/-- Given a partition `P` of `s`, as well as a proof that `a * m + b * (m + 1) = s.card`, we can
find a new partition `Q` of `s` where each part has size `m` or `m + 1`, every part of `P` is the
union of parts of `Q` plus at most `m` extra elements, there are `b` parts of size `m + 1` and
(provided `m > 0`, because a partition does not have parts of size `0`) there are `a` parts of size
`m` and hence `a + b` parts in total. -/
theorem equitabilise_aux (hs : a * m + b * (m + 1) = s.card) :
    ∃ Q : Finpartition s,
      (∀ x : Finset α, x ∈ Q.parts → x.card = m ∨ x.card = m + 1) ∧
        (∀ x, x ∈ P.parts → (x \ (Q.parts.filter fun y => y ⊆ x).biUnion id).card ≤ m) ∧
          (Q.parts.filter fun i => card i = m + 1).card = b := by
  -- Get rid of the easy case `m = 0`
  obtain rfl | m_pos := m.eq_zero_or_pos
  · refine' ⟨⊥, by simp, _, by simpa using hs.symm⟩
    simp only [le_zero_iff, card_eq_zero, mem_biUnion, exists_prop, mem_filter, id.def, and_assoc,
      sdiff_eq_empty_iff_subset, subset_iff]
    exact fun x hx a ha =>
      ⟨{a}, mem_map_of_mem _ (P.le hx ha), singleton_subset_iff.2 ha, mem_singleton_self _⟩
  -- Prove the case `m > 0` by strong induction on `s`
  induction' s using Finset.strongInduction with s ih generalizing a b
  -- If `a = b = 0`, then `s = ∅` and we can partition into zero parts
  by_cases hab : a = 0 ∧ b = 0
  · simp only [hab.1, hab.2, add_zero, zero_mul, eq_comm, card_eq_zero, Finset.bot_eq_empty] at hs
    subst hs
    -- Porting note: to synthesize `Finpartition ∅`, `have` is required
    have : P = Finpartition.empty _ := Unique.eq_default (α := Finpartition ⊥) P
    exact ⟨Finpartition.empty _, by simp, by simp [this], by simp [hab.2]⟩
  simp_rw [not_and_or, ← Ne.def, ← pos_iff_ne_zero] at hab
  -- `n` will be the size of the smallest part
  set n := if 0 < a then m else m + 1 with hn
  -- Some easy facts about it
  obtain ⟨hn₀, hn₁, hn₂, hn₃⟩ : 0 < n ∧ n ≤ m + 1 ∧ n ≤ a * m + b * (m + 1) ∧
      ite (0 < a) (a - 1) a * m + ite (0 < a) b (b - 1) * (m + 1) = s.card - n := by
    rw [hn, ← hs]
    split_ifs with h <;> rw [tsub_mul, one_mul]
    · refine' ⟨m_pos, le_succ _, le_add_right (le_mul_of_pos_left ‹0 < a›), _⟩
      rw [tsub_add_eq_add_tsub (le_mul_of_pos_left h)]
    · refine' ⟨succ_pos', le_rfl, le_add_left (le_mul_of_pos_left <| hab.resolve_left ‹¬0 < a›), _⟩
      rw [← add_tsub_assoc_of_le (le_mul_of_pos_left <| hab.resolve_left ‹¬0 < a›)]
  /- We will call the inductive hypothesis on a partition of `s \ t` for a carefully chosen `t ⊆ s`.
    To decide which, however, we must distinguish the case where all parts of `P` have size `m` (in
    which case we take `t` to be an arbitrary subset of `s` of size `n`) from the case where at
    least one part `u` of `P` has size `m + 1` (in which case we take `t` to be an arbitrary subset
    of `u` of size `n`). The rest of each branch is just tedious calculations to satisfy the
    induction hypothesis. -/
  by_cases h : ∀ u ∈ P.parts, card u < m + 1
  · obtain ⟨t, hts, htn⟩ := exists_smaller_set s n (hn₂.trans_eq hs)
    have ht : t.Nonempty := by rwa [← card_pos, htn]
    have hcard : ite (0 < a) (a - 1) a * m + ite (0 < a) b (b - 1) * (m + 1) = (s \ t).card := by
      

Proof state:
α : Type u_1
inst✝ : DecidableEq α
t✝ : Finset α
m n✝ : ℕ
m_pos : m > 0
s : Finset α
ih :
  ∀ t ⊂ s,
    ∀ {a b : ℕ} {P : Finpartition t},
      a * m + b * (m + 1) = card t →
        ∃ Q,
          (∀ x ∈ Q.parts, card x = m ∨ card x = m + 1) ∧
            (∀ x ∈ P.parts, card (x \ Finset.biUnion (filter (fun y => y ⊆ x) Q.parts) id) ≤ m) ∧
              card (filter (fun i => card i = m + 1) Q.parts) = b
a b : ℕ
P : Finpartition s
hs : a * m + b * (m + 1) = card s
hab : 0 < a ∨ 0 < b
n : ℕ := if 0 < a then m else m + 1
hn : n = if 0 < a then m else m + 1
hn₀ : 0 < n
hn₁ : n ≤ m + 1
hn₂ : n ≤ a * m + b * (m + 1)
hn₃ : (if 0 < a then a - 1 else a) * m + (if 0 < a then b else b - 1) * (m + 1) = card s - n
h : ∀ u ∈ P.parts, card u < m + 1
t : Finset α
hts : t ⊆ s
htn : card t = n
ht : Finset.Nonempty t
⊢ (if 0 < a then a - 1 else a) * m + (if 0 < a then b else b - 1) * (m + 1) = card (s \ t)

Next tactic:
rw [card_sdiff ‹t ⊆ s›, htn, hn₃]