dependent product type

**natural deduction** metalanguage, practical foundations

**type theory** (dependent, intensional, observational type theory, homotopy type theory)

**computational trinitarianism** = **propositions as types** +**programs as proofs** +**relation type theory/category theory**

In dependent type theory, a *dependent product type* $\prod_{x\colon A} B(x)$, for a dependent type $x\colon A\vdash B(x)\colon Type$ is the type of “dependently typed functions” assigning to each $x\colon A$ an element of $B(x)$.

In a model of the type theory in categorical semantics, this is a dependent product. In set theory, it is an element of an indexed product.

It includes function types as the special case when $B$ is not dependent on $A$. Note that a binary product type is rather different, being actually a special case of a dependent sum type.

type theory | category theory | |
---|---|---|

syntax | semantics | |

natural deduction | universal construction | |

dependent product type | dependent product | |

type formation | $\displaystyle\frac{\vdash\: X \colon Type \;\;\;\;\; x \colon X \;\vdash\; A(x)\colon Type}{\vdash \; \left(\prod_{x \colon X} A\left(x\right)\right) \colon Type}$ | |

term introduction | $\displaystyle\frac{x \colon X \;\vdash\; a\left(x\right) \colon A\left(x\right)}{\vdash (x \mapsto a\left(x\right)) \colon \prod_{x' \colon X} A\left(x'\right) }$ | |

term elimination | $\displaystyle\frac{\vdash\; f \colon \left(\prod_{x \colon X} A\left(x\right)\right)\;\;\;\; \vdash \; x \colon X}{x \colon X\;\vdash\; f(x) \colon A(x)}$ | |

computation rule | $(y \mapsto a(y))(x) = a(x)$ |

Like any type constructor in type theory, a dependent product type is specified by rules saying when we can introduce it as a type, how to construct terms of that type, how to use or “eliminate” terms of that type, and how to compute when we combine the constructors with the eliminators.

The type formation rule for dependent product type is:

$\frac{A\colon Type \qquad x\colon A \vdash B(x) \colon Type}{\prod_{x\colon A} B(x)\colon Type}$

Dependent product types are almost always defined as negative types. In this presentation, primacy is given to the eliminators. The natural eliminator of a dependent product type says that we can *apply* it to any input:

$\frac{f\colon \prod_{x\colon A} B(x) \qquad a\colon A}{f(a) \colon B(a)}$

The constructor is then determined as usual for a negative type: to construct a term of $\prod_{x\colon A} B(x)$, we have to specify how it behaves when applied to any input. In other words, we should have a term of type $B(x)$ containing a free variable $x\colon A$. This yields the usual “$\lambda$-abstraction” constructor:

$\frac{x\colon A\vdash b\colon B(x)}{\lambda x.b\colon \prod_{x\colon A} B(x)}$

The β-reduction rule is the obvious one, saying that when we evaluate a $\lambda$-abstraction, we do it by substituting for the bound variable.

$(\lambda x.b)(a) \;\to_\beta\; b[a/x]$

If we want an η-conversion rule, the natural one says that every dependently typed function is a $\lambda$-abstraction:

$\lambda x.f(x) \;\to_\eta\; f$

It is also possible to present dependent product types as a positive type. However, this requires a stronger metatheory, such as a logical framework. We use the same constructor ($\lambda$-abstraction), but now the eliminator says that to define an operation using a function, it suffices to say what to do in the case that that function is a lambda abstraction.

$\frac{(x\colon A \vdash b\colon B(x)) \vdash c\colon C \qquad f\colon \prod_{x\colon A} B(x)}{funsplit(c,f)\colon C}$

This rule cannot be formulated in the usual presentation of type theory, since it involves a “higher-order judgment”: the context of the term $c\colon C$ involves a “term of type $B(x)$ containing a free variable $x\colon A$”. However, it is possible to make sense of it. In dependent type theory, we need additionally to allow $C$ to depend on $\prod_{x\colon A} B(x)$.

The natural $\beta$-reduction rule for this eliminator is

$funsplit(c, \lambda x.g) \;\to_\beta c[g/b]$

and its $\eta$-conversion rule looks something like

$funsplit(c[\lambda x.b / g], f) \;\to_\eta\; c[f/g].$

Here $g\colon \prod_{x\colon A} B(x) \vdash c\colon C$ is a term containing a free variable of type $\prod_{x\colon A} B(x)$. By substituting $\lambda x.b$ for $g$, we obtain a term of type $C$ which depends on “a term $b\colon B(x)$ containing a free variable $x\colon A$”. We then apply the positive eliminator at $f\colon \prod_{x\colon A} B(x)$, and the $\eta$-rule says that this can be computed by just substituting $f$ for $g$ in $c$.

As usual, the positive and negative formulations are equivalent in a suitable sense. They have the same constructor, while we can formulate the eliminators in terms of each other:

$\begin{aligned}
f(a) &\coloneqq funsplit(b[a/x], f)\\
funsplit(c, f) &\coloneqq c[f(x)/b]
\end{aligned}$

The conversion rules also correspond.

In dependent type theory, this definition of $funsplit$ only gives us a properly typed dependent eliminator if the negative dependent product type satisfies $\eta$-conversion. As usual, if it satisfies propositional eta-conversion then we can transport along that instead—and conversely, the dependent eliminator allows us to prove propositional $\eta$-conversion. This is the content of Propositions 3.5, 3.6, and 3.7 in (Garner).

The standard rules for type-formation, term introduction/elimination and computation of dependent product type are listed for instance in part I of

- Nicola Gambino,
*Lectures on dependent type theory*(pdf)

See also

- Richard Garner,
*On the strength of dependent products in the type theory of Martin-Löf*, arXiv.

Revised on October 2, 2012 22:22:50
by Urs Schreiber
(82.169.65.155)