each valid sentence in the language of the axioms is then either true or false in the model. (It could be independent, too, but we're not concerned with that here). — fishfry

(1) I would avoid the word 'valid' there, since it could be misunderstood in the more ordinary sense of 'valid' meaning 'true in every model'. What you mean is 'well-formed'. But, by definition, every sentence is well-formed, so we only need to say 'sentence'.

(2) If by 'independent' you mean 'not determined to be true, and not determined to be false', then there are no such sentences. Per a given model, a sentence is either true in the model or false in the model, and not both.

the OP makes the point that there are more mathematical truths than there are symbol strings to express them. — fishfry

That depends on what things are truths.

If a truth is a true sentence, then there are exactly as many truths as there are true sentences, which is to say there are denumerably many

If a truth is "state-of-affairs", such as taken to be a relation on the universe, then, for an infinite universe, there are more "truths" then sentences.

I used to think that Carnap's theorem was the real culprit:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both." — Tarskian

Where did Carnap write that?

We know it is so because having both addition and multiplication entails incompleteness, so, since Presburger arithmetic is complete, it can't define multiplication. — TonesInDeepFreeze

And that was my basic question: why having both addition and multiplication entail incompleteness?
How does it entail incompleteness?

Is it that with both addition and multiplication you can make a diagonalization or what is the reason?

why having both addition and multiplication entail incompleteness?
How does it entail incompleteness? — ssu

See a proof Godel-Rosser.

Is it that with both addition and multiplication you can make a diagonalization or what is the reason? — ssu

Diagonalization is available in any case. But we need multiplication for Godel numbering. We also need exponentiation, but Godel proved that exponentiation is definable.

But we need multiplication for Godel numbering. — TonesInDeepFreeze

OK, I think you answered here my question.

One point though: Godel-numbering is in the meta-theory, but we want to know why we need multiplication in the object theory. But, if I'm not mistaken, we need that it is representable in the object theory; I'd have to study the proof again.

My favorite game on the internet is guessing the number of page views per day for math and other topics. I guessed 126 here, whereas it is 111. Close, but no cigar. — jgill

That's interesting. Which page views? I think you've mentioned in the past that you look at papers written or something like that. — fishfry

Daily pageview statistics on Wikipedia. And papers submitted to ArXiv.org

For example: True arithmetic (Talk)
And True arithmetic (pageviews)

Low priority in Mathematics in Wikipedia. About the same as my own low priority math article.

(The daily analysis can be misleading, however. The median is a better indicator of popularity. For example, I just checked my former sport, bouldering, and found a huge disparity with a daily average of 912, but a median of 351. It was running below 400 per day until one day only it shot up to nearly 12,000. I haven't a clue.)

For perspective, keep in mind that Skolem arithmetic and Presburger arithmetic are not fully analagous, since Skolem arithmetic has more detailed axioms about its operation symbol. — TonesInDeepFreeze

Qualification:

Presburger arithmetic is usually stated with a finite axiomatization. But it also can be finitely axiomatized.

On the other hand, the only axiomatization of Skolem arithmetic I can find is at Wikipedia. It seems to be a finite axiomatization (it doesn't have an induction schema), but I don't understand it because it includes exponentiation though exponentiation is not primitive. So I can't say whether that Wikipedia axiomatization makes sense.

My point is that if we compare a finite axiomatization of Presburger arithmetic with finite axiomatization of Skolem arithmetic, we may find that they are indeed "analogous" to some extent.

Where did Carnap write that? — TonesInDeepFreeze

The diagonal lemma:

https://en.m.wikipedia.org/wiki/Diagonal_lemma

Rudolf Carnap (1934) was the first to prove the general self-referential lemma,[6] which says that for any formula F in a theory T satisfying certain conditions, there exists a formula sentence ψ such that ψ ↔ F(°#(ψ)) is provable in T.

(By the way, there seems to be a mistake in the page: "formula" should be "sentence").

Equivalently replace F by ¬ F:

ψ ↔ ¬ F(°#(ψ))

It is used in this negative variant in Gödel's proof and Tarski's proof.

That is how you get:

For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both

With "formula" or well-formed formula replaced by "predicate" or "property".

ψ ↔ ¬ F(°#(ψ))

is not:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both."

Counterexample: Let P be the property: P(S) if and only if S is equivalent with S.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

Counterexample: Let P be the property: P(S) if and only if S is equivalent with S. — TonesInDeepFreeze

I guess you meant to write:

Let P be the property: P(S) if and only if S is equivalent with P(#S).

In that special case, P is actually Tarski's truth predicate, which is indeed not definable. The conclusion here is that truth is not a legitimate predicate.

No, I meant what I wrote, I showed you a property of sentences that every sentence has.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

And what you wrote doesn't even make sense. # S is a number not a sentence.

No, I meant what I wrote, I showed you a property of sentences that every sentence has.

And what you wrote doesn't even make sense. # S is a number not a sentence. — TonesInDeepFreeze

In arithmetic theory, the argument n in P(n) must be a natural number. You cannot apply the predicate to S. You can only apply it to its Godel number.

P(S) is simply not a predicate of PA.

No, I meant what I wrote, I showed you a property of sentences that every sentence has. — TonesInDeepFreeze

The identity predicate in PA is:

P(n) := n = n

It cannot be implemented as:

P(S) := S <-> S

You are trying to do something that is not supported in PA.

For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both. — TonesInDeepFreeze

You wrote:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both."

That doesn't mention PA. Rather, it a universal generalization over properties and sentences.

That doesn't mention PA. Rather, it a universal generalization over properties and sentences. — TonesInDeepFreeze

I wrote that about the diagonal lemma, i.e. Carnap's theorem. Of course, there are conditions for when it applies. The context required, is PA or equivalent.

Furthermore, the term P(S) is in and of itself ambiguous.

It is a predicate that seemingly applies to a truth value. At first glance, it always means P(true) or P(false). It's as if it were a predicate with a boolean argument.

The term P(S) only works if you do not distinguish between the source code of the sentence and its truth value. It requires judiciously swapping between both meanings and second-guessing what exactly S means: the source code of the expression or its truth value? Sometimes this and sometimes that.

It yields expressions that are in fact not computable. No compiler would ever be able to compile that kind of things.

('r' for 'the numeral for' and '#' for 'the Godel number of')

Let C be this theorem:

For certain theories T, for every formula F(x) there is a sentence S such that T |- S <-> F(r(#S)).

Let K be:

"For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both."

C is not correctly rendered as K.

(1) K doesn't qualify as to certain kinds of theories.

(2) C generalizes over formulas, not over properties.

(3) C doesn't say anything about 'true'.

(4) C doesn't say that for every property of sentences there is a true sentence that does not have the property. C doesn't say that for every property of sentences that there is a false sentence that does have the property.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

Moreover:

(5) I showed a counterexample to both prongs of K.

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

You said that my counterexample is not in PA. So what? It doesn't have to be in PA, it merely needs to be a counterexample to K. And, by the way, K is not in PA, especially since PA doesn't have a predicate 'true'. And C includes PA as one of the T's, but C itself is not in PA.

(6) And with the arithmetization of syntax, both 'is a sentence' and 'is equivalent with itself' are expressible in PA. But I didn't do that, because K doesn't specify any language or kinds of theories.

/

For certain theories T, for every formula F(x) there is a sentence S such that T |- S <-> F(r(#S)).

is not remotely anything like:

For any property of logic sentences, there always exists a true sentence that does not have it, or a false sentence that has it, or both.

For certain theories T, for every formula F(x) there is a sentence S such that T |- S <-> F(r(#S)). — TonesInDeepFreeze

First, we replace F by ¬F. If F is a property then its negation is also a property. So, the following is an equivalent statement:

For certain theories T, for every formula F(x) there is a sentence S such that T |- S <-> ¬F(r(#S)).

Next, we replace S <-> ¬F(r(#S)) by the equivalent expression:

(S ∧ ¬F(r(#S)) ∨ (¬S ∧ F(r(#S))

Meaning:

(S is true and F is false) or (S is false and F is true)

Since ∨ is an "inclusive or", we can add "or both":

(S is true and F is false) or (S is false and F is true) or both.

So, it means:

A true sentence that does not have the property, or a false sentence that has the property, or both.

(2) C generalizes over formulas, not over properties. — TonesInDeepFreeze

That is the same.

A formula that takes a sentence as argument is a property of that sentence.

(3) C doesn't say anything about 'true'. — TonesInDeepFreeze

That is just how logic works. Asserting:
.
S ∧ L

Means:

S is true and L is true.

You said that my counterexample is not in PA. — TonesInDeepFreeze

P(S) := S <-> S

Is indeed impossible in PA. However, you can implement it as:

P(n) := n=n

The diagonal lemma is still perfectly satisfied for the identity property:

There exists a true sentence that does not have the identity property or a false sentence that does have the identity property, or both.

All false sentences have the identity property. Hence, you can always find one to satisfy the lemma. So, in what way is the diagonal lemma not satisfied?

(1) Your quoted characterization did not have the specifications you are giving now. Your quoted characterization was a broad generalization about properties and sentences.

(2) PA doesn't say 'true' and 'false'.

(3) Inclusive 'or' allows 'both' but not regarding 'true' and 'false'.

We may have:

P is true and Q is false
or
P is false and Q is true

But we cannot have:

P is true and Q is false
and
P is false and Q is true

(4) There are properties not expressed by formulas, so the generalization should be over formulas, not properties.

(5) I did not say: "P(S) := S <-> S." I said: "with the arithmetization of syntax, both 'is a sentence' and 'is equivalent with itself' are expressible in PA." What is not expressible in PA are 'true' and 'false'.

Your quoted characterization did not have the specifications you are giving now. Your quoted characterization was a broad generalization about properties and sentences. — TonesInDeepFreeze

It has always been an explanation about the diagonal lemma:

S <-> ¬F(r(#S))

Meaning:
(S ∧ ¬F(r(#S)) ∨ (¬S ∧ F(r(#S))

Meaning:
(S is true and F is false) or (S is false and F is true)

Meaning:
A true sentence that does not have the property, or a false sentence that has the property, or both.

It was a choice not to provide these details because this kind of explanations quickly become impenetrable in a multidisciplinary environment.

(2) PA doesn't say 'true' and 'false'. — TonesInDeepFreeze

The meaning of the S above is "a true sentence". PA doesn't say it, but that is what it means, for reasons of first-order logic.

(4) There are properties not expressed by formulas, so the generalization should be over formulas, not properties. — TonesInDeepFreeze

In that case, it is not a property in PA, because that would require a predicate in PA. In fact, Tarski's truth is also a property but not one in PA.

It is possible to precisely state all the conditions that apply, but in that case, the explanation becomes impenetrable. Nobody would be interested in a multidisciplinary forum. In order to keep it readable, there is no other alternative than to leave things out.

(S ∧ ¬F(r(#S)) ∨ (¬S ∧ F(r(#S))

Meaning:
(S is true and F is false) or (S is false and F is true)

Meaning:
A true sentence that does not have the property, or a false sentence that has the property, or both. — Tarskian

(1) You skipped that I pointed out that:

(S is true and F is false) and (S is false and F is true)

is never the case.

(2)

"S ∧ ¬F(r(#S)" is not the same as "S & ~F".
"¬S ∧ F(r(#S)" is not the same as "~S & F".

F

does not have a truth value. What has a truth value is

F(r(#S))

Saying "F is false" is nonsense.

(3) I agree with this:

C entails that there is a sentence S such that T proves:

(S & ~F(r(#S))) v (~S & F(r(#S))).

F expresses a property. F(r(#S)) is true if and only if S has the property expressed by F.

But:

[EDIT CORRECTION: I misread the quote. The quote was a disjunction not a conjunction. Mine is not a counterexample. No true sentence is not equivalent with itself, but every false sentence is equivalent with itself.]

It is possible to precisely state all the conditions that apply, but in that case, the explanation becomes impenetrable. Nobody would be interested in a multidisciplinary forum. In order to keep it readable, there is no other alternative than to leave things out. — Tarskian

First, I made a mistake as I misread your disjunction for conjunction. I made edit notes for that in my posts now.

You don't know what every person is interested in. As far as posts thus far, the only person to comment on your remark is me, and I am interested in seeing the subject properly represented. Better not to post terribly poor renderings of technical matters than to mangle the subject. The fact that this is a philosophy forum doesn't entail that it is good to oversimplify to the point of foggy vagueness and/or substantive misstatement. And I don't know why you would suppose that people would care about your synopsis of Carnap if they didn't also grasp the mathematical basis. You think many (if any) people are going to read your one liner about the theorem and grasp anything about it without a clearly stated mathematical basis? Moreover, your one-liner is incorrect.

Your synopsis is poor:

It does not make clear that it pertains specifically to the mathematical theorem.

It should not say 'logic sentences' in general, since the theorem pertains to sentences in certain languages for certain theories.

You should generalize over formulas in those languages and not over properties (since there are properties not expressed by formulas).

Disjunction is inclusive, but that does not entail that it is ever the case that "P is true and Q is false" and "P is false and Q is true".

You conflate a predicate with a sentence.

"S ∧ ¬F(r(#S)" is not the same as "S & ~F".
"¬S ∧ F(r(#S)" is not the same as "~S & F". — TonesInDeepFreeze

I left out that detail because it is obvious. So, with the details:

(S is true and F(r(#S)) is false) or (S is false and F(r(#S)) is true)

It is more accurate but also much more impenetrable than:

(S is true and F is false) and (S is false and F is true)

The resulting syntactic noise detracts from understanding what exactly it is about. It muddies the explanation.

and lately, you confuse the predicate F with a sentence. — TonesInDeepFreeze

I simplified F(r(#S)) to just F, because I thought that it was obvious what it was about.

And I don't know why you would suppose that people would care about your synopsis of Carnap if they didn't also grasp the mathematical basis. — TonesInDeepFreeze

If that is truly the case, then the subject may not be suitable for a philosophy forum. I had hoped that it was, but you may be right.

The metaphysical implications do seem out of reach of philosophical investigation. Apparently, they have been for almost a century.

You still resist recognizing these points:

You should not say 'logic sentences' in general, since the theorem pertains to sentences in certain languages for certain theories.

You should generalize over formulas in those languages and not over properties (since there are properties not expressed by formulas).

Disjunction is inclusive, but it is never the case that both of these are true: "P is true and Q is false" and "P is false and Q is true".

In general, a disjunction 'phi or psi' might not allow 'phi and psi', depending on the content in phi and the content in psi.

/

People can decide for themselves what is too technical or not. Symbolic logic, some mathematical logic and set theory are included in some philosophy department programs. Symbolic logic, even in community colleges. And philosophy of mathematics is based on understanding the mathematics being philosophized about.

And my point was not that people wouldn't be interested or that there are not enough technically minded readers (we don't know who is reading now or who might read in the future), but rather that the subject deserves to not be mangled by oversimplification.

I have made the point that whether there are uncountably many truths or whether there are unexpressed truths depends on what is meant by 'a truth'.

In context of mathematical logic, I would take a truth to be a certain kind of sentence relative to a given model. So, for a countable language, there are only denumerably many truths (i.e. true sentences).

One might also say that truths are states-of-affairs, such as a certain tuple being in a certain relation is a truth, even though no sentence asserts that fact. In that sense, yes, for an infinite universe, there are uncountably many truths.

The truth of a sentence is per interpretation, not per axioms. — TonesInDeepFreeze

Hmm. You could fix the interpretation and change the axioms to show that truth depends on the axioms plus the interpretation. That's not the usual way of thinking about it but I believe it could be shown.

Some sentences are true in all models. Some sentences are true in no models. Some sentences are true in some models and not true in other models.

Axioms are sentences. Some axioms are true in all models (those are logical axioms). Some axioms are true in no models (those are logically false axioms, hence inconsistent, axioms). Some axioms are true in some models and not true in other models (those are typically mathematical axioms).

The key relationship between axioms and truth is: Every model in which the axioms are true is a model in which the theorems of the axioms are true. And every set of axioms induces the class of all and only those models in which the axioms are true. — TonesInDeepFreeze

Ok.

(1) I would avoid the word 'valid' there, since it could be misunderstood in the more ordinary sense of 'valid' meaning 'true in every model'. What you mean is 'well-formed'. But, by definition, every sentence is well-formed, so we only need to say 'sentence'. — TonesInDeepFreeze

Yes you're right.

(2) If by 'independent' you mean 'not determined to be true, and not determined to be false', then there are no such sentences. Per a given model, a sentence is either true in the model or false in the model, and not both. — TonesInDeepFreeze

Yes ok.

That depends on what things are truths.

If a truth is a true sentence, then there are exactly as many truths as there are true sentences, which is to say there are denumerably many

If a truth is "state-of-affairs", such as taken to be a relation on the universe, then, for an infinite universe, there are more "truths" then sentences. — TonesInDeepFreeze

Yes that's a bit of murkiness in the paper the OP linked. But Chaitin and the paper author seem to take truth in the latter sense.