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stuck at two places in proof of preservation of substitution

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wadler 2017-06-25 19:08:11 +01:00
parent 6ee9d8dd40
commit 500d56b3b0
1 changed files with 52 additions and 60 deletions
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112
src/StlcProp.lagda
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@ -280,25 +280,9 @@ postulate
contradiction : ∀ {A : Set} → ∀ {v : A} → ¬ (_≡_ {A = Maybe A} (just v) nothing)
contradiction ()
-- ∅⊢-closed′ : ∀ {M A} → ∅ ⊢ M ∈ A → closed M
-- ∅⊢-closed′ {M} {A} ⊢M {x} x∈M with freeLemma x∈M ⊢M
-- ... | (B , ∅x≡justB) = contradiction (trans (sym ∅x≡justB) apply-∅)
{-
∅⊢-closed′ : ∀ {t A} → ∅ ⊢ t ∶ A → Closed t
∅⊢-closed′ (var x ())
∅⊢-closed′ (app t₁∶A⇒B t₂∶A) (app1 x∈t₁) = ∅⊢-closed′ t₁∶A⇒B x∈t₁
∅⊢-closed′ (app t₁∶A⇒B t₂∶A) (app2 x∈t₂) = ∅⊢-closed′ t₂∶A x∈t₂
∅⊢-closed′ true ()
∅⊢-closed′ false ()
∅⊢-closed′ (if t₁∶bool then t₂∶A else t₃∶A) (if1 x∈t₁) = ∅⊢-closed′ t₁∶bool x∈t₁
∅⊢-closed′ (if t₁∶bool then t₂∶A else t₃∶A) (if2 x∈t₂) = ∅⊢-closed′ t₂∶A x∈t₂
∅⊢-closed′ (if t₁∶bool then t₂∶A else t₃∶A) (if3 x∈t₃) = ∅⊢-closed′ t₃∶A x∈t₃
∅⊢-closed′ (abs {x = x} t′∶A) {y} (abs x≠y y∈t′) with freeInCtxt y∈t′ t′∶A
∅⊢-closed′ (abs {x = x} t′∶A) {y} (abs x≠y y∈t′) | A , y∶A with x ≟ y
∅⊢-closed′ (abs {x = x} t′∶A) {y} (abs x≠y y∈t′) | A , y∶A | yes x=y = x≠y x=y
∅⊢-closed′ (abs {x = x} t′∶A) {y} (abs x≠y y∈t′) | A , () | no _
-}
∅⊢-closed′ : ∀ {M A} → ∅ ⊢ M ∈ A → closed M
∅⊢-closed′ {M} {A} ⊢M {x} x∈M with freeLemma x∈M ⊢M
... | (B , ∅x≡justB) = contradiction (trans (sym ∅x≡justB) (apply-∅ x))
\end{code}
</div>
@ -358,12 +342,6 @@ $$Γ ⊢ M ∈ A$$.
$$t_1$$ are also free in $$t_1\;t_2$$, and similarly for $$t_2$$;
hence the desired result follows from the induction hypotheses.
-- weaken : ∀ {Γ Γ′ M A}
-- → (∀ {x} → x FreeIn M → Γ x ≡ Γ′ x)
-- → Γ ⊢ M ∈ A
-- → Γ′ ⊢ M ∈ A
\begin{code}
weaken Γ⊆Γ′ (Ax Γx≡justA) rewrite (Γ⊆Γ′ free-varᵀ) = Ax Γx≡justA
weaken {Γ} {Γ′} {λᵀ x ∈ A ⇒ N} Γ⊆Γ′ (⇒-I ⊢N) = ⇒-I (weaken Γx⊆Γ′x ⊢N)
@ -417,10 +395,10 @@ _Lemma_: If $$\Gamma,x:U \vdash t : T$$ and $$\vdash v : U$$, then
$$\Gamma \vdash [x:=v]t : T$$.
\begin{code}
[:=]-preserves-⊢ : ∀ {Γ x A N P B}
[:=]-preserves-⊢ : ∀ {Γ y A P N B}
→ ∅ ⊢ P ∈ A
→ Γ , x ↦ A ⊢ N ∈ B
→ Γ , x ↦ A ⊢ N [ x := P ] ∈ B
→ (Γ , y ↦ A) ⊢ N ∈ B
→ (Γ , y ↦ A) ⊢ N [ y := P ] ∈ B
\end{code}
One technical subtlety in the statement of the lemma is that
@ -438,47 +416,47 @@ property. Intuitively, it says that substitution and typing can
be done in either order: we can either assign types to the terms
$$N$$ and $$P$$ separately (under suitable contexts) and then combine
them using substitution, or we can substitute first and then
assign a type to $$N [ x := P ]$$---the result is the same either
assign a type to $$N [ y := P ]$$---the result is the same either
way.
_Proof_: We show, by induction on $$N$$, that for all $$A$$ and
$$Γ$$, if $$Γ , x ↦ A \vdash N ∈ B$$ and $$∅ ⊢ P ∈ B$$, then
$$Γ \vdash N [ x := P ] ∈ B$$.
$$Γ$$, if $$Γ , y ↦ A \vdash N ∈ B$$ and $$∅ ⊢ P ∈ B$$, then
$$Γ \vdash N [ y := P ] ∈ B$$.
- If $$N$$ is a variable there are two cases to consider,
depending on whether $$N$$ is $$x$$ or some other variable.
depending on whether $$N$$ is $$y$$ or some other variable.
- If $$N = x$$, then from the fact that $$Γ , x ↦ A ⊢ N ∈ B$$
we conclude that $$A = B$$. We must show that $$x [ x := P] =
- If $$N = varᵀ y$$, then from the fact that $$Γ , y ↦ A ⊢ N ∈ B$$
we conclude that $$A = B$$. We must show that $$y [ y := P] =
P$$ has type $$A$$ under $$Γ$$, given the assumption that
$$P$$ has type $$A$$ under the empty context. This
follows from context invariance: if a closed term has type
$$A$$ in the empty context, it has that type in any context.
- If $$N$$ is some variable $$y$$ that is not equal to $$x$$, then
we need only note that $$y$$ has the same type under $$Γ , x ↦ A$$
- If $$N$$ is some variable $$x$$ that is not equal to $$y$$, then
we need only note that $$x$$ has the same type under $$Γ , y ↦ A$$
as under $$Γ$$.
- If $$N$$ is an abstraction $$λᵀ y ∈ A′ ⇒ N′$$, then the IH tells us,
for all $$Γ′$$́ and $$A′$$, that if $$Γ′ , x ↦ A ⊢ N′ ∈ B′$$
and $$∅ ⊢ P ∈ A$$, then $$Γ′ ⊢ N′ [ x := P ] ∈ B′$$.
- If $$N$$ is an abstraction $$λᵀ x ∈ A′ ⇒ N′$$, then the IH tells us,
for all $$Γ′$$́ and $$A′$$, that if $$Γ′ , y ↦ A ⊢ N′ ∈ B′$$
and $$∅ ⊢ P ∈ A$$, then $$Γ′ ⊢ N′ [ y := P ] ∈ B′$$.
The substitution in the conclusion behaves differently
depending on whether $$x$$ and $$y$$ are the same variable.
First, suppose $$x ≡ y$$. Then, by the definition of
substitution, $$N [ x := P] = N$$, so we just need to show $$Γ ⊢ N ∈ B$$.
But we know $$Γ , x ↦ A ⊢ N ∈ B$$ and, since $$x ≡ y$$
does not appear free in $$λᵀ y ∈ A′ ⇒ N′$$, the context invariance
substitution, $$N [ y := P] = N$$, so we just need to show $$Γ ⊢ N ∈ B$$.
But we know $$Γ , y ↦ A ⊢ N ∈ B$$ and, since $$x ≡ y$$
does not appear free in $$λᵀ x ∈ A′ ⇒ N′$$, the context invariance
lemma yields $$Γ ⊢ N ∈ B$$.
Second, suppose $$x ≢ y$$. We know $$Γ , x ↦ A , y ↦ A′ ⊢ N' ∈ B′$$
Second, suppose $$x ≢ y$$. We know $$Γ , y ↦ A , x ↦ A′ ⊢ N' ∈ B′$$
by inversion of the typing relation, from which
$$Γ , y ↦ A′ , x ↦ A ⊢ N′ ∈ B′$$ follows by update permute,
so the IH applies, giving us $$Γ , y ↦ A′ ⊢ N′ [ x := P ] ∈ B′$$
By $$⇒-I$$, we have $$Γ ⊢ λᵀ y ∈ A′ ⇒ (N′ [ x := P ]) ∈ A′ ⇒ B′$$
$$Γ , x ↦ A′ , y ↦ A ⊢ N′ ∈ B′$$ follows by update permute,
so the IH applies, giving us $$Γ , x ↦ A′ ⊢ N′ [ y := P ] ∈ B′$$
By $$⇒-I$$, we have $$Γ ⊢ λᵀ x ∈ A′ ⇒ (N′ [ y := P ]) ∈ A′ ⇒ B′$$
and the definition of substitution (noting $$x ≢ y$$) gives
$$Γ ⊢ (λᵀ y ∈ A′ ⇒ N′) [ x := P ] ∈ A′ ⇒ B′$$ as required.
$$Γ ⊢ (λᵀ x ∈ A′ ⇒ N′) [ y := P ] ∈ A′ ⇒ B′$$ as required.
- If $$N$$ is an application $$L ·ᵀ M$$, the result follows
straightforwardly from the definition of substitution and the
@ -486,24 +464,38 @@ $$Γ \vdash N [ x := P ] ∈ B$$.
- The remaining cases are similar to the application case.
One more technical note: This proof is a rare case where an
induction on terms, rather than typing derivations, yields a
simpler argument. The reason for this is that the assumption
$$Γ , x ↦ A ⊢ N ∈ B$$ is not completely generic, in the
sense that one of the "slots" in the typing relation---namely the
context---is not just a variable, and this means that Agda's
native induction tactic does not give us the induction hypothesis
that we want. It is possible to work around this, but the needed
generalization is a little tricky. The term $$t$$, on the other
hand, _is_ completely generic.
For one case, we need to know that weakening applies to any closed term.
\begin{code}
weaken-closed : ∀ {P A Γ} → ∅ ⊢ P ∈ A → Γ ⊢ P ∈ A
weaken-closed {P} {A} {Γ} ⊢P = weaken g ⊢P
where
g : ∀ {x} → x FreeIn P → ∅ x ≡ Γ x
g {x} x∈P = ⊥-elim (x∉P x∈P)
where
x∉P : ¬ (x FreeIn P)
x∉P = ∅⊢-closed ⊢P {x}
\end{code}
\begin{code}
[:=]-preserves-⊢ {Γ} {x} v∶A (var y y∈Γ) with x ≟ y
... | yes x=y = {!!}
... | no x≠y = {!!}
[:=]-preserves-⊢ {Γ} {y} {A} ⊢P (Ax {_} {x} {B} Γx≡justB) with x ≟ y
...| yes x≡y = {!!} -- weaken-closed ⊢P
...| no x≢y = Ax {_} {x} Γx≡justB
[:=]-preserves-⊢ {Γ} {y} {A} ⊢P (⇒-I {_} {x} ⊢N) with x ≟ y
...| yes x≡y = ⇒-I {_} {x} ⊢N
...| no x≢y = {!!} -- ⇒-I {_} {x} ([:=]-preserves-⊢ {_} {y} {A} ⊢P ⊢N)
[:=]-preserves-⊢ ⊢P (⇒-E ⊢L ⊢M) = ⇒-E ([:=]-preserves-⊢ ⊢P ⊢L) ([:=]-preserves-⊢ ⊢P ⊢M)
[:=]-preserves-⊢ ⊢P 𝔹-I₁ = 𝔹-I₁
[:=]-preserves-⊢ ⊢P 𝔹-I₂ = 𝔹-I₂
[:=]-preserves-⊢ ⊢P (𝔹-E ⊢L ⊢M ⊢N) = 𝔹-E ([:=]-preserves-⊢ ⊢P ⊢L) ([:=]-preserves-⊢ ⊢P ⊢M) ([:=]-preserves-⊢ ⊢P ⊢N)
{-
[:=]-preserves-⊢ {x} ⊢P (Ax {_} {y} Γy≡justB) with x ≟ y
... | yes x=y = ?
... | no x≠y = ?
[:=]-preserves-⊢ {Γ} {x} v∶A (var y y∈Γ) with x ≟ y
... | yes x=y = {!!}
... | no x≠y = {!!}