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You ellipsed ([...]) the "can assume" part of my statement. Every mathematician knows the different ways of constructing numbers. It really is a simple YES.

Someone 2y ago
No, it isn’t. Whether it is depends on what level the mathematician operates, and will differ on what they’re working on. That’s precisely the subject of this article: there’s an intuitive idea that, upon close scrutiny, turns out to before complex than it appears to be. So, if you want to start from the basis in a proof assistant, you’ll have to discriminate between them and either give them different names or introduce context from which the computer can infer what you mean. If, for example, you want to prove that two definitions of integers are equivalent, you need two different names.
- auggierose 2y ago
  It helps to state carefully what you are trying to achieve. I want a theorem prover where constructing numbers such that ℕ ⊂ ℤ is easy. Doing this in a theorem prover is not trivial, for example none of the type based theorem provers allow that (although, Isabelle/ZF might qualify; it is based on set theory on top of a weak intuitionistic type theory; but here ℕ and ℤ are not types, but sets of the same type). It is trivial though if you are a mathematician, no matter what you are working on. You know that these constructions can be done, and that as a result you can assume ℕ ⊂ ℤ (if you want to). So as long as your theorem prover doesn't support something as trivial as that, it is lacking. And yes, that is exactly what Buzzard is talking about. For the case I am talking about here though, that is just table stakes: make sure your prover can do ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ. If you can do that, you can start worrying about more advanced issues of equality, and you might find that you have already solved a significant amount of them.
  - robinzfc 2y ago
    I can confirm that in Isabelle/ZF one can set up a context (locale) with the meaning of the ℕ ℤ ℝ ℂ symbols defined so that ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ. However ℕ for example will not be equal in such case to the canonical set of ZF natural numbers where 1 = {0}, 2 = {0,1} etc.

Someone 2y ago

No, it isn’t. Whether it is depends on what level the mathematician operates, and will differ on what they’re working on. That’s precisely the subject of this article: there’s an intuitive idea that, upon close scrutiny, turns out to before complex than it appears to be. So, if you want to start from the basis in a proof assistant, you’ll have to discriminate between them and either give them different names or introduce context from which the computer can infer what you mean. If, for example, you want to prove that two definitions of integers are equivalent, you need two different names.

auggierose 2y ago
It helps to state carefully what you are trying to achieve. I want a theorem prover where constructing numbers such that ℕ ⊂ ℤ is easy. Doing this in a theorem prover is not trivial, for example none of the type based theorem provers allow that (although, Isabelle/ZF might qualify; it is based on set theory on top of a weak intuitionistic type theory; but here ℕ and ℤ are not types, but sets of the same type). It is trivial though if you are a mathematician, no matter what you are working on. You know that these constructions can be done, and that as a result you can assume ℕ ⊂ ℤ (if you want to). So as long as your theorem prover doesn't support something as trivial as that, it is lacking. And yes, that is exactly what Buzzard is talking about. For the case I am talking about here though, that is just table stakes: make sure your prover can do ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ. If you can do that, you can start worrying about more advanced issues of equality, and you might find that you have already solved a significant amount of them.
- robinzfc 2y ago
  I can confirm that in Isabelle/ZF one can set up a context (locale) with the meaning of the ℕ ℤ ℝ ℂ symbols defined so that ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ. However ℕ for example will not be equal in such case to the canonical set of ZF natural numbers where 1 = {0}, 2 = {0,1} etc.