/-- Over any ring `R`, if `b` is a basis for a module `M`,
and `s` is a maximal linearly independent set,
then the union of the supports of `x ∈ s` (when written out in the basis `b`) is all of `b`.
-/
theorem union_support_maximal_linearIndependent_eq_range_basis {ι : Type w} (b : Basis ι R M)
    {κ : Type w'} (v : κ → M) (i : LinearIndependent R v) (m : i.Maximal) :
    ⋃ k, ((b.repr (v k)).support : Set ι) = Set.univ := by
  -- If that's not the case,
  by_contra h
  simp only [← Ne.def, ne_univ_iff_exists_not_mem, mem_iUnion, not_exists_not,
    Finsupp.mem_support_iff, Finset.mem_coe] at h
  -- We have some basis element `b b'` which is not in the support of any of the `v i`.
  obtain ⟨b', w⟩ := h
  -- Using this, we'll construct a linearly independent family strictly larger than `v`,
  -- by also using this `b b'`.
  let v' : Option κ → M := fun o => o.elim (b b') v
  have r : range v ⊆ range v' := by
    rintro - ⟨k, rfl⟩
    use some k
    rfl
  have r' : b b' ∉ range v := by
    rintro ⟨k, p⟩
    simpa [w] using congr_arg (fun m => (b.repr m) b') p
  have r'' : range v ≠ range v' := by
    intro e
    have p : b b' ∈ range v' := by
      use none
      rfl
    rw [← e] at p
    exact r' p
  -- The key step in the proof is checking that this strictly larger family is linearly independent.
  have i' : LinearIndependent R ((↑) : range v' → M) := by
    apply LinearIndependent.to_subtype_range
    rw [linearIndependent_iff]
    intro l z
    rw [Finsupp.total_option] at z
    simp only [Option.elim'] at z
    change _ + Finsupp.total κ M R v l.some = 0 at z
    -- We have some linear combination of `b b'` and the `v i`, which we want to show is trivial.
    -- We'll first show the coefficient of `b b'` is zero,
    -- by expressing the `v i` in the basis `b`, and using that the `v i` have no `b b'` term.
    have l₀ : l none = 0 := by
      rw [← eq_neg_iff_add_eq_zero] at z
      replace z := neg_eq_iff_eq_neg.mpr z
      apply_fun fun x => b.repr x b' at z
      simp only [repr_self, LinearEquiv.map_smul, mul_one, Finsupp.single_eq_same, Pi.neg_apply,
        Finsupp.smul_single', LinearEquiv.map_neg, Finsupp.coe_neg] at z
      erw [FunLike.congr_fun (Finsupp.apply_total R (b.repr : M →ₗ[R] ι →₀ R) v l.some) b'] at z
      simpa [Finsupp.total_apply, w] using z
    -- Then all the other coefficients are zero, because `v` is linear independent.
    have l₁ : l.some = 0 := by
      rw [l₀, zero_smul, zero_add] at z
      exact linearIndependent_iff.mp i _ z
    -- Finally we put those facts together to show the linear combination is trivial.
    

Proof state:
case hf
K R : Type u
V V₁ V₂ V₃ : Type v
V' V'₁ : Type v'
V'' : Type v''
ι✝ : Type w
ι' : Type w'
η : Type u₁'
φ : η → Type u_1
inst✝⁷ : Ring R
M : Type v
inst✝⁶ : AddCommGroup M
inst✝⁵ : Module R M
M' : Type v'
inst✝⁴ : AddCommGroup M'
inst✝³ : Module R M'
M₁ : Type v
inst✝² : AddCommGroup M₁
inst✝¹ : Module R M₁
inst✝ : Nontrivial R
ι : Type w
b : Basis ι R M
κ : Type w'
v : κ → M
i : LinearIndependent R v
m : LinearIndependent.Maximal i
b' : ι
w : ∀ (x : κ), (b.repr (v x)) b' = 0
v' : Option κ → M := fun o => Option.elim o (b b') v
r : range v ⊆ range v'
r' : b b' ∉ range v
r'' : range v ≠ range v'
l : Option κ →₀ R
z : l none • Option.elim none (b b') v + (Finsupp.total κ M R v) (Finsupp.some l) = 0
l₀ : l none = 0
l₁ : Finsupp.some l = 0
⊢ l = 0

Next tactic:
ext (_ | a)