This notes review our development of the Girard-Reynolds polymorphic λ-calculus (also called System F), and the relationship of the GR calculus with the other λ-calculi developed during the semester.

The distinguishing feature of System F is that it provides *explicit* polymorphism, by
providing terms for type abstraction (type lambdas) and type application. The resulting system
is more verbose than implicit polymorphism, but also more expressive.

The types of System F are as follows. (Unlike in H-M polymorphism, we do not introduce a distinction between generic, or quantified, and ground types.)

Type |
Meaning |

`a` |
Type variables |

`t → u` |
Functions from `t` to `u` |

`Πa . t` |
Polymorphic type |

The type language is significantly reduced from earler calculi! Later, we will see how the remainder of our familiar types can be encoded in terms of the types of System F.

The term language is equally simple.

Term |
Meaning |

`x` |
Term variables |

`\ x : t → e` |
Function from `x` of type `t` to `e` |

`e` |
Application of `e` to `e` |

`Λ a. e` |
Type abstraction: `e` for any type `a` |

`e [t]` |
Type application: instantiation of outermost type parameter in `e` with `t` |

Functions have regrown their type annotations. Unlike H-M, we cannot compute the most general types for terms in the G-R calculus, so we will rely on the programmer to supply some type annotations.

Typing judgments take the form $\Delta; \Gamma \vdash e : t$, indicating that, with valid type variables in $\Delta$ and term variables as in $\Gamma$, expression $e$ has type $t$. For now, we only have one kind of type, and so $\Delta$ only needs to keep a list of variables.

Rule |
Name |

$$\frac{x:t \in \Gamma} {\Delta; \Gamma \vdash x : t}$$ | (var) |

$$\frac{\Delta; \Gamma, x : t \vdash e : u \quad \Delta \vdash t\;\mathsf{type}} {\Delta; \Gamma \vdash \backslash x : t \to e : t \to u}$$ | (→I) |

$$\frac{\Delta; \Gamma \vdash e_1 : t \to u \quad \Delta; \Gamma \vdash e_2 : t} {\Delta; \Gamma \vdash e_1\,e_2 : u}$$ | (→E) |

$$\frac{\Delta, a; \Gamma \vdash e : t \quad a \not\in \Delta} {\Delta; \Gamma \vdash \Lambda a. e : \Pi a. t}$$ | (ΠI) |

$$\frac{\Delta; \Gamma \vdash e : \Pi a. t \quad \Delta \vdash u \; \mathsf{type}} {\Delta; \Gamma \vdash e[u] : [a \mapsto u]t}$$ | (ΠE) |

The rules that mention types now rely on an auxiliary judgment $\Delta \vdash t \; \mathsf{type}$, meaning that with type variables in $\Delta$, $t$ is a valid type. This requires checking that any type variables mentioned in $t$ appear in $\Delta$.

$$ \frac{a \in \Delta} {\Delta \vdash a \; \mathsf{type}} \quad \frac{\Delta, a \vdash t \; \mathsf{type}} {\Delta \vdash \Pi a. t \; \mathsf{type}} \quad \frac{\Delta \vdash t \; \mathsf{type} \quad \Delta \vdash u \; \mathsf{type}} {\Delta \vdash t \to u \; \mathsf{type}} $$The identity function in System F is typed as follows

$$ \Lambda a. \backslash x : a \to x : \Pi a. a \to a $$The K combinator can be given either of the following typings

$$\begin{gather*} \Lambda a. \Lambda b. \backslash x : a \to \backslash y : b \to x : \Pi a. \Pi b. a \to b \to a \\ \Lambda a. \Lambda b. \backslash x : b \to \backslash y : a \to x : \Pi a. \Pi b. b \to a \to b \end{gather*}$$
*Optional exercise.* Show that the two types above are *isomorphic*. To do so, write two
terms, one with each of the following two types.

The evaluation rules for System F terms are as follows. Formally, we specify these rules in
terms of substitution of types into terms; however, in practice, all type manipulation can be
omitted from evaluation. Nevertheless, it is important to include *some* runtime
representation of type lambdas. Can you see why?)

Our central observation is that computation is character by elimination forms, rather than introduction forms. Consider the evaluation rules for products.

$$ \frac{e_1 \Downarrow v_1 \quad e_2 \Downarrow v_2} {(e_1, e_2) \Downarrow \langle v_1, v_2 \rangle} \quad \frac{e_1 \Downarrow \langle w_1, w_2 \rangle \quad [x_1 \mapsto w_1, x_2 \mapsto w_2]e_2 \Downarrow v} {\Let{(x_1,x_2)}{e_1}{e_2} \Downarrow v} $$The introduction form is just packaging the values $v_1$ and $v_2$; everything interesting happens in the elimination form. If we want to encode products, then, the key step is to be able to encode the elimination form. That is to say, we will encode the pair $(e_1, e_2)$ as something that can perform the corresponding elimination action.

Before we get to the translation itself, two observations on notation. First, we write
translation using double brackets (also called "Oxford bracket" or "semantic brackets"), so
$\llbracket X \rrbracket$ is notation for "the translation of $X$" (whatever kind of thing $X$
may happen to be). Second, our translation of terms relies on a small bit of typing
information; so, rather than defining a translation on terms alone $\llbracket e \rrbracket$, we
define a translation on terms *and their types* $\llbracket e : t \rrbracket$. Of
course, as we have claimed in class (and you are discovering in your homework) it is frequently
possible to compute the types of terms, so this is for convenience rather than illustrating a
fundamental weakness of the translation.

Now, we can have the encoding itself:

$$ \begin{align*} \Tr{t_1 \times t_2} &= \Pi a. (t_1 \to t_2 \to a) \to a \\ \Tr{(e_1, e_2) : t_1 \times t_2} &= \Lambda a. \backslash f : (t_1 \to t_2 \to a) \to f\, \Tr{e_1} \, \Tr{e_2} \\ \Tr{\Let{(x_1,x_2)}{e_1}{e_2} : u} &= \Tr{e_1} \, [\Tr u] \, (\backslash x_1 : t_1 \to \backslash x_2 : t_2 \to e_2) \end{align*} $$The translation is driven by the need to do elimination. Consider the second hypothesis of the elimination rule. Term $e_2$ has two free variables $x_1$ and $x_2$; given values for this variables, the evaluation of $e_2$ is the evaluation of the whole term. We make two steps. First, we can represent a term with holes as a function. So, rather than talking about $e_2$, of type $u$ directly, we will talk about a function of type $t_1 \to t_2 \to u$. Second, we encode a pair itself as a thing that performs the elimination form. That is, given a suitable $e_2$, wrapped as a function, it produces the resulting value.

Here is a simple example. For the purpose of this example, we will leave the constants untranslated.

$$\begin{align} & \Tr{\Let{(x,y)}{(4,5)}{x + y}} \\ &= \Tr{(4,5)} \, [\Tr\Int] \, (\backslash x : \Tr\Int \to \backslash y : \Tr\Int \to x \Tr{+} y) \\ &= (\Lambda a. \backslash f : (\Tr\Int \to \Tr\Int \to a) \to f\,\Tr 4\, \Tr 5)\, [\Tr\Int] (\backslash x : \Tr\Int \to \backslash y : \Tr\Int \to x \Tr{+} y) \\ &= (\backslash f : (\Tr\Int \to \Tr\Int \to \Tr\Int) \to f\,\Tr 4\, \Tr 5)\, (\backslash x : \Tr\Int \to \backslash y : \Tr\Int \to x \Tr{+} y) \\ &= (\backslash x : \Tr\Int \to \backslash y : \Tr\Int \to x \Tr{+} y) \,\Tr 4\, \Tr 5 \\ &= (\backslash y : \Tr\Int \to \Tr 5 \Tr{+} y)\,\Tr 5 \\ &= \Tr 4 \Tr + \Tr 5 \end{align}$$We perform translation in two steps: From line 1 to line 2, we translate the outer $\mathtt{let}$, and from line 2 to line 3 we translate the inner pair term. In line 2, observe that we have wrapped the body of the $\mathtt{let}$ expression in a function to abstract the variables $x$ and $y$. Line 3, then, gives the full translation of the original term. We are not at this point concerned with the translations of integers, so both the various integer constants and the integer type itself remain untranslated throughout this example. The remaining steps are all substitutions. Line 4 substites the type $\Tr\Int$ for type parameter $a$; line 5 substitutes the encoding of $e_2$ for $f$ in the encoding of the pair, and the remaining lines substitute the components of the pair into the body of $f$.

We can take a similar approach to encoding sums. Again, in the evaluation rules, everything interesting happens in the elimination form.

$$ \begin{gather*} \frac{e \Eval v}{\Inl e \Eval \Vinl v} \quad \frac{e \Eval \Vinl w \quad [x_1 \mapsto w]e_1 \Eval v} {\CCase e {x_1} {e_1} {x_2} {e_2} \Eval v} \\[5px] \frac{e \Eval v}{\Inr e \Eval \Vinr v} \quad \frac{e \Eval \Vinr w \quad [x_2 \mapsto w]e_2 \Eval v} {\CCase e {x_1} {e_1} {x_2} {e_2} \Eval v} \end{gather*} $$In the case of products, evaluating the elimination required substituting the components of the pair. For sums, we additionally choose between the two branches.

$$ \begin{align*} \Tr{t_1 + t_2} &= \Pi a. (\Tr{t_1} \to a) \to (\Tr{t_2} \to a) \to a \\ \Tr{\Inl{e} : t_1 + t_2} &= \Lambda a. \backslash f_1 : (t_1 \to a) \to \backslash f_2 : (t_2 \to a) \to f_1 \Tr{e : t_1} \\ \Tr{\Inr{e} : t_1 + t_2} &= \Lambda a. \backslash f_1 : (t_1 \to a) \to \backslash f_2 : (t_2 \to a) \to f_2 \Tr{e : t_2} \\ \Tr{\CCase e {x_1} {e_1} {x_2} {e_2} : u} &= \Tr{e : t_1 + t_2} \, [\Tr u] \, (\lambda x_1 : t_1 \to e_1) \, (\lambda x_2 : t_2 \to e_2) \end{align*} $$The encoding of the type $t_1 + t_2$ follows the pattern of the case statement: given two functions, one from $t_1$ to the result type, the other from $t_2$ to the result type, we produce a value of the result type. The encodings of $\mathtt{Inl}$ and $\mathtt{Inr}$ are reponsibile for making the choice: $\mathtt{Inl}$ calls its first argument, while $\mathtt{Inr}$ calls its second. Finally, the translation of $\texttt{case}$ just has to package the two branches as functions and pass them to the translation of the scrutinee.

Again, a simple example. We lave the constants untranslated.

$$ \begin{align} & \Tr{\CCase{\Inl{4}}{x}{\Isz{x}}{y}{y} : \Bool} \\ &= \Tr{\Inl 4 : \Tr\Int + \Tr\Bool} \, [\Tr\Bool] \, (\lambda x : \Tr\Int \to \Tr{\mathtt{isz}}\, x) \, (\backslash y : \Tr\Bool \to y) \\ &= \begin{split} (\Lambda a. \backslash (f_1 : \Tr\Int \to a) \to \backslash (f_2 : \Tr\Bool \to a) \to f_1 \Tr 4) \, [\Tr\Bool] \\ (\lambda x : \Tr\Int \to \Tr{\mathtt{isz}}\, x) \, (\backslash y : \Tr\Bool \to y) \end{split} \\ &= \begin{split} (\backslash (f_1 : \Tr\Int \to \Tr\Bool) \to \backslash (f_2 : \Tr\Bool \to \Tr\Bool) \to f_1 \Tr 4) \\ (\lambda x : \Tr\Int \to \Tr{\mathtt{isz}}\, x) \, (\backslash y : \Tr\Bool \to y) \end{split} \\ &= (\backslash (f_2 : \Tr\Bool \to \Tr\Bool) \to (\lambda x : \Tr\Int \to \Tr{\mathtt{isz}}\, x) \, \Tr 4) \, (\backslash y : \Tr\Bool \to y) \\ &= (\lambda x : \Tr\Int \to \Tr{\mathtt{isz}}\, x) \, \Tr 4 \\ &= \Tr{\mathtt{isz}}\, \Tr{4} \end{align} $$We split the translation of the source term across lines 8 to 10; from lines 8 to 9 we translate the case block, while from lines 9 to 10 we translate the scrutinee. The remaining lines are substitutions. The key step is from line 11 to 12, when we substitution for $f_1$, and thus set the behavior of the resulting term.

As a final note, we consider the unit type. By itself, the unit type has very little computational content of interest. However, by following the same pattern as we did for products and sums, we arrive at an interesting observation. Recall the evaluation rules for the unit.

$$ \frac{ }{() \Eval \langle \rangle} \quad \frac{e_1 \Eval \langle \rangle \quad e_2 \Eval v} {e_1 ; e_2 \Eval v} $$In evaluting the elimination form, there is no extra information to pass to the body ($e_2$), as the unit carries no other information. Following the pattern above, we have the following translation.

$$ \begin{align*} \Tr{1} &= \Pi a. a \to a \\ \Tr{()} &= \Lambda a. \backslash x : t \to x \\ \Tr{e_1 ; e_2 : u} &= \Tr{e_1} \, [\Tr u] \, \Tr{e_2} \end{align*} $$
The unit type has a unique inhabitant, the unit value. Now, consider the translation of the
unit type. As we have discussed (intuitively) in class, there is also only one possible value
of type $\Pi a. a \to a$. As we have no knowledge of type $a$, the only way we could produce an
$a$ value is to use the one we're given. So this value *must* be the identity function.
The pleasant result here being that the translation of the unit type does not just simulate its
behavior, but also captures its meaning.