Sure, if all you wanted to point out was my misunderstanding of Godel's Incompleteness Theorems then you made your point. Rest assured your mockery of me is not something I would have thought that you would personalize your own authority over my ineptitude over the matter; but, carry on if you wish with your tirade about how your educated on the matter and I'm not. :lol:

So, I'll try and restate the topic subject to the question as to why the reader thinks Wittgenstein called Godel's Incompleteness Theorems, a logical trick?

What's your own understanding over Wittgenstein's comment about Godel's Incompleteness Theorems?

My posts were about correcting misstatements about the theorem and about the lack of clarity in your errant arguments about reactions to the theorem. And deserved though quite slight sarcasm about your claim about Einstein.

Before my previous post, there were no overly personal comments about you. I responded sharply back only when you falsely and snidely insinuated regarding my acquaintanceship with the halting problem and the method of definition.

I don't claim any authority.

I did not mention our personal comparative knowledge until my previous post in reply to your false insinuations about my knowledge.

You seem not to know the meaning of the word 'tirade'.



I have no comment on Wittgenstein.

I have no comment on Wittgenstein.

I responded personally only when you falsely and snidely insinuated regarding my acquaintanceship with the halting problem and the method of definition. — TonesInDeepFreeze

Given the limitations of communication online, as I actually said that it was not meant to be a snide comment, then I rest my case as there's no further point in trying to change your perception of what I meant by what was said...

And deserved though quite slight sarcasm about your claim about Einstein. — TonesInDeepFreeze

To think that Einstein didn't have discussions about the import of the defining work of Godel, being his Incompleteness Theorems, would seem like a moot issue to profess skepticism over.

Anyway, this thread was mostly about why Wittgenstein or what Wittgenstein could have meant by claiming that Godel's Incompleteness Theorems are logical tricks.

I have no comment on Wittgenstein. — TonesInDeepFreeze

:up:

I actually said that it was not meant to be a snide comment — Shawn

I put that in the category of "Don't micturate on me and tell me it's precipitation."

It was a stupid comment: There's no reason to think I don't know what the halting problem is; and knowing what the halting problem is doesn't fill in the definitions I asked about.

To think that Einstein didn't have discussions about the import of the defining work of Godel, being his Incompleteness Theorems, would seem like a moot issue to profess skepticism over. — Shawn

Wow! What a great stawman! I will get decidedly personal at this point: You're a sneaky poster. The question was not about Einstein discussing incompleteness. The question was your claim that Einstein thought "nature was "incomplete," for lack of a better word".

this thread was mostly about why Wittgenstein or what Wittgenstein could have meant by claiming that Godel's Incompleteness Theorems are logical tricks. — Shawn

Your post was about different things. I comment on what I feel like commenting on.

I will get decidedly personal at this point — TonesInDeepFreeze

Then, I suppose there's no point in continuing this thread as is; but, if anyone wants to address Wittgenstein's disregard of Godels Incompleteness Theorems, then by all means.

You must mean that there's no point in you continuing. Others can choose for themselves.

You must mean that there's no point in you continuing. — TonesInDeepFreeze

No, you telling me at this point to simply 'shut up' won't happen, sorry.

Others can choose for themselves. — TonesInDeepFreeze

Yeah, so its a discussion forum, so if you don't like the thread, then there's no point in trying to cow down other members, which in this case is your personalization of the issue.

You must mean that there's no point in you continuing.
— TonesInDeepFreeze

No, you telling me at this point to simply 'shut up' won't happen, sorry. — Shawn

Amazing that you got that exactly backwards.

no point in trying to cow down other members, which in this case is your personalization of the issue. — Shawn

Like I said, I didn't get personal with you until you did with me.

Amazing that you got that exactly backwards. — TonesInDeepFreeze

I'm really not interested in continuing this.

Like I said, I didn't get personal with you until you did with me. — TonesInDeepFreeze

If you were offended by my comment about the halting problem, then I'm sorry about how you felt about it, which was bona fide-as stated-not meant to be a snide comment. That's the best I can offer and I hope not to participate in any future exchanges with you.

I accept your disclaimer.

But I point out still that your comment gratuitously shifted from you to me. I asked for definitions and your retort was to falsely insinuate that I don't know about something, even though knowledge of that thing doesn't provide the definitions. My point is not that I take umbrage at that; but rather to catch your rhetorical dodge.

As to what you participate in, suit yourself, of course.

Anyway, this thread was mostly about why Wittgenstein or what Wittgenstein could have meant by claiming that Godel's Incompleteness Theorems are logical tricks. — Shawn

((everything below is in the context of PA or similar))

Gödel's own witness is certainly a corner case, but so are all the witnesses for his incompleteness theorem. Goodstein's theorem, another example, is also a corner case.

It is actually not difficult to see why every example for Gödel's theorem will always be a weird corner case.

We usually know that a proposition is true because it is provable, aka, as a consequence of the soundness theorem ("provable implies true"). Otherwise, without proof, the proposition is not a theorem but just a hypothesis. Pure reason, and therefore, mathematics, is blind. Unlike in physical reality, we cannot discover truth in mathematics by somehow observing it. We have to discover truth by discovering its proof. So, at first glance, it even looks impossible to discover examples of true but unprovable propositions. It is still possible, though. But then again, it certainly means that the discovery of a Gödelian proposition can only be achieved by means of the one or the other contorted hack.

https://en.wikipedia.org/wiki/Remarks_on_the_Foundations_of_Mathematics

Particularly controversial in the Remarks was Wittgenstein's "notorious paragraph", which contained an unusual commentary on Gödel's incompleteness theorems. Multiple commentators read Wittgenstein as misunderstanding Gödel. In 2000 Juliet Floyd and Hilary Putnam suggested that the majority of commentary misunderstands Wittgenstein but their interpretation[8] has not been met with approval.[9][10]

Wittgenstein's notorious paragraph is certainly confused:

Wittgenstein wrote:

I imagine someone asking my advice; he says: "I have constructed a proposition (I will use 'P' to designate it) in Russell's symbolism, and by means of certain definitions and transformations it can be so interpreted that it says: 'P is not provable in Russell's system'. Must I not say that this proposition on the one hand is true, and on the other hand unprovable? For suppose it were false; then it is true that it is provable. And that surely cannot be! And if it is proved, then it is proved that it is not provable. Thus it can only be true, but unprovable." Just as we can ask, " 'Provable' in what system?," so we must also ask, "'True' in what system?" "True in Russell's system" means, as was said, proved in Russell's system, and "false" in Russell's system means the opposite has been proved in Russell's system.—Now, what does your "suppose it is false" mean? In the Russell sense it means, "suppose the opposite is proved in Russell's system"; if that is your assumption you will now presumably give up the interpretation that it is unprovable. And by "this interpretation" I understand the translation into this English sentence.—If you assume that the proposition is provable in Russell's system, that means it is true in the Russell sense, and the interpretation "P is not provable" again has to be given up. If you assume that the proposition is true in the Russell sense, the same thing follows. Further: if the proposition is supposed to be false in some other than the Russell sense, then it does not contradict this for it to be proved in Russell's system. (What is called "losing" in chess may constitute winning in another game.)[11]

In his "notorious paragraph", Wittgenstein over-complicates the matter. Concerning Gödel's canonical witness, "P is not provable":

If the proposition is true, then it is not provable.
If the proposition is false, then it is provable.

Hence, the proposition is (true and not provable) or (false and provable).

If you look at what exactly Gödel's theorem says, There exist propositions in Russell's system that are (true and not provable) or (false and provable), then you can see that Gödel's canonical witness is exactly such proposition.

Of course, it is a weird corner case.

As I have argued above, pure reason is blind, and therefore, discovering truth normally requires discovering proof. Hence, any true but unprovable proposition will necessarily be a weird corner case that was hard to discover.

Because of the difficulty of discovering Gödelian statements, the impression may arise that Gödelian statements are not the norm but the odd exception, i.e. that they are some kind of deviant abnormality. This is not the case. The overwhelmingly vast majority of true propositions in true arithmetic are Gödelian. True but unprovable statements vastly outnumber the true and provable ones (cfr. True But Unprovable by Noson S. Yanofsky). This does indeed mean that the overwhelmingly vast majority of mathematical truth is simply invisible to us. Pure reason is blind.

We usually know that a proposition is true because it is provable — Tarskian

There are two ways:

(1) Prove a theorem from axioms. Then the theorem is true in any model in which the axioms are true.

(2) Prove that the sentence is true in a given model or given class of models.

If you look at what exactly Gödel's theorem says, There exist propositions in Russell's system that are (true and not provable) or (false and provable) — Tarskian

But if the system is sound, then the second disjunct is precluded.

But if the system is sound, then the second disjunct is precluded. — TonesInDeepFreeze

Soundness implies consistency.

So, if you manage to prove soundness theorem from PA then you have also managed to prove PA's consistency from PA.

Gödel's second incompleteness theorem prevents PA from proving its own consistency. If PA proves its own consistency, then PA is necessarily inconsistent.

Hence, it is not possible to preclude the second disjunct. PA puts up a lot of resistance to doing that by making the attempt self-defeating.

The soundness theorem is "If a sentence P is provable from a set of sentences G, then all models of G are models of P". We don't prove that from PA.

What we do prove (in, for example, set theory) is that PA has a model thus PA is consistent.

It is possible to preclude the second disjunct if we assume or prove that PA is sound. Meanwhile, I know of only a rare few mathematicians who even seriously doubt that PA is sound. I didn't say that PA itself proves that PA is sound.

The incompleteness theorem is that any consistent, formal, arithmetically adequate theory is incomplete, thus there is a true but PA-unprovable sentence. And PA is formal and arithmetically adequate, and one may take PA to be consistent, thus to infer that there is a true sentence that is not a theorem of PA. Moreover one may take PA to be sound, thus to infer that there is not a false theorem of PA.

That does not violate that PA does not itself prove that there is a true but PA-unprovable sentence, nor that PA itself proves that there is not a false but PA provable sentence.

In any case, notice what I said: IF the system is sound then the second disjunct is precluded. That is a correct statement.

It is possible to preclude the second disjunct if we assume or prove that PA is sound. I didn't say that PA itself proves that PA is sound. Virtually every mathematician (including Godel) regards PA to be sound. — TonesInDeepFreeze

Well, if PA is not sound, then it is actually unusable. So, we have to assume that it is sound. We simply have no other choice.

However, proving soundness is even irrelevant.

Imagine that we prove soundness theorem. Does that make soundness theorem true? No, because the proposition that proof implies truth is exactly what we are trying to prove. So, that would just be a silly exercise in circular reasoning.

Proving soundness is therefore both irrelevant and self-defeating.

What we do prove (in, for example, set theory) is that PA has model thus PA is consistent. — TonesInDeepFreeze

Proving PA's soundness from set theory amounts to moving the goal post. How do you prove set theory's soundness? It is simply the same problem all over again.

if PA is not sound, then it is actually unusable — Tarskian

It wouldn't be used for arithmetic. But it would still have models if it is consistent.

However, proving soundness is even irrelevant. — Tarskian

Irrelevant to what? Irrelevant to whom?

It's relevant to whomever would like to see it proven for whatever reasons.

Imagine that we prove soundness theorem. — Tarskian

I guess by "the soundness theorem" you mean "PA is sound". That's not what "the soundness theorem" ordinarily means in mathematical logic. The soundness theorem is:

"If a sentence P is provable from a set of sentences G, then all models of G are models of P"

That is not hard to prove, though it requires many steps and details.

No, because the proposition that proof implies truth is exactly what we are trying to prove. — Tarskian

Who is trying to prove? Do you mean "If P is a theorem of PA then P is true in the standard model for the language of PA"?

If the standard model for the language of PA is a model of PA, then of course any theorem of PA is true true in the standard model.

The part that requires much proof is that the standard model is a model of PA.

There is no circularity there.

As to proving soundness in set theory, notice that I did not claim that that has epistemological clout.

Again, what I said is correct: IF the system is sound then the second disjunct is precluded.

Moreover, I don't think there are more than a rare few mathematicians who seriously doubt that PA is sound. Pretty much, we, along with Godel, live in world of mathematics in which the soundness of arithmetic is taken for granted.

/

And I should have mentioned that not only is PA proven consistent in set theory (and note that I didn't claim that that has epistemological clout) but the consistency of PA is proven by Gentzen from assumptions that are considered quasi-finitistic, which, if I am not mistaken, is regarded by a lot of people as having epistemological clout.

If you look at what exactly Gödel's theorem says, There exist propositions in Russell's system that are (true and not provable) or (false and provable) — Tarskian

Where did Godel say that?

Here's a breakdown:

'true' and 'false' here mean, respectively, 'true in the standard model for the language of PA' and 'false in the standard model for the language of PA'.

'sound' and 'unsound' here mean, respectively, 'every theorem is true' and 'it is not the case that every theorem is true'.

'provable' and unprovable here mean, respectively, 'theorem of PA' and 'not a theorem of PA'.

PA is formal and arithmetically adequate.

(1) If PA is consistent, then there is a true but unprovable sentence.

(2), From (1), by disjunction introduction, if PA is consistent, then there is a true but unprovable sentence or there is a false but provable sentence.

(3) If PA is sound, then PA is consistent.

(4) From (1) and (3), if PA is sound, then there is a true but unprovable sentence and there is not a false but provable sentence.

True but unprovable statements vastly outnumber the true and provable ones — Tarskian

There are denumerably many of each.

The part that requires much proof is that the standard model is a model of PA. — TonesInDeepFreeze

Yes, and once this part has been proven, there is no need to prove soundness theorem, because the model-theoretical construction already guarantees this.

"If a sentence P is provable from a set of sentences G, then all models of G are models of P" — TonesInDeepFreeze

This statement requires the use of model theory. Soundness can also be defined without using model theory:

https://en.wikipedia.org/wiki/Soundness

I am obviously not against using model-theoretical notions to define soundness, but it obviously raises the bar in terms of accessibility.

If you look at what exactly Gödel's theorem says, There exist propositions in Russell's system that are (true and not provable) or (false and provable) — Tarskian


Where did Godel say that? — TonesInDeepFreeze

If we start from Carnap's diagonal lemma:

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

Let T be a first-order theory in the language of arithmetic and capable of representing all computable functions, and F(y) be a formula in T with one free variable. Then there exists a sentence C such that

T ⊢ C ⇔ F (⌜C⌝)

Choose F(y) to be ¬ Bew(y), with Bew(y) the provability predicate. This will morph Carnap's diagonal lemma into Gödel's incompleteness theorem:

Let T be a first-order theory in the language of arithmetic and capable of representing all computable functions. Then there exists a sentence G such that

T ⊢ G ⇔ ¬ Bew(⌜G⌝)

We can see that:

T ⊢ G ⇔ ¬ Bew(⌜G⌝)

is equivalent to:

T ⊢ ( G ∧¬ Bew(⌜G⌝) ) ∨( ¬ G ∧ Bew(⌜G⌝)

In plain English:

G ∧¬ Bew(⌜G⌝) means G is true and not provable

¬ G ∧ Bew(⌜G⌝ means G is false and provable

Therefore, Gödel's incompleteness theorem can be written as:

Let T be a first-order theory in the language of arithmetic and capable of representing all computable functions. Then there exists a sentence G such that

G is (true and not provable) or G is (false and provable)

This is somewhat equivalent to the alternative phrasing in which we assume consistency:

https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems

First Incompleteness Theorem: "Any consistent formal system T within which a certain amount of elementary arithmetic can be carried out is incomplete; i.e. there are statements of the language of T which can neither be proved nor disproved in T." (Raatikainen 2020)

The result that follows straight out of Carnap's diagonal lemma does not assume consistency.That is why I do not see where the requirement comes from, to phrase it like that. I prefer to phrase it as the proof's verbatim output.

There are denumerably many of each. — TonesInDeepFreeze

In "True But Unprovable", Yanofsky insists that unprovably true statements vastly outnumber provably true ones:

http://www.sci.brooklyn.cuny.edu/~noson/True%20but%20Unprovable.pdf

We have come a long way since Gödel. A true but unprovable statement is not some strange, rare phenomenon. In fact, the opposite is correct. A fact that is true and provable is a rare phenomenon. The collection of mathematical facts is very large and what is expressible and true is a small part of it. Furthermore, what is provable is only a small part of those.

True and provable statements are denumerable while true and unprovable statements are non-denumerable.

(1) If PA is consistent, then there is a true but unprovable sentence. — TonesInDeepFreeze

Gödel's theorem can perfectly be phrased without assuming that PA is consistent.

In fact, by introducing the assumption "If PA is consistent", Gödel's theorem is no longer a theorem in PA. In that case, it is a theorem in PA + Cons(PA). That is not the same theory as PA.

Incompleteness as a theorem provable in PA proper must be phrased as:

There is a true but unprovable statement or a false but provable statement.

there is no need to prove soundness theorem — Tarskian

You keep using 'the soundness theorem' in a way that invites confusion.

For the third time, the soundness theorem is:

"If a sentence P is provable from a set of sentences G, then all models of G are models of P"

Soundness can also be defined without using model theory:

https://en.wikipedia.org/wiki/Soundness — Tarskian

You are very confused. That page relies on '|=' which is from model theory.

I am obviously not against using model-theoretical notions to define soundness — Tarskian

What's your definition?

If you look at what exactly Gödel's theorem says — Tarskian

https://en.wikipedia.org/wiki/Diagonal_lemma — Tarskian

I'm asking where Godel says, not what Wikipedia says.

G is (true and not provable) or G is (false and provable)

This is somewhat equivalent to the alternative phrasing in which we assume consistency:

https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems

First Incompleteness Theorem: "Any consistent formal system T within which a certain amount of elementary arithmetic can be carried out is incomplete; i.e. there are statements of the language of T which can neither be proved nor disproved in T." (Raatikainen 2020) — Tarskian

If PA is consistent, then there are true but unprovable sentences. So, trivially, by disjunction introduction, it follows that there are true but unprovable sentences or there are false but provable sentences.

Meanwhile, for the third time, my remark is correct: If we assume soundness, then the second disjunct is precluded.

not assume consistency — Tarskian

The theorem doesn't assume consistency. Rather, the theorem has consistency as a conjunct in the antecedent of a conditional. If you drop that conjunct, then you don't have the theorem that there is an unprovable sentence.

true and unprovable statements are non-denumerable. — Tarskian

There are not only finitely many of them, and there are not uncountably many of them (there are only countably many sentences in the language), so there are denumerably many.

In fact, by introducing the assumption "If PA is consistent", Gödel's theorem is no longer a theorem in PA. In that case, it is a theorem in PA + Cons(PA). That is not the same theory as PA. — Tarskian

You are very confused.

Incompleteness is not a theorem of PA, unless PA is inconsistent.

[EDIT: I'm not happy now with my wording above. I should have said, "there is a PA-unprovable sentence" is not a theorem of PA unless PA is inconsistent. And that actually accords with Tarskian, so I should not claim that he is confused on that point. However, I would need to think more whether "Con(PA) then PA is incomplete" (or "If PA is consistent then PA is incomplete") is a theorem of PA.]

Incompleteness is ordinarily proved informally or formally in a variety of theories (most signficantly, finitistically in, say, PRA). Proofs of incompleteness do not require Con(PA) as an assumption.

The theorem is: If PA is consistent, then there is a true but unprovable sentence.

I think you're saying, drop the antecedent, and we have: There is a true but unprovable sentence or there is false but provable sentence.

So what?

If PA is consistent, then there is a true but unprovable sentence. So, by disjunction introduction, there is true but unprovable sentence or there is a false but provable sentence.

If PA is inconsistent, then there is a false but provable sentence. So, by disjunction introduction, there is a true but unprovable sentence or there is a false but provable sentence.

For the fourth time: My statement is correct: If we assume soundness then we can preclude the second disjunct.

That page relies on '|=' which is from model theory. — TonesInDeepFreeze

The use of logical entailment predates model theory by decades:

https://en.wikipedia.org/wiki/Logical_consequence

The turnstile symbol ⊢ was originally introduced by Frege in 1879, but its current use only dates back to Rosser and Kleene (1934–1935).

The page uses it without distinguishing between theory T and its model M:

A1..An ⊢ S => A1..An ⊨ S with T equivalent to A1..An

Therefore, it is not a model-theoretic explanation. It just uses "T ⊢ S " as as a synonym for "T proves S" and T ⊨S as a synonym for "S is true in T".

If PA is consistent, then there are true but unprovable sentences. So, trivially, by disjunction introduction, it follows that there are true but unprovable sentences or there are false but provable sentences.

Meanwhile, for the third time, my remark is correct: If we assume soundness, then the second disjunct is precluded. — TonesInDeepFreeze

I did not say that your remark would be wrong or that the (Raatikainen 2020) characterization would be wrong. I just said that I prefer to introduce the disjunction and avoid assuming Cons(PA). This is just a personal preference.

There are not only finitely many of them, and there are not uncountably many of them (there are only countably many sentences in the language), so there are denumerably many. — TonesInDeepFreeze

Yanosky includes the Gödelian statements that cannot be expressed by language. There are uncountably many of those.