-- Lean 4 proof step from Mathlib.RingTheory.Ideal.Quotient.HasFiniteQuotients (goal)
Goal:
R : Type u_1
inst : CommRing R
inst_1 : HasFiniteQuotients R
B : ℕ
inl : Finite R → Set.Finite.{u_1} (α := Ideal R) {I : Ideal R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B}
inr inr₁ :
  Infinite R → Set.Finite.{u_1} (α := Ideal R) {I : Ideal R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B}
inr₂ :
  Infinite R →
    ∀ (s : Finset R) (hs : s.card = B + 1),
      Set.Finite.{u_1} (α := Ideal R) {I : Ideal R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B}
inr₃ :
  Infinite R →
    ∀ (s : Finset R) (hs : s.card = B + 1),
      Set.Finite.{u_1} (α := Ideal R) {I : Ideal R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B}
inr₄ :
  Infinite R →
    ∀ (s : Finset R) (hs : s.card = B + 1),
      Set.Finite (α := Ideal R) ({I : Ideal R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B} \ {⊥})
inr₅ :
  Infinite R →
    ∀ (s : Finset R) (hs : s.card = B + 1),
      let t := (s - s) \ {0};
      Subset (α := Set (Submodule R R))
        ({I : Submodule R R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B} \ {⊥})
        (⋃ x ∈ t, {I : Submodule R R | x ∈ I})
inr₆ :
  Infinite R →
    ∀ (s : Finset R) (hs : s.card = B + 1),
      let t := (s - s) \ {0};
      ∀ ⦃I : Submodule R R⦄
        (hI :
          Membership.mem (γ := Set (Submodule R R))
            ({I : Submodule R R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B} \ {⊥}) I),
        Membership.mem (γ := Set (Submodule R R)) (⋃ x ∈ t, {I : Submodule R R | x ∈ I}) I
⊢ Set.Finite.{u_1} (α := Ideal R) {I : Ideal R | Submodule.cardQuot.{u_1, u_1} (R := R) (M := R) I ≤ B}
have inr₇ : Infinite R →
  ∀ (s : Finset R) (hs : s.card = B + 1),
    let t := (s - s) \ {0};
    ∀ ⦃I : Submodule R R⦄ (hI : Nat.card.{u_1} (R ⧸ I) ≤ B ∧ I ∉ {⊥}),
      Membership.mem (γ := Set (Submodule R R)) (⋃ x ∈ t, {I : Submodule R R | x ∈ I}) I
Tactic: classical
  rcases finite_or_infinite R
  · apply Set.toFinite
  -- if `R` is infinite, then we can pick a finite set `s` of cardinality `B + 1`
  obtain ⟨s, hs⟩ := Infinite.exists_subset_card_eq R (B + 1)
  -- and consider the finite set `t` of nonzero differences
  let t := (s - s) \ {0}
  refine Set.Finite.of_diff ?_ (Set.finite_singleton ⊥)
  -- in a ring with finite quotients, each nonzero element is contained in only finitely many ideals
  -- so it is enough to show that each ideal `I` of norm at most `B` contains some element of `t`
  suffices {I | Submodule.cardQuot I ≤ B} \ {⊥} ⊆ ⋃ x ∈ t, {I | x ∈ I} from
    (t.finite_toSet.biUnion fun x hx ↦ finite_setOf_mem x (by grind)).subset this
  intro I hI
  rw [Set.mem_diff, Set.mem_setOf, Submodule.cardQuot_apply] at hI
  simp_rw [Set.mem_iUnion, exists_prop, Set.mem_setOf_eq]
  -- `s` has cardinality `B + 1`, but the quotient `R ⧸ I` has cardinality at most `B`
  replace hs : (s.image (Ideal.Quotient.mk I)).card < s.card := by
    have := finiteQuotient hI.2
    have := Fintype.ofFinite (R ⧸ I)
    grw [Finset.card_le_univ, Fintype.card_eq_nat_card, hI.1, hs, Nat.lt_add_one_iff]
  -- so we can find distinct `x, y ∈ s` with the desired collision `x - y ∈ I`
  obtain ⟨x, hx, y, hy, hxy, h⟩ := Finset.exists_ne_map_eq_of_card_image_lt hs
  refine ⟨x - y, ?_, (Submodule.Quotient.eq I).mp h⟩
  refine Finset.mem_sdiff.mpr ⟨Finset.mem_sub.mpr ⟨x, hx, y, hy, rfl⟩, ?_⟩
  rwa [Finset.notMem_singleton, sub_ne_zero]