ChatGPT just spit out nonsense in the first example. Look at the sum in step 6, over τ ≤ τ. Is τ the bound variable of the summation or a free variable? Put another way, how many terms are in this summation over "τ ≤ τ"? And how does this relate to establishing either the LHS or RHS of part 3, to then conclude the other side? Nothing actually coheres.
What happened here is that ChatGPT remembered the Möbius function for the partition lattice, regurgitated it* without providing proof, and then spit out some nonsense afterwards that looks superficially reasonable.
But establishing that Möbius function is essentially the whole ballgame! The question being asked is very nearly the same as just being asked to prove that the Möbius function has that form.
[*: The regurgitation also has a slight error, as μ(τ, σ) with τ a refinement of σ is actually (-1)^(|τ| - |σ|) * the product of (|τ restricted to c| - 1)! over each equivalence class c in σ. The formula given by ChatGPT matches this one when |σ| = 1, which is the important case for the proof at hand, but is incorrect more generally.]
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For what it's worth, the desired fact can be shown just by elementary induction, without explicitly invoking all the Möbius function machinery anyway:
Let N(t, e) be the number of partitions of {1, ..., e} into t many equivalence classes [we will always take e to be a non-negative integer, but allow for t to be an arbitrary integer, with N(t, e) = 0 when t is negative]. The sum in question is of (-1)^t * t! * N(t, e) over all t. Refer to this sum as Sum(e). We wish to show that Sum(e) comes out to (-1)^e. Since Sum(0) = 1 trivially, what we need to show is that Sum(e + 1) = -Sum(e).
There are two kinds of partitions on the set {1, ..., e + 1}: Those which put e + 1 in an equivalence class all on its own and those which put it in the same equivalence class as some value or values in {1, ..., e}. From this, we obtain the recurrence N(t, e + 1) = N(t - 1, e) + t * N(t, e).
Accordingly, Sum(e + 1) = Sum of (-1)^t * t! * N(t, e + 1) over all t = Sum of (-1)^t * t! * N(t - 1, e) over all t, plus sum of (-1)^t * t! * t * N(t, e) over all t. The first of these two sub-sums can be reparametrized as the sum of (-1)^(t + 1) * (t + 1)! * N(t, e) over all t. Now recombining the two sub-sums term-wise, and keeping in mind (-1)^(t + 1) * (t + 1)! = (-t - 1) * (-1)^t * t!, we get Sum(e + 1) = sum of (-1)^t * t! * (-t - 1 + t) * N(t, e) over all t. As (-t - 1 + t) = -1, this comes to -Sum(e) as desired, completing the proof.
The second ChatGPT example proof is mostly fine, except it failed to see that the claimed theorem isn't actually true at d = 2. (It writes "d ≥ 2" for a step in its reasoning where d ≥ 3 is actually required.)
But establishing that Möbius function is essentially the whole ballgame! The question being asked is very nearly the same as just being asked to prove that the Möbius function has that form.
[*: The regurgitation also has a slight error, as μ(τ, σ) with τ a refinement of σ is actually (-1)^(|τ| - |σ|) * the product of (|τ restricted to c| - 1)! over each equivalence class c in σ. The formula given by ChatGPT matches this one when |σ| = 1, which is the important case for the proof at hand, but is incorrect more generally.]
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For what it's worth, the desired fact can be shown just by elementary induction, without explicitly invoking all the Möbius function machinery anyway:
Let N(t, e) be the number of partitions of {1, ..., e} into t many equivalence classes [we will always take e to be a non-negative integer, but allow for t to be an arbitrary integer, with N(t, e) = 0 when t is negative]. The sum in question is of (-1)^t * t! * N(t, e) over all t. Refer to this sum as Sum(e). We wish to show that Sum(e) comes out to (-1)^e. Since Sum(0) = 1 trivially, what we need to show is that Sum(e + 1) = -Sum(e).
There are two kinds of partitions on the set {1, ..., e + 1}: Those which put e + 1 in an equivalence class all on its own and those which put it in the same equivalence class as some value or values in {1, ..., e}. From this, we obtain the recurrence N(t, e + 1) = N(t - 1, e) + t * N(t, e).
Accordingly, Sum(e + 1) = Sum of (-1)^t * t! * N(t, e + 1) over all t = Sum of (-1)^t * t! * N(t - 1, e) over all t, plus sum of (-1)^t * t! * t * N(t, e) over all t. The first of these two sub-sums can be reparametrized as the sum of (-1)^(t + 1) * (t + 1)! * N(t, e) over all t. Now recombining the two sub-sums term-wise, and keeping in mind (-1)^(t + 1) * (t + 1)! = (-t - 1) * (-1)^t * t!, we get Sum(e + 1) = sum of (-1)^t * t! * (-t - 1 + t) * N(t, e) over all t. As (-t - 1 + t) = -1, this comes to -Sum(e) as desired, completing the proof.