finished preservation, renamed substitution, substitution preservation still to do
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wadler
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2 changed files with 100 additions and 81 deletions
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@ -72,16 +72,16 @@ data value : Term → Set where
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\begin{code}
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\begin{code}
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_[_:=_] : Term → Id → Term → Term
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_[_:=_] : Term → Id → Term → Term
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(varᵀ x) [ y := P ] with x ≟ y
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(varᵀ x′) [ x := V ] with x ≟ x′
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... | yes _ = P
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... | yes _ = V
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... | no _ = varᵀ x
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... | no _ = varᵀ x′
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(λᵀ x ∈ A ⇒ N) [ y := P ] with x ≟ y
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(λᵀ x′ ∈ A′ ⇒ N′) [ x := V ] with x ≟ x′
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... | yes _ = λᵀ x ∈ A ⇒ N
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... | yes _ = λᵀ x′ ∈ A′ ⇒ N′
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... | no _ = λᵀ x ∈ A ⇒ (N [ y := P ])
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... | no _ = λᵀ x′ ∈ A′ ⇒ (N′ [ x := V ])
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(L ·ᵀ M) [ y := P ] = (L [ y := P ]) ·ᵀ (M [ y := P ])
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(L′ ·ᵀ M′) [ x := V ] = (L′ [ x := V ]) ·ᵀ (M′ [ x := V ])
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(trueᵀ) [ y := P ] = trueᵀ
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(trueᵀ) [ x := V ] = trueᵀ
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(falseᵀ) [ y := P ] = falseᵀ
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(falseᵀ) [ x := V ] = falseᵀ
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(ifᵀ L then M else N) [ y := P ] = ifᵀ (L [ y := P ]) then (M [ y := P ]) else (N [ y := P ])
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(ifᵀ L′ then M′ else N′) [ x := V ] = ifᵀ (L′ [ x := V ]) then (M′ [ x := V ]) else (N′ [ x := V ])
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\end{code}
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\end{code}
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## Reduction rules
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## Reduction rules
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@ -343,18 +343,18 @@ $$Γ ⊢ M ∈ A$$.
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hence the desired result follows from the induction hypotheses.
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hence the desired result follows from the induction hypotheses.
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\begin{code}
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\begin{code}
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weaken Γ⊆Γ′ (Ax Γx≡justA) rewrite (Γ⊆Γ′ free-varᵀ) = Ax Γx≡justA
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weaken Γ~Γ′ (Ax Γx≡justA) rewrite (Γ~Γ′ free-varᵀ) = Ax Γx≡justA
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weaken {Γ} {Γ′} {λᵀ x ∈ A ⇒ N} Γ⊆Γ′ (⇒-I ⊢N) = ⇒-I (weaken Γx⊆Γ′x ⊢N)
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weaken {Γ} {Γ′} {λᵀ x ∈ A ⇒ N} Γ~Γ′ (⇒-I ⊢N) = ⇒-I (weaken Γx⊆Γ′x ⊢N)
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where
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where
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Γx⊆Γ′x : ∀ {y} → y FreeIn N → (Γ , x ↦ A) y ≡ (Γ′ , x ↦ A) y
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Γx⊆Γ′x : ∀ {y} → y FreeIn N → (Γ , x ↦ A) y ≡ (Γ′ , x ↦ A) y
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Γx⊆Γ′x {y} y∈N with x ≟ y
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Γx⊆Γ′x {y} y∈N with x ≟ y
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... | yes refl = refl
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... | yes refl = refl
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... | no x≢y = Γ⊆Γ′ (free-λᵀ x≢y y∈N)
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... | no x≢y = Γ~Γ′ (free-λᵀ x≢y y∈N)
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weaken Γ⊆Γ′ (⇒-E ⊢L ⊢M) = ⇒-E (weaken (Γ⊆Γ′ ∘ free-·ᵀ₁) ⊢L) (weaken (Γ⊆Γ′ ∘ free-·ᵀ₂) ⊢M)
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weaken Γ~Γ′ (⇒-E ⊢L ⊢M) = ⇒-E (weaken (Γ~Γ′ ∘ free-·ᵀ₁) ⊢L) (weaken (Γ~Γ′ ∘ free-·ᵀ₂) ⊢M)
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weaken Γ⊆Γ′ 𝔹-I₁ = 𝔹-I₁
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weaken Γ~Γ′ 𝔹-I₁ = 𝔹-I₁
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weaken Γ⊆Γ′ 𝔹-I₂ = 𝔹-I₂
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weaken Γ~Γ′ 𝔹-I₂ = 𝔹-I₂
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weaken Γ⊆Γ′ (𝔹-E ⊢L ⊢M ⊢N)
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weaken Γ~Γ′ (𝔹-E ⊢L ⊢M ⊢N)
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= 𝔹-E (weaken (Γ⊆Γ′ ∘ free-ifᵀ₁) ⊢L) (weaken (Γ⊆Γ′ ∘ free-ifᵀ₂) ⊢M) (weaken (Γ⊆Γ′ ∘ free-ifᵀ₃) ⊢N)
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= 𝔹-E (weaken (Γ~Γ′ ∘ free-ifᵀ₁) ⊢L) (weaken (Γ~Γ′ ∘ free-ifᵀ₂) ⊢M) (weaken (Γ~Γ′ ∘ free-ifᵀ₃) ⊢N)
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{-
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{-
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replaceCtxt f (var x x∶A
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replaceCtxt f (var x x∶A
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@ -383,82 +383,82 @@ preserves types---namely, the observation that _substitution_
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preserves types.
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preserves types.
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Formally, the so-called _Substitution Lemma_ says this: Suppose we
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Formally, the so-called _Substitution Lemma_ says this: Suppose we
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have a term $$t$$ with a free variable $$x$$, and suppose we've been
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have a term $$N$$ with a free variable $$x$$, and suppose we've been
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able to assign a type $$T$$ to $$t$$ under the assumption that $$x$$ has
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able to assign a type $$B$$ to $$N$$ under the assumption that $$x$$ has
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some type $$U$$. Also, suppose that we have some other term $$v$$ and
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some type $$A$$. Also, suppose that we have some other term $$V$$ and
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that we've shown that $$v$$ has type $$U$$. Then, since $$v$$ satisfies
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that we've shown that $$V$$ has type $$A$$. Then, since $$V$$ satisfies
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the assumption we made about $$x$$ when typing $$t$$, we should be
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the assumption we made about $$x$$ when typing $$N$$, we should be
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able to substitute $$v$$ for each of the occurrences of $$x$$ in $$t$$
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able to substitute $$V$$ for each of the occurrences of $$x$$ in $$N$$
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and obtain a new term that still has type $$T$$.
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and obtain a new term that still has type $$B$$.
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_Lemma_: If $$\Gamma,x:U \vdash t : T$$ and $$\vdash v : U$$, then
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_Lemma_: If $$Γ , x ↦ A ⊢ N ∈ B$$ and $$∅ ⊢ V ∈ A$$, then
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$$\Gamma \vdash [x:=v]t : T$$.
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$$Γ ⊢ (N [ x := V ]) ∈ B$$.
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\begin{code}
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\begin{code}
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[:=]-preserves-⊢ : ∀ {Γ y A P N B}
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preservation-[:=] : ∀ {Γ x A N B V}
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→ ∅ ⊢ P ∈ A
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→ ∅ ⊢ V ∈ A
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→ (Γ , y ↦ A) ⊢ N ∈ B
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→ (Γ , x ↦ A) ⊢ N ∈ B
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→ (Γ , y ↦ A) ⊢ N [ y := P ] ∈ B
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→ Γ ⊢ (N [ x := V ]) ∈ B
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\end{code}
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\end{code}
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One technical subtlety in the statement of the lemma is that
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One technical subtlety in the statement of the lemma is that
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we assign $$P$$ the type $$A$$ in the _empty_ context---in other
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we assign $$V$$ the type $$A$$ in the _empty_ context---in other
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words, we assume $$P$$ is closed. This assumption considerably
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words, we assume $$V$$ is closed. This assumption considerably
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simplifies the $$λᵀ$$ case of the proof (compared to assuming
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simplifies the $$λᵀ$$ case of the proof (compared to assuming
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$$Γ ⊢ P ∈ A$$, which would be the other reasonable assumption
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$$Γ ⊢ V ∈ A$$, which would be the other reasonable assumption
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at this point) because the context invariance lemma then tells us
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at this point) because the context invariance lemma then tells us
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that $$P$$ has type $$A$$ in any context at all---we don't have to
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that $$V$$ has type $$A$$ in any context at all---we don't have to
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worry about free variables in $$P$$ clashing with the variable being
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worry about free variables in $$V$$ clashing with the variable being
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introduced into the context by $$λᵀ$$.
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introduced into the context by $$λᵀ$$.
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The substitution lemma can be viewed as a kind of "commutation"
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The substitution lemma can be viewed as a kind of "commutation"
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property. Intuitively, it says that substitution and typing can
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property. Intuitively, it says that substitution and typing can
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be done in either order: we can either assign types to the terms
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be done in either order: we can either assign types to the terms
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$$N$$ and $$P$$ separately (under suitable contexts) and then combine
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$$N$$ and $$V$$ separately (under suitable contexts) and then combine
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them using substitution, or we can substitute first and then
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them using substitution, or we can substitute first and then
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assign a type to $$N [ y := P ]$$---the result is the same either
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assign a type to $$N [ x := V ]$$---the result is the same either
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way.
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way.
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_Proof_: We show, by induction on $$N$$, that for all $$A$$ and
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_Proof_: We show, by induction on $$N$$, that for all $$A$$ and
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$$Γ$$, if $$Γ , y ↦ A \vdash N ∈ B$$ and $$∅ ⊢ P ∈ B$$, then
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$$Γ$$, if $$Γ , x ↦ A \vdash N ∈ B$$ and $$∅ ⊢ V ∈ A$$, then
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$$Γ \vdash N [ y := P ] ∈ B$$.
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$$Γ \vdash N [ x := V ] ∈ B$$.
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- If $$N$$ is a variable there are two cases to consider,
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- If $$N$$ is a variable there are two cases to consider,
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depending on whether $$N$$ is $$y$$ or some other variable.
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depending on whether $$N$$ is $$x$$ or some other variable.
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- If $$N = varᵀ y$$, then from the fact that $$Γ , y ↦ A ⊢ N ∈ B$$
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- If $$N = varᵀ x$$, then from the fact that $$Γ , x ↦ A ⊢ N ∈ B$$
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we conclude that $$A = B$$. We must show that $$y [ y := P] =
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we conclude that $$A = B$$. We must show that $$x [ x := V] =
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P$$ has type $$A$$ under $$Γ$$, given the assumption that
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V$$ has type $$A$$ under $$Γ$$, given the assumption that
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$$P$$ has type $$A$$ under the empty context. This
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$$V$$ has type $$A$$ under the empty context. This
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follows from context invariance: if a closed term has type
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follows from context invariance: if a closed term has type
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$$A$$ in the empty context, it has that type in any context.
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$$A$$ in the empty context, it has that type in any context.
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- If $$N$$ is some variable $$x$$ that is not equal to $$y$$, then
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- If $$N$$ is some variable $$x′$$ different from $$x$$, then
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we need only note that $$x$$ has the same type under $$Γ , y ↦ A$$
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we need only note that $$x′$$ has the same type under $$Γ , x ↦ A$$
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as under $$Γ$$.
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as under $$Γ$$.
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- If $$N$$ is an abstraction $$λᵀ x ∈ A′ ⇒ N′$$, then the IH tells us,
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- If $$N$$ is an abstraction $$λᵀ x′ ∈ A′ ⇒ N′$$, then the IH tells us,
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for all $$Γ′$$́ and $$A′$$, that if $$Γ′ , y ↦ A ⊢ N′ ∈ B′$$
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for all $$Γ′$$́ and $$B′$$, that if $$Γ′ , x ↦ A ⊢ N′ ∈ B′$$
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and $$∅ ⊢ P ∈ A$$, then $$Γ′ ⊢ N′ [ y := P ] ∈ B′$$.
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and $$∅ ⊢ V ∈ A$$, then $$Γ′ ⊢ N′ [ x := V ] ∈ B′$$.
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The substitution in the conclusion behaves differently
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The substitution in the conclusion behaves differently
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depending on whether $$x$$ and $$y$$ are the same variable.
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depending on whether $$x$$ and $$x′$$ are the same variable.
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First, suppose $$x ≡ y$$. Then, by the definition of
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First, suppose $$x ≡ x′$$. Then, by the definition of
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substitution, $$N [ y := P] = N$$, so we just need to show $$Γ ⊢ N ∈ B$$.
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substitution, $$N [ x := V] = N$$, so we just need to show $$Γ ⊢ N ∈ B$$.
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But we know $$Γ , y ↦ A ⊢ N ∈ B$$ and, since $$x ≡ y$$
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But we know $$Γ , x ↦ A ⊢ N ∈ B$$ and, since $$x ≡ x′$$
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does not appear free in $$λᵀ x ∈ A′ ⇒ N′$$, the context invariance
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does not appear free in $$λᵀ x′ ∈ A′ ⇒ N′$$, the context invariance
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lemma yields $$Γ ⊢ N ∈ B$$.
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lemma yields $$Γ ⊢ N ∈ B$$.
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Second, suppose $$x ≢ y$$. We know $$Γ , y ↦ A , x ↦ A′ ⊢ N' ∈ B′$$
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Second, suppose $$x ≢ x′$$. We know $$Γ , x ↦ A , x′ ↦ A′ ⊢ N′ ∈ B′$$
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by inversion of the typing relation, from which
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by inversion of the typing relation, from which
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$$Γ , x ↦ A′ , y ↦ A ⊢ N′ ∈ B′$$ follows by update permute,
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$$Γ , x′ ↦ A′ , x ↦ A ⊢ N′ ∈ B′$$ follows by update permute,
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so the IH applies, giving us $$Γ , x ↦ A′ ⊢ N′ [ y := P ] ∈ B′$$
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so the IH applies, giving us $$Γ , x′ ↦ A′ ⊢ N′ [ x := V ] ∈ B′$$
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By $$⇒-I$$, we have $$Γ ⊢ λᵀ x ∈ A′ ⇒ (N′ [ y := P ]) ∈ A′ ⇒ B′$$
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By $$⇒-I$$, we have $$Γ ⊢ λᵀ x′ ∈ A′ ⇒ (N′ [ x := V ]) ∈ A′ ⇒ B′$$
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and the definition of substitution (noting $$x ≢ y$$) gives
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and the definition of substitution (noting $$x ≢ x′$$) gives
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$$Γ ⊢ (λᵀ x ∈ A′ ⇒ N′) [ y := P ] ∈ A′ ⇒ B′$$ as required.
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$$Γ ⊢ (λᵀ x′ ∈ A′ ⇒ N′) [ x := V ] ∈ A′ ⇒ B′$$ as required.
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- If $$N$$ is an application $$L ·ᵀ M$$, the result follows
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- If $$N$$ is an application $$L′ ·ᵀ M′$$, the result follows
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straightforwardly from the definition of substitution and the
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straightforwardly from the definition of substitution and the
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induction hypotheses.
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induction hypotheses.
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@ -467,26 +467,26 @@ $$Γ \vdash N [ y := P ] ∈ B$$.
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For one case, we need to know that weakening applies to any closed term.
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For one case, we need to know that weakening applies to any closed term.
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\begin{code}
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\begin{code}
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weaken-closed : ∀ {P A Γ} → ∅ ⊢ P ∈ A → Γ ⊢ P ∈ A
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weaken-closed : ∀ {P A Γ} → ∅ ⊢ P ∈ A → Γ ⊢ P ∈ A
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weaken-closed {P} {A} {Γ} ⊢P = weaken g ⊢P
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weaken-closed {P} {A} {Γ} ⊢P = weaken Γ~Γ′ ⊢P
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where
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where
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g : ∀ {x} → x FreeIn P → ∅ x ≡ Γ x
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Γ~Γ′ : ∀ {x} → x FreeIn P → ∅ x ≡ Γ x
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g {x} x∈P = ⊥-elim (x∉P x∈P)
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Γ~Γ′ {x} x∈P = ⊥-elim (x∉P x∈P)
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where
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where
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x∉P : ¬ (x FreeIn P)
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x∉P : ¬ (x FreeIn P)
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x∉P = ∅⊢-closed ⊢P {x}
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x∉P = ∅⊢-closed ⊢P {x}
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\end{code}
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\end{code}
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\begin{code}
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\begin{code}
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[:=]-preserves-⊢ {Γ} {y} {A} ⊢P (Ax {_} {x} {B} Γx≡justB) with x ≟ y
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preservation-[:=] {Γ} {y} {A} ⊢P (Ax {_} {x} {B} Γx≡justB) with x ≟ y
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...| yes x≡y = {!!} -- weaken-closed ⊢P
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...| yes x≡y = {!!} -- weaken-closed ⊢P
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...| no x≢y = Ax {_} {x} Γx≡justB
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...| no x≢y = {!!} -- Ax {_} {x} Γx≡justB
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[:=]-preserves-⊢ {Γ} {y} {A} ⊢P (⇒-I {_} {x} ⊢N) with x ≟ y
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preservation-[:=] {Γ} {y} {A} ⊢P (⇒-I {_} {x} ⊢N) with x ≟ y
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...| yes x≡y = ⇒-I {_} {x} ⊢N
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...| yes x≡y = {!!} -- ⇒-I {_} {x} ⊢N
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...| no x≢y = {!!} -- ⇒-I {_} {x} ([:=]-preserves-⊢ {_} {y} {A} ⊢P ⊢N)
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...| no x≢y = {!!} -- ⇒-I {_} {x} (preservation-[:=] {_} {y} {A} ⊢P ⊢N)
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[:=]-preserves-⊢ ⊢P (⇒-E ⊢L ⊢M) = ⇒-E ([:=]-preserves-⊢ ⊢P ⊢L) ([:=]-preserves-⊢ ⊢P ⊢M)
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preservation-[:=] ⊢P (⇒-E ⊢L ⊢M) = ⇒-E (preservation-[:=] ⊢P ⊢L) (preservation-[:=] ⊢P ⊢M)
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[:=]-preserves-⊢ ⊢P 𝔹-I₁ = 𝔹-I₁
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preservation-[:=] ⊢P 𝔹-I₁ = 𝔹-I₁
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[:=]-preserves-⊢ ⊢P 𝔹-I₂ = 𝔹-I₂
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preservation-[:=] ⊢P 𝔹-I₂ = 𝔹-I₂
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[:=]-preserves-⊢ ⊢P (𝔹-E ⊢L ⊢M ⊢N) = 𝔹-E ([:=]-preserves-⊢ ⊢P ⊢L) ([:=]-preserves-⊢ ⊢P ⊢M) ([:=]-preserves-⊢ ⊢P ⊢N)
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preservation-[:=] ⊢P (𝔹-E ⊢L ⊢M ⊢N) = 𝔹-E (preservation-[:=] ⊢P ⊢L) (preservation-[:=] ⊢P ⊢M) (preservation-[:=] ⊢P ⊢N)
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{-
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{-
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@ -509,14 +509,13 @@ weaken-closed {P} {A} {Γ} ⊢P = weaken g ⊢P
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### Main Theorem
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### Main Theorem
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We now have the tools we need to prove preservation: if a closed
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We now have the tools we need to prove preservation: if a closed
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term $$t$$ has type $$T$$ and takes a step to $$t'$$, then $$t'$$
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term $$M$$ has type $$A$$ and takes a step to $$N$$, then $$N$$
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is also a closed term with type $$T$$. In other words, the small-step
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is also a closed term with type $$A$$. In other words, small-step
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reduction relation preserves types.
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reduction preserves types.
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Theorem preservation : forall t t' T,
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\begin{code}
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empty \vdash t : T →
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preservation : ∀ {M N A} → ∅ ⊢ M ∈ A → M ⟹ N → ∅ ⊢ N ∈ A
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t ==> t' →
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\end{code}
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empty \vdash t' : T.
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_Proof_: By induction on the derivation of $$\vdash t : T$$.
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_Proof_: By induction on the derivation of $$\vdash t : T$$.
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@ -549,6 +548,26 @@ _Proof_: By induction on the derivation of $$\vdash t : T$$.
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- Otherwise, $$t$$ steps by $$Sif$$, and the desired conclusion
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- Otherwise, $$t$$ steps by $$Sif$$, and the desired conclusion
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follows directly from the induction hypothesis.
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follows directly from the induction hypothesis.
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\begin{code}
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preservation (Ax x₁) ()
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preservation (⇒-I ⊢N) ()
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preservation (⇒-E (⇒-I ⊢N) ⊢V) (β⇒ valueV) = preservation-[:=] ⊢V ⊢N
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preservation (⇒-E ⊢L ⊢M) (γ⇒₁ L⟹L′) with preservation ⊢L L⟹L′
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...| ⊢L′ = ⇒-E ⊢L′ ⊢M
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preservation (⇒-E ⊢L ⊢M) (γ⇒₂ valueL M⟹M′) with preservation ⊢M M⟹M′
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...| ⊢M′ = ⇒-E ⊢L ⊢M′
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preservation 𝔹-I₁ ()
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preservation 𝔹-I₂ ()
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preservation (𝔹-E 𝔹-I₁ ⊢M ⊢N) β𝔹₁ = ⊢M
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preservation (𝔹-E 𝔹-I₂ ⊢M ⊢N) β𝔹₂ = ⊢N
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preservation (𝔹-E ⊢L ⊢M ⊢N) (γ𝔹 L⟹L′) with preservation ⊢L L⟹L′
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...| ⊢L′ = 𝔹-E ⊢L′ ⊢M ⊢N
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-- Writing out implicit parameters in full
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-- preservation (⇒-E {Γ} {λᵀ x ∈ A ⇒ N} {M} {.A} {B} (⇒-I {.Γ} {.x} {.N} {.A} {.B} ⊢N) ⊢M) (β⇒ {.x} {.A} {.N} {.M} valueM)
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-- = preservation-[:=] {Γ} {x} {A} {M} {N} {B} ⊢M ⊢N
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\end{code}
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Proof with eauto.
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Proof with eauto.
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remember (@empty ty) as Gamma.
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remember (@empty ty) as Gamma.
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intros t t' T HT. generalize dependent t'.
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intros t t' T HT. generalize dependent t'.
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|
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Reference in a new issue