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src/Maps.lagda
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@ -86,8 +86,6 @@ y = id "y"
z = id "z"
\end{code}
## Extensionality
## Total Maps
Our main job in this chapter will be to build a definition of
@ -112,8 +110,8 @@ TotalMap : Set → Set
TotalMap A = Id → A
\end{code}
Intuitively, a total map over anfi element type $$A$$ _is_ just a
function that can be used to look up ids, yielding $$A$$s.
Intuitively, a total map over anfi element type `A` _is_ just a
function that can be used to look up ids, yielding `A`s.
\begin{code}
module TotalMap where
@ -129,10 +127,12 @@ applied to any id.
\end{code}
More interesting is the update function, which (as before) takes
a map $$ρ$$, a key $$x$$, and a value $$v$$ and returns a new map that
takes $$x$$ to $$v$$ and takes every other key to whatever $$ρ$$ does.
a map `ρ`, a key `x`, and a value `v` and returns a new map that
takes `x` to `v` and takes every other key to whatever `ρ` does.
\begin{code}
infixl 100 _,_↦_
_,_↦_ : ∀ {A} → TotalMap A → Id → A → TotalMap A
(ρ , x ↦ v) y with x ≟ y
... | yes x=y = v
@ -140,30 +140,11 @@ takes $$x$$ to $$v$$ and takes every other key to whatever $$ρ$$ does.
\end{code}
This definition is a nice example of higher-order programming.
The update function takes a _function_ $$ρ$$ and yields a new
The update function takes a _function_ `ρ` and yields a new
function that behaves like the desired map.
We define handy abbreviations for updating a map two, three, or four times.
<div class="note hidden">
Wen: you don't actually need to define these, you can simply declare `_,_↦_` to
be a left-associative infix operator with an `infixl` statement, and then you'll
be able to just evaluate `M , x ↦ y , z ↦ w` as `(M , x ↦ y) , z ↦ w`.
</div>
\begin{code}
_,_↦_,_↦_ : ∀ {A} → TotalMap A → Id → A → Id → A → TotalMap A
ρ , x₁ ↦ v₁ , x₂ ↦ v₂ = (ρ , x₁ ↦ v₁), x₂ ↦ v₂
_,_↦_,_↦_,_↦_ : ∀ {A} → TotalMap A → Id → A → Id → A → Id → A → TotalMap A
ρ , x₁ ↦ v₁ , x₂ ↦ v₂ , x₃ ↦ v₃ = ((ρ , x₁ ↦ v₁), x₂ ↦ v₂), x₃ ↦ v₃
_,_↦_,_↦_,_↦_,_↦_ : ∀ {A} → TotalMap A → Id → A → Id → A → Id → A → Id → A → TotalMap A
ρ , x₁ ↦ v₁ , x₂ ↦ v₂ , x₃ ↦ v₃ , x₄ ↦ v₄ = (((ρ , x₁ ↦ v₁), x₂ ↦ v₂), x₃ ↦ v₃), x₄ ↦ v₄
\end{code}
For example, we can build a map taking ids to naturals, where $$x$$
maps to 42, $$y$$ maps to 69, and every other key maps to 0, as follows:
For example, we can build a map taking ids to naturals, where `x`
maps to 42, `y` maps to 69, and every other key maps to 0, as follows:
\begin{code}
ρ₀ : TotalMap ℕ
@ -206,9 +187,9 @@ The `always` map returns its default element for all keys:
</div>
#### Exercise: 2 stars, optional (update-eq)
Next, if we update a map $$ρ$$ at a key $$x$$ with a new value $$v$$
and then look up $$x$$ in the map resulting from the update, we get
back $$v$$:
Next, if we update a map `ρ` at a key `x` with a new value `v`
and then look up `x` in the map resulting from the update, we get
back `v`:
\begin{code}
postulate
@ -227,9 +208,9 @@ back $$v$$:
</div>
#### Exercise: 2 stars, optional (update-neq)
On the other hand, if we update a map $$m$$ at a key $$x$$ and
then look up a _different_ key $$y$$ in the resulting map, we get
the same result that $$m$$ would have given:
On the other hand, if we update a map `m` at a key `x` and
then look up a _different_ key `y` in the resulting map, we get
the same result that `m` would have given:
\begin{code}
update-neq : ∀ {A} (ρ : TotalMap A) (x : Id) (v : A) (y : Id)
@ -248,11 +229,11 @@ show two maps equal we will need to postulate extensionality.
\end{code}
#### Exercise: 2 stars, optional (update-shadow)
If we update a map $$ρ$$ at a key $$x$$ with a value $$v$$ and then
update again with the same key $$x$$ and another value $$w$$, the
If we update a map `ρ` at a key `x` with a value `v` and then
update again with the same key `x` and another value `w`, the
resulting map behaves the same (gives the same result when applied
to any key) as the simpler map obtained by performing just
the second update on $$ρ$$:
the second update on `ρ`:
\begin{code}
postulate
@ -274,9 +255,9 @@ the second update on $$ρ$$:
</div>
#### Exercise: 2 stars (update-same)
Prove the following theorem, which states that if we update a map $$ρ$$ to
assign key $$x$$ the same value as it already has in $$ρ$$, then the
result is equal to $$ρ$$:
Prove the following theorem, which states that if we update a map `ρ` to
assign key `x` the same value as it already has in `ρ`, then the
result is equal to `ρ`:
\begin{code}
postulate
@ -297,7 +278,7 @@ result is equal to $$ρ$$:
#### Exercise: 3 stars, recommended (update-permute)
Prove one final property of the `update` function: If we update a map
$$m$$ at two distinct keys, it doesn't matter in which order we do the
`m` at two distinct keys, it doesn't matter in which order we do the
updates.
\begin{code}
@ -354,21 +335,11 @@ module PartialMap where
\end{code}
\begin{code}
infixl 100 _,_↦_
_,_↦_ : ∀ {A} (ρ : PartialMap A) (x : Id) (v : A) → PartialMap A
ρ , x ↦ v = TotalMap._,_↦_ ρ x (just v)
\end{code}
As before, we define handy abbreviations for updating a map two, three, or four times.
\begin{code}
_,_↦_,_↦_ : ∀ {A} → PartialMap A → Id → A → Id → A → PartialMap A
ρ , x₁ ↦ v₁ , x₂ ↦ v₂ = (ρ , x₁ ↦ v₁), x₂ ↦ v₂
_,_↦_,_↦_,_↦_ : ∀ {A} → PartialMap A → Id → A → Id → A → Id → A → PartialMap A
ρ , x₁ ↦ v₁ , x₂ ↦ v₂ , x₃ ↦ v₃ = ((ρ , x₁ ↦ v₁), x₂ ↦ v₂), x₃ ↦ v₃
_,_↦_,_↦_,_↦_,_↦_ : ∀ {A} → PartialMap A → Id → A → Id → A → Id → A → Id → A → PartialMap A
ρ , x₁ ↦ v₁ , x₂ ↦ v₂ , x₃ ↦ v₃ , x₄ ↦ v₄ = (((ρ , x₁ ↦ v₁), x₂ ↦ v₂), x₃ ↦ v₃), x₄ ↦ v₄
\end{code}
We now lift all of the basic lemmas about total maps to partial maps.

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@ -168,6 +168,8 @@ example₁ = ⟨ step₀ ⟩ >> ⟨ step₁ ⟩ >> ⟨ step₂ ⟩ >> ⟨ step
Context : Set
Context = PartialMap Type
infix 50 _⊢_∈_
data _⊢_∈_ : Context → Term → Type → Set where
Ax : ∀ {Γ x A} →
Γ x ≡ just A →

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@ -28,8 +28,8 @@ theorem.
As we saw for the simple calculus in the [Stlc]({{ "Stlc" | relative_url }})
chapter, the first step in establishing basic properties of reduction and types
is to identify the possible _canonical forms_ (i.e., well-typed closed values)
belonging to each type. For $$bool$$, these are the boolean values $$true$$ and
$$false$$. For arrow types, the canonical forms are lambda-abstractions.
belonging to each type. For `bool`, these are the boolean values `true` and
`false`. For arrow types, the canonical forms are lambda-abstractions.
\begin{code}
data canonical_for_ : Term → Type → Set where
@ -63,43 +63,43 @@ first, then the formal version.
progress : ∀ {M A} → ∅ ⊢ M ∈ A → value M ⊎ ∃ λ N → M ⟹ N
\end{code}
_Proof_: By induction on the derivation of $$\vdash t : A$$.
_Proof_: By induction on the derivation of `\vdash t : A`.
- The last rule of the derivation cannot be `var`,
since a variable is never well typed in an empty context.
- The `true`, `false`, and `abs` cases are trivial, since in
each of these cases we can see by inspecting the rule that $$t$$
each of these cases we can see by inspecting the rule that `t`
is a value.
- If the last rule of the derivation is `app`, then $$t$$ has the
form $$t_1\;t_2$$ for som e$$t_1$$ and $$t_2$$, where we know that
$$t_1$$ and $$t_2$$ are also well typed in the empty context; in particular,
there exists a type $$B$$ such that $$\vdash t_1 : A\to T$$ and
$$\vdash t_2 : B$$. By the induction hypothesis, either $$t_1$$ is a
- If the last rule of the derivation is `app`, then `t` has the
form `t_1\;t_2` for som e`t_1` and `t_2`, where we know that
`t_1` and `t_2` are also well typed in the empty context; in particular,
there exists a type `B` such that `\vdash t_1 : A\to T` and
`\vdash t_2 : B`. By the induction hypothesis, either `t_1` is a
value or it can take a reduction step.
- If $$t_1$$ is a value, then consider $$t_2$$, which by the other
- If `t_1` is a value, then consider `t_2`, which by the other
induction hypothesis must also either be a value or take a step.
- Suppose $$t_2$$ is a value. Since $$t_1$$ is a value with an
arrow type, it must be a lambda abstraction; hence $$t_1\;t_2$$
- Suppose `t_2` is a value. Since `t_1` is a value with an
arrow type, it must be a lambda abstraction; hence `t_1\;t_2`
can take a step by `red`.
- Otherwise, $$t_2$$ can take a step, and hence so can $$t_1\;t_2$$
- Otherwise, `t_2` can take a step, and hence so can `t_1\;t_2`
by `app2`.
- If $$t_1$$ can take a step, then so can $$t_1 t_2$$ by `app1`.
- If `t_1` can take a step, then so can `t_1 t_2` by `app1`.
- If the last rule of the derivation is `if`, then $$t = \text{if }t_1
\text{ then }t_2\text{ else }t_3$$, where $$t_1$$ has type $$bool$$. By
the IH, $$t_1$$ either is a value or takes a step.
- If the last rule of the derivation is `if`, then `t = \text{if }t_1
\text{ then }t_2\text{ else }t_3`, where `t_1` has type `bool`. By
the IH, `t_1` either is a value or takes a step.
- If $$t_1$$ is a value, then since it has type $$bool$$ it must be
either $$true$$ or $$false$$. If it is $$true$$, then $$t$$ steps
to $$t_2$$; otherwise it steps to $$t_3$$.
- If `t_1` is a value, then since it has type `bool` it must be
either `true` or `false`. If it is `true`, then `t` steps
to `t_2`; otherwise it steps to `t_3`.
- Otherwise, $$t_1$$ takes a step, and therefore so does $$t$$ (by `if`).
- Otherwise, `t_1` takes a step, and therefore so does `t` (by `if`).
\begin{code}
progress (Ax ())
@ -141,23 +141,23 @@ interesting proofs), the story goes like this:
- The _preservation theorem_ is proved by induction on a typing
derivation, pretty much as we did in the [Stlc]({{ "Stlc" | relative_url }})
chapter. The one case that is significantly different is the one for the
$$red$$ rule, whose definition uses the substitution operation. To see that
`red` rule, whose definition uses the substitution operation. To see that
this step preserves typing, we need to know that the substitution itself
does. So we prove a...
- _substitution lemma_, stating that substituting a (closed)
term $$s$$ for a variable $$x$$ in a term $$t$$ preserves the type
of $$t$$. The proof goes by induction on the form of $$t$$ and
term `s` for a variable `x` in a term `t` preserves the type
of `t`. The proof goes by induction on the form of `t` and
requires looking at all the different cases in the definition
of substitition. This time, the tricky cases are the ones for
variables and for function abstractions. In both cases, we
discover that we need to take a term $$s$$ that has been shown
to be well-typed in some context $$\Gamma$$ and consider the same
term $$s$$ in a slightly different context $$\Gamma'$$. For this
discover that we need to take a term `s` that has been shown
to be well-typed in some context `\Gamma` and consider the same
term `s` in a slightly different context `\Gamma'`. For this
we prove a...
- _context invariance_ lemma, showing that typing is preserved
under "inessential changes" to the context $$\Gamma$$---in
under "inessential changes" to the context `\Gamma`---in
particular, changes that do not affect any of the free
variables of the term. And finally, for this, we need a
careful definition of...
@ -172,13 +172,13 @@ order...
### Free Occurrences
A variable $$x$$ _appears free in_ a term $$M$$ if $$M$$ contains some
occurrence of $$x$$ that is not under an abstraction over $$x$$.
A variable `x` _appears free in_ a term `M` if `M` contains some
occurrence of `x` that is not under an abstraction over `x`.
For example:
- $$y$$ appears free, but $$x$$ does not, in $$λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y$$
- both $$x$$ and $$y$$ appear free in $$(λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y) ·ᵀ x$$
- no variables appear free in $$λᵀ x ∈ (A ⇒ B) ⇒ (λᵀ y ∈ A ⇒ x ·ᵀ y)$$
- `y` appears free, but `x` does not, in `λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y`
- both `x` and `y` appear free in `(λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y) ·ᵀ x`
- no variables appear free in `λᵀ x ∈ (A ⇒ B) ⇒ (λᵀ y ∈ A ⇒ x ·ᵀ y)`
Formally:
@ -211,42 +211,42 @@ are really the crux of the lambda-calculus.)
### Substitution
To prove that substitution preserves typing, we first need a
technical lemma connecting free variables and typing contexts: If
a variable $$x$$ appears free in a term $$M$$, and if we know $$M$$ is
well typed in context $$Γ$$, then it must be the case that
$$Γ$$ assigns a type to $$x$$.
a variable `x` appears free in a term `M`, and if we know `M` is
well typed in context `Γ`, then it must be the case that
`Γ` assigns a type to `x`.
\begin{code}
freeLemma : ∀ {x M A Γ} → x FreeIn M → Γ ⊢ M ∈ A → ∃ λ B → Γ x ≡ just B
\end{code}
_Proof_: We show, by induction on the proof that $$x$$ appears
free in $$P$$, that, for all contexts $$Γ$$, if $$P$$ is well
typed under $$Γ$$, then $$Γ$$ assigns some type to $$x$$.
_Proof_: We show, by induction on the proof that `x` appears
free in `P`, that, for all contexts `Γ`, if `P` is well
typed under `Γ`, then `Γ` assigns some type to `x`.
- If the last rule used was `free-varᵀ`, then $$P = x$$, and from
the assumption that $$M$$ is well typed under $$Γ$$ we have
immediately that $$Γ$$ assigns a type to $$x$$.
- If the last rule used was `free-varᵀ`, then `P = x`, and from
the assumption that `M` is well typed under `Γ` we have
immediately that `Γ` assigns a type to `x`.
- If the last rule used was `free-·₁`, then $$P = L ·ᵀ M$$ and $$x$$
appears free in $$L$$. Since $$L$$ is well typed under $$\Gamma$$,
we can see from the typing rules that $$L$$ must also be, and
the IH then tells us that $$Γ$$ assigns $$x$$ a type.
- If the last rule used was `free-·₁`, then `P = L ·ᵀ M` and `x`
appears free in `L`. Since `L` is well typed under `\Gamma`,
we can see from the typing rules that `L` must also be, and
the IH then tells us that `Γ` assigns `x` a type.
- Almost all the other cases are similar: $$x$$ appears free in a
subterm of $$P$$, and since $$P$$ is well typed under $$Γ$$, we
know the subterm of $$M$$ in which $$x$$ appears is well typed
under $$Γ$$ as well, and the IH gives us exactly the
- Almost all the other cases are similar: `x` appears free in a
subterm of `P`, and since `P` is well typed under `Γ`, we
know the subterm of `M` in which `x` appears is well typed
under `Γ` as well, and the IH gives us exactly the
conclusion we want.
- The only remaining case is `free-λᵀ`. In this case $$P =
λᵀ y ∈ A ⇒ N$$, and $$x$$ appears free in $$N$$; we also know that
$$x$$ is different from $$y$$. The difference from the previous
cases is that whereas $$P$$ is well typed under $$\Gamma$$, its
body $$N$$ is well typed under $$(Γ , y ↦ A)$$, so the IH
allows us to conclude that $$x$$ is assigned some type by the
extended context $$(Γ , y ↦ A)$$. To conclude that $$Γ$$
assigns a type to $$x$$, we appeal the decidable equality for names
`_≟_`, noting that $$x$$ and $$y$$ are different variables.
- The only remaining case is `free-λᵀ`. In this case `P =
λᵀ y ∈ A ⇒ N`, and `x` appears free in `N`; we also know that
`x` is different from `y`. The difference from the previous
cases is that whereas `P` is well typed under `\Gamma`, its
body `N` is well typed under `(Γ , y ↦ A)`, so the IH
allows us to conclude that `x` is assigned some type by the
extended context `(Γ , y ↦ A)`. To conclude that `Γ`
assigns a type to `x`, we appeal the decidable equality for names
`_≟_`, noting that `x` and `y` are different variables.
\begin{code}
freeLemma free-varᵀ (Ax Γx≡justA) = (_ , Γx≡justA)
@ -261,11 +261,11 @@ freeLemma (free-λᵀ {x} {y} y≢x x∈N) (⇒-I ⊢N) with freeLemma x∈N ⊢
... | no _ = Γx=justC
\end{code}
[A subtle point: if the first argument of $$free-λᵀ$$ was of type
$$x ≢ y$$ rather than of type $$y ≢ x$$, then the type of the
term $$Γx=justC$$ would not simplify properly.]
[A subtle point: if the first argument of `free-λᵀ` was of type
`x ≢ y` rather than of type `y ≢ x`, then the type of the
term `Γx=justC` would not simplify properly.]
Next, we'll need the fact that any term $$M$$ which is well typed in
Next, we'll need the fact that any term `M` which is well typed in
the empty context is closed (it has no free variables).
#### Exercise: 2 stars, optional (∅⊢-closed)
@ -286,11 +286,11 @@ contradiction ()
\end{code}
</div>
Sometimes, when we have a proof $$Γ ⊢ M ∈ A$$, we will need to
replace $$Γ$$ by a different context $$Γ′$$. When is it safe
Sometimes, when we have a proof `Γ ⊢ M ∈ A`, we will need to
replace `Γ` by a different context `Γ′`. When is it safe
to do this? Intuitively, it must at least be the case that
$$Γ′$$ assigns the same types as $$Γ$$ to all the variables
that appear free in $$M$$. In fact, this is the only condition that
`Γ′` assigns the same types as `Γ` to all the variables
that appear free in `M`. In fact, this is the only condition that
is needed.
\begin{code}
@ -301,45 +301,45 @@ weaken : ∀ {Γ Γ′ M A}
\end{code}
_Proof_: By induction on the derivation of
$$Γ ⊢ M ∈ A$$.
`Γ ⊢ M ∈ A`.
- If the last rule in the derivation was `var`, then $$t = x$$
and $$\Gamma x = T$$. By assumption, $$\Gamma' x = T$$ as well, and
hence $$\Gamma' \vdash t : T$$ by `var`.
- If the last rule in the derivation was `var`, then `t = x`
and `\Gamma x = T`. By assumption, `\Gamma' x = T` as well, and
hence `\Gamma' \vdash t : T` by `var`.
- If the last rule was `abs`, then $$t = \lambda y:A. t'$$, with
$$T = A\to B$$ and $$\Gamma, y : A \vdash t' : B$$. The
induction hypothesis is that, for any context $$\Gamma''$$, if
$$\Gamma, y:A$$ and $$\Gamma''$$ assign the same types to all the
free variables in $$t'$$, then $$t'$$ has type $$B$$ under
$$\Gamma''$$. Let $$\Gamma'$$ be a context which agrees with
$$\Gamma$$ on the free variables in $$t$$; we must show
$$\Gamma' \vdash \lambda y:A. t' : A\to B$$.
- If the last rule was `abs`, then `t = \lambda y:A. t'`, with
`T = A\to B` and `\Gamma, y : A \vdash t' : B`. The
induction hypothesis is that, for any context `\Gamma''`, if
`\Gamma, y:A` and `\Gamma''` assign the same types to all the
free variables in `t'`, then `t'` has type `B` under
`\Gamma''`. Let `\Gamma'` be a context which agrees with
`\Gamma` on the free variables in `t`; we must show
`\Gamma' \vdash \lambda y:A. t' : A\to B`.
By $$abs$$, it suffices to show that $$\Gamma', y:A \vdash t' : t'$$.
By the IH (setting $$\Gamma'' = \Gamma', y:A$$), it suffices to show
that $$\Gamma, y:A$$ and $$\Gamma', y:A$$ agree on all the variables
that appear free in $$t'$$.
By `abs`, it suffices to show that `\Gamma', y:A \vdash t' : t'`.
By the IH (setting `\Gamma'' = \Gamma', y:A`), it suffices to show
that `\Gamma, y:A` and `\Gamma', y:A` agree on all the variables
that appear free in `t'`.
Any variable occurring free in $$t'$$ must be either $$y$$ or
some other variable. $$\Gamma, y:A$$ and $$\Gamma', y:A$$
clearly agree on $$y$$. Otherwise, note that any variable other
than $$y$$ that occurs free in $$t'$$ also occurs free in
$$t = \lambda y:A. t'$$, and by assumption $$\Gamma$$ and
$$\Gamma'$$ agree on all such variables; hence so do $$\Gamma, y:A$$ and
$$\Gamma', y:A$$.
Any variable occurring free in `t'` must be either `y` or
some other variable. `\Gamma, y:A` and `\Gamma', y:A`
clearly agree on `y`. Otherwise, note that any variable other
than `y` that occurs free in `t'` also occurs free in
`t = \lambda y:A. t'`, and by assumption `\Gamma` and
`\Gamma'` agree on all such variables; hence so do `\Gamma, y:A` and
`\Gamma', y:A`.
- If the last rule was `app`, then $$t = t_1\;t_2$$, with
$$\Gamma \vdash t_1:A\to T$$ and $$\Gamma \vdash t_2:A$$.
One induction hypothesis states that for all contexts $$\Gamma'$$,
if $$\Gamma'$$ agrees with $$\Gamma$$ on the free variables in $$t_1$$,
then $$t_1$$ has type $$A\to T$$ under $$\Gamma'$$; there is a similar IH
for $$t_2$$. We must show that $$t_1\;t_2$$ also has type $$T$$ under
$$\Gamma'$$, given the assumption that $$\Gamma'$$ agrees with
$$\Gamma$$ on all the free variables in $$t_1\;t_2$$. By `app`, it
suffices to show that $$t_1$$ and $$t_2$$ each have the same type
under $$\Gamma'$$ as under $$\Gamma$$. But all free variables in
$$t_1$$ are also free in $$t_1\;t_2$$, and similarly for $$t_2$$;
- If the last rule was `app`, then `t = t_1\;t_2`, with
`\Gamma \vdash t_1:A\to T` and `\Gamma \vdash t_2:A`.
One induction hypothesis states that for all contexts `\Gamma'`,
if `\Gamma'` agrees with `\Gamma` on the free variables in `t_1`,
then `t_1` has type `A\to T` under `\Gamma'`; there is a similar IH
for `t_2`. We must show that `t_1\;t_2` also has type `T` under
`\Gamma'`, given the assumption that `\Gamma'` agrees with
`\Gamma` on all the free variables in `t_1\;t_2`. By `app`, it
suffices to show that `t_1` and `t_2` each have the same type
under `\Gamma'` as under `\Gamma`. But all free variables in
`t_1` are also free in `t_1\;t_2`, and similarly for `t_2`;
hence the desired result follows from the induction hypotheses.
\begin{code}
@ -383,16 +383,16 @@ preserves types---namely, the observation that _substitution_
preserves types.
Formally, the so-called _Substitution Lemma_ says this: Suppose we
have a term $$N$$ with a free variable $$x$$, and suppose we've been
able to assign a type $$B$$ to $$N$$ under the assumption that $$x$$ has
some type $$A$$. Also, suppose that we have some other term $$V$$ and
that we've shown that $$V$$ has type $$A$$. Then, since $$V$$ satisfies
the assumption we made about $$x$$ when typing $$N$$, we should be
able to substitute $$V$$ for each of the occurrences of $$x$$ in $$N$$
and obtain a new term that still has type $$B$$.
have a term `N` with a free variable `x`, and suppose we've been
able to assign a type `B` to `N` under the assumption that `x` has
some type `A`. Also, suppose that we have some other term `V` and
that we've shown that `V` has type `A`. Then, since `V` satisfies
the assumption we made about `x` when typing `N`, we should be
able to substitute `V` for each of the occurrences of `x` in `N`
and obtain a new term that still has type `B`.
_Lemma_: If $$Γ , x ↦ A ⊢ N ∈ B$$ and $$∅ ⊢ V ∈ A$$, then
$$Γ ⊢ (N [ x := V ]) ∈ B$$.
_Lemma_: If `Γ , x ↦ A ⊢ N ∈ B` and `∅ ⊢ V ∈ A`, then
`Γ ⊢ (N [ x := V ]) ∈ B`.
\begin{code}
preservation-[:=] : ∀ {Γ x A N B V}
@ -402,63 +402,63 @@ preservation-[:=] : ∀ {Γ x A N B V}
\end{code}
One technical subtlety in the statement of the lemma is that
we assign $$V$$ the type $$A$$ in the _empty_ context---in other
words, we assume $$V$$ is closed. This assumption considerably
simplifies the $$λᵀ$$ case of the proof (compared to assuming
$$Γ ⊢ V ∈ A$$, which would be the other reasonable assumption
we assign `V` the type `A` in the _empty_ context---in other
words, we assume `V` is closed. This assumption considerably
simplifies the `λᵀ` case of the proof (compared to assuming
`Γ ⊢ V ∈ A`, which would be the other reasonable assumption
at this point) because the context invariance lemma then tells us
that $$V$$ has type $$A$$ in any context at all---we don't have to
worry about free variables in $$V$$ clashing with the variable being
introduced into the context by $$λᵀ$$.
that `V` has type `A` in any context at all---we don't have to
worry about free variables in `V` clashing with the variable being
introduced into the context by `λᵀ`.
The substitution lemma can be viewed as a kind of "commutation"
property. Intuitively, it says that substitution and typing can
be done in either order: we can either assign types to the terms
$$N$$ and $$V$$ separately (under suitable contexts) and then combine
`N` and `V` separately (under suitable contexts) and then combine
them using substitution, or we can substitute first and then
assign a type to $$N [ x := V ]$$---the result is the same either
assign a type to `N [ x := V ]`---the result is the same either
way.
_Proof_: We show, by induction on $$N$$, that for all $$A$$ and
$$Γ$$, if $$Γ , x ↦ A \vdash N ∈ B$$ and $$∅ ⊢ V ∈ A$$, then
$$Γ \vdash N [ x := V ] ∈ B$$.
_Proof_: We show, by induction on `N`, that for all `A` and
`Γ`, if `Γ , x ↦ A \vdash N ∈ B` and `∅ ⊢ V ∈ A`, then
`Γ \vdash N [ x := V ] ∈ B`.
- If $$N$$ is a variable there are two cases to consider,
depending on whether $$N$$ is $$x$$ or some other variable.
- If `N` is a variable there are two cases to consider,
depending on whether `N` is `x` or some other variable.
- If $$N = varᵀ x$$, then from the fact that $$Γ , x ↦ A ⊢ N ∈ B$$
we conclude that $$A = B$$. We must show that $$x [ x := V] =
V$$ has type $$A$$ under $$Γ$$, given the assumption that
$$V$$ has type $$A$$ under the empty context. This
- If `N = varᵀ x`, then from the fact that `Γ , x ↦ A ⊢ N ∈ B`
we conclude that `A = B`. We must show that `x [ x := V] =
V` has type `A` under `Γ`, given the assumption that
`V` has type `A` under the empty context. This
follows from context invariance: if a closed term has type
$$A$$ in the empty context, it has that type in any context.
`A` in the empty context, it has that type in any context.
- If $$N$$ is some variable $$x′$$ different from $$x$$, then
we need only note that $$x′$$ has the same type under $$Γ , x ↦ A$$
as under $$Γ$$.
- If `N` is some variable `x′` different from `x`, then
we need only note that `x′` has the same type under `Γ , x ↦ A`
as under `Γ`.
- If $$N$$ is an abstraction $$λᵀ x′ ∈ A′ ⇒ N′$$, then the IH tells us,
for all $$Γ′$$́ and $$B′$$, that if $$Γ′ , x ↦ A ⊢ N′ ∈ B′$$
and $$∅ ⊢ V ∈ A$$, then $$Γ′ ⊢ N′ [ x := V ] ∈ B′$$.
- If `N` is an abstraction `λᵀ x′ ∈ A′ ⇒ N′`, then the IH tells us,
for all `Γ′`́ and `B′`, that if `Γ′ , x ↦ A ⊢ N′ ∈ B′`
and `∅ ⊢ V ∈ A`, then `Γ′ ⊢ N′ [ x := V ] ∈ B′`.
The substitution in the conclusion behaves differently
depending on whether $$x$$ and $$x′$$ are the same variable.
depending on whether `x` and `x′` are the same variable.
First, suppose $$x ≡ x′$$. Then, by the definition of
substitution, $$N [ x := V] = N$$, so we just need to show $$Γ ⊢ N ∈ B$$.
But we know $$Γ , x ↦ A ⊢ N ∈ B$$ and, since $$x ≡ x′$$
does not appear free in $$λᵀ x′ ∈ A′ ⇒ N′$$, the context invariance
lemma yields $$Γ ⊢ N ∈ B$$.
First, suppose `x ≡ x′`. Then, by the definition of
substitution, `N [ x := V] = N`, so we just need to show `Γ ⊢ N ∈ B`.
But we know `Γ , x ↦ A ⊢ N ∈ B` and, since `x ≡ x′`
does not appear free in `λᵀ x′ ∈ A′ ⇒ N′`, the context invariance
lemma yields `Γ ⊢ N ∈ B`.
Second, suppose $$x ≢ x′$$. We know $$Γ , x ↦ A , x′ ↦ A′ ⊢ N′ ∈ B′$$
Second, suppose `x ≢ x′`. We know `Γ , x ↦ A , x′ ↦ A′ ⊢ N′ ∈ B′`
by inversion of the typing relation, from which
$$Γ , x′ ↦ A′ , x ↦ A ⊢ N′ ∈ B′$$ follows by update permute,
so the IH applies, giving us $$Γ , x′ ↦ A′ ⊢ N′ [ x := V ] ∈ B′$$
By $$⇒-I$$, we have $$Γ ⊢ λᵀ x′ ∈ A′ ⇒ (N′ [ x := V ]) ∈ A′ ⇒ B′$$
and the definition of substitution (noting $$x ≢ x′$$) gives
$$Γ ⊢ (λᵀ x′ ∈ A′ ⇒ N′) [ x := V ] ∈ A′ ⇒ B′$$ as required.
`Γ , x′ ↦ A′ , x ↦ A ⊢ N′ ∈ B′` follows by update permute,
so the IH applies, giving us `Γ , x′ ↦ A′ ⊢ N′ [ x := V ] ∈ B′`
By `⇒-I`, we have `Γ ⊢ λᵀ x′ ∈ A′ ⇒ (N′ [ x := V ]) ∈ A′ ⇒ B′`
and the definition of substitution (noting `x ≢ x′`) gives
`Γ ⊢ (λᵀ x′ ∈ A′ ⇒ N′) [ x := V ] ∈ A′ ⇒ B′` as required.
- If $$N$$ is an application $$L′ ·ᵀ M′$$, the result follows
- If `N` is an application `L′ ·ᵀ M′`, the result follows
straightforwardly from the definition of substitution and the
induction hypotheses.
@ -492,7 +492,7 @@ preservation-[:=] {Γ} {x} {A} {λᵀ x′ ∈ A′ ⇒ N′} {.A′ ⇒ B′} {
...| no _ = refl
...| no x≢x′ = ⇒-I ⊢N′V
where
x′x⊢N′ : (Γ , x′ ↦ A′ , x ↦ A) ⊢ N′ ∈ B′
x′x⊢N′ : Γ , x′ ↦ A′ , x ↦ A ⊢ N′ ∈ B′
x′x⊢N′ rewrite update-permute Γ x A x′ A′ x≢x′ = ⊢N′
⊢N′V : (Γ , x′ ↦ A′) ⊢ N′ [ x := V ] ∈ B′
⊢N′V = preservation-[:=] x′x⊢N′ ⊢V
@ -507,43 +507,43 @@ preservation-[:=] (𝔹-E ⊢L ⊢M ⊢N) ⊢V =
### Main Theorem
We now have the tools we need to prove preservation: if a closed
term $$M$$ has type $$A$$ and takes a step to $$N$$, then $$N$$
is also a closed term with type $$A$$. In other words, small-step
term `M` has type `A` and takes a step to `N`, then `N`
is also a closed term with type `A`. In other words, small-step
reduction preserves types.
\begin{code}
preservation : ∀ {M N A} → ∅ ⊢ M ∈ A → M ⟹ N → ∅ ⊢ N ∈ A
\end{code}
_Proof_: By induction on the derivation of $$\vdash t : T$$.
_Proof_: By induction on the derivation of `\vdash t : T`.
- We can immediately rule out $$var$$, $$abs$$, $$T_True$$, and
$$T_False$$ as the final rules in the derivation, since in each of
these cases $$t$$ cannot take a step.
- We can immediately rule out `var`, `abs`, `T_True`, and
`T_False` as the final rules in the derivation, since in each of
these cases `t` cannot take a step.
- If the last rule in the derivation was $$app$$, then $$t = t_1
t_2$$. There are three cases to consider, one for each rule that
could have been used to show that $$t_1 t_2$$ takes a step to $$t'$$.
- If the last rule in the derivation was `app`, then `t = t_1
t_2`. There are three cases to consider, one for each rule that
could have been used to show that `t_1 t_2` takes a step to `t'`.
- If $$t_1 t_2$$ takes a step by $$Sapp1$$, with $$t_1$$ stepping to
$$t_1'$$, then by the IH $$t_1'$$ has the same type as $$t_1$$, and
hence $$t_1' t_2$$ has the same type as $$t_1 t_2$$.
- If `t_1 t_2` takes a step by `Sapp1`, with `t_1` stepping to
`t_1'`, then by the IH `t_1'` has the same type as `t_1`, and
hence `t_1' t_2` has the same type as `t_1 t_2`.
- The $$Sapp2$$ case is similar.
- The `Sapp2` case is similar.
- If $$t_1 t_2$$ takes a step by $$Sred$$, then $$t_1 =
\lambda x:t_{11}.t_{12}$$ and $$t_1 t_2$$ steps to $$$$x:=t_2$$t_{12}$$; the
- If `t_1 t_2` takes a step by `Sred`, then `t_1 =
\lambda x:t_{11}.t_{12}` and `t_1 t_2` steps to ``x:=t_2`t_{12}`; the
desired result now follows from the fact that substitution
preserves types.
- If the last rule in the derivation was $$if$$, then $$t = if t_1
then t_2 else t_3$$, and there are again three cases depending on
how $$t$$ steps.
- If the last rule in the derivation was `if`, then `t = if t_1
then t_2 else t_3`, and there are again three cases depending on
how `t` steps.
- If $$t$$ steps to $$t_2$$ or $$t_3$$, the result is immediate, since
$$t_2$$ and $$t_3$$ have the same type as $$t$$.
- If `t` steps to `t_2` or `t_3`, the result is immediate, since
`t_2` and `t_3` have the same type as `t`.
- Otherwise, $$t$$ steps by $$Sif$$, and the desired conclusion
- Otherwise, `t` steps by `Sif`, and the desired conclusion
follows directly from the induction hypothesis.
\begin{code}
@ -571,11 +571,11 @@ Proof with eauto.
intros t t' T HT. generalize dependent t'.
induction HT;
intros t' HE; subst Gamma; subst;
try solve $$inversion HE; subst; auto$$.
try solve `inversion HE; subst; auto`.
- (* app
inversion HE; subst...
(* Most of the cases are immediate by induction,
and $$eauto$$ takes care of them
and `eauto` takes care of them
+ (* Sred
apply substitution_preserves_typing with t_{11}...
inversion HT_1...
@ -585,7 +585,7 @@ Qed.
An exercise in the [Stlc]({{ "Stlc" | relative_url }}) chapter asked about the
subject expansion property for the simple language of arithmetic and boolean
expressions. Does this property hold for STLC? That is, is it always the case
that, if $$t ==> t'$$ and $$has_type t' T$$, then $$empty \vdash t : T$$? If
that, if `t ==> t'` and `has_type t' T`, then `empty \vdash t : T`? If
so, prove it. If not, give a counter-example not involving conditionals.
@ -604,7 +604,7 @@ Corollary soundness : forall t t' T,
~(stuck t').
Proof.
intros t t' T Hhas_type Hmulti. unfold stuck.
intros $$Hnf Hnot_val$$. unfold normal_form in Hnf.
intros `Hnf Hnot_val`. unfold normal_form in Hnf.
induction Hmulti.
@ -621,10 +621,10 @@ Formalize this statement and prove it.
#### Exercise: 1 star (progress_preservation_statement)
Without peeking at their statements above, write down the progress
and preservation theorems for the simply typed lambda-calculus.
$$$$
``
#### Exercise: 2 stars (stlc_variation1)
Suppose we add a new term $$zap$$ with the following reduction rule
Suppose we add a new term `zap` with the following reduction rule
--------- (ST_Zap)
t ==> zap
@ -639,7 +639,7 @@ the presence of these rules? For each property, write either
"remains true" or "becomes false." If a property becomes
false, give a counterexample.
- Determinism of $$step$$
- Determinism of `step`
- Progress
@ -647,7 +647,7 @@ false, give a counterexample.
#### Exercise: 2 stars (stlc_variation2)
Suppose instead that we add a new term $$foo$$ with the following
Suppose instead that we add a new term `foo` with the following
reduction rules:
----------------- (ST_Foo1)
@ -661,20 +661,20 @@ the presence of this rule? For each one, write either
"remains true" or else "becomes false." If a property becomes
false, give a counterexample.
- Determinism of $$step$$
- Determinism of `step`
- Progress
- Preservation
#### Exercise: 2 stars (stlc_variation3)
Suppose instead that we remove the rule $$Sapp1$$ from the $$step$$
Suppose instead that we remove the rule `Sapp1` from the `step`
relation. Which of the following properties of the STLC remain
true in the presence of this rule? For each one, write either
"remains true" or else "becomes false." If a property becomes
false, give a counterexample.
- Determinism of $$step$$
- Determinism of `step`
- Progress
@ -692,7 +692,7 @@ the presence of this rule? For each one, write either
"remains true" or else "becomes false." If a property becomes
false, give a counterexample.
- Determinism of $$step$$
- Determinism of `step`
- Progress
@ -714,7 +714,7 @@ the presence of this rule? For each one, write either
"remains true" or else "becomes false." If a property becomes
false, give a counterexample.
- Determinism of $$step$$
- Determinism of `step`
- Progress
@ -736,7 +736,7 @@ the presence of this rule? For each one, write either
"remains true" or else "becomes false." If a property becomes
false, give a counterexample.
- Determinism of $$step$$
- Determinism of `step`
- Progress
@ -756,7 +756,7 @@ the presence of this rule? For each one, write either
"remains true" or else "becomes false." If a property becomes
false, give a counterexample.
- Determinism of $$step$$
- Determinism of `step`
- Progress
@ -798,10 +798,10 @@ with arithmetic. Specifically:
the definition of values through the Type Soundness theorem), and
paste it into the file at this point.
- Extend the definitions of the $$subst$$ operation and the $$step$$
- Extend the definitions of the `subst` operation and the `step`
relation to include appropriate clauses for the arithmetic operators.
- Extend the proofs of all the properties (up to $$soundness$$) of
- Extend the proofs of all the properties (up to `soundness`) of
the original STLC to deal with the new syntactic forms. Make
sure Agda accepts the whole file.