CIS6700-Spring2025/Homework 3.md

Definition (negative formulation of hereditary termination):

• 🆕 HT{T₁ × T₂}(M) ≝ HT{T₁}(M.1) ∧ HT{T₂}(M.2)
• 🆕 HT{T₁ → T₂}(M) ≝ ∀M₁, HT{T₁}(M₁) ⊃ HT{T₂}(M M₁)
• HT{1}(M) ≝ M →* ⟨⟩
• HT{Ans}(M) ≝ M →* yes ∨ M →* no

Lemma (head expansion): If M' → M and HT{A}(M) then HT{A}(M').

Theorem (fundamental): If Γ ⊢ M : A then Γ ≫ M ∈ A. Proof: By induction on the typing derivation. Assume γ and HT{Γ}(γ); the goal is to show HT{A}(γ̂(M)).

• Case Lam: The premise is Γ, x : A₁ ⊢ M₂ : A₂, with conclusion Γ ⊢ λx. M₂ : A₁ → A₂. Assuming M₁ and HT{A₁}(M₁), the goal is to show HT{A₂}((λx. γ̂(M₂)) M₁). Using the assumptions, we have HT{Γ, x : A₁}(γ, x ↦ M₁), which we apply to the induction hypothesis on the premise to get HT{A₂}(γ̂(M₂)[x ↦ M₁]). By head expansion on (λx. γ̂(M₂)) M₁ → γ̂(M₂)[x ↦ M₁], we obtain our goal.
• Case App: The premises are Γ ⊢ M : A₁ → A₂ and Γ ⊢ M₁ : A₁, with conclusion is Γ ⊢ M M₁ : A₂; the goal is to show HT{A₂}(γ̂(M) γ̂(M₁)). By the induction hypothesis on the first premise specialized to γ̂(M₁), we have that HT{T₁}(γ̂(M₁)) implies HT{T₂}(γ̂(M) γ̂(M₁)), our goal. It suffices to show the antecedent, which holds by the induction hypothesis on the second premise.
• Case Pair: The premises are Γ ⊢ M₁ : A₁ and Γ ⊢ M₂ : A₂, with conclusion Γ ⊢ ⟨M₁, M₂⟩ : A₁ × A₂. The goal is to show HT{A₁}(⟨γ̂(M₁), γ̂(M₂)⟩.1) and HT{A₂}(⟨γ̂(M₁), γ̂(M₂)⟩.2). By the induction hypotheses on the two premises, we have HT{A₁}(γ̂(M₁)) and HT{A₂}(γ̂(M₂)). By head expansion on ⟨γ̂(M₁), γ̂(M₂)⟩.i → γ̂(Mᵢ), we have our goal.
• Case Left/Right: The premise is Γ ⊢ M : A₁ × A₂, and the conclusion is Γ ⊢ M.i : Aᵢ. The goal is to show HT{Aᵢ}(γ̂(M).i). By the induction hypothesis on the premise, we have HT{Aᵢ}(γ̂(M).i) as needed.
• Cases Var, Yes, No, Unit: Unchanged.

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Lemma: If M₁ and M₂[x ↦ M₁] are strongly normalizing, then M₂ is strongly normalizing. Proof: By induction on M₂. In the variable case for x, we use strong normalization of M₂.

Claim: The largest saturated set is SN itself. Obviously there is no larger set, since all saturated sets are subsets of SN. To check saturation:

• If N... ∈ SN then x N... ∈ SN: By congruence of application, the only possible reduction paths are through reduction of N..., which are strongly normalizing.
• If M₁, N... ∈ SN and M₂[x ↦ M₁] N... ∈ SN then (λx. M₂) M₁ N... ∈ SN: The possible reduction paths are by congruence through M₁, M₂, or N..., or by β-reduction of the head redex.
  • β-reduction: (λx. M₂) M₁ N... → M₂[x ↦ M₁] N..., which was assumed to be strongly normalizing.
  • Congruence (M₁ and N...): They were assumed to be strongly normalizing.
  • Congruence (M₂): By assumption, M₂[x ↦ M₁] N... is strongly normalizing, so M₂[x ↦ M₁] must be strongly normalizing.† Then by the above lemma, M₂ is strongly normalizing. † I'm fairly certain this is true, but I'm now doubting that it's simple to prove rigorously...