Properties: Progress and Preservation
module plfa.part2.Properties where
This chapter covers properties of the simplytyped 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 simplytyped lambda calculus, including the notions of closed terms, terms that are values, reducing one term to another, and welltyped 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, welltyped 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, welltyped term.
Preservation: If ∅ ⊢ M ⦂ A
and M —→ N
then ∅ ⊢ N ⦂ A
.
This gives us a recipe for automating evaluation. Start with a closed
and welltyped 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¬—→ Vƛ () V¬—→ Vzero () V¬—→ (Vsuc VM) (ξsuc 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
ξsuc
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
Welltyped 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 Cƛ : ∀ {x A N B} → ∅ , x ⦂ A ⊢ N ⦂ B  → Canonical (ƛ x ⇒ N) ⦂ (A ⇒ B) Czero :  Canonical `zero ⦂ `ℕ Csuc : ∀ {V} → Canonical V ⦂ `ℕ  → Canonical `suc V ⦂ `ℕ
Every closed, welltyped value is canonical:
canonical : ∀ {V A} → ∅ ⊢ V ⦂ A → Value V  → Canonical V ⦂ A canonical (⊢` ()) () canonical (⊢ƛ ⊢N) Vƛ = Cƛ ⊢N canonical (⊢L · ⊢M) () canonical ⊢zero Vzero = Czero canonical (⊢suc ⊢V) (Vsuc VV) = Csuc (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 welltyping of the term guarantees welltyping 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 (Cƛ ⊢N) = Vƛ value Czero = Vzero value (Csuc CM) = Vsuc (value CM) typed : ∀ {M A} → Canonical M ⦂ A  → ∅ ⊢ M ⦂ A typed (Cƛ ⊢N) = ⊢ƛ ⊢N typed Czero = ⊢zero typed (Csuc 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 Vƛ 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 ...  Cƛ _ = step (βƛ VM) progress ⊢zero = done Vzero progress (⊢suc ⊢M) with progress ⊢M ...  step M—→M′ = step (ξsuc M—→M′) ...  done VM = done (Vsuc VM) progress (⊢case ⊢L ⊢M ⊢N) with progress ⊢L ...  step L—→L′ = step (ξcase L—→L′) ...  done VL with canonical ⊢L VL ...  Czero = step βzero ...  Csuc CL = step (βsuc (value CL)) progress (⊢μ ⊢M) = step βμ
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 thatL
is well typed:
If the term steps, we have evidence that
L —→ L′
, which byξ·₁
means that our original term steps toL′ · M

If the term is done, we have evidence that
L
is a value. Recursively apply progress to the derivation thatM
is well typed:
If the term steps, we have evidence that
M —→ M′
, which byξ·₂
means that our original term steps toL · M′
. Stepξ·₂
applies only if we have evidence thatL
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 thatL
is well typed and a value, which since we are in an application leads to the conclusion thatL
must be a lambda abstraction. We also have evidence thatM
is a value, so our original term steps byβƛ
.


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 βƛ
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 welltyped 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 threestep 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 asy
, we usedZ
to access the last variable in the extendedΓ
; and can similarly useZ
to access the last variable in the extendedΔ
. 
If
x
differs fromy
, then we usedS
to skip over the last variable in the extendedΓ
, wherex≢y
is evidence thatx
andy
differ, and∋x
is the evidence thatx
appears inΓ
; and we can similarly useS
to skip over the last variable in the extendedΔ
, applyingρ
to find the evidence thatx
appears inΔ
.
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 inΓ
yields the corresponding evidence that the variable appears inΔ
. 
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 z≢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
andy
are necessarily identical, as areA
andB
. Nonetheless, we must evaluatex ≟ 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
andy
are necessarily distinct. Nonetheless, we must again evaluatex ≟ 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 ⊢ 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
andM
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.
For Agda it makes a difference whether we write x ≟ y
or
y ≟ x
. In an interactive proof, Agda will show which residual with
clauses in the definition of _[_:=_]
need to be simplified, and the
with
clauses in subst
need to match these exactly. The guideline is
that Agda knows nothing about symmetry or commutativity, which require
invoking appropriate lemmas, so it is important to think about order of
arguments and to be consistent.
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) (βƛ VV) = subst ⊢V ⊢N preserve ⊢zero () preserve (⊢suc ⊢M) (ξsuc M—→M′) = ⊢suc (preserve ⊢M M—→M′) preserve (⊢case ⊢L ⊢M ⊢N) (ξcase L—→L′) = ⊢case (preserve ⊢L L—→L′) ⊢M ⊢N preserve (⊢case ⊢zero ⊢M ⊢N) (βzero) = ⊢M preserve (⊢case (⊢suc ⊢V) ⊢M ⊢N) (βsuc VV) = subst ⊢V ⊢N preserve (⊢μ ⊢M) (βμ) = 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 haveL —→ L′  L · M —→ L′ · M
where the lefthand 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 righthand side follows immediately.

Rule
βƛ
. We haveValue V  (ƛ x ⇒ N) · V —→ N [ x := V ]
where the lefthand 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 righthand 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 welltyped term. In this section, we will present an Agda function that computes the reduction sequence from any given closed, welltyped 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 —→⟨ βμ ⟩ `suc sucμ —→⟨ ξsuc βμ ⟩ `suc `suc sucμ —→⟨ ξsuc (ξsuc βμ) ⟩ `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 longrunning or nonterminating 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:
record Gas : Set where constructor gas field amount : ℕ
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 outofgas :  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 ∎) outofgas eval {L} (gas (suc m)) ⊢L with progress ⊢L ...  done VL = steps (L ∎) (done VL) ...  step {M} 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 thatL
is well typed, and an indication that we are out of gas. 
It is nonzero and after the next step we have
m
gas remaining. Apply progress to the evidence that termL
is well typed. There are two possibilities:
Term
L
is a value, so we are done. We return the trivial reduction sequenceL —↠ L
, evidence thatL
is well typed, and the evidence thatL
is a value. 
Term
L
steps to another termM
. Preservation provides evidence thatM
is also well typed, and we recursively invokeeval
on the remaining gas. The result is evidence thatM —↠ N
, together with evidence thatN
is well typed and an indication of whether reduction finished. We combine the evidence thatL —→ M
andM —↠ N
to return evidence thatL —↠ N
, together with the other relevant evidence.

Examples
We can now use Agda to compute the nonterminating 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" —→⟨ βμ ⟩ `suc (μ "x" ⇒ `suc ` "x") —→⟨ ξsuc βμ ⟩ `suc (`suc (μ "x" ⇒ `suc ` "x")) —→⟨ ξsuc (ξsuc βμ) ⟩ `suc (`suc (`suc (μ "x" ⇒ `suc ` "x"))) ∎) outofgas _ = refl
Similarly, we can use Agda to compute the reduction 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 (βƛ Vƛ) ⟩ (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) · `zero —→⟨ βƛ Vzero ⟩ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · `zero) —→⟨ ξ·₂ Vƛ (βƛ Vzero) ⟩ (ƛ "n" ⇒ `suc ` "n") · `suc `zero —→⟨ βƛ (Vsuc Vzero) ⟩ `suc (`suc `zero) ∎) (done (Vsuc (Vsuc Vzero))) _ = refl
The example above was generated by using Cc Cn
to normalise the
lefthand side of the equation and pasting in the result as the
righthand 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 βμ) ⟩ (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 (βƛ (Vsuc (Vsuc Vzero))) ⟩ (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) —→⟨ βƛ (Vsuc (Vsuc Vzero)) ⟩ 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)) ] —→⟨ βsuc (Vsuc Vzero) ⟩ `suc ((μ "+" ⇒ (ƛ "m" ⇒ (ƛ "n" ⇒ case ` "m" [zero⇒ ` "n" suc "m" ⇒ `suc (` "+" · ` "m" · ` "n") ]))) · `suc `zero · `suc (`suc `zero)) —→⟨ ξsuc (ξ·₁ (ξ·₁ βμ)) ⟩ `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)) —→⟨ ξsuc (ξ·₁ (βƛ (Vsuc Vzero))) ⟩ `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)) —→⟨ ξsuc (βƛ (Vsuc (Vsuc Vzero))) ⟩ `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)) ] —→⟨ ξsuc (βsuc Vzero) ⟩ `suc (`suc ((μ "+" ⇒ (ƛ "m" ⇒ (ƛ "n" ⇒ case ` "m" [zero⇒ ` "n" suc "m" ⇒ `suc (` "+" · ` "m" · ` "n") ]))) · `zero · `suc (`suc `zero))) —→⟨ ξsuc (ξsuc (ξ·₁ (ξ·₁ βμ))) ⟩ `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))) —→⟨ ξsuc (ξsuc (ξ·₁ (βƛ Vzero))) ⟩ `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))) —→⟨ ξsuc (ξsuc (βƛ (Vsuc (Vsuc Vzero)))) ⟩ `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)) ]) —→⟨ ξsuc (ξsuc βzero) ⟩ `suc (`suc (`suc (`suc `zero))) ∎) (done (Vsuc (Vsuc (Vsuc (Vsuc Vzero))))) _ = 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 (βƛ Vƛ))) ⟩ (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 (βƛ Vƛ)) ⟩ (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 (βƛ Vƛ) ⟩ (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 —→⟨ βƛ Vzero ⟩ (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 (βƛ Vƛ) ⟩ (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 Vƛ (ξ·₁ (βƛ Vƛ)) ⟩ (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 Vƛ (βƛ Vzero) ⟩ (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) · ((ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · `zero)) —→⟨ ξ·₂ Vƛ (ξ·₂ Vƛ (βƛ Vzero)) ⟩ (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) · ((ƛ "n" ⇒ `suc ` "n") · `suc `zero) —→⟨ ξ·₂ Vƛ (βƛ (Vsuc Vzero)) ⟩ (ƛ "z" ⇒ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · ` "z")) · `suc (`suc `zero) —→⟨ βƛ (Vsuc (Vsuc Vzero)) ⟩ (ƛ "n" ⇒ `suc ` "n") · ((ƛ "n" ⇒ `suc ` "n") · `suc (`suc `zero)) —→⟨ ξ·₂ Vƛ (βƛ (Vsuc (Vsuc Vzero))) ⟩ (ƛ "n" ⇒ `suc ` "n") · `suc (`suc (`suc `zero)) —→⟨ βƛ (Vsuc (Vsuc (Vsuc Vzero))) ⟩ `suc (`suc (`suc (`suc `zero))) ∎) (done (Vsuc (Vsuc (Vsuc (Vsuc Vzero))))) _ = refl
And again, the example in the previous section was derived by editing the above.
Exercise muleval
(recommended)
Using the evaluator, confirm that two times two is four.
 Your code goes here
Exercise: progresspreservation
(practice)
Without peeking at their statements above, write down the progress and preservation theorems for the simply typed lambdacalculus.
 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 counterexamples to subject expansion, one
with case expressions and one not involving case expressions.
 Your code goes here
Welltyped 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 welltyped 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 welltyped term remains well typed:
postulate preserves : ∀ {M N A} → ∅ ⊢ M ⦂ A → M —↠ N  → ∅ ⊢ N ⦂ A
An easy consequence is that starting from a welltyped 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 welltyped 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 welltyped terms don’t go wrong.)
Exercise stuck
(practice)
Give an example of an illtyped 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.
Our proof will need a variant
of congruence to deal with functions of four arguments
(to deal with case_[zero⇒_suc_⇒_]
). 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′) (βƛ _) = ⊥elim (V¬—→ Vƛ 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′) (βƛ VM) = ⊥elim (V¬—→ VM M—→M′) det (βƛ _) (ξ·₁ L—→L″) = ⊥elim (V¬—→ Vƛ L—→L″) det (βƛ VM) (ξ·₂ _ M—→M″) = ⊥elim (V¬—→ VM M—→M″) det (βƛ _) (βƛ _) = refl det (ξsuc M—→M′) (ξsuc M—→M″) = cong `suc_ (det M—→M′ M—→M″) det (ξcase L—→L′) (ξcase L—→L″) = cong₄ case_[zero⇒_suc_⇒_] (det L—→L′ L—→L″) refl refl refl det (ξcase L—→L′) βzero = ⊥elim (V¬—→ Vzero L—→L′) det (ξcase L—→L′) (βsuc VL) = ⊥elim (V¬—→ (Vsuc VL) L—→L′) det βzero (ξcase M—→M″) = ⊥elim (V¬—→ Vzero M—→M″) det βzero βzero = refl det (βsuc VL) (ξcase L—→L″) = ⊥elim (V¬—→ (Vsuc VL) L—→L″) det (βsuc _) (βsuc _) = refl det βμ βμ = 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 congruenceL′ · M ≡ L″ · M
. 
An instance of
ξ·₁
and an instance ofξ·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 requiresL
to be a value. This is a contradiction since values do not reduce. If the value constraint was removed fromξ·₂
, or from one of the other reduction rules, then determinism would no longer hold. 
Two instances of
βƛ
:Value V Value V  βƛ  βƛ (ƛ x ⇒ N) · V —→ N [ x := V ] (ƛ x ⇒ N) · V —→ N [ x := V ]
Since the lefthand sides are identical, the righthand 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
 βzap
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:
 βfoo₁
(λ x ⇒ ` x) —→ foo
 βfoo₂
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
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
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 (\:)