sets that are not elements of themselves
does not have to exist I do not know why the set of ~(n el D) must
exist.

And I would argue that the set D is not meaningfull
analogically to "this sentence is true."

I think that the disconnet in your argument is in "every recursively
enumerable set is definable." How do you know what every set is before
you defined it? D is an ill defined concept, such a set does not
exist, hence it is NOT an r.e. set.

Yes, it certainly does. You agree that T does not prove
Phi_n(n). The formalization in T of the statement
"T does not prove Phi_x(x)" is Phi_n(x). So the
formalization of the statement "T does not prove
Phi_n(n)" is Phi_n(n).

Here is your error. There are many formalizations of "T does not prove
Phi_n(n)."

The situation does not seem very different from PA but you
could argue that D is not well defined.

No, you could not. If T is a well-defined axiomatizable
theory, then the set of naturals n such that T proves
Phi_n(n) is well-defined. How could it be otherwise?

Affirmative.

If the set of all sets that are not elements of themselves
does not have to exist I do not know why the set of ~(n el D) must
exist.

Because we can recursively enumerate all the elements
of D. So D is well-defined. If a set of naturals is
well-defined, then so is its complement.

So where is the problem?

I think that the disconnet in your argument is in "every recursively
enumerable set is definable." How do you know what every set is before
you defined it? D is an ill defined concept, such a set does not
exist, hence it is NOT an r.e. set.

As I said, D is as well-defined as T. I can certainly
believe that T is ill-defined, but if we substitute an
actualy theory, such as PA or ZF, then the corresponding
D is not only well-defined, but we can write a computer
program that will generate all the elements of D.

If D is ill defined in T you still have to show that T cannot
represent all p.r. functions.

That something analogical to D exists in
PA is not a proof. There are infinitely
many ways to express each p.r. function.
We have eliminated only one.

Newberry says...

Yes, it certainly does. You agree that T does not prove
Phi_n(n). The formalization in T of the statement
"T does not prove Phi_x(x)" is Phi_n(x). So the
formalization of the statement "T does not prove
Phi_n(n)" is Phi_n(n).

Here is your error. There are many formalizations of "T does not prove
Phi_n(n)."

Pick *any* formalization of "T does not prove Phi_x(x)".
That will be some formula Phi_n(x). Then use *that*
formalization.

The situation does not seem very different from PA but you
could argue that D is not well defined.

No, you could not. If T is a well-defined axiomatizable
theory, then the set of naturals n such that T proves
Phi_n(n) is well-defined. How could it be otherwise?

Affirmative.

If the set of all sets that are not elements of themselves
does not have to exist I do not know why the set of ~(n el D) must
exist.

Because we can recursively enumerate all the elements
of D. So D is well-defined. If a set of naturals is
well-defined, then so is its complement.

So where is the problem?

There is no problem. It's just that assuming
that
(1) T is well-defined enough that its theorems form
an r.e. set, and
(2) T is expressive enough to define every r.e. set, and
(3) T is consistent.

it follows that (1) there is a well-defined co-r.e. set D-complement
and (2) there is a natural number n that is in D-complement, and
(3) T can express the statement "n is an element of D-complement",
but (4) T cannot prove that statement. So T is semantically
incomplete.

I think that the disconnet in your argument is in "every recursively
enumerable set is definable." How do you know what every set is before
you defined it? D is an ill defined concept, such a set does not
exist, hence it is NOT an r.e. set.

As I said, D is as well-defined as T. I can certainly
believe that T is ill-defined, but if we substitute an
actualy theory, such as PA or ZF, then the corresponding
D is not only well-defined, but we can write a computer
program that will generate all the elements of D.

If D is ill defined in T you still have to show that T cannot
represent all p.r. functions.

I just said, D is well-defined if T is. D is an r.e.
set.

That something analogical to D exists in
PA is not a proof. There are infinitely
many ways to express each p.r. function.
We have eliminated only one.

I've already been through this. It doesn't
matter that there are multiple ways to define
a primitive recursive function. The set D *is*
an r.e. set. Therefore, it is the image of some
primitive recursive function. Therefore, if T
can define every primitive recursive function,
then it can define D.

We will say that a sentence is neither true nor false if it and its
negation are not provable or if the proof is infinitely long.

On Apr 29, 7:37 pm, Newberry <newberr...@gmail.com> wrote:



We will say that a sentence is neither true nor false if it and its
negation are not provable or if the proof is infinitely long.

It seems to me that all you've done is (re)define the word
"true" to mean "provable." In which case, we should not
be surprised to find that many systems are semantically
complete!

Rewording your following sentence accordingly:

  According to this definition G is neither provable nor disprovable
  The system is semantically complete because there is no provable
  but unprovable sentence.

Well, sure. You've made "complete" mean the same thing as
"consistent."

Marshall

PS. Consistency is a piddling correctness condition.

Let's break it down.
Step 1:
We will say that a sentence is neither true nor false if it and its
negation are not provable or if the proof is infinitely long.

According to this definition G is neither true nor false. The system
is semantically complete because there is no true but unprovable
sentence. The system is still capable of expressing all p.r. functions
and all r.e. sers.

Do you find any problem with this approach?

Let's break it down.
Step 1:
We will say that a sentence is neither true nor false if it and its
negation are not provable or if the proof is infinitely long.
According to this definition G is neither true nor false. The system
is semantically complete because there is no true but unprovable
senetnce. The system is still capable of expressing all p.r. functions
and all r.e. sers.

Do you find any problem with this approach?