216 Logic as a Tool This property is equivalent (exercise) to the following: a first-order theory Γ is D-consistent iff every finite subset of Γ is D-consistent Theorem 176 (Lindenbaum–Henkin Theorem) Let Γ be D-consistent theory in a countable language L and let L+ be the extension of L obtained by adding countably many new constants c1 , , cn , Then Γ can be extended to a maximal Henkin theory H (Γ) in L+ Proof Let A0 , A1 , be a list of all sentences in L+ (NB: these are still only countably many!) We define a chain by inclusion of theories Γ0 , Γ1 , by recursion on n as follows • Γ0 := Γ • Γn+1 ⎧ Γn ∪ {An } if Γn D An and An is not of the type ∃xB ; ⎪ ⎪ ⎪ ⎪ ⎪ ∪ { A } ∪ { B [ c/x ]} if Γn D An and An = ∃xB, where c is the Γ n ⎨ n first new constant that does not occur in := ⎪ ⎪ ⎪ Γn ∪ {An }; ⎪ ⎪ ⎩ Γn ∪ {¬An } if Γn D An Note that every Γn is an D-consistent theory (Exercise: prove this by induction on n, using Corollary 173.) We now define H (Γ) := Γn n∈N Clearly, Γ ⊆ H (Γ) We claim that H (Γ) is a maximal D-consistent Henkin theory Indeed: • H (Γ) is D-consistent To prove this, suppose otherwise Then H (Γ) D A and H (Γ) D ¬A Since the deductive consequence is compact by Lemma 175, it follows that Γn D A and Γn D ¬A for some large enough index n (exercise: complete the details here), which contradicts the consistency of Γn • H (Γ) is maximal D-consistent Indeed, take any sentence A Let A = Am Then Am ∈ Γm+1 or ¬Am ∈ Γm+1 , hence A ∈ H (Γ) or ¬A ∈ H (Γ), hence H (Γ) cannot be extended to a larger D-consistent theory • H (Γ) is a Henkin theory The proof of this is left as an exercise The theory H (Γ) defined above is called a maximal Henkin extension of Γ Remark: The result above also applies to uncountable languages, by using Zorn’s Lemma