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Remove kth-x lemma #1242

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23 changes: 7 additions & 16 deletions src/plfa/part3/Adequacy.lagda.md
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Expand Up @@ -454,23 +454,12 @@ function, then it terminates via call-by-name. More formally, if
`γ ⊢ M ↓ v` and `𝔾 γ γ'`, then `𝔼 v (clos M γ')`. The proof is by
induction on the derivation of `γ ⊢ M ↓ v` we discuss each case below.

The following lemma, kth-x, is used in the case for the `var` rule.

```agda
kth-x : ∀{Γ}{γ' : ClosEnv Γ}{x : Γ ∋ ★}
→ Σ[ Δ ∈ Context ] Σ[ δ ∈ ClosEnv Δ ] Σ[ M ∈ Δ ⊢ ★ ]
γ' x ≡ clos M δ
kth-x{γ' = γ'}{x = x} with γ' x
... | clos{Γ = Δ} M δ = ⟨ Δ , ⟨ δ , ⟨ M , refl ⟩ ⟩ ⟩
```

```agda
↓→𝔼 : ∀{Γ}{γ : Env Γ}{γ' : ClosEnv Γ}{M : Γ ⊢ ★ }{v}
→ 𝔾 γ γ' → γ ⊢ M ↓ v → 𝔼 v (clos M γ')
↓→𝔼 {Γ} {γ} {γ'} 𝔾γγ' (var{x = x}) fγx
with kth-x{Γ}{γ'}{x} | 𝔾γγ'{x = x}
... | ⟨ Δ , ⟨ δ , ⟨ M' , eq ⟩ ⟩ ⟩ | 𝔾γγ'x rewrite eq
with 𝔾γγ'x fγx
↓→𝔼 {Γ} {γ} {γ'} 𝔾γγ' (var{x = x}) fγx with γ' x in eq | 𝔾γγ'{x = x}
... | clos M' δ | 𝔾γγ'x
with 𝔾γγ'x fγx
... | ⟨ c , ⟨ M'⇓c , 𝕍γx ⟩ ⟩ =
⟨ c , ⟨ (⇓-var eq M'⇓c) , 𝕍γx ⟩ ⟩
↓→𝔼 {Γ} {γ} {γ'} 𝔾γγ' (↦-elim{L = L}{M = M}{v = v₁}{w = v} d₁ d₂) fv
Expand Down Expand Up @@ -522,8 +511,8 @@ kth-x{γ' = γ'}{x = x} with γ' x
and from `𝔾 γ γ'` we have `𝔼 (γ x) (clos M' δ)`. With the premise
`above-fun (γ x)`, we obtain a closure `c` such that `δ ⊢ M' ⇓ c`
and `𝕍 (γ x) c`. To conclude `γ' ⊢ x ⇓ c` via `⇓-var`, we
need `γ' x ≡ clos M' δ`, which is obvious, but it requires some
Agda shananigans via the `kth-x` lemma to get our hands on it.
need `γ' x ≡ clos M' δ`, which is given to us via
[With-abstraction equality](with-abstraction-equality).

* Case `↦-elim`. We have `γ ⊢ L · M ↓ v`.
The induction hypothesis for `γ ⊢ L ↓ v₁ ↦ v`
Expand Down Expand Up @@ -654,3 +643,5 @@ This chapter uses the following unicode:
𝔼 U+1D53C MATHEMATICAL DOUBLE-STRUCK CAPITAL E (\bE)
𝔾 U+1D53E MATHEMATICAL DOUBLE-STRUCK CAPITAL G (\bG)
𝕍 U+1D53E MATHEMATICAL DOUBLE-STRUCK CAPITAL V (\bV)

[with-abstraction-equality]: https://agda.readthedocs.io/en/v2.7.0/language/with-abstraction.html#with-abstraction-equality