Started rewriting with Reasoning notation

src/Maps.lagda

@@ -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
+Intuitively, a total map over anfi element type `A` _is_ just a
-function that can be used to look up ids, yielding $$A$$s.
+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
+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.
+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.
+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:
-<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:
 
 \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 $$v$$
+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
+and then look up `x` in the map resulting from the update, we get
-back $$v$$:
+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
+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
+then look up a _different_ key `y` in the resulting map, we get
-the same result that $$m$$ would have given:
+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
+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
+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
+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
+assign key `x` the same value as it already has in `ρ`, then the
-result is equal to $$ρ$$:
+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.
 

src/Stlc.lagda

@@ -199,6 +199,8 @@ 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 →

src/StlcProp.lagda

@@ -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
+belonging to each type.  For `bool`, these are the boolean values `true` and
-$$false$$.  For arrow types, the canonical forms are lambda-abstractions. 
+`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
+  - 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
+    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,
+    `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
+    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
+    `\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
+    - Suppose `t_2` is a value.  Since `t_1` is a value with an
-      arrow type, it must be a lambda abstraction; hence $$t_1\;t_2$$
+      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 $$t = \text{if }t_1
+  - If the last rule of the derivation is `if`, then `t = \text{if }t_1
-    \text{ then }t_2\text{ else }t_3$$, where $$t_1$$ has type $$bool$$.  By
+    \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.
+    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
+  - 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
+    either `true` or `false`.  If it is `true`, then `t` steps
-    to $$t_2$$; otherwise it steps to $$t_3$$.
+    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
+    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
+    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
+    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
+    to be well-typed in some context `\Gamma` and consider the same
-    term $$s$$ in a slightly different context $$\Gamma'$$.  For this
+    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
+A variable `x` _appears free in_ a term `M` if `M` contains some
-occurrence of $$x$$ that is not under an abstraction over $$x$$.
+occurrence of `x` that is not under an abstraction over `x`.
 For example:
 
-  - $$y$$ appears free, but $$x$$ does not, in $$λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y$$
+  - `y` appears free, but `x` does not, in `λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y`
-  - both $$x$$ and $$y$$ appear free in $$(λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y) ·ᵀ x$$
+  - both `x` and `y` appear free in `(λᵀ x ∈ (A ⇒ B) ⇒ x ·ᵀ y) ·ᵀ x`
-  - no variables appear free in $$λᵀ x ∈ (A ⇒ B) ⇒ (λᵀ y ∈ A ⇒ x ·ᵀ y)$$
+  - 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
+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
+well typed in context `Γ`, then it must be the case that
-$$Γ$$ assigns a type to $$x$$.
+`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
+_Proof_: We show, by induction on the proof that `x` appears
-  free in $$P$$, that, for all contexts $$Γ$$, if $$P$$ is well
+  free in `P`, that, for all contexts `Γ`, if `P` is well
-  typed under $$Γ$$, then $$Γ$$ assigns some type to $$x$$.
+  typed under `Γ`, then `Γ` assigns some type to `x`.
 
-  - If the last rule used was `free-varᵀ`, then $$P = x$$, and from
+  - If the last rule used was `free-varᵀ`, then `P = x`, and from
-    the assumption that $$M$$ is well typed under $$Γ$$ we have
+    the assumption that `M` is well typed under `Γ` we have
-    immediately that $$Γ$$ assigns a type to $$x$$.
+    immediately that `Γ` assigns a type to `x`.
 
-  - If the last rule used was `free-·₁`, then $$P = L ·ᵀ M$$ and $$x$$
+  - 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 $$\Gamma$$,
+    appears free in `L`.  Since `L` is well typed under `\Gamma`,
-    we can see from the typing rules that $$L$$ must also be, and
+    we can see from the typing rules that `L` must also be, and
-    the IH then tells us that $$Γ$$ assigns $$x$$ a type.
+    the IH then tells us that `Γ` assigns `x` a type.
 
-  - Almost all the other cases are similar: $$x$$ appears free in a
+  - Almost all the other cases are similar: `x` appears free in a
-    subterm of $$P$$, and since $$P$$ is well typed under $$Γ$$, we
+    subterm of `P`, and since `P` is well typed under `Γ`, we
-    know the subterm of $$M$$ in which $$x$$ appears is well typed
+    know the subterm of `M` in which `x` appears is well typed
-    under $$Γ$$ as well, and the IH gives us exactly the
+    under `Γ` as well, and the IH gives us exactly the
     conclusion we want.
 
-  - The only remaining case is `free-λᵀ`.  In this case $$P =
+  - The only remaining case is `free-λᵀ`.  In this case `P =
-    λᵀ y ∈ A ⇒ N$$, and $$x$$ appears free in $$N$$; we also know that
+    λᵀ y ∈ A ⇒ N`, and `x` appears free in `N`; we also know that
-    $$x$$ is different from $$y$$.  The difference from the previous
+    `x` is different from `y`.  The difference from the previous
-    cases is that whereas $$P$$ is well typed under $$\Gamma$$, its
+    cases is that whereas `P` is well typed under `\Gamma`, its
-    body $$N$$ is well typed under $$(Γ , y ↦ A)$$, so the IH
+    body `N` is well typed under `(Γ , y ↦ A)`, so the IH
-    allows us to conclude that $$x$$ is assigned some type by the
+    allows us to conclude that `x` is assigned some type by the
-    extended context $$(Γ , y ↦ A)$$.  To conclude that $$Γ$$
+    extended context `(Γ , y ↦ A)`.  To conclude that `Γ`
-    assigns a type to $$x$$, we appeal the decidable equality for names
+    assigns a type to `x`, we appeal the decidable equality for names
-    `_≟_`, noting that $$x$$ and $$y$$ are different variables.
+    `_≟_`, 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
+[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
+`x ≢ y` rather than of type `y ≢ x`, then the type of the
-term $$Γx=justC$$ would not simplify properly.]
+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
+Sometimes, when we have a proof `Γ ⊢ M ∈ A`, we will need to
-replace $$Γ$$ by a different context $$Γ′$$.  When is it safe
+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
+`Γ′` assigns the same types as `Γ` to all the variables
-that appear free in $$M$$. In fact, this is the only condition that
+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 $$t = x$$
+  - If the last rule in the derivation was `var`, then `t = x`
-    and $$\Gamma x = T$$.  By assumption, $$\Gamma' x = T$$ as well, and
+    and `\Gamma x = T`.  By assumption, `\Gamma' x = T` as well, and
-    hence $$\Gamma' \vdash t : T$$ by `var`.
+    hence `\Gamma' \vdash t : T` by `var`.
 
-  - If the last rule was `abs`, then $$t = \lambda y:A. t'$$, with
+  - 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
+    `T = A\to B` and `\Gamma, y : A \vdash t' : B`.  The
-    induction hypothesis is that, for any context $$\Gamma''$$, if
+    induction hypothesis is that, for any context `\Gamma''`, if
-    $$\Gamma, y:A$$ and $$\Gamma''$$ assign the same types to all the
+    `\Gamma, y:A` and `\Gamma''` assign the same types to all the
-    free variables in $$t'$$, then $$t'$$ has type $$B$$ under
+    free variables in `t'`, then `t'` has type `B` under
-    $$\Gamma''$$.  Let $$\Gamma'$$ be a context which agrees with
+    `\Gamma''`.  Let `\Gamma'` be a context which agrees with
-    $$\Gamma$$ on the free variables in $$t$$; we must show
+    `\Gamma` on the free variables in `t`; we must show
-    $$\Gamma' \vdash \lambda y:A. t' : A\to B$$.
+    `\Gamma' \vdash \lambda y:A. t' : A\to B`.
 
-    By $$abs$$, it suffices to show that $$\Gamma', y:A \vdash t' : t'$$.
+    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
+    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 `\Gamma, y:A` and `\Gamma', y:A` agree on all the variables
-    that appear free in $$t'$$.
+    that appear free in `t'`.
 
-    Any variable occurring free in $$t'$$ must be either $$y$$ or
+    Any variable occurring free in `t'` must be either `y` or
-    some other variable.  $$\Gamma, y:A$$ and $$\Gamma', y:A$$
+    some other variable.  `\Gamma, y:A` and `\Gamma', y:A`
-    clearly agree on $$y$$.  Otherwise, note that any variable other
+    clearly agree on `y`.  Otherwise, note that any variable other
-    than $$y$$ that occurs free in $$t'$$ also occurs free in
+    than `y` that occurs free in `t'` also occurs free in
-    $$t = \lambda y:A. t'$$, and by assumption $$\Gamma$$ and
+    `t = \lambda y:A. t'`, and by assumption `\Gamma` and
-    $$\Gamma'$$ agree on all such variables; hence so do $$\Gamma, y:A$$ and
+    `\Gamma'` agree on all such variables; hence so do `\Gamma, y:A` and
-    $$\Gamma', y:A$$.
+    `\Gamma', y:A`.
 
-  - If the last rule was `app`, then $$t = t_1\;t_2$$, with
+  - If the last rule was `app`, then `t = t_1\;t_2`, with
-    $$\Gamma \vdash t_1:A\to T$$ and $$\Gamma \vdash t_2:A$$.
+    `\Gamma \vdash t_1:A\to T` and `\Gamma \vdash t_2:A`.
-    One induction hypothesis states that for all contexts $$\Gamma'$$,
+    One induction hypothesis states that for all contexts `\Gamma'`,
-    if $$\Gamma'$$ agrees with $$\Gamma$$ on the free variables in $$t_1$$,
+    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
+    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
+    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'`, given the assumption that `\Gamma'` agrees with
-    $$\Gamma$$ on all the free variables in $$t_1\;t_2$$.  By `app`, it
+    `\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
+    suffices to show that `t_1` and `t_2` each have the same type
-    under $$\Gamma'$$ as under $$\Gamma$$.  But all free variables in
+    under `\Gamma'` as under `\Gamma`.  But all free variables in
-    $$t_1$$ are also free in $$t_1\;t_2$$, and similarly for $$t_2$$;
+    `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
+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
+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
+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
+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
+the assumption we made about `x` when typing `N`, we should be
-able to substitute $$V$$ for each of the occurrences of $$x$$ in $$N$$
+able to substitute `V` for each of the occurrences of `x` in `N`
-and obtain a new term that still has type $$B$$.
+and obtain a new term that still has type `B`.
 
-_Lemma_: If $$Γ , x ↦ A ⊢ N ∈ B$$ and $$∅ ⊢ V ∈ A$$, then
+_Lemma_: If `Γ , x ↦ A ⊢ N ∈ B` and `∅ ⊢ V ∈ A`, then
-$$Γ ⊢ (N [ x := V ]) ∈ B$$.
+`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
+we assign `V` the type `A` in the _empty_ context---in other
-words, we assume $$V$$ is closed.  This assumption considerably
+words, we assume `V` is closed.  This assumption considerably
-simplifies the $$λᵀ$$ case of the proof (compared to assuming
+simplifies the `λᵀ` case of the proof (compared to assuming
-$$Γ ⊢ V ∈ A$$, which would be the other reasonable assumption
+`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
+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
+worry about free variables in `V` clashing with the variable being
-introduced into the context by $$λᵀ$$.
+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
+_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`, if `Γ , x ↦ A \vdash N ∈ B` and `∅ ⊢ V ∈ A`, then
-$$Γ \vdash N [ x := V ] ∈ B$$.
+`Γ \vdash N [ x := V ] ∈ B`.
 
-  - If $$N$$ is a variable there are two cases to consider,
+  - If `N` is a variable there are two cases to consider,
-    depending on whether $$N$$ is $$x$$ or some other variable.
+    depending on whether `N` is `x` or some other variable.
 
-      - If $$N = varᵀ x$$, then from the fact that $$Γ , x ↦ A ⊢ N ∈ B$$
+      - If `N = varᵀ x`, then from the fact that `Γ , x ↦ A ⊢ N ∈ B`
-        we conclude that $$A = B$$.  We must show that $$x [ x := V] =
+        we conclude that `A = B`.  We must show that `x [ x := V] =
-        V$$ has type $$A$$ under $$Γ$$, given the assumption that
+        V` has type `A` under `Γ`, given the assumption that
-        $$V$$ has type $$A$$ under the empty context.  This
+        `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
+      - If `N` is some variable `x′` different from `x`, then
-        we need only note that $$x′$$ has the same type under $$Γ , x ↦ A$$
+        we need only note that `x′` has the same type under `Γ , x ↦ A`
-        as under $$Γ$$.
+        as under `Γ`.
 
-  - If $$N$$ is an abstraction $$λᵀ x′ ∈ A′ ⇒ N′$$, then the IH tells us,
+  - If `N` is an abstraction `λᵀ x′ ∈ A′ ⇒ N′`, then the IH tells us,
-    for all $$Γ′$$́ and $$B′$$, that if $$Γ′ , x ↦ A ⊢ N′ ∈ B′$$
+    for all `Γ′`́ and `B′`, that if `Γ′ , x ↦ A ⊢ N′ ∈ B′`
-    and $$∅ ⊢ V ∈ A$$, then $$Γ′ ⊢ N′ [ x := V ] ∈ B′$$.
+    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
+    First, suppose `x ≡ x′`.  Then, by the definition of
-    substitution, $$N [ x := V] = N$$, so we just need to show $$Γ ⊢ N ∈ B$$.
+    substitution, `N [ x := V] = N`, so we just need to show `Γ ⊢ N ∈ B`.
-    But we know $$Γ , x ↦ A ⊢ N ∈ B$$ and, since $$x ≡ x′$$
+    But we know `Γ , x ↦ A ⊢ N ∈ B` and, since `x ≡ x′`
-    does not appear free in $$λᵀ x′ ∈ A′ ⇒ N′$$, the context invariance
+    does not appear free in `λᵀ x′ ∈ A′ ⇒ N′`, the context invariance
-    lemma yields $$Γ ⊢ N ∈ B$$.
+    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,
+    `Γ , x′ ↦ A′ , x ↦ A ⊢ N′ ∈ B′` follows by update permute,
-    so the IH applies, giving us $$Γ , x′ ↦ A′ ⊢ N′ [ x := V ] ∈ B′$$
+    so the IH applies, giving us `Γ , x′ ↦ A′ ⊢ N′ [ x := V ] ∈ B′`
-    By $$⇒-I$$, we have $$Γ ⊢ λᵀ x′ ∈ A′ ⇒ (N′ [ x := V ]) ∈ A′ ⇒ B′$$
+    By `⇒-I`, we have `Γ ⊢ λᵀ x′ ∈ A′ ⇒ (N′ [ x := V ]) ∈ A′ ⇒ B′`
-    and the definition of substitution (noting $$x ≢ x′$$) gives
+    and the definition of substitution (noting `x ≢ x′`) gives
-    $$Γ ⊢ (λᵀ x′ ∈ A′ ⇒ N′) [ x := V ] ∈ A′ ⇒ B′$$ as required.
+    `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.
 
@@ -483,12 +483,7 @@ just-injective refl = refl
 preservation-[:=] {_} {x} (Ax {_} {x′} [Γ,x↦A]x′≡B) ⊢V with x ≟ x′
 ...| yes x≡x′ rewrite just-injective [Γ,x↦A]x′≡B  =  weaken-closed ⊢V
 ...| no  x≢x′  =  Ax [Γ,x↦A]x′≡B
-{-
+preservation-[:=] {Γ} {x} {A} {λᵀ x′ ∈ A′ ⇒ N′} {.A′ ⇒ B′} {V} (⇒-I ⊢N′) ⊢V with x ≟ x′
-preservation-[:=] {Γ} {x} {A} {varᵀ x′} {B} {V} (Ax {.(Γ , x ↦ A)} {.x′} {.B} Γx′≡B) ⊢V with x ≟ x′
-...| yes x≡x′ rewrite just-injective Γx′≡B  =  weaken-closed ⊢V
-...| no  x≢x′  =  Ax {Γ} {x′} {B} Γx′≡B
--}
-preservation-[:=] {Γ} {x} {A} {λᵀ x′ ∈ A′ ⇒ N′} {.A′ ⇒ B′} {V} (⇒-I {.(Γ , x ↦ A)} {.x′} {.N′} {.A′} {.B′} ⊢N′) ⊢V with x ≟ x′
 ...| yes x≡x′ rewrite x≡x′ = weaken Γ′~Γ (⇒-I ⊢N′)
   where
   Γ′~Γ : ∀ {y} → y FreeIn (λᵀ x′ ∈ A′ ⇒ N′) → (Γ , x′ ↦ A) y ≡ Γ y
@@ -497,79 +492,58 @@ 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′!}
+  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
-{-
-...| yes x′≡x rewrite x′≡x | update-shadow Γ x A A′  =  {!!}
-  -- ⇒-I ⊢N′
-...| no  x′≢x rewrite update-permute Γ x′ A′ x A x′≢x  =  {!!}
-  --  ⇒-I {Γ} {x′} {N′} {A′} {B′} (preservation-[:=] {(Γ , x′ ↦ A′)} {x} {A} ⊢N′ ⊢V)
--}
 preservation-[:=] (⇒-E ⊢L ⊢M) ⊢V = ⇒-E (preservation-[:=] ⊢L ⊢V) (preservation-[:=] ⊢M ⊢V)
 preservation-[:=] 𝔹-I₁ ⊢V = 𝔹-I₁
 preservation-[:=] 𝔹-I₂ ⊢V = 𝔹-I₂
 preservation-[:=] (𝔹-E ⊢L ⊢M ⊢N) ⊢V =
   𝔹-E (preservation-[:=] ⊢L ⊢V) (preservation-[:=] ⊢M ⊢V) (preservation-[:=] ⊢N ⊢V)
-
-{-
-[:=]-preserves-⊢ {Γ} {x} v∶A (var y y∈Γ) with x ≟ y
-... | yes x=y = {!!}
-... | no  x≠y = {!!}
-[:=]-preserves-⊢ v∶A (abs t′∶B) = {!!}
-[:=]-preserves-⊢ v∶A (app t₁∶A⇒B t₂∶A) =
-  app ([:=]-preserves-⊢ v∶A t₁∶A⇒B) ([:=]-preserves-⊢ v∶A t₂∶A)
-[:=]-preserves-⊢ v∶A true  = true
-[:=]-preserves-⊢ v∶A false = false
-[:=]-preserves-⊢ v∶A (if t₁∶bool then t₂∶B else t₃∶B) =
-  if   [:=]-preserves-⊢ v∶A t₁∶bool
-  then [:=]-preserves-⊢ v∶A t₂∶B
-  else [:=]-preserves-⊢ v∶A t₃∶B
--}
 \end{code}
 
 
 ### 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 $$N$$
+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
+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
+- We can immediately rule out `var`, `abs`, `T_True`, and
-  $$T_False$$ as the final rules in the derivation, since in each of
+  `T_False` as the final rules in the derivation, since in each of
-  these cases $$t$$ cannot take a step.
+  these cases `t` cannot take a step.
 
-- If the last rule in the derivation was $$app$$, then $$t = t_1
+- If the last rule in the derivation was `app`, then `t = t_1
-  t_2$$.  There are three cases to consider, one for each rule that
+  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 $$t'$$.
+  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
+    - 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
+      `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 $$t_1 t_2$$.
+      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 $$t_1 =
+    - If `t_1 t_2` takes a step by `Sred`, then `t_1 =
-      \lambda x:t_{11}.t_{12}$$ and $$t_1 t_2$$ steps to $$$$x:=t_2$$t_{12}$$; the
+      \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 $$t = if t_1
+    - If the last rule in the derivation was `if`, then `t = if t_1
-      then t_2 else t_3$$, and there are again three cases depending on
+      then t_2 else t_3`, and there are again three cases depending on
-      how $$t$$ steps.
+      how `t` steps.
 
-    - If $$t$$ steps to $$t_2$$ or $$t_3$$, the result is immediate, since
+    - If `t` steps to `t_2` or `t_3`, the result is immediate, since
-      $$t_2$$ and $$t_3$$ have the same type as $$t$$.
+      `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}
@@ -597,11 +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...
@@ -611,7 +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. 
 
 
@@ -630,7 +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.
 
 
@@ -647,10 +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
@@ -665,7 +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
 
@@ -673,7 +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)
@@ -687,20 +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
 
@@ -718,7 +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
 
@@ -740,7 +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
 
@@ -762,7 +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
 
@@ -782,7 +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
 
@@ -824,10 +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.