Resolution of the question as to whether math is discovered or invented

[Impenitent]

I have just pointed out what the fundamental difference is between mathematics, science, and philosophy. I don't think that it is necessary to give examples. Everybody is supposed to know examples already. There are millions of machinists out there.

Where is the machinist in the Platonic realm?

in a cave

-Imp

[Skepdick]

I have just pointed out what the fundamental difference is between mathematics, science, and philosophy. I don't think that it is necessary to give examples. Everybody is supposed to know examples already. There are millions of machinists out there.

Where is the machinist in the Platonic realm?

in a cave

-Imp

Is that cave inside; or outside the Platonic realm?

Which way's the escaping?

[godelian]

Where is the machinist in the Platonic realm?

ChatGPT: Is it possible to define the ontology of mathematics solely by making deductive claims within mathematics itself?

This is a deep and subtle question that touches on the philosophy of mathematics, specifically concerning mathematical ontology (what kinds of things exist in mathematics) and epistemology (how we come to know them).

Short Answer:

No, it is not possible to fully define the ontology of mathematics solely by making deductive claims within mathematics itself. Ontology inherently involves meta-mathematical and philosophical considerations that go beyond purely deductive reasoning within a formal system.

Deductive reasoning within mathematics is essential for developing mathematical theories, but it cannot on its own answer ontological questions like what kinds of mathematical entities exist or what is the nature of mathematical truth. Those are philosophical questions that require stepping outside the deductive system.

There is no mathematical answer to the question what mathematics is. That is a philosophical matter. You won't get anywhere by insisting on theory A versus theory B.

[Skepdick]

There is no mathematical answer to the question what mathematics is. That is a philosophical matter.

Just as well. I don't give a fuck about philosophical questions. But I do give a fuck about well-formed Mathematical questions.

You know...a subtle difference between a query encoded in a formal query language; and a query encoded in English.

You won't get anywhere by insisting on theory A versus theory B.

Ironic. because that is a well-formed question, not a philosophical one! And it is therefore decidable in principle if Platonism is true.

So weird that you keep retreating into vacuousness.

[godelian]

Ironic. because that is a well-formed question, not a philosophical one! And it is therefore decidable in principle if Platonism is true.

So weird that you keep retreating into vacuousness.

Even that is not true. The question is not decidable.

DeepSeek: Is it possible to determine for any arbitrary theory A and B, if the following is true: A ⊢ ∃0, B ⊢ ¬∃0? If a zero-like entity is definable in them?

The question asks whether it is possible to determine, for any arbitrary theories \(A\) and \(B\), whether \(A \vdash \exists 0\) and \(B \vdash \neg \exists 0\), where "\(\exists 0\)" refers to the existence of a zero-like entity definable in both theories. The answer is **no**, and here's a detailed explanation:

### 1. **Definability of "Zero-Like Entity"**
- A "zero-like entity" must be defined by a specific formula \(\phi(x)\) in the language of the theories. For example:
- In arithmetic, \(\phi(x)\) might be \(x = 0\) (if \(0\) is a constant) or a formula defining an additive identity: \(\forall y\, (y + x = y \land x + y = y)\).
- In set theory, \(\phi(x)\) might define the empty set: \(\forall y\, \neg (y \in x)\).
- The formula \(\phi(x)\) must be fixed and shared between \(A\) and \(B\) for "\(\exists 0\)" (i.e., \(\exists x \, \phi(x)\)) to be meaningful. Without a fixed \(\phi(x)\), the concept of "zero-like" is ambiguous.

### 2. **Undecidability of Provability**
- For an arbitrary theory \(A\), determining whether \(A \vdash \exists x \, \phi(x)\) is **undecidable** in general. This follows from Church's theorem: the decision problem for first-order logic (i.e., whether \(A \vdash \psi\) for an arbitrary sentence \(\psi\)) is recursively unsolvable.
- Similarly, determining whether \(B \vdash \neg \exists x \, \phi(x)\) is also undecidable.
- Thus, even with a fixed \(\phi(x)\), there is no algorithm that can decide these provability relations for arbitrary \(A\) and \(B\).

### 3. **Dependence on Specific Theories and \(\phi(x)\)**
- **Example where it holds**:
- Let \(\phi(x) := \forall y\, (x + y = y \land y + x = y)\) (additive identity).
- Let \(A\) be group theory. Then \(A \vdash \exists x \, \phi(x)\) (groups have an identity element).
- Let \(B\) be the theory of semigroups with no identity axiom. Then \(B \not\vdash \exists x \, \phi(x)\), but it may not prove \(\neg \exists x \, \phi(x)\). To force \(B \vdash \neg \exists x \, \phi(x)\), we could define \(B\) as the theory of non-unital semigroups with an added axiom \(\forall x \, \neg \phi(x)\). However, this is constructed specifically.
- **Counterexample where it fails**:
- Consider \(A\) and \(B\) both as consistent extensions of Peano Arithmetic (PA). PA proves \(\exists x\, (x = 0)\), so \(A \vdash \exists 0\). If \(B\) is consistent, it cannot prove \(\neg \exists 0\) without being inconsistent (since PA \(\vdash \exists 0\)). Thus, \(B \vdash \neg \exists 0\) is impossible unless \(B\) is inconsistent.
- If \(A\) and \(B\) are inconsistent, both prove every sentence, so \(A \vdash \exists 0\) and \(B \vdash \neg \exists 0\) trivially hold. However, inconsistency is undecidable in general.

### 4. **Role of Language and Consistency**
- **Language Mismatch**: If \(A\) and \(B\) use different languages, \(\phi(x)\) may not be expressible in both. Even if expressible, provability depends on axioms.
- **Consistency Assumption**: Without assuming consistency, \(A\) and \(B\) could prove anything. But consistency is undecidable (by Gödel's second incompleteness theorem).

### 5. **General Impossibility**
- For arbitrary \(A\), \(B\), and \(\phi(x)\):
- There is no procedure to verify \(A \vdash \exists x \, \phi(x)\) (provability is r.e. but not decidable).
- There is no procedure to verify \(B \vdash \neg \exists x \, \phi(x)\).
- Combining these, the joint condition is **algorithmically undecidable**.
- Even if "zero-like" is definable (i.e., \(\phi(x)\) exists), this does not mitigate the undecidability.

### Conclusion
It is **not possible** to determine for arbitrary theories \(A\) and \(B\) whether \(A \vdash \exists 0\) and \(B \vdash \neg \exists 0\), even if a zero-like entity is definable in both. The provability of such statements is undecidable in general due to fundamental limitations in logic (Church's theorem). Specific cases may be resolvable, but no universal method exists.

It is not possible to know if map/theory T indirectly supports addition -and multiplication-like operations for which there exists an object that is simultaneously neutral to the addition while absorbing the multiplication.

For example, a collection of NAND gates that operate on the presence or absence of electric current physically support this, but only for a finite Galois field. You would need unbounded expansion of that collection to simulate PA. There is no universal method to figure that out for any possible such construction.

Furthermore, it still won't say what such collection of logic gates fundamentally tries to represent, because that is the real question with regards to Platonism.

[Skepdick]

Ironic. because that is a well-formed question, not a philosophical one! And it is therefore decidable in principle if Platonism is true.

So weird that you keep retreating into vacuousness.

Even that is not true. The question is not decidable.

Do you want me to hand you the gold medal in irony?

The question is well-formed. It is just an instantiation of LEM. It is formulated as a trivial disjunction.

To assert that the question is NOT decidable is to provide a constructive proof that no procedure exists as to which disjunct holds.

It is to assert that NO objective fact exists on the matter.

That contradicts Platonism. Because on Platonism the question is IN PRINCIPLE decidable. Even if you don't know the answer.

[Skepdick]

Furthermore, it still won't say what such collection of logic gates fundamentally tries to represent, because that is the real question with regards to Platonism.

yeah, but I am going to tell you precisely what the symbol ∃ represents. It represents the Platonic realm.

To say ∃x is to assert access to Platonic truth. The realm is otherwise vacuous.

When you assert ∃0 you are putting it there.All you are doing is saying "the symbol 0 is now bound; and the representation evaluates to itself".
You create a token; you bind it to itself and then you pretend you've discovered some objective truth.

That's why ∃ is just another quantifier in intuitionistic logic. Axiomatically - you can assert anything exist and place it into the Platonic realm.

But lets just call it what it really is - The Construct. It's pure computational nominalism.

[godelian]

To assert that the question is NOT decidable is to provide a constructive proof that no procedure exists as to which disjunct holds.
It is to assert that NO objective fact exists on the matter.

The map won't give an answer.

That contradicts Platonism. Because on Platonism the question is IN PRINCIPLE decidable. Even if you don't know the answer.

You confuse classical logic with Platonism.

[godelian]

That's why ∃ is just another quantifier in intuitionistic logic. Axiomatically - you can assert anything exist and place it into the Platonic realm.

You can place anything you want on the map. You cannot put in on the territory.

[Skepdick]

You confuse classical logic with Platonism.

No, I don't. Classical logic is metaphysically identical with Platonism.

It entails that every well-defined disjunction (e.g ∃0 ∧ ¬∃0) is decidable. This is a re-statement of LEM.
It is NOT possible that ∃0 ∧ ¬∃0. This is a re-statement of non-contradiction

[Skepdick]

You can place anything you want on the map. You cannot put in on the territory.

Great!

So when I place 0 on the map it exists on the map.

Is the map true before; or after I placed 0 on it?

[godelian]

No, I don't. Classical logic is metaphysically identical with Platonism.

They are not identical.

ChatGPT: Is classical logic is metaphysically identical with Platonism?

No, classical logic is not metaphysically identical with Platonism, although the two are historically and philosophically related in certain ways. Classical logic does not commit to a specific metaphysical view by itself. It can be interpreted within different metaphysical or epistemological frameworks. Some philosophers (like Gödel) explicitly linked their defense of classical logic with Platonist views.

You can have:
Classical logic without Platonism: A nominalist or formalist might use classical logic purely as a practical system with no commitment to abstract entities.
Platonism without classical logic: Some Platonists might endorse intuitionistic logic if they believe truth is grounded in provability, not bivalence.

So when I place 0 on the map it exists on the map.
Is the map true before; or after I placed 0 on it?

The map would be rather "more incomplete" versus "more complete". The notion of "true but not provable" means that it is not on the map but it is in the territory. The first incompleteness theorem basically says that the map is incomplete (true but not provable) or inconsistent (false but provable).

[Skepdick]

They are not identical

Yeah.. they are.

LEM entails all propositions are decidable.

The Platonic realm is where all the answers live. The mind of the Omniscient.

The One in Neoplatonism.

[godelian]

They are not identical

Yeah.. they are.

LEM entails all propositions are decidable.

The Platonic realm is where all the answers live. The mind of the Omniscient.

The One in Neoplatonism.

We only see the map. There is an awareness of the fact that it cannot be the territory, but that's about it. You are seeing too much in it

[Skepdick]

We only see the map. There is an awareness of the fact that it cannot be the territory, but that's about it. You are seeing too much in it

Yeah, bullshit. I am not seeing anything. Is that "too much"?

All I have on my "map" is a question? Does 0 exist; or not?

That's why I am asking the Platonists. So they can have a look and let me know.