The calculus of fig. 1.1 corresponds to a set of models. Models are used here and elsewhere as a tool. They have merely instrumental and no explanatory value. As is discussed in chapter 2 they can in principle be dispensed with for philosophical purposes.
The notion of a model is expanded throughout this work just as the notion of a position is expanded to capture different aspects of the practice of offering a descrip- tion of the way things are. A model in this case is a function v :L ÝÑ tT, F u such that
• If ϕ is atomic then vpϕq P tT, F u • If ϕ is ψ then vpϕq “ T iff vpψq “ F .
• If ϕ is ψ ^ θ then vpϕq “ T iff vpψq “ vpθq “ T .
A model is a counter-example to a position Γ ñ Σ iff for all γ P Γ vpγq “ T and for all σ P Σ vpσq “ F . If there are no counter-examples to Γ ñ Σ this is written as Γ |ù Σ.
Theorem 1.6.1 (Soundness). If $ Γ ñ Σ then Γ |ù Σ.
Proof. This is proved by induction on the length deductions in fig. 1.1. Let δ be a deduction whose last inference is I. Since the smallest deductions are instances of Id that is the base case. There are no counter-examples to ϕ ñ ϕ because all models are functions. The inductive hypothesis (IH) is that if δ1 is a deduction smaller than δ then there are no counter-examples to the end sequent of δ1. There are seven other cases to consider.
Case 1 (I is TL). In this case δ has the form δ .. . Γ ñ Σ TL Γ, ϕ ñ Σ
Suppose that there is a model v that is a counter-example to Γ, ϕ ñ Σ. For all γ P Γ Y tϕu vpγq “ T . It follows that for all γ P Γ vpγq “ T . Since vpσq “ F for all σ P Σ v is a counter-example to Γ ñ Σ. This contradicts IH which entails that there is no counter-example to Γ ñ Σ.
Case 2 (I is TR). This case is similar to the case where I is TL.
Case 3 (I is cut). For this case it is necessary to establish that for any v and ϕ vpϕq “ T or vpϕq “ F . This is proved by induction on the complexity of ϕ. If ϕ
is atomic this is built into the definition of a model. Suppose that ϕ is ψ. By the inner inductive hypothesis (IIH) vpψq “ T or vpψq “ F . In the first case vpϕq “ T and in the second vpϕq “ F . Suppose that ϕ is ψ ^ θ. By IIH either vpψq “ T or vpψq “ F and either vpθq “ T or vpθq “ F . This generates four sub-cases to consider.
c1 vpψq “ T and vpθq “ T . c2 vpψq “ T and vpθq “ F . c3 vpψq “ F and vpθq “ T . c4 vpψq “ F and vpθq “ F .
In c2 – c4 it is not the case that vpψq “ vpθq “ T . It follows that vpϕq “ F and so vpϕq “ F or vpϕq “ T . In c1 vpψq “ vpθq “ T . So vpϕq “ T and thus that vpϕq “ T or vpϕq “ F .
In this case δ has the form
δ1 .. . Γ ñ ϕ, Σ δ2 .. . Γ, ϕ ñ Σ cut Γ ñ Σ
Applying IH to δ1 and δ2 yields that there are no counter-examples to Γ ñ ϕ, Σ and there are no counter-examples to Γ, ϕ ñ Σ. Suppose for reductio that there is a counter-example v to Γ ñ Σ . By the above proof either vpϕq “ T or vpϕq “ F . In the first case v is a counter-example to Γ, ϕ ñ Σ which is impossible. In the second case v is a counter-example to Γ ñ ϕ, Σ.
Case 4 (I is L ). In this case δ has the form δ .. . Γ ñ ϕ, Σ L Γ, ϕ ñ Σ
By IH there is no counter-example to Γ ñ ϕ, Σ. For reductio let v be a counter- example to Γ, ϕ ñ Σ. In particular, vp ϕq “ T and so vpϕq “ F . It follows that v is a counter-example to Γ ñ ϕ, Σ.
Case 5 (I is R ). This case is similar to the case where I is L . Case 6 (I is L^). In this case δ has the form
δ .. . Γ, ϕ, ψ ñ Σ L^ Γ, ϕ ^ ψ ñ Σ
By IH there is no counter-example to Γ, ϕ, ψ ñ Σ. Suppose for reductio that there is a counter-example v to Γ, ϕ^ψ ñ Σ. In particular vpϕ^ψq “ T so vpϕq “ vpψq “ T . v is also a counter-example to Γ, ϕ, ψ ñ Σ.
Case 7 (I is R^). In this case δ has the form
δ1 .. . Γ ñ ϕ, Σ δ2 .. . Γ ñ ψ, Σ R^ Γ ñ ϕ ^ ψ, Σ
By IH there are no counter-examples to Γ ñ ϕ, Σ and Γ ñ ψ, Σ. Suppose for reductio that v is a counter-example to Γ ñ ϕ ^ ψ, Σ. In particular, vpϕ ^ ψq “ F . Either vpϕq “ F or vpψq “ F . In the first case v is a counter-example to Γ ñ ϕ, Σ
which is impossible. In the second case v is a counter-example to Γ ñ ψ, Σ which is also impossible.
Theorem 1.6.2 requires the definition of several structures about which some important lemmas are proved. The first such is a tree τ constructed from a position Γ ñ Σ written τ pΓ ñ Σq. At the root of the tree is Γ ñ Σ. The tree leaves of the tree at any stage are positions. A leaf Γ ñ Σ is open iff Γ X Σ “ H. A leaf is called ‘closed’ if it is not open. A branch is open when it contains no closed leaves. Let ϕ1, ϕ2, . . . be a list of the sentences ofL . At each stage of construction i do the following for each open leaf Γ ñ Σ in the construction:
1. If ϕi is atomic then do nothing. 2. If ϕi is ψ then
• if ϕi P Γ then expand the branch of the tree under consideration by Γ ñ ψ, Σ
Γ ñ Σ
• If ϕi P Σ then expand the branch of the tree under consideration by Γ, ψ ñ Σ
Γ ñ Σ 3. If ϕi is ψ ^ θ then
• If ϕi P Γ then expand the branch of the tree under consideration by Γ, ψ, θ ñ Σ
Γ ñ Σ 27
• If ϕi P Σ then expand the branch of the tree under consideration by Γ ñ ψ, Σ Γ ñ θ, Σ
Γ ñ Σ 4. Repeat steps (1-3) for ϕj where j ă i.
Let β “ Γ ñ Σ, Γ1 ñ Σ1, . . . be an open branch in τ pΓ ñ Σq. f pβq is the position
YiΓi ñ YiΣi
If β is an open branch where ∆j ñ Λj occurs before ∆k ñ Λk then ∆j Ď ∆k and Λj Ď Λk. The set
ą
pΓ ñ Σq is the set of all f pβq for open branches β in τ pΓ ñ Σq. If a position Γ ñ Σ is such that &cf Γ ñ Σ then
ą
pΓ ñ Σq ‰ H where $cf Γ ñ Σ indicates that Γ ñ Σ is provable in the calculus of fig. 1.1 without the rule of cut. Ifą pΓ ñ Σq “ H then there are no branches in τ pΓ ñ Σq that are open but in that case τ pΓ ñ Σq is a cut-free deduction of Γ ñ Σ.
Lemma 1.2. For any position ∆ ñ Λ P ą pΓ ñ Σq if ϕ P ∆ then ϕ P Λ and if ϕ P Λ then ϕ P ∆.
Proof. Let β be a branch such that f pβq “ ∆ ñ Λ. Let ϕj be ϕ. At stage j (or later, without loss of generality let it be j) there is a position ∆1 ñ Λ1 at which ϕ
j is considered. Let ∆2
ñ Λ2 be the position in β considered at stage j ` 1. At that stage if ϕ P ∆1then ϕ P Λ2 and if ϕ P Λ1 then ϕ P ∆1 by case 2 of the construction of τ pΓ ñ Σq. Since ∆2
Ď ∆ and Λ2 Ď Λ those facts hold for ∆ ñ Λ too.
Lemma 1.3. For any position ∆ ñ Λ P ą pΓ ñ Σq if ϕ ^ ψ P ∆ then ϕ P ∆ and ψ P ∆ and if ϕ ^ ψ P Λ then either ϕ P Λ or ψ P Λ.
Proof. Let β be a branch such that f pβq “ ∆ ñ Λ. Let ϕj be ϕ. At stage j (or later, without loss of generality let it be j) there is a position ∆1 ñ Λ1 at which ϕ
j is considered. Let ∆2 ñ Λ2 be the position in β considered at stage j ` 1. At that stage if ϕj P ∆1 then ϕ P ∆2 and ψ P ∆2 by case 3 of the construction of τ pΓ ñ Σq. Similarly, if ϕj P Λ1 then at stage j by clause 3 the τ pΓ ñ Σq branches. β either passes through the left branch or the right branch. In the first case ϕ P Λ2. In the second case ψ P Λ2.
Theorem 1.6.2 (Completeness). If Γ |ù Σ then $cf Γ ñ Σ.
Proof. Suppose that & Γ ñ Σ. Let ∆ ñ Λ P ą pΓ ñ Σq. ∆ ñ Λ is guaranteed to exist because otherwise Γ ñ Σ would have a deduction. Let v be the model that assigns all the atomic sentences in Γ T and all other atomic sentences F .
Claim 1. For all ϕ P ∆ vpϕq “ T and for all ϕ P Λ vpϕq “ F .
Proof of Claim 1. This is proved by induction on the complexity of ϕ. The base case is trivial. The inductive hypothesis (IH) is that for all formula ψ of complexity less than ϕ if ψ P ∆ then vpψq “ T and if ψ P Λ then vpψq “ F . There are four cases to consider
Case 1 (ϕ is ψ and ϕ P ∆). By lemma1.2 ψ P Λ. By IH, vpψq “ F . It follows that vpϕq “ T .
Case 2 (ϕ is ψ and ϕ P Λ). This case is similar to the first one.
Case 3 (ϕ is ψ ^ θ and ϕ P ∆). By lemma 1.3 ψ P ∆ and θ P ∆. By IH vpψq “ T and vpθq “ T . So vpϕq “ T .
Case 4 (ϕ is ψ ^ θ and ϕ P Λ). By lemma1.3 either ψ P Λ or θ P Λ. In the first case by IH vpψq “ F so vpϕq “ F . In the second case vpθq “ F so vpϕq “ F .
It follows from claim 1 that v is a counter-example to ∆ ñ Λ. Since Γ Ď ∆ and Σ Ď Λ v is also a counter-example to Γ ñ Σ.
Theorem 1.6.3 (Cut Admissibility). If $ Γ ñ Σ then $cf Γ ñ Σ.
Proof. Suppose that $ Γ ñ Σ. By theorem1.6.1 Γ |ù Σ. By theorem1.6.2$cf Γ ñ Σ.