Aristotle SDK bug reports

Stream: mathlib4

[Nick_adfor]

solution_path = await project.prove_from_file"" show me Too many positional arguments for classmethod call

[Nick_adfor]

How can this work before and fail now?

[Yury G. Kudryashov]

Nick_adfor said:

solution_path = await project.prove_from_file"" show me Too many positional arguments for classmethod call

Is it from your own file, or from aristotlelib? If it's from your file, then please post a Python #mwe (with imports).

[Nick_adfor]

Yury G. Kudryashov said:

Nick_adfor said:

solution_path = await project.prove_from_file"" show me Too many positional arguments for classmethod call

Is it from your own file, or from aristotlelib? If it's from your file, then please post a Python #mwe (with imports).

See in dm

[Nick_adfor]

import os
import warnings

import asyncio
import aristotlelib

os.environ['HTTP_PROXY'] = ''
os.environ['HTTPS_PROXY'] = ''
os.environ['ALL_PROXY'] = ''
os.environ['NO_PROXY'] = '*'

for key in list(os.environ.keys()):
    if 'proxy' in key.lower() or 'PROXY' in key:
        os.environ[key] = ''

warnings.filterwarnings("ignore", category=UserWarning)

aristotlelib.set_api_key("")

async def main():
    # Prove a theorem from a Lean file
    solution_path = await aristotlelib.Project.prove_from_file("")
    print(f"Solution saved to: {solution_path}")

asyncio.run(main())

[Nick_adfor]

TypeError: Project.prove_from_file() takes 1 positional argument but 2 were given (base) where can I see the 2 were given??????

[Nick_adfor]

There's only one in solution_path = await aristotlelib.Project.prove_from_file("")

[Nick_adfor]

This problem arises after the upgrade of Aristotle. I mean that now I have no way to get back to the old version and also have no way to check what happened to the code.

[Nick_adfor]

Why does my entire career as a programmer have to be spent solving compatibility issues caused by version updates???

[Nick_adfor]

async def main():
    # Prove a theorem from a Lean file
    solution_path = await aristotlelib.Project.prove_from_file("path/to/your/file.lean")
    print(f"Solution saved to: {solution_path}")

asyncio.run(main())

I've changed the former one into the following one:

async def main():
    solution_path = await aristotlelib.Project.prove_from_file(input_file_path="path/to/your/file.lean")
    print(f"Solution saved to: {solution_path}")

asyncio.run(main())

But I should say that the former one in in https://pypi.org/project/aristotlelib/. Maybe the website should change.

[Kevin Buzzard]

Nick_adfor said:

Why does my entire career as a programmer have to be spent solving compatibility issues caused by version updates???

Because Lean/mathlib is a very fast-moving project with no backward compatibility guarantees at this point, and language models are unable to keep up right now.

[Nick_adfor]

Kevin Buzzard said:

Nick_adfor said:

Why does my entire career as a programmer have to be spent solving compatibility issues caused by version updates???

Because Lean/mathlib is a very fast-moving project with no backward compatibility guarantees at this point, and language models are unable to keep up right now.

Today's problem is because of Aristotle or python update though

[Notification Bot]

3 messages were moved here from #mathlib4 > Discussion thread for Aristotle API by Yury G. Kudryashov.

[Nick_adfor]

/-
This file was edited by Aristotle.

Lean version: leanprover/lean4:v4.24.0
Mathlib version: f897ebcf72cd16f89ab4577d0c826cd14afaafc7

Aristotle failed to load this code into its environment. Double check that the syntax is correct.
You are importing a file that is unknown to Aristotle, Aristotle supports importing user projects, but files must be uploaded as project context, please see the aristotlelib docs or help menu.
Details:
Lean error:
Errors during import; aborting. Details:
input.lean:1:0: error: unknown module prefix 'Init'

No directory 'Init' or file 'Init.olean' in the search path entries:
/code/harmonic-lean/.lake/packages/batteries/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/Qq/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/aesop/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/proofwidgets/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/importGraph/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/LeanSearchClient/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/plausible/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/MD4Lean/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/BibtexQuery/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/UnicodeBasic/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/Cli/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/mathlib/.lake/build/lib/lean
/code/harmonic-lean/.lake/packages/doc-gen4/.lake/build/lib/lean
/code/harmonic-lean/.lake/build/lib/lean
/root/.elan/toolchains/leanprover--lean4---v4.24.0/lib/lean
/root/.elan/toolchains/leanprover--lean4---v4.24.0/lib/lean


-/

import Mathlib

variable {p : ℕ} [Fact (Nat.Prime p)] {k : ℕ}

open MvPolynomial

/--
\textbf{Theorem 2.1.} Let \( p \) be a prime and let \( h = h(x_0, \ldots, x_k) \) be a polynomial over \( \mathbb{Z}_p \). Let \( A_0, A_1, \ldots, A_k \) be nonempty subsets of \( \mathbb{Z}_p \), where \( |A_i| = c_i + 1 \) and define \( m = \sum_{i=0}^k c_i - \deg(h) \). If the coefficient of \( \prod_{i=0}^k x_i^{c_i} \) in
\[
(x_0 + x_1 + \cdots + x_k)^m h(x_0, x_1, \ldots, x_k)
\]
is nonzero (in \( \mathbb{Z}_p \)) then
\[
\left| \oplus_h \sum_{i=0}^k A_i \right| \geq m + 1
\]
(and hence \( m < p \)).

To prove this theorem we need the following simple and well known result. Since the argument is very short we reproduce it here.

\textbf{Lemma.} Let \( P = P(x_0, x_1, \ldots, x_k) \) be a polynomial in \( k+1 \) variables over an arbitrary field \( F \). Suppose that the degree of \( P \) as a polynomial in \( x_i \) is at most \( c_i \) for \( 0 \leq i \leq k \), and let \( A_i \subset F \) be a set of cardinality \( c_i + 1 \). If \( P(x_0, x_1, \ldots, x_k) = 0 \) for all \( (k+1) \)-tuples \( (x_0, \ldots, x_k) \in A_0 \times A_1 \times \cdots \times A_k \), then \( P \equiv 0 \), that is: all the coefficients in \( P \) are zeros.

\textit{Proof.} We apply induction on \( k \). For \( k = 0 \), the lemma is simply the assertion that a non-zero polynomial of degree \( c_0 \) in one variable can have at most \( c_0 \) distinct zeros. Assuming that the lemma holds for \( k-1 \), we prove it for \( k \) (\( k \geq 1 \)). Given a polynomial \( P = P(x_0, \ldots, x_k) \) and sets \( A_i \) satisfying the hypotheses of the lemma, let us write \( P \) as a polynomial in \( x_k \), that is,
\[
P = \sum_{i=0}^{c_k} P_i(x_0, \ldots, x_{k-1}) x_k^i,
\]
where each \( P_i \) is a polynomial with \( x_j \)-degree bounded by \( c_j \). For each fixed \( k \)-tuple \( (x_0, \ldots, x_{k-1}) \in A_0 \times A_1 \times \cdots \times A_{k-1} \), the polynomial in \( x_k \) obtained from \( P \) by substituting the values of \( x_0, \ldots, x_{k-1} \) vanishes for all \( x_k \in A_k \), and is thus identically \( 0 \). Thus \( P_i(x_0, \ldots, x_{k-1}) = 0 \) for all \( (x_0, \ldots, x_{k-1}) \in A_0 \times \cdots \times A_{k-1} \). Hence, by the induction hypothesis, \( P_i \equiv 0 \) for all \( i \), implying that \( P \equiv 0 \). This completes the induction and the proof of the lemma. \( \square \)

\textbf{Proof of Theorem 2.1.} Suppose the assertion is false, and let \( E \) be a (multi-) set of \( m \) (not necessarily distinct) elements of \( \mathbb{Z}_p \) that contains the set \( \oplus_h \sum_{i=0}^k A_i \). Let \( Q = Q(x_0, \ldots, x_k) \) be the polynomial defined as follows:
\[
Q(x_0, \ldots, x_k) = h(x_0, x_1, \ldots, x_k) \cdot \prod_{e \in E} (x_0 + \ldots + x_k - e).
\]
Note that
\[
Q(x_0, \ldots, x_k) = 0 \quad \text{for all } (x_0, \ldots, x_k) \in (A_0, \ldots, A_k). \tag{1}
\]
This is because for each such \( (x_0, \ldots, x_k) \) either \( h(x_0, \ldots, x_k) = 0 \) or \( x_0 + \ldots + x_k \in \oplus_h \sum_{i=0}^k A_i \subset E \). Note also that \( \deg(Q) = m + \deg(h) = \sum_{i=0}^k c_i \) and hence the coefficient of the monomial \( x_0^{c_0} \cdots x_k^{c_k} \) in \( Q \) is the same as that of this monomial in the polynomial \( (x_0 + \ldots + x_k)^m h(x_0, \ldots, x_k) \), which is nonzero, by assumption.

For each \( i \), \( 0 \leq i \leq k \), define
\[
g_i(x_i) = \prod_{a \in A_i} (x_i - a) = x_i^{c_i + 1} - \sum_{j=0}^{c_i} b_{ij} x_i^j.
\]
Let \( \overline{Q} = \overline{Q}(x_0, \ldots, x_k) \) be the polynomial obtained from the standard representation of \( Q \) as a linear combination of monomials by replacing, repeatedly, each occurrence of \( x_i^{c_i + 1} \) by \( \sum_{j=0}^{c_i} b_{ij} x_i^j \). Note that since for every \( x_i \in A_i \), \( x_i^{c_i + 1} \) is equal to this sum, equation (1) holds for \( \overline{Q} \) as well. However, the \( x_i \)-degree of \( \overline{Q} \) is at most \( c_i \) and hence, by the lemma it is identically zero. To obtain a contradiction, we claim that the coefficient of the monomial \( \prod_{i=0}^k x_i^{c_i} \) in \( \overline{Q} \) is not \( 0 \) (in \( \mathbb{Z}_p \)). To see this note that the coefficient of this monomial in \( Q \) is nonzero modulo \( p \) by assumption. The crucial observation is that the coefficient of this monomial in \( \overline{Q} \) is equal to its coefficient in \( Q \). This is because the process of replacing each of the expressions \( x_i^{c_i + 1} \) by \( \sum_{j=0}^{c_i} b_{ij} x_i^j \) does not affect the above monomial itself. Moreover, since the total degree of \( Q \) is \( \sum_{i=0}^k c_i \) and the process of replacing the expressions as above strictly reduces degrees, this process cannot create any additional scalar multiples of this monomial, proving the claim.

It thus follows that \( \overline{Q} \) is not identically zero, supplying the desired contradiction and completing the proof. \( \square \)
-/
theorem ANR_polynomial_method (h : MvPolynomial (Fin (k + 1)) (ZMod p))
    (A : Fin (k + 1) → Finset (ZMod p))
    (c : Fin (k + 1) → ℕ)
    (hA : ∀ i, (A i).card = c i + 1)
    (m : ℕ) (hm : m = (∑ i, c i) - h.totalDegree)
    (h_coeff : MvPolynomial.coeff (Finsupp.equivFunOnFinite.symm c)
    ((∑ i : Fin (k + 1), MvPolynomial.X i) ^ m * h) ≠ 0) :
    let S : Finset (ZMod p) :=
      (Fintype.piFinset A).filter (fun f => h.eval f ≠ 0) |>.image (fun f => ∑ i, f i)
    S.card ≥ m + 1 := by sorry

[Nick_adfor]

Aristotle told me that there's an unknown Init, but I do not think that my code need Init

[Aaron Liu]

I am 100% sure your code needs Init

[Aaron Liu]

It does important things like define what equality is and sets up tactics like sorry

[Alex J. Best]

Apologies this was a temporary issue on our end, retrying the same request should succeed if you didn't already!

[Nick_adfor]

Aaron Liu said:

I am 100% sure your code needs Init

Okay. My habit is only import Mathlib

[Nick_adfor]

After I import Init, it is

Cannot find import 'Init' from ... Have you tried running `lake update` ?
Could not resolve import 'Init' from ...

[Nick_adfor]

I may think that if a lean statement can work when import Mathlib only then the solution of the lean statement can also just import Mathlib and do not need others. But this time my thought seems wrong

[Julian Berman]

My interpretation of the last few messages is -- you said Aristotle complained Init was missing, you said you didn't think you needed it, Aaron assured you you did need it (but didn't mean for you to add import Init, it's already implicitly imported), and then Alex was telling you you should try your same request to Aristotle again and you won't get the error about Init this time.

[Nick_adfor]

My last submit includes import Mathlib and import Init both. Though Aristotle gives me "Could not resolve...", it is still working on it.

[Aaron Liu]

So the problem is that whatever Aristotle is doing can't find the imports

[Vikram Shanker]

Hi Aaron, Nick, and others,

Thanks for the bug report here. I pushed a fix last night (in v0.5.1). Please pip install --upgrade aristotlelib to pick it up, and let me know if the issue persists.

Vikram

[Nick_adfor]

It is really effective that Aristotle can even upgrade in half a day :)

[Nick_adfor]

I do something silly: I submit three times into Aristotle :(

[Nick_adfor]

But there's no way to delete some. Can my API check my three files in the same time?

[Yury G. Kudryashov]

If you just start aristotle, you'll find a way to view all your requests.

[Yury G. Kudryashov]

You can also write a Python script that does it. I suggest that you either read the sources of aristotlelib (docs can be outdated), or ask some AI to do it for you.

[Nick_adfor]

raise api_request.AristotleAPIError(
aristotlelib.api_request.AristotleAPIError: Project failed due to an internal error. The team at Harmonic has been notified; please try again.
(base)

[Yury G. Kudryashov]

Exactly as it says, we've been notified about the error, and we'll work on fixing it when we can, but we can't share the details without disclosing secret information about our system.

[Boris Alexeev]

Elsewhere, I see "We're just rolling out an upgrade". Isn't that likely to be the underlying reason?

[Yury G. Kudryashov]

That's possible. @Nick_adfor , you can retry your request now and see if it works.

[Vikram Shanker]

Sorry for the interruption, we just upgraded Aristotle to lean v4.24!

[Nick_adfor]

All right... Aristotle always upgrades in half a day.