-- Lean 4 proof step from Mathlib.Combinatorics.Configuration (goal)
Goal:
P : Type u_1
L : Type u_2
inst : Membership P L
inst_1 : HasLines P L
inst_2 : Fintype P
inst_3 : Fintype L
h : Fintype.card P = Fintype.card L
l₁ l₂ : L
hl : l₁ ≠ l₂
obtain : ∀ (f : L → P), Function.Bijective f → ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)), ∃ p ∈ l₁, p ∈ l₂
obtain₁ :
  ∀ (f : L → P),
    Function.Bijective f →
      ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)) (this : Nontrivial L), ∃ p ∈ l₁, p ∈ l₂
obtain₂ :
  ∀ (f : L → P),
    Function.Bijective f →
      ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)) (this : Nontrivial L) (this : Nontrivial P),
        ∃ p ∈ l₁, p ∈ l₂
obtain₃ :
  ∀ (f : L → P),
    Function.Bijective f →
      ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)) (this : Nontrivial L) (this : Nontrivial P)
        (h₁ : ∀ (p : P), 0 < lineCount L p), ∃ p ∈ l₁, p ∈ l₂
obtain₄ :
  ∀ (f : L → P),
    Function.Bijective f →
      ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)) (this : Nontrivial L) (this : Nontrivial P)
        (h₁ : ∀ (p : P), 0 < lineCount L p) (h₂ : ∀ (l : L), 0 < pointCount P l), ∃ p ∈ l₁, p ∈ l₂
obtain₅ :
  ∀ (f : L → P),
    Function.Bijective f →
      ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)) (this : Nontrivial L) (this : Nontrivial P)
        (h₁ : ∀ (p : P), 0 < lineCount L p) (h₂ : ∀ (l : L), 0 < pointCount P l) (p : P) (hl₁ : p ∈ l₁),
        ∃ p ∈ l₁, p ∈ l₂
pos :
  ∀ (f : L → P),
    Function.Bijective f →
      ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)) (this : Nontrivial L) (this : Nontrivial P)
        (h₁ : ∀ (p : P), 0 < lineCount L p) (h₂ : ∀ (l : L), 0 < pointCount P l) (p : P) (hl₁ : p ∈ l₁) (hl₂ : p ∈ l₂),
        ∃ p ∈ l₁, p ∈ l₂
⊢ ∃ p ∈ l₁, p ∈ l₂
have neg : ∀ (f : L → P),
  Function.Bijective f →
    ∀ (hf2 : ∀ (l : L), pointCount P l = lineCount L (f l)) (this : Nontrivial L) (this : Nontrivial P)
      (h₁ : ∀ (p : P), 0 < lineCount L p) (h₂ : ∀ (l : L), 0 < pointCount P l) (p : P) (hl₁ : p ∈ l₁) (hl₂ : p ∉ l₂),
      ∃ p ∈ l₁, p ∈ l₂
Tactic: classical
      obtain ⟨f, _, hf2⟩ := HasLines.exists_bijective_of_card_eq h
      haveI : Nontrivial L := ⟨⟨l₁, l₂, hl⟩⟩
      haveI := Fintype.one_lt_card_iff_nontrivial.mp ((congr_arg _ h).mpr Fintype.one_lt_card)
      have h₁ : ∀ p : P, 0 < lineCount L p := fun p =>
        Exists.elim (exists_ne p) fun q hq =>
          (congr_arg _ Nat.card_eq_fintype_card).mpr
            (Fintype.card_pos_iff.mpr ⟨⟨mkLine hq, (mkLine_ax hq).2⟩⟩)
      have h₂ : ∀ l : L, 0 < pointCount P l := fun l => (congr_arg _ (hf2 l)).mpr (h₁ (f l))
      obtain ⟨p, hl₁⟩ := Fintype.card_pos_iff.mp ((congr_arg _ Nat.card_eq_fintype_card).mp (h₂ l₁))
      by_cases hl₂ : p ∈ l₂
      · exact ⟨p, hl₁, hl₂⟩
      have key' : Fintype.card { q : P // q ∈ l₂ } = Fintype.card { l : L // p ∈ l } :=
        ((HasLines.lineCount_eq_pointCount h hl₂).trans Nat.card_eq_fintype_card).symm.trans
          Nat.card_eq_fintype_card
      have : ∀ q : { q // q ∈ l₂ }, p ≠ q := fun q hq => hl₂ ((congr_arg (· ∈ l₂) hq).mpr q.2)
      let f : { q : P // q ∈ l₂ } → { l : L // p ∈ l } := fun q =>
        ⟨mkLine (this q), (mkLine_ax (this q)).1⟩
      have hf : Function.Injective f := fun q₁ q₂ hq =>
        Subtype.ext ((eq_or_eq q₁.2 q₂.2 (mkLine_ax (this q₁)).2
            ((congr_arg (_ ∈ ·) (Subtype.ext_iff.mp hq)).mpr (mkLine_ax (this q₂)).2)).resolve_right
            fun h => (congr_arg (p ∉ ·) h).mp hl₂ (mkLine_ax (this q₁)).1)
      have key' := ((Fintype.bijective_iff_injective_and_card f).mpr ⟨hf, key'⟩).2
      obtain ⟨q, hq⟩ := key' ⟨l₁, hl₁⟩
      exact ⟨q, (congr_arg (_ ∈ ·) (Subtype.ext_iff.mp hq)).mp (mkLine_ax (this q)).2, q.2⟩