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[ re #473 ] Takahashi's complete development (#493)

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@ -401,14 +401,92 @@ sub-par : ∀{Γ A B} {N N′ : Γ , A ⊢ B} {M M′ : Γ ⊢ A}
sub-par pn pm = subst-par (par-subst-zero pm) pn sub-par pn pm = subst-par (par-subst-zero pm) pn
``` ```
## Parallel reduction satisfies the diamond property ## Parallel reduction satisfies the diamond property
The heart of the confluence proof is made of stone, or rather, of The heart of the confluence proof is made of stone, or rather, of
diamond! We show that parallel reduction satisfies the diamond diamond! We show that parallel reduction satisfies the diamond
property: that if `M ⇛ N` and `M ⇛ N′`, then `N ⇛ L` and `N′ ⇛ L` for property: that if `M ⇛ N` and `M ⇛ N′`, then `N ⇛ L` and `N′ ⇛ L` for
some `L`. The proof is relatively easy; it is parallel reduction's some `L`. The typical proof is an induction on `M ⇛ N` and `M ⇛ N′`
_raison d'etre_. so that every possible pair gives rise to a witeness `L` given by
performing enough beta reductions in parallel.
However, a simpler approach is to perform as many beta reductions in
parallel as possible on `M`, say `M ⁺`, and then show that `N` also
parallel reduces to `M ⁺`. This is the idea of Takahashi's _complete
development_. The desired property may be illustrated as
M
/|
/ |
/ |
N 2
\ |
\ |
\|
M⁺
where downward lines are instances of `⇛`, so we call it the _triangle
property_.
```
_⁺ : ∀ {Γ A}
→ Γ ⊢ A → Γ ⊢ A
(` x) ⁺ = ` x
(ƛ M) ⁺ = ƛ (M ⁺)
((ƛ N) · M) ⁺ = N ⁺ [ M ⁺ ]
(L · M) ⁺ = L ⁺ · (M ⁺)
par-triangle : ∀ {Γ A} {M N : Γ ⊢ A}
→ M ⇛ N
-------
→ N ⇛ M ⁺
par-triangle pvar = pvar
par-triangle (pabs p) = pabs (par-triangle p)
par-triangle (pbeta p1 p2) = sub-par (par-triangle p1) (par-triangle p2)
par-triangle (papp {L = ƛ _ } (pabs p1) p2) =
pbeta (par-triangle p1) (par-triangle p2)
par-triangle (papp {L = ` _} p1 p2) = papp (par-triangle p1) (par-triangle p2)
par-triangle (papp {L = _ · _} p1 p2) = papp (par-triangle p1) (par-triangle p2)
```
The proof of the triangle property is an induction on `M ⇛ N`.
* Suppose `x ⇛ x`. Clearly `x ⁺ = x`, so `x ⇛ x`.
* Suppose `ƛ M ⇛ ƛ N`. By the induction hypothesis we have `N ⇛ M ⁺`
and by definition `(λ M) ⁺ = λ (M ⁺)`, so we conclude that `λ N ⇛ λ
(M ⁺)`.
* Suppose `(λ N) · M ⇛ N′ [ M′ ]`. By the induction hypothesis, we have
`N′ ⇛ N ⁺` and `M′ ⇛ M ⁺`. Since substitution respects parallel reduction,
it follows that `N′ [ M′ ] ⇛ N ⁺ [ M ⁺ ]`, but the right hand side
is exactly `((λ N) · M) ⁺`, hence `N′ [ M′ ] ⇛ ((λ N) · M) ⁺`.
* Suppose `(λ L) · M ⇛ (λ L′) · M′`. By the induction hypothesis
we have `L′ ⇛ L ⁺` and `M′ ⇛ M ⁺`; by definition `((λ L) · M) ⁺ = L ⁺ [ M ⁺ ]`.
It follows `(λ L′) · M′ ⇛ L ⁺ [ M ⁺ ]`.
* Suppose `x · M ⇛ x · M′`. By the induction hypothesis we have `M′ ⇛ M ⁺`
and `x ⇛ x ⁺` so that `x · M′ ⇛ x · M ⁺`.
The remaining case is proved in the same way, so we ignore it. (As
there is currently no way in Agda to expand the catch-all pattern in
the definition of `_⁺` for us before checking the right-hand side,
we have to write down the remaining case explicitly.)
The diamond property then follows by halving the diamond into two triangles.
M
/|\
/ | \
/ | \
N 2 N′
\ | /
\ | /
\|/
M⁺
That is, the diamond property is proved by applying the
triangle property on each side with the same confluent term `M ⁺`.
``` ```
par-diamond : ∀{Γ A} {M N N′ : Γ ⊢ A} par-diamond : ∀{Γ A} {M N N′ : Γ ⊢ A}
@ -416,90 +494,24 @@ par-diamond : ∀{Γ A} {M N N′ : Γ ⊢ A}
→ M ⇛ N′ → M ⇛ N′
--------------------------------- ---------------------------------
→ Σ[ L ∈ Γ ⊢ A ] (N ⇛ L) × (N′ ⇛ L) → Σ[ L ∈ Γ ⊢ A ] (N ⇛ L) × (N′ ⇛ L)
par-diamond (pvar{x = x}) pvar = ⟨ ` x , ⟨ pvar , pvar ⟩ ⟩ par-diamond {M = M} p1 p2 = ⟨ M ⁺ , ⟨ par-triangle p1 , par-triangle p2 ⟩ ⟩
par-diamond (pabs p1) (pabs p2)
with par-diamond p1 p2
... | ⟨ L′ , ⟨ p3 , p4 ⟩ ⟩ =
⟨ ƛ L′ , ⟨ pabs p3 , pabs p4 ⟩ ⟩
par-diamond{Γ}{A}{L · M}{N}{N′} (papp{Γ}{L}{L₁}{M}{M₁} p1 p3)
(papp{Γ}{L}{L₂}{M}{M₂} p2 p4)
with par-diamond p1 p2
... | ⟨ L₃ , ⟨ p5 , p6 ⟩ ⟩
with par-diamond p3 p4
... | ⟨ M₃ , ⟨ p7 , p8 ⟩ ⟩ =
⟨ (L₃ · M₃) , ⟨ (papp p5 p7) , (papp p6 p8) ⟩ ⟩
par-diamond (papp (pabs p1) p3) (pbeta p2 p4)
with par-diamond p1 p2
... | ⟨ N₃ , ⟨ p5 , p6 ⟩ ⟩
with par-diamond p3 p4
... | ⟨ M₃ , ⟨ p7 , p8 ⟩ ⟩ =
⟨ N₃ [ M₃ ] , ⟨ pbeta p5 p7 , sub-par p6 p8 ⟩ ⟩
par-diamond (pbeta p1 p3) (papp (pabs p2) p4)
with par-diamond p1 p2
... | ⟨ N₃ , ⟨ p5 , p6 ⟩ ⟩
with par-diamond p3 p4
... | ⟨ M₃ , ⟨ p7 , p8 ⟩ ⟩ =
⟨ (N₃ [ M₃ ]) , ⟨ sub-par p5 p7 , pbeta p6 p8 ⟩ ⟩
par-diamond {Γ}{A} (pbeta p1 p3) (pbeta p2 p4)
with par-diamond p1 p2
... | ⟨ N₃ , ⟨ p5 , p6 ⟩ ⟩
with par-diamond p3 p4
... | ⟨ M₃ , ⟨ p7 , p8 ⟩ ⟩ =
⟨ N₃ [ M₃ ] , ⟨ sub-par p5 p7 , sub-par p6 p8 ⟩ ⟩
``` ```
The proof is by induction on both premises. This step is optional, though, in the presence of triangle property.
* Suppose `x ⇛ x` and `x ⇛ x`.
We choose `L = x` and immediately have `x ⇛ x` and `x ⇛ x`.
* Suppose `ƛ N ⇛ ƛ N₁` and `ƛ N ⇛ ƛ N₂`.
By the induction hypothesis, there exists `L′` such that
`N₁ ⇛ L′` and `N₂ ⇛ L′`. We choose `L = ƛ L′` and
by `pabs` conclude that `ƛ N₁ ⇛ ƛ L′` and `ƛ N₂ ⇛ ƛ L′.
* Suppose that `L · M ⇛ L₁ · M₁` and `L · M ⇛ L₂ · M₂`.
By the induction hypothesis we have
`L₁ ⇛ L₃` and `L₂ ⇛ L₃` for some `L₃`.
Likewise, we have
`M₁ ⇛ M₃` and `M₂ ⇛ M₃` for some `M₃`.
We choose `L = L₃ · M₃` and conclude with two uses of `papp`.
* Suppose that `(ƛ N) · M ⇛ (ƛ N₁) · M₁` and `(ƛ N) · M ⇛ N₂ [ M₂ ]`
By the induction hypothesis we have
`N₁ ⇛ N₃` and `N₂ ⇛ N₃` for some `N₃`.
Likewise, we have
`M₁ ⇛ M₃` and `M₂ ⇛ M₃` for some `M₃`.
We choose `L = N₃ [ M₃ ]`.
We have `(ƛ N₁) · M₁ ⇛ N₃ [ M₃ ]` by rule `pbeta`
and conclude that `N₂ [ M₂ ] ⇛ N₃ [ M₃ ]` because
substitution respects parallel reduction.
* Suppose that `(ƛ N) · M ⇛ N₁ [ M₁ ]` and `(ƛ N) · M ⇛ (ƛ N₂) · M₂`.
The proof of this case is the mirror image of the last one.
* Suppose that `(ƛ N) · M ⇛ N₁ [ M₁ ]` and `(ƛ N) · M ⇛ N₂ [ M₂ ]`.
By the induction hypothesis we have
`N₁ ⇛ N₃` and `N₂ ⇛ N₃` for some `N₃`.
Likewise, we have
`M₁ ⇛ M₃` and `M₂ ⇛ M₃` for some `M₃`.
We choose `L = N₃ [ M₃ ]`.
We have both `(ƛ N₁) · M₁ ⇛ N₃ [ M₃ ]`
and `(ƛ N₂) · M₂ ⇛ N₃ [ M₃ ]`
by rule `pbeta`
#### Exercise (practice) #### Exercise (practice)
Draw pictures that represent the proofs of each of the six cases in * Prove the diamond property `par-diamond` directly by induction on `M ⇛ N` and `M ⇛ N′`.
the above proof of `par-diamond`. The pictures should consist of nodes
and directed edges, where each node is labeled with a term and each
edge represents parallel reduction.
* Draw pictures that represent the proofs of each of the six cases in
the direct proof of `par-diamond`. The pictures should consist of nodes
and directed edges, where each node is labeled with a term and each
edge represents parallel reduction.
## Proof of confluence for parallel reduction ## Proof of confluence for parallel reduction
As promised at the beginning, the proof that parallel reduction is As promised at the beginning, the proof that parallel reduction is
confluent is easy now that we know it satisfies the diamond property. confluent is easy now that we know it satisfies the triangle property.
We just need to prove the strip lemma, which states that We just need to prove the strip lemma, which states that
if `M ⇛ N` and `M ⇛* N′`, then if `M ⇛ N` and `M ⇛* N′`, then
`N ⇛* L` and `N′ ⇛ L` for some `L`. `N ⇛* L` and `N′ ⇛ L` for some `L`.
@ -519,7 +531,7 @@ where downward lines are instances of `⇛` or `⇛*`, depending on how
they are marked. they are marked.
The proof of the strip lemma is a straightforward induction on `M ⇛* N′`, The proof of the strip lemma is a straightforward induction on `M ⇛* N′`,
using the diamond property in the induction step. using the triangle property in the induction step.
``` ```
strip : ∀{Γ A} {M N N′ : Γ ⊢ A} strip : ∀{Γ A} {M N N′ : Γ ⊢ A}
@ -529,11 +541,8 @@ strip : ∀{Γ A} {M N N′ : Γ ⊢ A}
→ Σ[ L ∈ Γ ⊢ A ] (N ⇛* L) × (N′ ⇛ L) → Σ[ L ∈ Γ ⊢ A ] (N ⇛* L) × (N′ ⇛ L)
strip{Γ}{A}{M}{N}{N′} mn (M ∎) = ⟨ N , ⟨ N ∎ , mn ⟩ ⟩ strip{Γ}{A}{M}{N}{N′} mn (M ∎) = ⟨ N , ⟨ N ∎ , mn ⟩ ⟩
strip{Γ}{A}{M}{N}{N′} mn (M ⇛⟨ mm' ⟩ m'n') strip{Γ}{A}{M}{N}{N′} mn (M ⇛⟨ mm' ⟩ m'n')
with par-diamond mn mm' with strip (par-triangle mm') m'n'
... | ⟨ L , ⟨ nl , m'l ⟩ ⟩ ... | ⟨ L , ⟨ ll' , n'l' ⟩ ⟩ = ⟨ L , ⟨ N ⇛⟨ par-triangle mn ⟩ ll' , n'l' ⟩ ⟩
with strip m'l m'n'
... | ⟨ L′ , ⟨ ll' , n'l' ⟩ ⟩ =
⟨ L′ , ⟨ (N ⇛⟨ nl ⟩ ll') , n'l' ⟩ ⟩
``` ```
The proof of confluence for parallel reduction is now proved by The proof of confluence for parallel reduction is now proved by
@ -605,19 +614,16 @@ Broadly speaking, this proof of confluence, based on parallel
reduction, is due to W. Tait and P. Martin-Lof (see Barendredgt 1984, reduction, is due to W. Tait and P. Martin-Lof (see Barendredgt 1984,
Section 3.2). Details of the mechanization come from several sources. Section 3.2). Details of the mechanization come from several sources.
The `subst-par` lemma is the "strong substitutivity" lemma of Shafer, The `subst-par` lemma is the "strong substitutivity" lemma of Shafer,
Tebbi, and Smolka (ITP 2015). The proofs of `par-diamond`, `strip`, Tebbi, and Smolka (ITP 2015). The proofs of `par-triangle`, `strip`,
and `par-confluence` are based on Pfenning's 1992 technical report and `par-confluence` are based on the notion of complete development
about the Church-Rosser theorem. In addition, we consulted Nipkow and by Takahashi (1995) and Pfenning's 1992 technical report about the
Church-Rosser theorem. In addition, we consulted Nipkow and
Berghofer's mechanization in Isabelle, which is based on an earlier Berghofer's mechanization in Isabelle, which is based on an earlier
article by Nipkow (JAR 1996). We opted not to use the "complete article by Nipkow (JAR 1996).
developments" approach of Takahashi (1995) because we felt that the
proof was simple enough based solely on parallel reduction. There are
many more mechanizations of the Church-Rosser theorem that we have not
yet had the time to read, including Shankar's (J. ACM 1988) and
Homeier's (TPHOLs 2001).
## Unicode ## Unicode
This chapter uses the following unicode: This chapter uses the following unicode:
⇛ U+3015 RIGHTWARDS TRIPLE ARROW (\r== or \Rrightarrow) ⇛ U+21DB RIGHTWARDS TRIPLE ARROW (\r== or \Rrightarrow)
⁺ U+207A SUPERSCRIPT PLUS SIGN (\^+)