/-- A maximal consistent set contains every sentence or its negation, provided consistency splits.
Hint: apply `hsplit Δ φ hΔ.prop` and, in each case, `Maximal.mem_of_prop_insert`. -/
theorem maximal_mem_or_neg_mem {S : Type u} (Con : Set S → Prop) (neg : S → S)
(hsplit : ∀ Γ φ, Con Γ → Con (insert φ Γ) ∨ Con (insert (neg φ) Γ))
{Δ : Set S} (hΔ : Maximal Con Δ) (φ : S) : φ ∈ Δ ∨ neg φ ∈ Δ := by
sorryShow the compiled solution
/-- **A maximal consistent set decides every sentence**, provided consistency *splits*: whenever
`Γ` is consistent, so is `Γ ∪ {φ}` or `Γ ∪ {¬φ}`. (For provability this holds because
`Γ ∪ {φ} ⊢ ⊥` and `Γ ∪ {¬φ} ⊢ ⊥` together give `Γ ⊢ ⊥`.) By maximality, whichever of the two
enlarged sets is consistent is `Γ` itself. -/
theorem maximal_mem_or_neg_mem {S : Type u} (Con : Set S → Prop) (neg : S → S)
(hsplit : ∀ Γ φ, Con Γ → Con (insert φ Γ) ∨ Con (insert (neg φ) Γ))
{Δ : Set S} (hΔ : Maximal Con Δ) (φ : S) : φ ∈ Δ ∨ neg φ ∈ Δ := by
rcases hsplit Δ φ hΔ.prop with h | h
· exact Or.inl (hΔ.mem_of_prop_insert h)
· exact Or.inr (hΔ.mem_of_prop_insert h)Axioms used: propext, Classical.choice, Quot.soundformal/Lumenary/CompletenessTheorem.leanTry it in the Lean web editor