Properties: Progress and Preservation

module plfa.part2.Properties where

This chapter covers properties of the simply-typed lambda calculus, as introduced in the previous chapter. The most important of these properties are progress and preservation. We introduce these below, and show how to combine them to get Agda to compute reduction sequences for us.

Imports

open import Relation.Binary.PropositionalEquality
  using (_≡_; _≢_; refl; sym; cong; cong₂)
open import Data.String using (String; _≟_)
open import Data.Nat using (ℕ; zero; suc)
open import Data.Empty using (⊥; ⊥-elim)
open import Data.Product
  using (_×_; proj₁; proj₂; ∃; ∃-syntax)
  renaming (_,_ to ⟨_,_⟩)
open import Data.Sum using (_⊎_; inj₁; inj₂)
open import Relation.Nullary using (¬_; Dec; yes; no)
open import Function using (_∘_)
open import plfa.part1.Isomorphism
open import plfa.part2.Lambda

Introduction

The last chapter introduced simply-typed lambda calculus, including the notions of closed terms, terms that are values, reducing one term to another, and well-typed terms.

Ultimately, we would like to show that we can keep reducing a term until we reach a value. For instance, in the last chapter we showed that two plus two is four,

plus · two · two  —↠  `suc `suc `suc `suc `zero

which was proved by a long chain of reductions, ending in the value on the right. Every term in the chain had the same type, `ℕ. We also saw a second, similar example involving Church numerals.

What we might expect is that every term is either a value or can take a reduction step. As we will see, this property does not hold for every term, but it does hold for every closed, well-typed term.

Progress: If ∅ ⊢ M ⦂ A then either M is a value or there is an N such that M —→ N.

So, either we have a value, and we are done, or we can take a reduction step. In the latter case, we would like to apply progress again. But to do so we need to know that the term yielded by the reduction is itself closed and well typed. It turns out that this property holds whenever we start with a closed, well-typed term.

Preservation: If ∅ ⊢ M ⦂ A and M —→ N then ∅ ⊢ N ⦂ A.

This gives us a recipe for automating evaluation. Start with a closed and well-typed term. By progress, it is either a value, in which case we are done, or it reduces to some other term. By preservation, that other term will itself be closed and well typed. Repeat. We will either loop forever, in which case evaluation does not terminate, or we will eventually reach a value, which is guaranteed to be closed and of the same type as the original term. We will turn this recipe into Agda code that can compute for us the reduction sequence of plus · two · two, and its Church numeral variant.

(The development in this chapter was inspired by the corresponding development in Software Foundations, Volume Programming Language Foundations, Chapter StlcProp. It will turn out that one of our technical choices — to introduce an explicit judgment Γ ∋ x ⦂ A in place of treating a context as a function from identifiers to types — permits a simpler development. In particular, we can prove substitution preserves types without needing to develop a separate inductive definition of the appears_free_in relation.)

Values do not reduce

We start with an easy observation. Values do not reduce:

V¬—→ : ∀ {M N}
  → Value M
    ----------
  → ¬ (M —→ N)
V¬—→ V-ƛ        ()
V¬—→ V-zero     ()
V¬—→ (V-suc VM) (ξ-suc M—→N) = V¬—→ VM M—→N

We consider the three possibilities for values:

If it is an abstraction then no reduction applies
If it is zero then no reduction applies
If it is a successor then rule ξ-suc may apply, but in that case the successor is itself of a value that reduces, which by induction cannot occur.

As a corollary, terms that reduce are not values:

—→¬V : ∀ {M N}
  → M —→ N
    ---------
  → ¬ Value M
—→¬V M—→N VM  =  V¬—→ VM M—→N

If we expand out the negations, we have

V¬—→ : ∀ {M N} → Value M → M —→ N → ⊥
—→¬V : ∀ {M N} → M —→ N → Value M → ⊥

which are the same function with the arguments swapped.

Canonical Forms

Well-typed values must take one of a small number of canonical forms, which provide an analogue of the Value relation that relates values to their types. A lambda expression must have a function type, and a zero or successor expression must be a natural. Further, the body of a function must be well typed in a context containing only its bound variable, and the argument of successor must itself be canonical:

infix  4 Canonical_⦂_

data Canonical_⦂_ : Term → Type → Set where

  C-ƛ : ∀ {x A N B}
    → ∅ , x ⦂ A ⊢ N ⦂ B
      -----------------------------
    → Canonical (ƛ x ⇒ N) ⦂ (A ⇒ B)

  C-zero :
      --------------------
      Canonical `zero ⦂ `ℕ

  C-suc : ∀ {V}
    → Canonical V ⦂ `ℕ
      ---------------------
    → Canonical `suc V ⦂ `ℕ

Every closed, well-typed value is canonical:

canonical : ∀ {V A}
  → ∅ ⊢ V ⦂ A
  → Value V
    -----------
  → Canonical V ⦂ A
canonical (⊢` ())          ()
canonical (⊢ƛ ⊢N)          V-ƛ         =  C-ƛ ⊢N
canonical (⊢L · ⊢M)        ()
canonical ⊢zero            V-zero      =  C-zero
canonical (⊢suc ⊢V)        (V-suc VV)  =  C-suc (canonical ⊢V VV)
canonical (⊢case ⊢L ⊢M ⊢N) ()
canonical (⊢μ ⊢M)          ()

There are only three interesting cases to consider:

If the term is a lambda abstraction, then well-typing of the term guarantees well-typing of the body.
If the term is zero then it is canonical trivially.
If the term is a successor then since it is well typed its argument is well typed, and since it is a value its argument is a value. Hence, by induction its argument is also canonical.

The variable case is thrown out because a closed term has no free variables and because a variable is not a value. The cases for application, case expression, and fixpoint are thrown out because they are not values.

Conversely, if a term is canonical then it is a value and it is well typed in the empty context:

value : ∀ {M A}
  → Canonical M ⦂ A
    ----------------
  → Value M
value (C-ƛ ⊢N)    =  V-ƛ
value C-zero      =  V-zero
value (C-suc CM)  =  V-suc (value CM)

typed : ∀ {M A}
  → Canonical M ⦂ A
    ---------------
  → ∅ ⊢ M ⦂ A
typed (C-ƛ ⊢N)    =  ⊢ƛ ⊢N
typed C-zero      =  ⊢zero
typed (C-suc CM)  =  ⊢suc (typed CM)

The proofs are straightforward, and again use induction in the case of successor.

Progress

We would like to show that every term is either a value or takes a reduction step. However, this is not true in general. The term

`zero · `suc `zero

is neither a value nor can take a reduction step. And if s ⦂ `ℕ ⇒ `ℕ then the term

 s · `zero

cannot reduce because we do not know which function is bound to the free variable s. The first of those terms is ill typed, and the second has a free variable. Every term that is well typed and closed has the desired property.

Progress: If ∅ ⊢ M ⦂ A then either M is a value or there is an N such that M —→ N.

To formulate this property, we first introduce a relation that captures what it means for a term M to make progress:

data Progress (M : Term) : Set where

  step : ∀ {N}
    → M —→ N
      ----------
    → Progress M

  done :
      Value M
      ----------
    → Progress M

A term M makes progress if either it can take a step, meaning there exists a term N such that M —→ N, or if it is done, meaning that M is a value.

If a term is well typed in the empty context then it satisfies progress:

progress : ∀ {M A}
  → ∅ ⊢ M ⦂ A
    ----------
  → Progress M
progress (⊢` ())
progress (⊢ƛ ⊢N)                            =  done V-ƛ
progress (⊢L · ⊢M) with progress ⊢L
... | step L—→L′                            =  step (ξ-·₁ L—→L′)
... | done VL with progress ⊢M
...   | step M—→M′                          =  step (ξ-·₂ VL M—→M′)
...   | done VM with canonical ⊢L VL
...     | C-ƛ _                             =  step (β-ƛ VM)
progress ⊢zero                              =  done V-zero
progress (⊢suc ⊢M) with progress ⊢M
...  | step M—→M′                           =  step (ξ-suc M—→M′)
...  | done VM                              =  done (V-suc VM)
progress (⊢case ⊢L ⊢M ⊢N) with progress ⊢L
... | step L—→L′                            =  step (ξ-case L—→L′)
... | done VL with canonical ⊢L VL
...   | C-zero                              =  step β-zero
...   | C-suc CL                            =  step (β-suc (value CL))
progress (⊢μ ⊢M)                            =  step β-μ

We induct on the evidence that the term is well typed. Let’s unpack the first three cases:

The term cannot be a variable, since no variable is well typed in the empty context.
If the term is a lambda abstraction then it is a value.
If the term is an application L · M, recursively apply progress to the derivation that L is well typed:
- If the term steps, we have evidence that L —→ L′, which by ξ-·₁ means that our original term steps to L′ · M
- If the term is done, we have evidence that L is a value. Recursively apply progress to the derivation that M is well typed:
  - If the term steps, we have evidence that M —→ M′, which by ξ-·₂ means that our original term steps to L · M′. Step ξ-·₂ applies only if we have evidence that L is a value, but progress on that subterm has already supplied the required evidence.
  - If the term is done, we have evidence that M is a value. We apply the canonical forms lemma to the evidence that L is well typed and a value, which since we are in an application leads to the conclusion that L must be a lambda abstraction. We also have evidence that M is a value, so our original term steps by β-ƛ.

The remaining cases are similar. If by induction we have a step case we apply a ξ rule, and if we have a done case then either we have a value or apply a β rule. For fixpoint, no induction is required as the β rule applies immediately.

Our code reads neatly in part because we consider the step option before the done option. We could, of course, do it the other way around, but then the ... abbreviation no longer works, and we will need to write out all the arguments in full. In general, the rule of thumb is to consider the easy case (here step) before the hard case (here done). If you have two hard cases, you will have to expand out ... or introduce subsidiary functions.

Instead of defining a data type for Progress M, we could have formulated progress using disjunction and existentials:

postulate
  progress′ : ∀ M {A} → ∅ ⊢ M ⦂ A → Value M ⊎ ∃[ N ](M —→ N)

This leads to a less perspicuous proof. Instead of the mnemonic done and step we use inj₁ and inj₂, and the term N is no longer implicit and so must be written out in full. In the case for β-ƛ this requires that we match against the lambda expression L to determine its bound variable and body, ƛ x ⇒ N, so we can show that L · M reduces to N [ x := M ].

Exercise `Progress-≃` (practice)

Show that Progress M is isomorphic to Value M ⊎ ∃[ N ](M —→ N).

-- Your code goes here

Exercise `progress′` (practice)

Write out the proof of progress′ in full, and compare it to the proof of progress above.

-- Your code goes here

Exercise `value?` (practice)

Combine progress and —→¬V to write a program that decides whether a well-typed term is a value:

postulate
  value? : ∀ {A M} → ∅ ⊢ M ⦂ A → Dec (Value M)

Prelude to preservation

The other property we wish to prove, preservation of typing under reduction, turns out to require considerably more work. The proof has three key steps.

The first step is to show that types are preserved by renaming.

Renaming: Let Γ and Δ be two contexts such that every variable that appears in Γ also appears with the same type in Δ. Then if any term is typeable under Γ, it has the same type under Δ.

In symbols:

∀ {x A} → Γ ∋ x ⦂ A  →  Δ ∋ x ⦂ A
---------------------------------
∀ {M A} → Γ ⊢ M ⦂ A  →  Δ ⊢ M ⦂ A

Three important corollaries follow. The weaken lemma asserts that a term which is well typed in the empty context is also well typed in an arbitrary context. The drop lemma asserts that a term which is well typed in a context where the same variable appears twice remains well typed if we drop the shadowed occurrence. The swap lemma asserts that a term which is well typed in a context remains well typed if we swap two variables.

(Renaming is similar to the context invariance lemma in Software Foundations, but it does not require the definition of appears_free_in nor the free_in_context lemma.)

The second step is to show that types are preserved by substitution.

Substitution: Say we have a closed term V of type A, and under the assumption that x has type A the term N has type B. Then substituting V for x in N yields a term that also has type B.

In symbols:

∅ ⊢ V ⦂ A
Γ , x ⦂ A ⊢ N ⦂ B
--------------------
Γ ⊢ N [ x := V ] ⦂ B

The result does not depend on V being a value, but it does require that V be closed; recall that we restricted our attention to substitution by closed terms in order to avoid the need to rename bound variables. The term into which we are substituting is typed in an arbitrary context Γ, extended by the variable x for which we are substituting; and the result term is typed in Γ.

The lemma establishes that substitution composes well with typing: typing the components separately guarantees that the result of combining them is also well typed.

The third step is to show preservation.

Preservation: If ∅ ⊢ M ⦂ A and M —→ N then ∅ ⊢ N ⦂ A.

The proof is by induction over the possible reductions, and the substitution lemma is crucial in showing that each of the β rules that uses substitution preserves types.

We now proceed with our three-step programme.

Renaming

We often need to “rebase” a type derivation, replacing a derivation Γ ⊢ M ⦂ A by a related derivation Δ ⊢ M ⦂ A. We may do so as long as every variable that appears in Γ also appears in Δ, and with the same type.

Three of the rules for typing (lambda abstraction, case on naturals, and fixpoint) have hypotheses that extend the context to include a bound variable. In each of these rules, Γ appears in the conclusion and Γ , x ⦂ A appears in a hypothesis. Thus:

Γ , x ⦂ A ⊢ N ⦂ B
------------------- ⊢ƛ
Γ ⊢ ƛ x ⇒ N ⦂ A ⇒ B

for lambda expressions, and similarly for case and fixpoint. To deal with this situation, we first prove a lemma showing that if one context maps to another, this is still true after adding the same variable to both contexts:

ext : ∀ {Γ Δ}
  → (∀ {x A}     →         Γ ∋ x ⦂ A →         Δ ∋ x ⦂ A)
    -----------------------------------------------------
  → (∀ {x y A B} → Γ , y ⦂ B ∋ x ⦂ A → Δ , y ⦂ B ∋ x ⦂ A)
ext ρ Z           =  Z
ext ρ (S x≢y ∋x)  =  S x≢y (ρ ∋x)

Let ρ be the name of the map that takes evidence that x appears in Γ to evidence that x appears in Δ. The proof is by case analysis of the evidence that x appears in the extended map Γ , y ⦂ B:

If x is the same as y, we used Z to access the last variable in the extended Γ; and can similarly use Z to access the last variable in the extended Δ.
If x differs from y, then we used S to skip over the last variable in the extended Γ, where x≢y is evidence that x and y differ, and ∋x is the evidence that x appears in Γ; and we can similarly use S to skip over the last variable in the extended Δ, applying ρ to find the evidence that x appears in Δ.

With the extension lemma under our belts, it is straightforward to prove renaming preserves types:

rename : ∀ {Γ Δ}
  → (∀ {x A} → Γ ∋ x ⦂ A → Δ ∋ x ⦂ A)
    ----------------------------------
  → (∀ {M A} → Γ ⊢ M ⦂ A → Δ ⊢ M ⦂ A)
rename ρ (⊢` ∋w)           =  ⊢` (ρ ∋w)
rename ρ (⊢ƛ ⊢N)           =  ⊢ƛ (rename (ext ρ) ⊢N)
rename ρ (⊢L · ⊢M)         =  (rename ρ ⊢L) · (rename ρ ⊢M)
rename ρ ⊢zero             =  ⊢zero
rename ρ (⊢suc ⊢M)         =  ⊢suc (rename ρ ⊢M)
rename ρ (⊢case ⊢L ⊢M ⊢N)  =  ⊢case (rename ρ ⊢L) (rename ρ ⊢M) (rename (ext ρ) ⊢N)
rename ρ (⊢μ ⊢M)           =  ⊢μ (rename (ext ρ) ⊢M)

As before, let ρ be the name of the map that takes evidence that x appears in Γ to evidence that x appears in Δ. We induct on the evidence that M is well typed in Γ. Let’s unpack the first three cases:

If the term is a variable, then applying ρ to the evidence that the variable appears in Γ yields the corresponding evidence that the variable appears in Δ.
If the term is a lambda abstraction, use the previous lemma to extend the map ρ suitably and use induction to rename the body of the abstraction.
If the term is an application, use induction to rename both the function and the argument.

The remaining cases are similar, using induction for each subterm, and extending the map whenever the construct introduces a bound variable.

The induction is over the derivation that the term is well typed, so extending the context doesn’t invalidate the inductive hypothesis. Equivalently, the recursion terminates because the second argument always grows smaller, even though the first argument sometimes grows larger.

We have three important corollaries, each proved by constructing a suitable map between contexts.

First, a closed term can be weakened to any context:

weaken : ∀ {Γ M A}
  → ∅ ⊢ M ⦂ A
    ----------
  → Γ ⊢ M ⦂ A
weaken {Γ} ⊢M = rename ρ ⊢M
  where
  ρ : ∀ {z C}
    → ∅ ∋ z ⦂ C
      ---------
    → Γ ∋ z ⦂ C
  ρ ()

Here the map ρ is trivial, since there are no possible arguments in the empty context ∅.

Second, if the last two variables in a context are equal then we can drop the shadowed one:

drop : ∀ {Γ x M A B C}
  → Γ , x ⦂ A , x ⦂ B ⊢ M ⦂ C
    --------------------------
  → Γ , x ⦂ B ⊢ M ⦂ C
drop {Γ} {x} {M} {A} {B} {C} ⊢M = rename ρ ⊢M
  where
  ρ : ∀ {z C}
    → Γ , x ⦂ A , x ⦂ B ∋ z ⦂ C
      -------------------------
    → Γ , x ⦂ B ∋ z ⦂ C
  ρ Z                 =  Z
  ρ (S x≢x Z)         =  ⊥-elim (x≢x refl)
  ρ (S z≢x (S _ ∋z))  =  S z≢x ∋z

Here map ρ can never be invoked on the inner occurrence of x since it is masked by the outer occurrence. Skipping over the x in the first position can only happen if the variable looked for differs from x (the evidence for which is x≢x or z≢x) but if the variable is found in the second position, which also contains x, this leads to a contradiction (evidenced by x≢x refl).

Third, if the last two variables in a context differ then we can swap them:

swap : ∀ {Γ x y M A B C}
  → x ≢ y
  → Γ , y ⦂ B , x ⦂ A ⊢ M ⦂ C
    --------------------------
  → Γ , x ⦂ A , y ⦂ B ⊢ M ⦂ C
swap {Γ} {x} {y} {M} {A} {B} {C} x≢y ⊢M = rename ρ ⊢M
  where
  ρ : ∀ {z C}
    → Γ , y ⦂ B , x ⦂ A ∋ z ⦂ C
      --------------------------
    → Γ , x ⦂ A , y ⦂ B ∋ z ⦂ C
  ρ Z                   =  S x≢y Z
  ρ (S y≢x Z)           =  Z
  ρ (S z≢x (S z≢y ∋z))  =  S z≢y (S z≢x ∋z)

Here the renaming map takes a variable at the end into a variable one from the end, and vice versa. The first line is responsible for moving x from a position at the end to a position one from the end with y at the end, and requires the provided evidence that x ≢ y.

Substitution

The key to preservation – and the trickiest bit of the proof – is the lemma establishing that substitution preserves types.

Recall that in order to avoid renaming bound variables, substitution is restricted to be by closed terms only. This restriction was not enforced by our definition of substitution, but it is captured by our lemma to assert that substitution preserves typing.

Our concern is with reducing closed terms, which means that when we apply β reduction, the term substituted in contains a single free variable (the bound variable of the lambda abstraction, or similarly for case or fixpoint). However, substitution is defined by recursion, and as we descend into terms with bound variables the context grows. So for the induction to go through, we require an arbitrary context Γ, as in the statement of the lemma.

Here is the formal statement and proof that substitution preserves types:

subst : ∀ {Γ x N V A B}
  → ∅ ⊢ V ⦂ A
  → Γ , x ⦂ A ⊢ N ⦂ B
    --------------------
  → Γ ⊢ N [ x := V ] ⦂ B
subst {x = y} ⊢V (⊢` {x = x} Z) with x ≟ y
... | yes _           =  weaken ⊢V
... | no  x≢y         =  ⊥-elim (x≢y refl)
subst {x = y} ⊢V (⊢` {x = x} (S x≢y ∋x)) with x ≟ y
... | yes refl        =  ⊥-elim (x≢y refl)
... | no  _           =  ⊢` ∋x
subst {x = y} ⊢V (⊢ƛ {x = x} ⊢N) with x ≟ y
... | yes refl        =  ⊢ƛ (drop ⊢N)
... | no  x≢y         =  ⊢ƛ (subst ⊢V (swap x≢y ⊢N))
subst ⊢V (⊢L · ⊢M)    =  (subst ⊢V ⊢L) · (subst ⊢V ⊢M)
subst ⊢V ⊢zero        =  ⊢zero
subst ⊢V (⊢suc ⊢M)    =  ⊢suc (subst ⊢V ⊢M)
subst {x = y} ⊢V (⊢case {x = x} ⊢L ⊢M ⊢N) with x ≟ y
... | yes refl        =  ⊢case (subst ⊢V ⊢L) (subst ⊢V ⊢M) (drop ⊢N)
... | no  x≢y         =  ⊢case (subst ⊢V ⊢L) (subst ⊢V ⊢M) (subst ⊢V (swap x≢y ⊢N))
subst {x = y} ⊢V (⊢μ {x = x} ⊢M) with x ≟ y
... | yes refl        =  ⊢μ (drop ⊢M)
... | no  x≢y         =  ⊢μ (subst ⊢V (swap x≢y ⊢M))

We induct on the evidence that N is well typed in the context Γ extended by x.

First, we note a wee issue with naming. In the lemma statement, the variable x is an implicit parameter for the variable substituted, while in the type rules for variables, abstractions, cases, and fixpoints, the variable x is an implicit parameter for the relevant variable. We are going to need to get hold of both variables, so we use the syntax {x = y} to bind y to the substituted variable and the syntax {x = x} to bind x to the relevant variable in the patterns for ⊢`, ⊢ƛ, ⊢case, and ⊢μ. Using the name y here is consistent with the naming in the original definition of substitution in the previous chapter. The proof never mentions the types of x, y, V, or N, so in what follows we choose type names as convenient.

Now that naming is resolved, let’s unpack the first three cases:

In the variable case, we must show
```
∅ ⊢ V ⦂ B
Γ , y ⦂ B ⊢ ` x ⦂ A
------------------------
Γ ⊢ ` x [ y := V ] ⦂ A
```
where the second hypothesis follows from:
```
Γ , y ⦂ B ∋ x ⦂ A
```
There are two subcases, depending on the evidence for this judgment:
- The lookup judgment is evidenced by rule Z:
```
----------------
Γ , x ⦂ A ∋ x ⦂ A
```
  In this case, x and y are necessarily identical, as are A and B. Nonetheless, we must evaluate x ≟ y in order to allow the definition of substitution to simplify:
  - If the variables are equal, then after simplification we must show
    ∅ ⊢ V ⦂ A --------- Γ ⊢ V ⦂ A
    which follows by weakening.
  - If the variables are unequal we have a contradiction.
- The lookup judgment is evidenced by rule S:
```
x ≢ y
Γ ∋ x ⦂ A
-----------------
Γ , y ⦂ B ∋ x ⦂ A
```
  In this case, x and y are necessarily distinct. Nonetheless, we must again evaluate x ≟ y in order to allow the definition of substitution to simplify:
  - If the variables are equal we have a contradiction.
  - If the variables are unequal, then after simplification we must show
    ∅ ⊢ V ⦂ B x ≢ y Γ ∋ x ⦂ A ------------- Γ ⊢ ` x ⦂ A
    which follows by the typing rule for variables.

In the abstraction case, we must show

∅ ⊢ V ⦂ B
Γ , y ⦂ B ⊢ (ƛ x ⇒ N) ⦂ A ⇒ C
--------------------------------
Γ ⊢ (ƛ x ⇒ N) [ y := V ] ⦂ A ⇒ C

where the second hypothesis follows from

Γ , y ⦂ B , x ⦂ A ⊢ N ⦂ C

We evaluate x ≟ y in order to allow the definition of substitution to simplify:

If the variables are equal then after simplification we must show:

∅ ⊢ V ⦂ B
Γ , x ⦂ B , x ⦂ A ⊢ N ⦂ C
-------------------------
Γ ⊢ ƛ x ⇒ N ⦂ A ⇒ C

From the drop lemma, drop, we may conclude:

Γ , x ⦂ B , x ⦂ A ⊢ N ⦂ C
-------------------------
Γ , x ⦂ A ⊢ N ⦂ C

The typing rule for abstractions then yields the required conclusion.

If the variables are distinct then after simplification we must show:

∅ ⊢ V ⦂ B
Γ , y ⦂ B , x ⦂ A ⊢ N ⦂ C
--------------------------------
Γ ⊢ ƛ x ⇒ (N [ y := V ]) ⦂ A ⇒ C

From the swap lemma we may conclude:

Γ , y ⦂ B , x ⦂ A ⊢ N ⦂ C
-------------------------
Γ , x ⦂ A , y ⦂ B ⊢ N ⦂ C

The inductive hypothesis gives us:

∅ ⊢ V ⦂ B
Γ , x ⦂ A , y ⦂ B ⊢ N ⦂ C
------------------------------------
Γ , x ⦂ A , y ⦂ B ⊢ N [ y := V ] ⦂ C

The typing rule for abstractions then yields the required conclusion.

In the application case, we must show

∅ ⊢ V ⦂ C
Γ , y ⦂ C ⊢ L · M ⦂ B
--------------------------
Γ ⊢ (L · M) [ y := V ] ⦂ B

where the second hypothesis follows from the two judgments

Γ , y ⦂ C ⊢ L ⦂ A ⇒ B
Γ , y ⦂ C ⊢ M ⦂ A

By the definition of substitution, we must show:

∅ ⊢ V ⦂ C
Γ , y ⦂ C ⊢ L ⦂ A ⇒ B
Γ , y ⦂ C ⊢ M ⦂ A
---------------------------------------
Γ ⊢ (L [ y := V ]) · (M [ y := V ]) ⦂ B

Applying the induction hypothesis for L and M and the typing rule for applications yields the required conclusion.

The remaining cases are similar, using induction for each subterm. Where the construct introduces a bound variable we need to compare it with the substituted variable, applying the drop lemma if they are equal and the swap lemma if they are distinct.

Exercise `subst′` (stretch)

Rewrite subst to work with the modified definition _[_:=_]′ from the exercise in the previous chapter. As before, this should factor dealing with bound variables into a single function, defined by mutual recursion with the proof that substitution preserves types.

-- Your code goes here

Preservation

Once we have shown that substitution preserves types, showing that reduction preserves types is straightforward:

preserve : ∀ {M N A}
  → ∅ ⊢ M ⦂ A
  → M —→ N
    ----------
  → ∅ ⊢ N ⦂ A
preserve (⊢` ())
preserve (⊢ƛ ⊢N)                 ()
preserve (⊢L · ⊢M)               (ξ-·₁ L—→L′)     =  (preserve ⊢L L—→L′) · ⊢M
preserve (⊢L · ⊢M)               (ξ-·₂ VL M—→M′)  =  ⊢L · (preserve ⊢M M—→M′)
preserve ((⊢ƛ ⊢N) · ⊢V)          (β-ƛ VV)         =  subst ⊢V ⊢N
preserve ⊢zero                   ()
preserve (⊢suc ⊢M)               (ξ-suc M—→M′)    =  ⊢suc (preserve ⊢M M—→M′)
preserve (⊢case ⊢L ⊢M ⊢N)        (ξ-case L—→L′)   =  ⊢case (preserve ⊢L L—→L′) ⊢M ⊢N
preserve (⊢case ⊢zero ⊢M ⊢N)     β-zero           =  ⊢M
preserve (⊢case (⊢suc ⊢V) ⊢M ⊢N) (β-suc VV)       =  subst ⊢V ⊢N
preserve (⊢μ ⊢M)                 (β-μ)            =  subst (⊢μ ⊢M) ⊢M

The proof never mentions the types of M or N, so in what follows we choose type name as convenient.

Let’s unpack the cases for two of the reduction rules:

Rule ξ-·₁. We have

L —→ L′
----------------
L · M —→ L′ · M

where the left-hand side is typed by

Γ ⊢ L ⦂ A ⇒ B
Γ ⊢ M ⦂ A
-------------
Γ ⊢ L · M ⦂ B

By induction, we have

Γ ⊢ L ⦂ A ⇒ B
L —→ L′
--------------
Γ ⊢ L′ ⦂ A ⇒ B

from which the typing of the right-hand side follows immediately.

Rule β-ƛ. We have

Value V
----------------------------
(ƛ x ⇒ N) · V ⊢ N [ x := V ]

where the left-hand side is typed by

Γ , x ⦂ A ⊢ N ⦂ B
-------------------
Γ ⊢ ƛ x ⇒ N ⦂ A ⇒ B    Γ ⊢ V ⦂ A
--------------------------------
Γ ⊢ (ƛ x ⇒ N) · V ⦂ B

By the substitution lemma, we have

Γ ⊢ V ⦂ A
Γ , x ⦂ A ⊢ N ⦂ B
--------------------
Γ ⊢ N [ x := V ] ⦂ B

from which the typing of the right-hand side follows immediately.

The remaining cases are similar. Each ξ rule follows by induction, and each β rule follows by the substitution lemma.

Evaluation

By repeated application of progress and preservation, we can evaluate any well-typed term. In this section, we will present an Agda function that computes the reduction sequence from any given closed, well-typed term to its value, if it has one.

Some terms may reduce forever. Here is a simple example:

sucμ  =  μ "x" ⇒ `suc (` "x")

_ =
  begin
    sucμ
  —→⟨ β-μ ⟩
    `suc sucμ
  —→⟨ ξ-suc β-μ ⟩
    `suc `suc sucμ
  —→⟨ ξ-suc (ξ-suc β-μ) ⟩
    `suc `suc `suc sucμ
  --  ...
  ∎

Since every Agda computation must terminate, we cannot simply ask Agda to reduce a term to a value. Instead, we will provide a natural number to Agda, and permit it to stop short of a value if the term requires more than the given number of reduction steps.

A similar issue arises with cryptocurrencies. Systems which use smart contracts require the miners that maintain the blockchain to evaluate the program which embodies the contract. For instance, validating a transaction on Ethereum may require executing a program for the Ethereum Virtual Machine (EVM). A long-running or non-terminating program might cause the miner to invest arbitrary effort in validating a contract for little or no return. To avoid this situation, each transaction is accompanied by an amount of gas available for computation. Each step executed on the EVM is charged an advertised amount of gas, and the transaction pays for the gas at a published rate: a given number of Ethers (the currency of Ethereum) per unit of gas.

By analogy, we will use the name gas for the parameter which puts a bound on the number of reduction steps. Gas is specified by a natural number:

data Gas : Set where
  gas : ℕ → Gas

When our evaluator returns a term N, it will either give evidence that N is a value or indicate that it ran out of gas:

data Finished (N : Term) : Set where

  done :
      Value N
      ----------
    → Finished N

  out-of-gas :
      ----------
      Finished N

Given a term L of type A, the evaluator will, for some N, return a reduction sequence from L to N and an indication of whether reduction finished:

data Steps (L : Term) : Set where

  steps : ∀ {N}
    → L —↠ N
    → Finished N
      ----------
    → Steps L

The evaluator takes gas and evidence that a term is well typed, and returns the corresponding steps:

eval : ∀ {L A}
  → Gas
  → ∅ ⊢ L ⦂ A
    ---------
  → Steps L
eval {L} (gas zero)    ⊢L                             =  steps (L ∎) out-of-gas
eval {L} (gas (suc m)) ⊢L with progress ⊢L
... | done VL                                         =  steps (L ∎) (done VL)
... | step L—→M with eval (gas m) (preserve ⊢L L—→M)
...    | steps M—↠N fin                               =  steps (L —→⟨ L—→M ⟩ M—↠N) fin

Let L be the name of the term we are reducing, and ⊢L be the evidence that L is well typed. We consider the amount of gas remaining. There are two possibilities:

It is zero, so we stop early. We return the trivial reduction sequence L —↠ L, evidence that L is well typed, and an indication that we are out of gas.
It is non-zero and after the next step we have m gas remaining. Apply progress to the evidence that term L is well typed. There are two possibilities:
- Term L is a value, so we are done. We return the trivial reduction sequence L —↠ L, evidence that L is well-typed, and the evidence that L is a value.
- Term L steps to another term M. Preservation provides evidence that M is also well typed, and we recursively invoke eval on the remaining gas. The result is evidence that M —↠ N, together with evidence that N is well typed and an indication of whether reduction finished. We combine the evidence that L —→ M and M —↠ N to return evidence that L —↠ N, together with the other relevant evidence.

Examples

We can now use Agda to compute the non-terminating reduction sequence given earlier. First, we show that the term sucμ is well typed:

⊢sucμ : ∅ ⊢ μ "x" ⇒ `suc ` "x" ⦂ `ℕ
⊢sucμ = ⊢μ (⊢suc (⊢` ∋x))
  where
  ∋x = Z

To show the first three steps of the infinite reduction sequence, we evaluate with three steps worth of gas:

_ : eval (gas 3) ⊢sucμ ≡
  steps
   (μ "x" ⇒ `suc ` "x"
   —→⟨ β-μ ⟩
    `suc (μ "x" ⇒ `suc ` "x")
   —→⟨ ξ-suc β-μ ⟩
    `suc (`suc (μ "x" ⇒ `suc ` "x"))
   —→⟨ ξ-suc (ξ-suc β-μ) ⟩
    `suc (`suc (`suc (μ "x" ⇒ `suc ` "x")))
   ∎)
   out-of-gas
_ = refl

Similarly, we can use Agda to compute the reductions sequences given in the previous chapter. We start with the Church numeral two applied to successor and zero. Supplying 100 steps of gas is more than enough:

_ : eval (gas 100) (⊢twoᶜ · ⊢sucᶜ · ⊢zero) ≡
  steps
   ((ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · (ƛ "n" ⇒ `suc ` "n")
   · `zero
   —→⟨ ξ-·₁ (β-ƛ V-ƛ) ⟩
    (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) ·
     `zero
   —→⟨ β-ƛ V-zero ⟩
    (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · `zero)
   —→⟨ ξ-·₂ V-ƛ (β-ƛ V-zero) ⟩
    (ƛ "n" ⇒ `suc ` "n") · `suc `zero
   —→⟨ β-ƛ (V-suc V-zero) ⟩
    `suc (`suc `zero)
   ∎)
   (done (V-suc (V-suc V-zero)))
_ = refl

The example above was generated by using C-c C-n to normalise the left-hand side of the equation and pasting in the result as the right-hand side of the equation. The example reduction of the previous chapter was derived from this result, reformatting and writing twoᶜ and sucᶜ in place of their expansions.

Next, we show two plus two is four:

_ : eval (gas 100) ⊢2+2 ≡
  steps
   ((μ "+" ⇒
     (ƛ "m" ⇒
      (ƛ "n" ⇒
       case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
       ])))
    · `suc (`suc `zero)
    · `suc (`suc `zero)
   —→⟨ ξ-·₁ (ξ-·₁ β-μ) ⟩
    (ƛ "m" ⇒
     (ƛ "n" ⇒
      case ` "m" [zero⇒ ` "n" |suc "m" ⇒
      `suc
      ((μ "+" ⇒
        (ƛ "m" ⇒
         (ƛ "n" ⇒
          case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
          ])))
       · ` "m"
       · ` "n")
      ]))
    · `suc (`suc `zero)
    · `suc (`suc `zero)
   —→⟨ ξ-·₁ (β-ƛ (V-suc (V-suc V-zero))) ⟩
    (ƛ "n" ⇒
     case `suc (`suc `zero) [zero⇒ ` "n" |suc "m" ⇒
     `suc
     ((μ "+" ⇒
       (ƛ "m" ⇒
        (ƛ "n" ⇒
         case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
         ])))
      · ` "m"
      · ` "n")
     ])
    · `suc (`suc `zero)
   —→⟨ β-ƛ (V-suc (V-suc V-zero)) ⟩
    case `suc (`suc `zero) [zero⇒ `suc (`suc `zero) |suc "m" ⇒
    `suc
    ((μ "+" ⇒
      (ƛ "m" ⇒
       (ƛ "n" ⇒
        case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
        ])))
     · ` "m"
     · `suc (`suc `zero))
    ]
   —→⟨ β-suc (V-suc V-zero) ⟩
    `suc
    ((μ "+" ⇒
      (ƛ "m" ⇒
       (ƛ "n" ⇒
        case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
        ])))
     · `suc `zero
     · `suc (`suc `zero))
   —→⟨ ξ-suc (ξ-·₁ (ξ-·₁ β-μ)) ⟩
    `suc
    ((ƛ "m" ⇒
      (ƛ "n" ⇒
       case ` "m" [zero⇒ ` "n" |suc "m" ⇒
       `suc
       ((μ "+" ⇒
         (ƛ "m" ⇒
          (ƛ "n" ⇒
           case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
           ])))
        · ` "m"
        · ` "n")
       ]))
     · `suc `zero
     · `suc (`suc `zero))
   —→⟨ ξ-suc (ξ-·₁ (β-ƛ (V-suc V-zero))) ⟩
    `suc
    ((ƛ "n" ⇒
      case `suc `zero [zero⇒ ` "n" |suc "m" ⇒
      `suc
      ((μ "+" ⇒
        (ƛ "m" ⇒
         (ƛ "n" ⇒
          case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
          ])))
       · ` "m"
       · ` "n")
      ])
     · `suc (`suc `zero))
   —→⟨ ξ-suc (β-ƛ (V-suc (V-suc V-zero))) ⟩
    `suc
    case `suc `zero [zero⇒ `suc (`suc `zero) |suc "m" ⇒
    `suc
    ((μ "+" ⇒
      (ƛ "m" ⇒
       (ƛ "n" ⇒
        case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
        ])))
     · ` "m"
     · `suc (`suc `zero))
    ]
   —→⟨ ξ-suc (β-suc V-zero) ⟩
    `suc
    (`suc
     ((μ "+" ⇒
       (ƛ "m" ⇒
        (ƛ "n" ⇒
         case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
         ])))
      · `zero
      · `suc (`suc `zero)))
   —→⟨ ξ-suc (ξ-suc (ξ-·₁ (ξ-·₁ β-μ))) ⟩
    `suc
    (`suc
     ((ƛ "m" ⇒
       (ƛ "n" ⇒
        case ` "m" [zero⇒ ` "n" |suc "m" ⇒
        `suc
        ((μ "+" ⇒
          (ƛ "m" ⇒
           (ƛ "n" ⇒
            case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
            ])))
         · ` "m"
         · ` "n")
        ]))
      · `zero
      · `suc (`suc `zero)))
   —→⟨ ξ-suc (ξ-suc (ξ-·₁ (β-ƛ V-zero))) ⟩
    `suc
    (`suc
     ((ƛ "n" ⇒
       case `zero [zero⇒ ` "n" |suc "m" ⇒
       `suc
       ((μ "+" ⇒
         (ƛ "m" ⇒
          (ƛ "n" ⇒
           case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
           ])))
        · ` "m"
        · ` "n")
       ])
      · `suc (`suc `zero)))
   —→⟨ ξ-suc (ξ-suc (β-ƛ (V-suc (V-suc V-zero)))) ⟩
    `suc
    (`suc
     case `zero [zero⇒ `suc (`suc `zero) |suc "m" ⇒
     `suc
     ((μ "+" ⇒
       (ƛ "m" ⇒
        (ƛ "n" ⇒
         case ` "m" [zero⇒ ` "n" |suc "m" ⇒ `suc (` "+" · ` "m" · ` "n")
         ])))
      · ` "m"
      · `suc (`suc `zero))
     ])
   —→⟨ ξ-suc (ξ-suc β-zero) ⟩
    `suc (`suc (`suc (`suc `zero)))
   ∎)
   (done (V-suc (V-suc (V-suc (V-suc V-zero)))))
_ = refl

Again, the derivation in the previous chapter was derived by editing the above.

Similarly, we can evaluate the corresponding term for Church numerals:

_ : eval (gas 100) ⊢2+2ᶜ ≡
  steps
   ((ƛ "m" ⇒
     (ƛ "n" ⇒
      (ƛ "s" ⇒ (ƛ "z" ⇒ ` "m" · ` "s" · (` "n" · ` "s" · ` "z")))))
    · (ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z")))
    · (ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z")))
    · (ƛ "n" ⇒ `suc ` "n")
    · `zero
   —→⟨ ξ-·₁ (ξ-·₁ (ξ-·₁ (β-ƛ V-ƛ))) ⟩
    (ƛ "n" ⇒
     (ƛ "s" ⇒
      (ƛ "z" ⇒
       (ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · ` "s" ·
       (` "n" · ` "s" · ` "z"))))
    · (ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z")))
    · (ƛ "n" ⇒ `suc ` "n")
    · `zero
   —→⟨ ξ-·₁ (ξ-·₁ (β-ƛ V-ƛ)) ⟩
    (ƛ "s" ⇒
     (ƛ "z" ⇒
      (ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · ` "s" ·
      ((ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · ` "s" · ` "z")))
    · (ƛ "n" ⇒ `suc ` "n")
    · `zero
   —→⟨ ξ-·₁ (β-ƛ V-ƛ) ⟩
    (ƛ "z" ⇒
     (ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · (ƛ "n" ⇒ `suc ` "n")
     ·
     ((ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · (ƛ "n" ⇒ `suc ` "n")
      · ` "z"))
    · `zero
   —→⟨ β-ƛ V-zero ⟩
    (ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · (ƛ "n" ⇒ `suc ` "n")
    ·
    ((ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · (ƛ "n" ⇒ `suc ` "n")
     · `zero)
   —→⟨ ξ-·₁ (β-ƛ V-ƛ) ⟩
    (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) ·
    ((ƛ "s" ⇒ (ƛ "z" ⇒ ` "s" · (` "s" · ` "z"))) · (ƛ "n" ⇒ `suc ` "n")
     · `zero)
   —→⟨ ξ-·₂ V-ƛ (ξ-·₁ (β-ƛ V-ƛ)) ⟩
    (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) ·
    ((ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) ·
     `zero)
   —→⟨ ξ-·₂ V-ƛ (β-ƛ V-zero) ⟩
    (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) ·
    ((ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · `zero))
   —→⟨ ξ-·₂ V-ƛ (ξ-·₂ V-ƛ (β-ƛ V-zero)) ⟩
    (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) ·
    ((ƛ "n" ⇒ `suc ` "n") · `suc `zero)
   —→⟨ ξ-·₂ V-ƛ (β-ƛ (V-suc V-zero)) ⟩
    (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) ·
    `suc (`suc `zero)
   —→⟨ β-ƛ (V-suc (V-suc V-zero)) ⟩
    (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · `suc (`suc `zero))
   —→⟨ ξ-·₂ V-ƛ (β-ƛ (V-suc (V-suc V-zero))) ⟩
    (ƛ "n" ⇒ `suc ` "n") · `suc (`suc (`suc `zero))
   —→⟨ β-ƛ (V-suc (V-suc (V-suc V-zero))) ⟩
    `suc (`suc (`suc (`suc `zero)))
   ∎)
   (done (V-suc (V-suc (V-suc (V-suc V-zero)))))
_ = refl

And again, the example in the previous section was derived by editing the above.

Exercise `mul-eval` (recommended)

Using the evaluator, confirm that two times two is four.

-- Your code goes here

Exercise: `progress-preservation` (practice)

Without peeking at their statements above, write down the progress and preservation theorems for the simply typed lambda-calculus.

-- Your code goes here

Exercise `subject_expansion` (practice)

We say that M reduces to N if M —→ N, but we can also describe the same situation by saying that N expands to M. The preservation property is sometimes called subject reduction. Its opposite is subject expansion, which holds if M —→ N and ∅ ⊢ N ⦂ A imply ∅ ⊢ M ⦂ A. Find two counter-examples to subject expansion, one with case expressions and one not involving case expressions.

-- Your code goes here

Well-typed terms don’t get stuck

A term is normal if it cannot reduce:

Normal : Term → Set
Normal M  =  ∀ {N} → ¬ (M —→ N)

A term is stuck if it is normal yet not a value:

Stuck : Term → Set
Stuck M  =  Normal M × ¬ Value M

Using progress, it is easy to show that no well-typed term is stuck:

postulate
  unstuck : ∀ {M A}
    → ∅ ⊢ M ⦂ A
      -----------
    → ¬ (Stuck M)

Using preservation, it is easy to show that after any number of steps, a well-typed term remains well typed:

postulate
  preserves : ∀ {M N A}
    → ∅ ⊢ M ⦂ A
    → M —↠ N
      ---------
    → ∅ ⊢ N ⦂ A

An easy consequence is that starting from a well-typed term, taking any number of reduction steps leads to a term that is not stuck:

postulate
  wttdgs : ∀ {M N A}
    → ∅ ⊢ M ⦂ A
    → M —↠ N
      -----------
    → ¬ (Stuck N)

Felleisen and Wright, who introduced proofs via progress and preservation, summarised this result with the slogan well-typed terms don’t get stuck. (They were referring to earlier work by Robin Milner, who used denotational rather than operational semantics. He introduced wrong as the denotation of a term with a type error, and showed well-typed terms don’t go wrong.)

Exercise `stuck` (practice)

Give an example of an ill-typed term that does get stuck.

-- Your code goes here

Exercise `unstuck` (recommended)

Provide proofs of the three postulates, unstuck, preserves, and wttdgs above.

-- Your code goes here

Reduction is deterministic

When we introduced reduction, we claimed it was deterministic. For completeness, we present a formal proof here.

A case term takes four arguments (three subterms and a bound variable), and our proof will need a variant of congruence to deal with functions of four arguments. It is exactly analogous to cong and cong₂ as defined previously:

cong₄ : ∀ {A B C D E : Set} (f : A → B → C → D → E)
  {s w : A} {t x : B} {u y : C} {v z : D}
  → s ≡ w → t ≡ x → u ≡ y → v ≡ z → f s t u v ≡ f w x y z
cong₄ f refl refl refl refl = refl

It is now straightforward to show that reduction is deterministic:

det : ∀ {M M′ M″}
  → (M —→ M′)
  → (M —→ M″)
    --------
  → M′ ≡ M″
det (ξ-·₁ L—→L′)   (ξ-·₁ L—→L″)     =  cong₂ _·_ (det L—→L′ L—→L″) refl
det (ξ-·₁ L—→L′)   (ξ-·₂ VL M—→M″)  =  ⊥-elim (V¬—→ VL L—→L′)
det (ξ-·₁ L—→L′)   (β-ƛ _)          =  ⊥-elim (V¬—→ V-ƛ L—→L′)
det (ξ-·₂ VL _)    (ξ-·₁ L—→L″)     =  ⊥-elim (V¬—→ VL L—→L″)
det (ξ-·₂ _ M—→M′) (ξ-·₂ _ M—→M″)   =  cong₂ _·_ refl (det M—→M′ M—→M″)
det (ξ-·₂ _ M—→M′) (β-ƛ VM)         =  ⊥-elim (V¬—→ VM M—→M′)
det (β-ƛ _)        (ξ-·₁ L—→L″)     =  ⊥-elim (V¬—→ V-ƛ L—→L″)
det (β-ƛ VM)       (ξ-·₂ _ M—→M″)   =  ⊥-elim (V¬—→ VM M—→M″)
det (β-ƛ _)        (β-ƛ _)          =  refl
det (ξ-suc M—→M′)  (ξ-suc M—→M″)    =  cong `suc_ (det M—→M′ M—→M″)
det (ξ-case L—→L′) (ξ-case L—→L″)   =  cong₄ case_[zero⇒_|suc_⇒_]
                                         (det L—→L′ L—→L″) refl refl refl
det (ξ-case L—→L′) β-zero           =  ⊥-elim (V¬—→ V-zero L—→L′)
det (ξ-case L—→L′) (β-suc VL)       =  ⊥-elim (V¬—→ (V-suc VL) L—→L′)
det β-zero         (ξ-case M—→M″)   =  ⊥-elim (V¬—→ V-zero M—→M″)
det β-zero         β-zero           =  refl
det (β-suc VL)     (ξ-case L—→L″)   =  ⊥-elim (V¬—→ (V-suc VL) L—→L″)
det (β-suc _)      (β-suc _)        =  refl
det β-μ            β-μ              =  refl

The proof is by induction over possible reductions. We consider three typical cases:

Two instances of ξ-·₁:

L —→ L′                 L —→ L″
--------------- ξ-·₁    --------------- ξ-·₁
L · M —→ L′ · M         L · M —→ L″ · M

By induction we have L′ ≡ L″, and hence by congruence L′ · M ≡ L″ · M.

An instance of ξ-·₁ and an instance of ξ-·₂:
```
                        Value L
L —→ L′                 M —→ M″
--------------- ξ-·₁    --------------- ξ-·₂
L · M —→ L′ · M         L · M —→ L · M″
```
The rule on the left requires L to reduce, but the rule on the right requires L to be a value. This is a contradiction since values do not reduce. If the value constraint was removed from ξ-·₂, or from one of the other reduction rules, then determinism would no longer hold.

Two instances of β-ƛ:

Value V                              Value V
----------------------------- β-ƛ    ----------------------------- β-ƛ
(ƛ x ⇒ N) · V —→ N [ x := V ]        (ƛ x ⇒ N) · V —→ N [ x := V ]

Since the left-hand sides are identical, the right-hand sides are also identical. The formal proof simply invokes refl.

Five of the 18 lines in the above proof are redundant, e.g., the case when one rule is ξ-·₁ and the other is ξ-·₂ is considered twice, once with ξ-·₁ first and ξ-·₂ second, and the other time with the two swapped. What we might like to do is delete the redundant lines and add

det M—→M′ M—→M″ = sym (det M—→M″ M—→M′)

to the bottom of the proof. But this does not work: the termination checker complains, because the arguments have merely switched order and neither is smaller.

Quiz

Suppose we add a new term zap with the following reduction rule

-------- β-zap
M —→ zap

and the following typing rule:

----------- ⊢zap
Γ ⊢ zap ⦂ A

Which of the following properties remain true in the presence of these rules? For each property, write either “remains true” or “becomes false.” If a property becomes false, give a counterexample:

Determinism of step
Progress
Preservation

Quiz

Suppose instead that we add a new term foo with the following reduction rules:

------------------ β-foo₁
(λ x ⇒ ` x) —→ foo

----------- β-foo₂
foo —→ zero

Which of the following properties remain true in the presence of this rule? For each one, write either “remains true” or else “becomes false.” If a property becomes false, give a counterexample:

Determinism of step
Progress
Preservation

Quiz

Suppose instead that we remove the rule ξ·₁ from the step relation. Which of the following properties remain true in the absence of this rule? For each one, write either “remains true” or else “becomes false.” If a property becomes false, give a counterexample:

Determinism of step
Progress
Preservation

Quiz

We can enumerate all the computable function from naturals to naturals, by writing out all programs of type `ℕ ⇒ `ℕ in lexical order. Write fᵢ for the i‘th function in this list.

Say we add a typing rule that applies the above enumeration to interpret a natural as a function from naturals to naturals:

Γ ⊢ L ⦂ `ℕ
Γ ⊢ M ⦂ `ℕ
-------------- _·ℕ_
Γ ⊢ L · M ⦂ `ℕ

And that we add the corresponding reduction rule:

fᵢ(m) —→ n
---------- δ
i · m —→ n

Determinism of step
Progress
Preservation

Are all properties preserved in this case? Are there any other alterations we would wish to make to the system?

Unicode

This chapter uses the following unicode:

ƛ  U+019B  LATIN SMALL LETTER LAMBDA WITH STROKE (\Gl-)
Δ  U+0394  GREEK CAPITAL LETTER DELTA (\GD or \Delta)
β  U+03B2  GREEK SMALL LETTER BETA (\Gb or \beta)
δ  U+03B4  GREEK SMALL LETTER DELTA (\Gd or \delta)
μ  U+03BC  GREEK SMALL LETTER MU (\Gm or \mu)
ξ  U+03BE  GREEK SMALL LETTER XI (\Gx or \xi)
ρ  U+03B4  GREEK SMALL LETTER RHO (\Gr or \rho)
ᵢ  U+1D62  LATIN SUBSCRIPT SMALL LETTER I (\_i)
ᶜ  U+1D9C  MODIFIER LETTER SMALL C (\^c)
–  U+2013  EM DASH (\em)
₄  U+2084  SUBSCRIPT FOUR (\_4)
↠  U+21A0  RIGHTWARDS TWO HEADED ARROW (\rr-)
⇒  U+21D2  RIGHTWARDS DOUBLE ARROW (\=>)
∅  U+2205  EMPTY SET (\0)
∋  U+220B  CONTAINS AS MEMBER (\ni)
≟  U+225F  QUESTIONED EQUAL TO (\?=)
⊢  U+22A2  RIGHT TACK (\vdash or \|-)
⦂  U+2982  Z NOTATION TYPE COLON (\:)

Imports

Introduction

Values do not reduce

Canonical Forms

Progress

Exercise Progress-≃ (practice)

Exercise progress′ (practice)

Exercise value? (practice)

Prelude to preservation

Renaming

Substitution

Exercise subst′ (stretch)

Preservation

Evaluation

Examples

Exercise mul-eval (recommended)

Exercise: progress-preservation (practice)

Exercise subject_expansion (practice)

Well-typed terms don’t get stuck

Exercise stuck (practice)

Exercise unstuck (recommended)

Reduction is deterministic

Quiz

Quiz

Quiz

Quiz

Unicode

Exercise `Progress-≃` (practice)

Exercise `progress′` (practice)

Exercise `value?` (practice)

Exercise `subst′` (stretch)

Exercise `mul-eval` (recommended)

Exercise: `progress-preservation` (practice)

Exercise `subject_expansion` (practice)

Exercise `stuck` (practice)

Exercise `unstuck` (recommended)