One of my goals this summer is to re-read Bob Carpenter’s Type-Logical Grammar and to actually do (at least some of) the exercises. If I manage to follow through with that goal, I’ll write up my solutions (or attempts at solutions) to some of the more interesting or difficult exercises. I’ll start that off with the last exercise from Chapter 2, which says to add to the simply typed lambda calculus the type $\sigma^*$ for lists of objects of type $\sigma$. First, let me review Chapter 2 to set the stage.

## Review

Chapter 2 deals with the syntax, semantics, and proof theory of the simply typed lambda calculus. I’ll review here the syntax and semantics, but not the proof theory.

### Syntax

Carpenter builds up the set $\mathbf{Typ}$ of types inductively from a set $\mathbf{BasTyp}$ of basic types, using just one type constructor: $\to$.

Definition (types). The set of types is the smallest set $\mathbf{Typ}$ such that

1. $\mathbf{BasTyp} \subseteq \mathbf{Typ}$.
2. If $\sigma, \tau \in \mathbf{Typ}$, then $(\sigma \to \tau) \in \mathbf{Typ}$.

A type of the form $(\sigma \to \tau)$ is called a functional type.

For each type $\tau$, there is a set $\mathbf{Var}_\tau$ of variables of type $\tau$ and a set $\mathbf{Con}_\tau$ of constants of type $\tau$. The set of lambda terms of type $\tau$ is defined as follows.

Definition (terms). The set of (lambda) terms of type $\tau$ is the smallest set $\mathbf{Term}_\tau$ such that

1. $\mathbf{Var}_\tau \subseteq \mathbf{Term}_\tau$.
2. $\mathbf{Con_\tau} \subseteq \mathbf{Term}_\tau$.
3. If $\alpha \in \mathbf{Term}_{\sigma \to \tau}$ and $\beta \in \mathbf{Term}_\sigma$, then $(\alpha(\beta)) \in \mathbf{Term}_\tau$.
4. If $\tau = \sigma \to \rho$ and $x \in \mathbf{Var}_\sigma$ and $\alpha \in \mathbf{Term}_\rho$, then $\lambda x . (\alpha) \in \mathbf{Term}_\tau$.

A term of the form $(\alpha(\beta))$, usually abbreviated $\alpha(\beta)$ or $\alpha\ \beta$, is a functional application of $\alpha$ to $\beta$. However, a term $\alpha$ with a functional type $(\sigma \to \tau)$ is not a function per se—$\alpha(\beta)$ is (the result of) a syntactic operation more akin to term concatenation than functional application. Nevertheless, $\alpha$ is called functional (it has a functional type) because of how it is interpreted semantically (model theoretically).

### Semantics

In the model theory, functional terms denote functions in the normal mathematical sense. Before defining denotations, I need to define frames and models.

Definition (frame). A frame for the set $\mathbf{BasTyp}$ is a collection $\mathbf{Dom} = \bigcup\mathbf{Dom}_\tau$ ($\tau \in \mathbf{BasTyp}$) of basic domains.

Definition (model). A model is a pair $\langle \mathbf{Dom},I \rangle$ in which $\mathbf{Dom}$ is a frame, and $I$ is the interpretation function mapping constants to the appropriate domain, i.e. $I(\alpha) \in \mathbf{Dom}_\tau$ if $\alpha \in \textbf{Con}_\tau$.

Definition (denotation). Given a model $\mathcal{M} = \langle \textbf{Dom}, I \rangle$ and a variable assignment $\theta$, the denotation $[\![\alpha]\!]^\theta_\mathcal{M}$ of a term $\alpha$ with respect to $\mathcal{M}$ and $\theta$ is given by:

1. If $\alpha \in \mathbf{Var}_\tau$, then $[\![\alpha]\!]^\theta_\mathcal{M} = \theta(\alpha)$.
2. If $\alpha \in \mathbf{Con}_\tau$, then $[\![\alpha]\!]^\theta_\mathcal{M} = I(\alpha)$.
3. $[\![\alpha(\beta)]\!]^\theta_\mathcal{M} = [\![\alpha]\!]^\theta_\mathcal{M} ([\![\beta]\!]^\theta_\mathcal{M}]\!])$, assuming $\alpha$ and $\beta$ are appropriately typed.
4. $[\![\lambda x . \alpha]\!]^\theta_\mathcal{M} = f$ such that $f(a) = [\![\alpha]\!]^{\theta[x:=a]}_\mathcal{M}$, where $\theta[x:=a]$ is that variable assignment $\theta'$ such that $\theta'(x) = a$ and $\theta'(y) = \theta(y)$ for each $y \neq x$.

Note that the left-hand side of the equality statement in (3) involves term concatenation (functional application), whereas the right-hand side involves real function application in the strict sense.

In other words, $\alpha$ of type $(\sigma \to \tau)$ is a lambda term whose denotation is a function that maps (elements of the domain of) $\sigma$-type things to (elements of the domain of) $\tau$-type things.

Carpenter goes on to give the usual axioms for the simply typed lamda calculus ($\alpha$-, $\beta$-, and $\eta$-reduction schemata) and the usual rules of inference (reflexivity, transitivity, congruence, and equivalence) and to prove that the resulting system is both sound and complete.

### Product and Sum Types

The latter part of the chapter adds product and sum types to the system. Product types allow functional terms to take multiple arguments at one time (in a certain sense), while sum types introduce functional terms that can act polymorphically.

## Exercise: List Types

Let me now turn to the actual exercise.

Consider adding $\sigma^*$ for lists of objects of type $\sigma$. What would the terms look like in this scheme and how are they interpreted and treated proof-theoretically? Is there any gain in expressive power by admitting arbitrary lists? What might we do to interpret infinite sequences?

In this post, I’ll only address the first two questions: what do list terms look like, and how are they interpreted. For my answer, I’ll be borrowing from the functional programming language Haskell.

In Haskell, lists are things like [3, 5, 0], ["hello", "world"], etc. The order of list elements is important, and list elements can repeat. In that sense, lists are more like tuples (sequences) than sets, except that whereas a tuple (both in Haskell and in Carpenter’s section on product types) can contain objects of all different types, list elements must all be of the same type. For example, in Haskell, (3, "hello") is a possible pair, consisting of an integer and a string, but [3, "hello"] is not a possible list.

In Haskell, the list type is written as [a], where a is any type. For example, [3, 5, 0] has type [Int], while ["hello", "world"] has type [String]. (NB: In Haskell a string is itself a list of characters, i.e. String is just a type synonym for [Char].)

Note that Haskell overloads the symbols []: they are used to construct both terms (Haskell expressions) and types.

I’ll follow Haskell’s lead by writing $[\sigma]$ instead of Carpenter’s $\sigma^*$ for the type of lists containing $\sigma$-type objects, and I’ll similarly overload the symbols $[\ ]$.

### Syntax

I begin by adding $[\sigma]$ to $\mathbf{Typ}$. I do so by adding the following clause to the definition of types:

1. If $\sigma \in \mathbf{Typ}$, then $[\sigma] \in \mathbf{Typ}$.

So $[\ ]$ is a unary type constructor (it takes a single type and returns a new type), whereas $\to$ is a binary type constructor (it takes two types and returns a new type).

As for terms, I assume that for each list type $\tau = [\sigma]$ there is a set $\mathbf{Var}_\tau$ of variables of type $\tau$ and a set $\mathbf{Con}_\tau$ of constants of type $\tau$. We now need a way to construct arbitrary list terms (similar to how abstraction lets us construct functional terms).

In Haskell, there are two list constructors: [], or nil (the empty list), and (:), or cons. The cons operator has the following type signature:

(:) :: a -> [a] -> [a]


This means it takes something of type a and something of type [a] and returns something of type [a]. More precisely, it takes an a-type object—call it x— and a list of a-type objects—call it list—and it returns a new list, list', whose elements are x followed by the elements of list. In other words, cons prepends its first argument (an object of type a) to the beginning of its second argument (a list of objects of type a). (Of course, in Haskell a pure function always returns a new object; it doesn’t modify objects given as arguments, so cons does not literally prepend anything to a given list, unlike, say, list.insert(0, x) in Python.)

For example, to construct a list of integers (objects of type Int), we start with [] of type [Int], and using (:) we add to the front of [] an integer like 0 to obtain (:) 0 [], or using infix notation, 0 : []. The result is of type [Int] and is normally written as [0], which hides the cons operator. From [0] we can obtain 5 : [0], written [5, 0], and from that we can obtain 3 : [5, 0], or [3, 5, 0], and so on.

More generally, [x1, x2, ..., xN] is syntactic sugar for x1 : ( x2 : ( ... ( xN : [] ) ... ) ).

#### Back to Terms

Returning to our lambda terms, we want, for each type $\sigma$, a nil term $[\ ]$ of type $[\sigma]$, and a term constructor for adding an object of type $\sigma$ to a list of type $[\sigma]$ to produce a new such list. I do so by adding the following clauses to the definition of the set $\mathbf{Term}_\tau$ of terms of type $\tau$.

1. If $\tau = [\sigma]$, then $[\ ] \in \mathbf{Term}_\tau$.
2. If $\tau = [\sigma]$ and $\alpha \in \mathbf{Term}_\sigma$ and $\beta \in \mathbf{Term}_\tau$, then $(\alpha : \beta) \in \mathbf{Term}_\tau$.

Like functional application, $:$ is a binary term contructor: it takes two terms to produce a new term.

Following Haskell, I make the following conventions:

• $[\alpha] =_\mbox{def} (\alpha : [\ ])$.
• $[\alpha, \beta, \dots] =_\mbox{def} (\alpha : [\beta, \dots])$.

The latter schema can be applied recursively for any list of arbitrary length. For example

Since there is a one-to-one correspondence between $[\alpha, \beta, \dots, \gamma]$-looking things and $(\alpha : (\beta : (\dots (\gamma : [\ ]) \dots)))$-looking things, one might wonder if we can instead reformulate (6) as follows (in two steps).

1. If $\tau = [\sigma]$ and $\alpha \in \mathbf{Term}_\sigma$, then $[\alpha] \in \mathbf{Term}_\tau$.
2. If $\tau = [\sigma]$ and $\alpha \in \mathbf{Term}_\sigma$ and $[\beta] \in \mathbf{Term}_\tau$, then $[\alpha, \beta] \in \mathbf{Term}_\tau$.

(In other words, we can construct a singleton list from a single object, and a 2-element list from two objects.)

The problem is that (7) cannot be applied recursively. For example, from $0$ we can obtain $[0]$ by appyling (6), and from $5$ and $[0]$ we can obtain $[5,0]$ by applying (7). But we cannot apply (7) to, say, $3$ and $[5,0]$ to obtain $[3,5,0]$ because, although $[5,0]$ is of type $[\sigma]$, $5,0$ is not a term of any kind, let alone one of type $\sigma$.

For this reason, and for the sake of making cons explicit, I’ll stick with the first formulation of (6) above.

### Semantics

Now that I’ve added list types and list terms to the syntax, I need to add their denotations to the model theory, and this is where I’m a bit stuck, primarily because there are several options and I don’t know which is best/normal, or which one Carpenter has in mind.

If $\alpha$ is of a list type, say $[\sigma]$, then the denotation of $\alpha$ is in $\mathbf{Dom}_{[\sigma]}$, but what precisely is the structure of $\mathbf{Dom}_{[\sigma]}$?

For example, functional terms denote functions, so the denotation of $\alpha$ of type $(\sigma \to \tau)$ is a member of $\mathbf{Dom}_{\tau}^{\mathbf{Dom}_\sigma}$, i.e. a function from $\mathbf{Dom}_\sigma$ to $\mathbf{Dom}_\tau$. Product terms denote pairs (and more generally, tuples), so the denotation of $\langle \alpha, \beta \rangle$ of type $\sigma \times \tau$ is a member of $\mathbf{Dom}_\sigma \times \mathbf{Dom}_\tau$. But what does/should a list $\alpha$ of type $[\sigma]$ denote, set theoretically? A flat set, i.e. a member of the power set of $\mathbf{Dom}_\sigma$? Probably not, because sets don’t care about order or repetition, whereas lists (as I’ve defined them) do. What about tuples, i.e. members of $\mathbf{Dom}_\sigma \times \dots \times \mathbf{Dom}_\sigma$? Maybe, but product terms already do that. Something else? I’m not sure.

A concrete example might help decide. Suppose we have the type $\mathbf{Ind}$ for individuals and that $\mathbf{john}, \mathbf{bill}, \mathbf{sue} \in \mathbf{Term}_\mathbf{Ind}$. Then we can create a list of these individuals, $[\mathbf{john}, \mathbf{bill}, \mathbf{sue}]$, which is of type $[\mathbf{Ind}]$. Suppose also that in some model $\mathcal{M}$, $\mathbf{john}$ denotes $j$, $\mathbf{bill}$ denotes $b$, and $\mathbf{sue}$ denotes $s$, where $j,b,s \in \mathbf{Dom}_\mathbf{Ind}$. Then what should the list of these three individuals denote? Probably not the flat set $\{j,b,s\} \in \mbox{Pow}(\mathbf{Dom}_\mathbf{Ind})$, because then $[\mathbf{john}, \mathbf{bill}, \mathbf{sue}]$ and $[\mathbf{bill}, \mathbf{john}, \mathbf{sue}]$ and $[\mathbf{john}, \mathbf{bill}, \mathbf{bill}, \mathbf{sue}]$ would all denote the same set (at least if we apply the most obvious denotation mapping), which is probably undesirable. One reason for introducing lists in the first place is, I assume, so they can serve as meaning representations for natural language expressions that care about order and repetition.

Assuming, then, that we want these three lists to denote different things, the only reasonable possibility I can think of is to map lists to tuples. For example, $[\mathbf{john}, \mathbf{bill}, \mathbf{sue}]$ will denote $\langle j,b,s \rangle$, which is an element of $\mathbf{Ind} \times \mathbf{Ind} \times \mathbf{Ind}$. In this way, list terms are essentially like product terms, but with two main differences: (i) both products and lists denote tuples, but lists are composed of objects of the same type, meaning that the tuples they denote will contain objects from the same domain; and (ii) empty and singleton lists are possible, whereas empty and singleton tuples are in general (at least for Carpenter) not, as far as I can tell.

The following clauses get added to the definition of denotation.

1. If $\tau = [\sigma]$ and $\alpha = [\ ] \in \mathbf{Term}_\tau$, then $[\![\alpha]\!]^\theta_\mathcal{M} = \emptyset$.
2. $[\![(\alpha : \beta)]\!]^\theta_\mathcal{M} = \langle [\![\alpha]\!]^\theta_\mathcal{M}, [\![\beta]\!]^\theta_\beta \rangle$, assuming $\alpha$ and $\beta$ are appropriately typed.

Continuing the example above, $[\mathbf{john}] = (\mathbf{john} : [\ ])$ denotes $\langle j, \emptyset \rangle \in \mathbf{Dom}_\mathbf{[Ind]}$. And $[\mathbf{john}, \mathbf{bill}, \mathbf{sue}]$ denotes $\langle j, \langle b, \langle s,\emptyset \rangle \rangle \rangle \in \mathbf{Dom}_\mathbf{[Ind]}$. Rewriting tuples in the usual way (analogous to lists above), where $\langle a,b,c, \dots \rangle =_\mbox{def} \langle a, \langle b,c, \dots \rangle \rangle$, the latter denotation becomes $\langle j,b,s,\emptyset \rangle$.

So what is the structure of $\mathbf{Dom}_\mathbf{[Ind]}$? It’s a set whose members are tuples, of varying size, consisting of individuals (except the last tuple member, which is always $\emptyset$ except in the case of the empty tuple).

More generally, $\mathbf{Dom}_{[\sigma]}$ is a set whose members are tuples, of varying size, consisting of elements from $\mathbf{Dom}_\sigma$.

This is different from the case of products: the domain of intepretation for, say, a pair of terms $\langle \alpha,\beta \rangle$ is $\mathbf{Dom}_\sigma \times \mathbf{Dom}_\tau$, which contains only pairs, not tuples of any other size; and the domain of interpretation for, say, a triple of terms $\langle \alpha,\beta,\gamma \rangle$ is $\mathbf{Dom}_\sigma \times \mathbf{Dom}_\tau \times \mathbf{Dom}_\rho$, which contains only triples, not tuples of any other size. Hence, whereas, say, $[\alpha, \beta]$ and $[\alpha, \beta, \gamma]$ have denotations in the same domain, $\langle \alpha,\beta \rangle$ and $\langle \alpha,\beta,\gamma \rangle$ do not.

I’ll conclude by mentioning that it’s kind of weird to use $\emptyset$ in the denotation of lists, but this is necessary because the denotation function is a total function, and since the empty list, $[\ ]$, is a term, it requires a denotation. If $[\ ]$ is of type $[\sigma]$, it doesn’t make sense for its denotation to include anything from $\mathbf{Dom}_\sigma$, so we’re stuck with having it denote something like $\emptyset$.

One way to avoid this would be to redefine lists: empty lists are simply not possible terms (objects). The smallest list would then be a singleton list, which would denote a $1$-tuple, so $[\mathbf{john}]$ would denote $\langle j \rangle$ rather than $\langle j,\emptyset \rangle$, and $[\mathbf{john}, \mathbf{bill}, \mathbf{sue}]$ would denote $\langle j,b,s \rangle$ rather than $\langle j,b,s,\emptyset \rangle$.