-- Lean 4 proof step from Mathlib.Algebra.Lie.CartanCriterion (goal)
Goal:
L : Type u_2
M : Type u_3
inst : LieRing L
inst_1 : AddCommGroup M
inst_2 : LieRingModule L M
K : Type u_4
inst_3 : Field K
inst_4 : CharZero K
inst_5 : IsAlgClosed K
inst_6 : LieAlgebra K L
inst_7 : Module K M
inst_8 : LieModule K L M
inst_9 : FiniteDimensional K M
h : Eq (α := LinearMap.BilinForm K L) (traceForm K L M) 0
φ : LieHom.{u_4, u_2, u_3} K L (End K M) := toEnd K L M
x : L
hx : Membership.mem (γ := LieIdeal K L) (derivedSeries K L 1) x
n : End K M
hn_adj : Membership.mem (γ := Subalgebra K (End K M)) K[((φ : (a : L) → End K M) x : End K M)] n
s : End K M
hn_nil : _root_.IsNilpotent n
hs_ss : s.IsSemisimple
hX_ns : (φ : (a : L) → End K M) x = n + s
hns : Commute n s
Decompose Decompose₁ Decompose₂ Decompose₃ Decompose₄ : s = 0
Decompose₅ Decompose₆ :
  let I := (ν : K) × Fin (finrank.{u_4, u_3} K ↥(s.eigenspace ν));
  ∀ (this : Fintype I), s = 0
Decompose₇ Decompose₈ :
  let eigenDecomp : DirectSum.IsInternal s.eigenspace := ⋯;
  let I := (ν : K) × Fin (finrank.{u_4, u_3} K ↥(s.eigenspace ν));
  let v : Basis I K M :=
    eigenDecomp.collectedBasis (α := fun (ν : K) ↦ Fin (finrank.{u_4, u_3} K ↥(s.eigenspace ν))) fun (μ : K) ↦
      finBasis K ↥(s.eigenspace μ);
  ∀ (this : Fintype I),
    let μ : I → K := Sigma.fst (β := fun (ν : K) ↦ Fin (finrank.{u_4, u_3} K ↥(s.eigenspace ν)));
    ∀
      (hsv :
        ∀ (i : I), (s : (a : M) → M) ((v : (a : I) → M) i : M) = HSMul.hSMul (β := M) (μ i) ((v : (a : I) → M) i : M)),
      s = 0
⊢ s = 0
have Decompose₉ : let eigenDecomp : DirectSum.IsInternal s.eigenspace := sorry;
let I := (ν : K) × Fin (finrank.{u_4, u_3} K ↥(s.eigenspace ν));
let v : Basis I K M :=
  eigenDecomp.collectedBasis (α := fun (ν : K) ↦ Fin (finrank.{u_4, u_3} K ↥(s.eigenspace ν))) fun (μ : K) ↦
    finBasis K ↥(s.eigenspace μ);
∀ (this : Fintype I),
  let μ : I → K := Sigma.fst (β := fun (ν : K) ↦ Fin (finrank.{u_4, u_3} K ↥(s.eigenspace ν)));
  ∀
    (hsv :
      ∀ (i : I), (s : (a : M) → M) ((v : (a : I) → M) i : M) = HSMul.hSMul (β := M) (μ i) ((v : (a : I) → M) i : M)),
    let E : Submodule ℚ K := Submodule.span ℚ (Set.range μ);
    ∀ (hμ : ∀ (i : I), μ i ∈ E), s = 0
Tactic: classical
  /- Decompose `M` as a direct sum of eigenspaces of `s`. -/
  let eigenDecomp := DirectSum.isInternal_submodule_of_iSupIndep_of_iSup_eq_top
    s.eigenspaces_iSupIndep hs_ss.iSup_eigenspace_eq_top
  let I := (ν : K) × Fin (finrank K (s.eigenspace ν))
  let v : Basis I K M := eigenDecomp.collectedBasis fun μ ↦ finBasis K (s.eigenspace μ)
  have : Fintype I := FiniteDimensional.fintypeBasisIndex v
  let μ : I → K := Sigma.fst
  have hsv (i : I) : s (v i) = μ i • v i :=
    mem_eigenspace_iff.mp (eigenDecomp.collectedBasis_mem _ i)
  /- Let `E ⊆ K` be the `ℚ`-submodule of scalars spanned by the eigenvalues of `s`. -/
  let E : Submodule ℚ K := Submodule.span ℚ (Set.range μ)
  have hμ (i : I) : μ i ∈ E := Submodule.subset_span (Set.mem_range_self i)
  /- It suffices to prove that the `ℚ`-dual of `E` is trivial.  This can be regarded as a trick to
     handle the fact that our scalars `K` are not ordered. -/
  suffices ∀ f : Dual ℚ E, f = 0 by
    suffices ∀ ν, s.HasEigenvalue ν → ν = 0 from hs_ss.eq_zero_iff_forall_eigenvalue.mpr this
    intro ν hν
    have : Nontrivial (s.eigenspace ν) :=
      Submodule.nontrivial_iff_ne_bot.mpr (hasEigenvalue_iff.mp hν)
    replace hν : ν ∈ E := Submodule.subset_span ⟨⟨ν, ⟨0, finrank_pos⟩⟩, rfl⟩
    have : Subsingleton E := (subsingleton_dual_iff ℚ).mp ⟨by aesop⟩
    simpa using Subsingleton.elim (⟨ν, hν⟩ : E) 0
  intro f
  /- It suffices to show that any `f : Dual ℚ E` vanishes on all the eigenvalues of `s`. -/
  suffices ∀ i, f ⟨μ i, hμ i⟩ = 0 by
    rw [Submodule.linearMap_eq_zero_iff_of_eq_span f rfl]
    rintro ⟨-, ⟨i, rfl⟩⟩
    exact this i
  /- We will deduce this by proving that the sum of the squares of all such values vanishes. -/
  suffices ∑ i, f ⟨μ i, hμ i⟩ ^ 2 = 0 from fun i ↦ eq_zero_of_pow_eq_zero <|
    (Finset.sum_eq_zero_iff_of_nonneg (fun _ _ ↦ sq_nonneg _)).mp this i (Finset.mem_univ _)
  /- Which will follow from the fact that the following `f`-linear expression vanishes. -/
  suffices ∑ i, (f ⟨μ i, hμ i⟩) • (⟨μ i, hμ i⟩ : E) = 0 by
    simpa only [map_sum, map_zero, map_smul, sq] using f.congr_arg this
  let fμ (i : I) : K := f ⟨μ i, hμ i⟩
  /- Defining `fμ i = f ⟨μ i, hμ i⟩`, we can restate our goal as `∑ i, fμ i * μ i = 0`. -/
  suffices ∑ i, fμ i * μ i = 0 by simp [Subtype.ext_iff, fμ, ← this, smul_def]
  /- We will do this by constructing endomorphism `y` such that `trace K M (φ x * y) = 0` and also
     `trace K M (φ x * y) = ∑ i, fμ i * μ i`. -/
  suffices ∃ y : End K M, trace K M (φ x * y) = 0 ∧ trace K M (φ x * y) = ∑ i, fμ i * μ i by grind
  /- We define `y` diagonal wrt our basis `v` and takes the values `fμ` on the diagonal. -/
  let y : End K M := (Matrix.diagonal fμ).toLin v v
  have hyv (i : I) : y (v i) = fμ i • v i :=
    mem_eigenspace_iff.mp (hasEigenvector_toLin_diagonal fμ i v).1
  /- Using Lagrange interpolation, we can show that the representation is stable under `y`. -/
  have hy_range (z : End K M) (hz : z ∈ LieHom.range φ) : ⁅y, z⁆ ∈ LieHom.range φ := by
    obtain ⟨q, hq⟩ : ∃ q : K[X], q.aeval (ad K _ (φ x)) = ad K _ y := by
      obtain ⟨r, hr⟩ : ∃ r : K[X], r.aeval (ad K _ s) = ad K _ y := by
        have h₁ (i j : I) : ⁅s, v.end (i, j)⁆ = (μ i - μ j) • v.end (i, j) := by
          rw [instLieRingModule_eq, v.lie_end_of_apply_eq_smul μ s hsv]
        have h₂ (i j : I) : ⁅y, v.end (i, j)⁆ = (fμ i - fμ j) • v.end (i, j) := by
          rw [instLieRingModule_eq, v.lie_end_of_apply_eq_smul fμ y hyv]
        obtain ⟨r, hr⟩ := exists_polynomial_eval_sub_aux μ hμ f
        refine ⟨r, v.end.ext fun ⟨i, j⟩ ↦ ?_⟩
        rw [ad_apply, ← instLieRingModule_eq, aeval_apply_of_mem_apply_eq_smul (h₁ i j), hr, h₂]
        rfl
      obtain ⟨p, hp⟩ : ∃ p : K[X], p.aeval (ad K _ (φ x)) = ad K _ s :=
        adjoin_mem_exists_aeval K _ <| by
          simpa only [← hX_ns] using ad_mem_adjoin_of_isSemisimple hns hn_nil hs_ss
      exact ⟨r.comp p, by rw [aeval_comp, hp, hr]⟩
    rw [instLieRingModule_eq, ← ad_apply K, ← hq]
    apply q.aeval_apply_smul_mem_of_le_comap hz _ ?_
    rintro - ⟨w, rfl⟩
    exact ⟨⁅x, w⁆, LieHom.map_lie φ x w⟩
  /- Using Lagrange interpolation again we can show that `n` and `y` commute. -/
  have hny_comm : Commute n y := by
    suffices y ∈ K[s] from commute_of_mem_adjoin_singleton_of_commute this hns
    rw [adjoin_singleton_eq_range_aeval]
    obtain ⟨q, hq⟩ : ∃ q : K[X], ∀ i, q.eval (μ i) = fμ i :=
      (Polynomial.exists_eval_eq_iff μ fμ).mpr <| by aesop
    refine ⟨q, v.ext fun i ↦ ?_⟩
    rw [AlgHom.toRingHom_eq_coe, RingHom.coe_coe, aeval_apply_of_mem_apply_eq_smul (hsv i), hq, hyv]
  /- By general results we need only prove `trace K M (φ x * y) = ∑ i, fμ i * μ i`. -/
  suffices trace K M (φ x * y) = ∑ i, fμ i * μ i from
    ⟨y, trace_toEnd_mul_eq_zero_of_traceForm_eq_zero h y hy_range x hx, this⟩
  /- And this is an easy calculation. -/
  have htr_n : trace K M (n * y) = 0 :=
    (isNilpotent_trace_of_isNilpotent (hny_comm.isNilpotent_mul_right hn_nil)).eq_zero
  have htr_s : trace K M (s * y) = ∑ i, fμ i * μ i := by
    rw [trace_eq_matrix_trace _ v, Matrix.trace]
    exact Finset.sum_congr rfl <| by simp [toMatrix_apply, hyv, hsv]
  rw [hX_ns, add_mul, map_add, htr_n, htr_s, zero_add]