awhecmu/canonical-drafter-extract-mathlib huggingface dataset:
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₄ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₅ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₆ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₁₇ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₄ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₅ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₆ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₇ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₁₈ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₄ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₅ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₆ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₇ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₈ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₁₉ : a_2 = in₀ a_1 → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₄ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₅ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₆ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₇ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₈ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₉ : a_2 = in₀ a_1 → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₂₀ h₂₁ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₂₂ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → ¬a_2 = in₂ c → a_2 = in₀ a_1
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₄ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₅ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₆ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₇ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₈ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₉ : a_2 = in₀ a_1 → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₂₀ h₂₁ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₂ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → ¬a_2 = in₂ c → a_2 = in₀ a_1
h₂₃ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₄ h₂₅ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₂₆ : a_2 = in₂ c → ¬a_2 = in₁ b → ¬a_2 = in₂ c → a_2 = in₀ a_1
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₄ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₅ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₆ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₇ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₈ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₉ : a_2 = in₀ a_1 → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₂₀ h₂₁ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₂ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → ¬a_2 = in₂ c → a_2 = in₀ a_1
h₂₃ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₄ h₂₅ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₆ : a_2 = in₂ c → ¬a_2 = in₁ b → ¬a_2 = in₂ c → a_2 = in₀ a_1
h₂₇ h₂₈ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₂₉ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h_1 : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
b_1 : β
h_3 : (a_1, b_1, c) ∈ t
a_2 : α ⊕ β ⊕ γ
h : a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁ : a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₂ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₃ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₄ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₅ : a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₆ : a_2 = in₁ b → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₇ : a_2 = in₂ c → a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
h₈ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₉ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₀ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₁₁ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₂ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₃ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₄ : a_2 = in₁ b → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₅ : a_2 = in₂ c → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₆ : a_2 = in₀ a_1 → a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c
h₁₇ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₈ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₁₉ : a_2 = in₀ a_1 → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₂₀ h₂₁ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₂ : a_2 = in₀ a_1 → ¬a_2 = in₁ b → ¬a_2 = in₂ c → a_2 = in₀ a_1
h₂₃ : a_2 = in₁ b → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₄ h₂₅ : a_2 = in₂ c → ¬a_2 = in₁ b → a_2 = in₂ c ∨ a_2 = in₀ a_1
h₂₆ : a_2 = in₂ c → ¬a_2 = in₁ b → ¬a_2 = in₂ c → a_2 = in₀ a_1
h₂₇ h₂₈ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
h₂₉ : a_2 = in₁ b → ¬a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c
h₃₀ h₃₁ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → a_2 = in₁ b ∨ a_2 = in₂ c
⊢ a_2 = in₀ a_1 ∨ a_2 = in₁ b ∨ a_2 = in₂ c ↔ a_2 = in₁ b ∨ a_2 = in₂ c ∨ a_2 = in₀ a_1
have h₃₂ : a_2 = in₂ c → ¬a_2 = in₀ a_1 → ¬a_2 = in₁ b → a_2 = in₂ c
Tactic: try tauto
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_1 : (a_1, b, c_1) ∈ t
b_1 : β
h_2 : (a_1, b_1, c) ∈ t
⊢ ∃ (a : α) (b_2 : β) (c_2 : γ),
    Eq (α := Finset (α ⊕ β ⊕ γ)) {in₀ a, in₁ b_2, in₂ c_2} {in₁ b, in₂ c, in₀ a_1} ∧
      (graph t).Adj (in₀ a) (in₁ b_2) ∧ (graph t).Adj (in₀ a) (in₂ c_2) ∧ (graph t).Adj (in₁ b_2) (in₂ c_2)
have in₁₂ : (graph t).Adj (in₀ a_1) (in₁ b)
Tactic: refine ⟨_, _, _, by ext; simp only [Finset.mem_insert, Finset.mem_singleton]; try tauto,
      ?_, ?_, ?_⟩
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_1 : (a_1, b, c_1) ∈ t
b_1 : β
h_2 : (a_1, b_1, c) ∈ t
in₁₂ : (graph t).Adj (in₀ a_1) (in₁ b)
⊢ ∃ (a : α) (b_2 : β) (c_2 : γ),
    Eq (α := Finset (α ⊕ β ⊕ γ)) {in₀ a, in₁ b_2, in₂ c_2} {in₁ b, in₂ c, in₀ a_1} ∧
      (graph t).Adj (in₀ a) (in₁ b_2) ∧ (graph t).Adj (in₀ a) (in₂ c_2) ∧ (graph t).Adj (in₁ b_2) (in₂ c_2)
have in₁₃ : (graph t).Adj (in₀ a_1) (in₂ c)
Tactic: refine ⟨_, _, _, by ext; simp only [Finset.mem_insert, Finset.mem_singleton]; try tauto,
      ?_, ?_, ?_⟩
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h : (a, b, c) ∈ t
a_1 : α
c_1 : γ
h_1 : (a_1, b, c_1) ∈ t
b_1 : β
h_2 : (a_1, b_1, c) ∈ t
in₁₂ : (graph t).Adj (in₀ a_1) (in₁ b)
in₁₃ : (graph t).Adj (in₀ a_1) (in₂ c)
⊢ ∃ (a : α) (b_2 : β) (c_2 : γ),
    Eq (α := Finset (α ⊕ β ⊕ γ)) {in₀ a, in₁ b_2, in₂ c_2} {in₁ b, in₂ c, in₀ a_1} ∧
      (graph t).Adj (in₀ a) (in₁ b_2) ∧ (graph t).Adj (in₀ a) (in₂ c_2) ∧ (graph t).Adj (in₁ b_2) (in₂ c_2)
have in₁₄ : (graph t).Adj (in₁ b) (in₂ c)
Tactic: refine ⟨_, _, _, by ext; simp only [Finset.mem_insert, Finset.mem_singleton]; try tauto,
      ?_, ?_, ?_⟩
-- Lean 4 proof step from Mathlib.Combinatorics.SimpleGraph.Triangle.Tripartite (goal)
Goal:
α : Type u_1
β : Type u_2
γ : Type u_3
t : Finset (α × β × γ)
inst : DecidableEq α
inst_1 : DecidableEq β
inst_2 : DecidableEq γ
a : α
b : β
c : γ
h : (a, b, c) ∈ t
a_1 : α
b_1 : β
h_1 : (a_1, b_1, c) ∈ t
c_1 : γ
h_2 : (a_1, b, c_1) ∈ t
⊢ Eq (α := Finset (α ⊕ β ⊕ γ)) {in₀ a_1, in₁ b, in₂ c} {in₂ c, in₁ b, in₀ a_1}
have h : ∀ (a : α ⊕ β ⊕ γ),
  Membership.mem (γ := Finset (α ⊕ β ⊕ γ)) {in₀ a_1, in₁ b, in₂ c} a ↔
    Membership.mem (γ := Finset (α ⊕ β ⊕ γ)) {in₂ c, in₁ b, in₀ a_1} a
Tactic: ext