Chapter 1. Once 〈X,α〉 is in 〈U,α〉 where X ⊂ Y and 〈U,α〉⊆ 〈V,α〉, then 〈X,α〉 is also in 〈V,α〉. Proof Suppose the contrary, that is, 〈X,α〉 is not in 〈V,α〉. Then we can expand the proof tree by adding a clause of the form〈〈Y,β〉,α〉. By hypothesis, 〈X,α〉 is in 〈U,α〉, so Y ⊂ X. Because 〈Y,β〉 is in 〈U,β〉 and 〈X,α〉 is in 〈U,α〉, then also 〈Y,β〉 is in 〈U,β〉. And because 〈Y,β〉 is in 〈U,β〉 and 〈X,α〉 is not in 〈U,α〉, then also 〈X,α〉 is not in 〈U,α〉, which is a contradiction. □ Hence, X ≠ α means 〈X,α〉 ∈ 〈U,α〉, so 〈X,α〉 ∈ 〈U,α〉, X ⊂ Y , and 〈〈Y,α〉,α〉 is in the proof tree. This corresponds to the fact that X ⊂ Y in the natural deduction proof of AX (i.e. that Y is a proper subset of X). 〈X,α〉 is in 〈U,α〉 means 〈X,α〉 is not in 〈V,α〉. Also, 〈〈X,α〉,α〉 is in the proof tree. The proof-irrelevance of A and H, on the other hand, allows us to move any of 〈X,α〉 or 〈Y,α〉 from 〈U,α〉 to 〈V,α〉. More precisely, A and H are symmetric in the following sense: A is the only symbol in H with a possible type of α, and vice versa. The symbol A cannot contain type information. It can only capture the position of an atom, so its type is fully determined by its index. H on the other hand can only be evaluated once its types are fully known, which is when it is being applied to the whole axiom tree. At the end of the axiom tree, the types of the symbols are fixed by their indices, and so the order of the typechecks does not matter. Therefore, A and H are symmetric in the sense that each can be applied to the other. Now, we can prove that the type-structure of the natural deduction proof mirrors the index-structure of the axiom tree. Consider the case where there are no assumptions in the proof; that is, there are no X ⊂ Y , so 〈X,α〉 is never added to the proof tree. Then the natural deduction proof of AX reduces to the presentation tree in which the axioms and inference rules are listed with the same order. The same applies to 〈X,α〉 and 〈Y,α〉 when X = Y . This means that if we reverse the order of the natural deduction proof of AX, we get the same presentation tree. More formally, we can define two maps Γ (the type-structure) and Θ (the index-structure), so that 〈X,α〉 ∈ Γ (X, α) = Θ (α, X) and 〈〈Y,α〉,α〉 ∈ Γ (Y,α) = Θ (α, Y). To prove Γ = Θ , we use the fact that the type-structure 〈U,α〉 and index-structure 〈V,α〉, which are used for reasoning about whole proofs, are defined as the union of the types and indices, respectively, of the proof-structure 〈〈U,α〉,α〉 and proof-structure 〈〈V,α〉,α〉, that is, 〈U,α〉 = 〈V,α〉. Also, Γ and Θ are symmetric, in the sense that if 〈〈X,α〉,α〉 ∈ Γ, then Γ 〈〈X,α〉,α〉 = 〈〈X,α〉,α〉. Then we prove that if Γ(X, α) = Γ(Y, α) then Γ(X, α) = Γ(Y, α). Suppose X = Y, which means that 〈X,α〉 is in the proof tree and in the proof-structure of the proof. Therefore, Γ(X, α) = Γ(Y, α) ≡ Γ(〈〈X,α〉,α〉). Now, 〈〈X,α〉,α〉 is in the proof-structure 〈V,α〉, but then 〈X,α〉 is also in 〈V,α〉. Because Γ(〈X,α〉, α) = Γ(〈〈X,α〉,α〉), this means that Γ(〈〈X,α〉,α〉) = Γ(〈V,α〉, α). So if 〈〈X,α〉,α〉 ∈ Γ(〈V,α〉, α), then 〈〈X,α〉,α〉 ∈ Γ(〈〈V,α〉,α〉), which means that X = Y is a type error in the context of the proof. It is easy to prove that this argument holds for X ⊂ Y . So Γ = Θ . The lemma shows that natural deduction proofs (when they exist) are the type-structure of the corresponding axiom tree (that is, they are consistent with respect to the given semantics). But now we have found that the type-structure of the axiom tree cannot be used to identify the axiom. This is because the axioms of an axiom system correspond to types in the context of the proof. The axiom cannot be identified with the unique type-structure that is associated with it by the natural deduction proof. Moreover, we have to take into account the semantics of the type information. In the axiom case, the associated type information refers to types that are assumed to be in the context of the proof, but we cannot simply assume that they are. This means that we cannot simply replace A with X, in the axioms, since it would be just as well to say that A ∈ X, or Y ∈ X, in the axioms, instead. But, at the same time, we cannot simply add type information that is assumed to be in the context of the proof, as this would result in a type-structure that is inconsistent with respect to the semantics of the type information. The axioms and derivations capture two different aspects of the axiom system. The axioms reflect the ways that the type information can be used to prove new theorems, and the derivations capture how the type information is used in the proofs. It is only after we combine the axiom system and the proof that the type-structures, and hence the corresponding axiom trees, coincide. 3.3 Substitution of Equivalent Terms The axioms of system T (without type information) have been used to construct a simple typed λ-calculus by Substitution. First, we translate each A into a function fA, where fA is a function symbol. For example, fA ∈ fA ∈ λx : ⊥.A x : ⊥.A. Then, we expand our axioms from axiom trees to derivations. For example, the derivation of A ∈ B ⇒ A ∈ B gets replaced with the typed derivation ruleSubstitution. Then, we can apply typechecking and type-structure construction. For example, Substitution is a proof-irrelevant derivable rule. Proof. We can use the axioms of T and the substitution rule to construct derivations of the form: Substitution ▶ fA(x) : ⊥.A ∈ B,A ∈ B ⇒ A ∈ B (1) So B is the only axiom that can be used to get the substitution rule to apply to fA. But then we can use Substitution again to construct the derivations: Substitution ▶ fA(x) : ⊥.A ∈ B,A ∈ B ⇒ A ∈ B (2) fA ∈ λx : ⊥.B.