[tptacek]

That's not what happened here. This isn't a proof; it's a counterexample. The model was perfectly capable of verifying its correctness. You could have verified it by hand if you wanted; the verification is trivial. Finding it was the hard part.

[YeGoblynQueenne]

>> The model was perfectly capable of verifying its correctness.

It's an LLM. It can't do that.

[tptacek]

Are you doing the "LLMs don't know how many R's are in 'Raspberry'" thing here?

A bunch of people on the original thread about the Conjecture were like "we need to wait and see if real mathematicians verify this proof it's probably just LLM psychosis", because they don't understand that the counterexample is a trivial calculation. Checking it isn't hard; any AP calc student can do it quickly. It's finding the counterexample that's the challenge.

It's as if someone presented a SHA-2 collision, which anyone could just feed to `openssl sha256` to see, and then naysayers were like "we need to wait for independent verification because the LLM can't know if that collision was real".

[YeGoblynQueenne]

>> Are you doing the "LLMs don't know how many R's are in 'Raspberry'" thing here?

I don't know. Are you doing the thing were you're arguing against something you don't like that the other person never said?

[tptacek]

The model is very capable of checking this counterexample, was my point.

[YeGoblynQueenne]

Checking, it how? I mean can you describe the mechanism? How does an LLM "check" that something is (or isn't) a counterexample? Does that involve pattern matching, formal verification or model checking, some other approach?

[akalin]

In this case, the same way I would -- calling out to SymPy.

Once you verify (with SymPy or another CAS, like Mathematica) that the given function has the claimed Jacobian determinant (which involves only taking partial derivatives and taking a determinant) and the given inputs map to the same output (which involves only evaluating polynomials with some inputs), you're done.