One may consult introductory textbooks in mathematics to see how we can prove undefinability from incompleteness or prove incompleteness from undefinability.

The definition of 'incomplete' is simple:

A theory T is incomplete if and only if there is a sentence S in the language for T such that neither S nor its negation are a theorem of T.

It is trivial to prove that there are incomplete theories, and not trivial, though pretty easy, is proving the soundness theorem that then trivially proves the incompleteness of certain theories. What is interesting about Godel-Rosser is that there are incomplete theories of a particular kind (consistent, recursively axiomatizable and arithmetically sufficient).

If we define 'true' as 'provable', then of course all bets are off regarding these theorems as they are stated. And if in baseball we define 'hit' as 'home run', then we would throw away all the baseball statistics books. Yeah, we know all that.

But to accommodate someone who insists that 'true' means 'provable', then we could simply say that wherever we have written the word 'true', it is to be replaced by 'gorue'. Then read all the proofs and discussions about them with that change. It matters not toward understanding the substance of them.

There is no proof of G in F.

That's the point.

Too miss that point is to utterly not know what the theorem is about.

"Why" is not a technical term, more a heuristic matter, and could mean different things to different people. In the most bare sense, "Why is T a theorem?" is answered by showing the proof of T . But, heuristically, there is a massive amount of discussion in the literature giving insight into the theorem and its proof; and insight is given in Godel's own paper.

With identity theory, '=' is primitive and not defined, and the axiom of extensionality merely provides a sufficient basis for equality that is not in identity theory. Without identity theory, for a definition of '=' we need not just the axiom of extensionality but also the 'xez <-> yez' clause.

As to manipulation of symbols, the incompleteness theorem can be be done in mere primitive recursive arithmetic, so the assumptions and means of reasoning are well within the scope of the methods of finite arithmetical calculations.

Regarding Tarski's undefinablity theorem, Tarski proved that in certain systems, there does not even exist such a sentence. Not only did Tarski not use such sentences as a basis, he actually proved that such sentences don't even exist in the relevant systems. To not understand that is to not understand what the theorem is even about.

/

I know the context in which interrogatory sentences were mentioned lately. But the matter of interrogatories has been brought into other posts in this forum as part of incorrect attempts to refute the theorems.

Again, as has been explained several times in this forum:

G asserts that G is not provable in system P.

But P does not prove G, and P does not prove that it does not prove G.

/

Proofs don't "hide" things. From fully declared axioms and rules of inference, we may prove Godel-Rosser. We may prove versions that do not mention semantics. And we may prove versions that mention both syntax and semantics. This is all famous and understood by reading an introductory textbook in mathematical logic.

The incompleteness theorem requires no notion or terminology 'True(L, x)' where L is a set of axioms or system.

Rather, using the above style of notation, we have:

True(M x) where M is a model and x is a sentence. Read as "x is true in M".

and

Theorem(L x) where L is a set of axioms and x is a sentence. Read as "x is a theorem from L".

And we prove about certain systems:

Theorem(L x) implies that for every model M, if True(M y) for every y in L, then True(M x). (This is the soundness theorem).

And we prove about certain axiom sets:

If True(M x) in every model M, then Theorem(L x).

And incompleteness proves that there are M and L such that:

(1) For every y such that Theorem(M y), we have True(M y); and (2) True(M x); but (3) it is not the case that Theorem(L x).

"Did you lie?" doesn't have a truth value, because it is not a declarative sentence. Indeed, interrogatory sentences do not appear as lines in proofs.

Contrary to a claim made in this thread (and made by the same poster several other times in this forum), it is not the case the Godel sentence requires that there is a sequence of inference steps that prove that they don't exist (as has been explained several other times in this forum).

More generally, Godel's and Tarski's proofs do not have the defects claimed in this thread (and claimed by the same poster several other times in this forum). That can be verified by reading an introductory textbook on mathematical logic in which the groundwork and proofs of Godel-Rosser incompleteness and Tarski undefinability are provided.

I don't prefer Wikipedia as a reference on such matters, but it was asked where in the Wikipedia article on the 'Axiom of extensionality' is it said that 'equals' means 'the same'.

The article states that the axiom of extensionality uses '=' with regard to predicate logic, with a link to an article on 'First-order logic'. And that article correctly states that the most common convention is that 'equals' means 'the same'. Moreover, the article on 'Equality (mathematics)' defines equality as sameness, and the article on the equals sign refers to equality, and the article on 'Identity (mathematics)' refers to equality.

In ordinary contexts in mathematics, including mathematical logic, including set theory, 'equals' means 'is the same as', which means the same as 'is identical with'. This is formalized by identity theory, which extends first order logic without identity, and adds a primitive binary predicate symbol '=' with axioms and a semantics.

More specifically:

/

Identity theory is first order logic plus:

Axiom: Ax x=x

Axiom schema:
For all formulas P,
Axy((x=y & P(x)) -> P(y))

Semantics:

For every model M, for all terms T and S,
T = S
is true if and only if M assigns T and S to the same member of the universe.

/

Set theory can be developed in at least two ways:

(1) First adopt identity theory. This gives us:

Theorem: Axy(x=y -> Az((xez <-> yez) & (zex <-> zey)))

Then add the axiom of extensionality:

Axiom: Axy(Az(zex <-> zey) -> x=y)

This gives us:

Theorem: Axy(x=y <-> Az(zex -> zey))

Thus, with identity theory and the axiom of extensionality, every model of
Az(zeT <-> zeS)
is a model that assigns T and S to the same member of the universe.

(2) Don't adopt identity theory. Instead:

Definition: Axy(x=y <-> Az((xez <-> yez) & (zex <-> zey)))

That gives us as theorems all the axioms of identity theory.

However, if we don't also stipulate the semantics of identity theory, the axioms of identity theory along with the axiom of extensionality do not provide that every model in which S=T is true is a model that assigns S and T to same member of the universe.

I don't think they're stupid. Rather, I find that there is complacency and sloppiness in the writing of certain articles, sometimes to the extent that there are plain falsehoods in them. But in the case of the article being discussed, I'm not pointing out falsehoods, but rather the confusions the article opens.

Before the reply to my post, I deleted "To see that, you just need to read the article that you yourself say is "clear and accurate"", as I thought it would be better not to invite referencing that article again.

But to address that article, here are the proofs without ""liar"" (scare quotes in original), "ask", "truth bearer" or anything else extraneous to the mathematical proofs:

In this context, 'formula' and 'sentence' mean 'formula in the language of first order arithmetic' and 'sentence in the language of first order arithmetic'.

In this context, 'true' and 'false' mean 'true in the standard model for the language of first order arithmetic' and 'false in in the standard model for the language of first order arithmetic'.

For every formula M, let g(M) be the numeral for the Godel number of M.

/

Theorem: There is no formula T(x) such that for every sentence S, T(g(S)) is true if and only if S is true.

Proof:

Toward a contradiction, suppose there is such a T(x).

So, there is a formula D(x) such that for every numeral m, D(m) is true if and only if m is the numeral for the Godel number of a formula P(x) such that P(m) is false. (The steps in obtaining this line from the previous line are not included in the article.)

D(g(D(x))) is true
if and only if
g(D(x)) is the numeral for the Godel number of a formula P(x) such that P(g(D(x))) is false.

Toward a contradiction, suppose D(g(D(x))) is true.
So g(D(x)) is the numeral for the Godel number of a formula P(x) such that P(g(D(x))) is false.
g(D(x)) is g(P(x)), so D(x) is P(x), so D(g(S(x))) is P(g(S(x))), so D(g(S(x))) is false. Contradiction.

Toward a contradiction, suppose D(g(D(x))) is false.
So it is not the case that g(D(x)) is the numeral for the Godel number of a formula P(x) such that P(g(D(x))) is false.
So D(g(D(x))) is true. Contradiction.

So there is no formula T(x) such that for every sentence S, T(g(S)) is true if and only if S is true.

/

Theorem: There is no formula T(x) such that for every sentence S, S is true if and only if T(g(S)) is true.

Proof:

Lemma: For every formula P(x) there is a sentence D such that D <-> P(g(D)) is true.

Toward a contradiction, suppose there is a formula T(x) such that for every sentence S, S is true if and only if T(g(S)) is true.

So, for every sentence S, S <-> T(g(S)) is true.

By the lemma, there is a sentence D such that D <-> ~T(g(D)) is true. But also, D <-> T(g(D)) is true. Contradiction.

So there is no formula T(x) such that for every sentence S, S is true if and only if T(g(S)) is true.

It's not a question of what was relevant to your point. I cited faults in the article, whether or not those faults bear on your point.

Tarski's proof doesn't work the way you describe it.

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For any sentence T, set of axioms S and set of rules R, we may ask the question "Is T derivable from S with R?" That fact doesn't entail the counterfactual that there are questions in proofs. Tarski's proof does not have questions in it.

Yet I showed exactly what is amiss in the Wikipedia article recently cited.

"Is there a proof of T?" is a question.

But a proof of T does not have questions in it.

One can couch things as questions. But the proofs themselves do not have questions in them.

All steps in proofs are statements, not questions.

Whatever the case may be with your characterization of the subject, at least we know that disallowing sets to be members of themselves does not avoid Russell's paradox but rather, as to first order set theory, Russell's paradox is avoided by not having the schema of unrestricted comprehension.

we'll just go around in circles, and then perhaps we'll come back to the p and q point again. — Philosopher19

Yes, going back to your p and q would be going back full circle yet again.

To break the circle requires that you give serious consideration to the fact that no one understands your phraseology but you. That would lead to you learning more about the subject so that you could communicate your ideas about it with other people.

What is incorrect is the assumption that there is a barber who shaves all and only those who do not shave themselves.

And we don't even need any set theory to prove that, for any property P, there is no x such that, for all y, Pxy if and only if ~Pyy. We prove that by pure logic alone. So, perforce, we prove it in set theory too.

/

The hierarchy of sets is not type theory nor higher order logic. ZFC is a first order theory. The context here is not type theory nor a higher order logic, but rather first order set theory:

The main point here could not be more clear: The logic is monotonic, so adding an axiom can't block inconsistency.

Said another way:

Let G and H be sets of formulas, and P a formula:

If G proves P, then G union H proves P.

So, if set theory without the axiom of regularity proves a contradiction, then set theory with the axiom of regularity proves a contradiction.

The schema of unrestricted comprehension proves a contradiction. So the schema of unrestricted comprehension with also the axiom of regularity proves a contradiction.

So the contradiction can't be avoided by adding the axiom of regularity to any set of axioms, but rather the contradiction can be avoided only by not having the schema of unrestricted comprehension.

No one knows what you mean by such locutions as "x is a member of itself only in its own set". You have not defined what it might mean.

Someone might as well say "x is not a member of itself not only outside itself" or similar nonsense and then require you to understand it though it is impossible to understand.

./

The assertion that there exists a set of which every set is a member allows Russell's paradox.

I did not contradict myself.

And, again, as I just explained, disallowing sets from being members of themselves does not avoid inconsistency. Again, as I just explained, inconsistency can be avoided only by deleting axioms (such as unrestricted comprehension) and not by adding them (such as regularity).

The fact that ZFC avoids inconsistency by not having unrestricted comprehension does not contradict that ZFC also has the axiom of regularity that disallows sets from being members of themselves. Moreover, the purpose of the axiom of regularity is not to avoid inconsistency but rather to facilitate the study of sets as in a hierarchy indexed by the ordinals. It is a nice feature of the axiom of regularity that it disallows sets being members of themselves, as many people regard it counter-intuitive or against the basic concept of 'set' that there are sets that are members of themselves. But the axiom of regularity, even with that feature, is not how set theory avoids inconsistency.

Usage may vary. One prominent definition of 'theory' is that a theory is a set of sentences closed under derivability. Then, any set of axioms determines a theory.

It is a common misconception on Internet forums that ZFC avoids inconsistency by disallowing sets to be members of themselves.

Yes, the axiom of regularity, which is adopted in ZFC, disallows that a set can be a member of itself. But the axiom of regularity does not block inconsistency. What avoids inconsistency is not having the axiom schema of unrestricted comprehension. If we have the axiom schema of unrestricted comprehension, then we get inconsistency, no matter whether we also have the axiom of regularity or not, and no matter whether there may be sets that are members of themselves or not.

Indeed, since the logic is monotonic, adding an axiom cannot avoid inconsistency. The only way to avoid inconsistency is to delete the axioms that provide inconsistency.

ZFC is undecidable. That means that there is no decision procedure to determine whether a given sentence in the language of ZFC is or is not a theorem of ZFC.

S might not be countable.

But, yes, we do have that either S in S or S not in S.

Either S in S or S not in S.

Suppose S in S. Then S not in S. So S not in S.

Suppose S not in S. Then S in S. So S in S.

So both S in S and S not in S.

I'm not in the practice of editing Wikipedia articles. Meanwhile, my points about the article stand. More generally, a good amount of caution is warranted when referencing Wikipedia.

Maybe I'm missing something, but that seems to rely on the premise that either S = 0 or S = {S}, which is a premise we are not allowed.

V = the set of all v
Z = the set of all z — Philosopher19

That doesn't mean anything.

We don't say "the set of all z". We say "the set of all z such that [fill in some property here]"

Examples:

the set of all z such that z is an even number
which is
{z | z is an even number}

the set of all z such that z is a subset of the set of natural numbers
which is
{z | z is a subset of the set of natural numbers}

/

'domain' and 'range' are defined:

domain(S) = {x | Ey <x y> in S}

range(S) = {y | Ex <x y> in S}

That is most applicable when S is a relation (a relation is a set of ordered pairs).

And every function is a relation.

The notion of 'list' is captured by a certain kind of function.

{<1 John> <2 Paul> <3 George> <4 Ringo>} is a list. It maps

1 to John
2 to Paul
3 to George
4 to Ringo

It tells you not just the members of the set {John Paul George Ringo} but also a particular ordering of that set.

The set of objects that the function maps to is the range of the function. In this case

the domain is {1 2 3 4} and the range is {John Paul George Ringo}.

Think of it this way:

Suppose I tell you that B is the set whose members are the Beatles. I haven't told you an order of the musicians in the Beatles, just that B is the set of them no matter in what order.

There are 24 ways to list the Beatles.

Set theory conveys the notion of a list with functions:

{<1 John> <2 Paul> <3 George> <4 Ringo>} is one list.

{<1 George> <2 John> <3 Paul> <4 Ringo>} is another list.

and there are 22 more.

In other words, we exhibit the order of the list by putting the number of the item in the list before the item.

Again:

the members of {John Paul George Ringo} are John, Paul, George and Ringo, in no particular order.

the members of one LIST of the Beatles are <1 John>, <2 Paul>, <3 George> and <4 Ringo>.

the members of another LIST of the Beatles are <1 George>, <2 John>, <3 Paul> and <4 Ringo>.

but all those lists have the same RANGE, which is {John Paul George Ringo}, whose members are John, Paul, George and Ringo.

/

So, if L is the list of all lists, then L is a function. And the items listed by L are the range of L. In this case, the range of L is the set of all lists. So, since L is itself a list, L is in the range of L. But L is not in L, since L is not an ordered pair.

Whether an item in a set is a member of itself or not is dependent on what it is and in what set it is in. — Philosopher19

If x in x, then x in x no matter what other sets x is or is not in.

If you don't agree or don't understand, then either you need to state your alternative system and terminology or you need to learn basic set theory. I recommend starting with the latter.

As I said, you have a highly idiosyncratic notion of sets and a highly idiosyncratic terminology to go with that. If you wish to be understood by other people, then you need to either state your system of notions and terminology, or show how they can be defined from notions and terminology that other people do already understand.

/

You can start at the very beginning:

'is an element of'
'is a member of'
'is in'
'in'
'e'

are just variants of the same primitive relation of set theory.

The symbol 'e' itself is a primitive relation symbol. It is not defined. Rather, there are axioms that determine what theorems are derived with 'e' in them. However, from 'e' we define a number of other symbols that intuitively stand for various concepts, including 'list'. With those definitions we formulate yet more theorems (though they all are in principle reducible to just the primitive 'e').

Your notions though, as so far stated, don't resolve to that process. Therefore, if you want to be understood, then you need to either show how your notions do resolve to those of set theory, or you need to state your own primitives, axioms and definitions that do explicate your own notions, or at least give some coherent outline about that.

/

Meanwhile, at least you should pay attention to the fact that there is a difference between 'set' and 'list' as a list is a certain kind of set. For example:

{John Paul George Ringo} = {George Paul Ringo John}

but

{<1 John> <2 Paul> <3 George> <4 Ringo>} not= {<1 George> <2 Paul> <3 Ringo> <4 John>}

and

<John Paul George Ringo> not= <George Paul Ringo John>

{John Paul George Ringo} is the set whose members are John, Paul, George and Ringo, in whatever order you want to mention them.

{<1 John> <2 Paul> <3 George> <4 Ringo>} is a set theoretic list, which is itself a set whose members are <1 John>, <2 Paul>, <3 George> and <4 Ringo>, in whatever order you want to mention them.

The range of {<1 John> <2 Paul> <3 George> <4 Ringo>} is {John Paul George Ringo}. Or, we can say that the entries of the list are John, Paul, George and Ringo.

<John Paul George Ringo> is the ordered 4-tuple that corresponds to the above list. (It actually unpacks to nested ordered pairs that unpack to certain unordered pairs, but we don't need to spell the details of that now.) Or we can say that the coordinates of <John Paul George Ringo> are John, Paul, George and Ringo.

Crucial takeway: A set is not in any particular order, except a list is a special kind of set that does convey a particular order. And the members of a list are ordered pairs, not the members of the RANGE of the set.

So, I've given you information that is at least a start for you to use common mathematical terminology, so that you may be understood by people other than yourself.

You didn't answer the question. I won't bother to post it yet again.

If you can find even one mathematician who can say what you mean by "L is a member of itself in L" then I'd welcome hearing about it.

And if you can define 'where', in whatever sense you mean by it, from the terminology of set theory, then I'd welcome hearing about it.

(1) The article conflates a language with a theory.

(2) The proof in the article handwaves past the crucial lemma, thus appearing to commit a serious non sequitur.

Set theory does not have a "where".

My question remains:

What exactly would need to be the case for "L is a member of itself" to be true while "L is a member of itself in L" is false? And what exactly would need to be the case for "L is a member of itself" to be false while "L is a member of itself in L" is true? — TonesInDeepFreeze

You are using your own personal terminology for an unclear notion that no one other than you can make sense of.

You need to define your terminology in already understood mathematical phrasing. Otherwise, whatever you have in mind won't be understood by others.

That article doesn't properly state the subject matter.

The terminology in this context needs to be exact.

(I use 'K' instead of 'LL' because it is not good notation to use a letter 'L' as a standalone constant and also concatenated with itself to form another constant.)

L = the list of all lists
K = the list of all lists that list themselves

I hope it is recognized that these are equivalent:

x is in S
x in S
x is an element of S
x is a member of S
xeS {read 'e' as the letter epsilon)

So, these are equivalent:

L is in L
L in L
L is an element of L
L is a member of L
LeL

These are equivalent:

T lists y
T is a list & y in range(T)

These are equivalent:

T lists T
T is a list & T in range(T)
T lists itself

So, these are equivalent:

L lists L
L is a list & L in range(L)
L lists itself

And these are equivalent:

K lists K
K is a list & K in range(K)
K lists itself

/

Does L list itself in L? — Philosopher19

"L lists itself" is sensical.

"L lists itself in L". What does that mean other than "L lists itself"?

In other words, what exactly would need to be the case for "L lists itself" to be true while "L lists itself in L" is false? And what exactly would need to be the case for "L lists itself" to be false while "L lists itself in L" is true?

More generally, what exactly would need to be the case for "S lists y" to be true while "S lists y in x" is false? And what exactly would need to be the case for "S lists y" to be false while "S lists y in x" is true?

Even more generally, what exactly would need to be the case for "S has property P" to be true while "S has property P in x" is false? And what exactly would need to be the case for "S has property P" to be false while "S has property P in x" is true?

Is L a member of itself in L? — Philosopher19

"L is a member of itself" is sensical.

"L is a member of itself in L". What does that mean other than "L is a member of itself"?

In other words, what exactly would need to be the case for "L is a member of itself" to be true while "L is a member of itself in L" is false? And what exactly would need to be the case for "L is a member of itself" to be false while "L is a member of itself in L" is true?

More generally, what exactly would need to be the case for "S is a member of y" to be true while "S is a member of y in x" is false? And what exactly would need to be the case for "S is a member of y" to be false while "S is a member of y in x" is true?

Again, even more generally, what exactly would need to be the case for "S has property P" to be true while "S has property P in x" is false? And what exactly would need to be the case for "S has property P" to be false while "S has property P in x" is true?

LL/not-L — Philosopher19

What does that mean?

What does 'not-L' stand for? Does it stand for the set of all things not in L? That is, {x | x not in L}. Or does it mean the set of all things that are not L? That is, {x | x not= L}.

What does '/' stand for?

What is the difference between asking if L lists itself and asking if L is a member of itself? — Michael

A list is a sequence. S lists x if and only if x is in the range of S. In that case, x is not a member of S, but rather it is a member of the range of S.

For example:

B = {John Paul George Ringo} is just a set.

T = {<1 John> <2 Paul> <3 George> <4 Ringo>} is a list of the members of B. Sometimes represented as the equivalent ordered 4-tuple:

<John Paul George Ringo>

or

1 John
2 Paul
3 George
4 Ringo

or

John
Paul
George
Ringo

B = range(T)

So John is not a member of T, but John is a member of the range of T = B.

I believe you are in essence removing/ignoring the semantical implications of "member of self" with the way you engage. — Philosopher19

Saying 'semantical' adds nothing substantive in this context.

The notion of "member of self" is not more than "x is a member of itself if and only if x is a member of x".

And I addressed exactly whether L is a member of L. It is not. Rather, L is a member of the range of L.

"L is a member of itself in L" has no apparent meaning if it does not simply mean "L is a member of L". Nothing is added by saying "in L". If L is a member of L then L is in L. Nothing is qualified by saying "in L" again.

There are two matters:

Whether L is a member of L. It is not, since the members of L are ordered pairs and L is not an ordered pair.

Whether L is a member of the range of L. It is.
I.e. whether L is one of the items listed by L. It is.

I think I've seen this one before:

"Is the correct answer to this question 'no'?"

If 'no' is the correct answer, then 'no' is not the correct answer.

If 'yes' is the correct answer, then 'yes' is not the correct answer.

A question is not rhetorical or not. An utterance of a question is rhetorical or not. An utterance of a question is rhetorical if and only if the utterer does not intend to provoke an answer.

"Is this a rhetorical question?"

It the utterer intends to provoke an answer, then the utterance is not a rhetorical question. If the utterer does not intend to provoke an answer, then the utterance is a rhetorical question.

My guess is that the poster intended to provoke answers, in which case the utterance of the question was not a rhetorical question and the correct answer regarding that utterance is 'no'.

I don't see a paradox here.

It has everything to do with what you said.

And I said that the exact answer to "Does L list itself?" is yes.

Progress will begin upon you paying attention to the replies you've received.