nLab
natural numbers type

Idea

In type theory: the the natural numbers type is the type of natural numbers.

Definition

Definition

The type of natural numbers $\mathbb{N}$ is the inductive type defined as follows.

1. type formation rule

   $$\frac{}{\mathbb{N} : Type}$$

2. term introduction rules

   $$\frac{}{0 : \mathbb{N}}$$

   and

   $$\frac{n : \mathbb{N}}{s n : \mathbb{N}}$$

3. term elimination rule

   $$\frac{ x : \mathbb{N} \vdash P(x) : Type; \;\; \vdash p_0 : P(0); \;\; x : \mathbb{N}, p : P(x) \vdash p_s(x,p) : P(s x); \;\; \vdash n : \mathbb{N} }{ rec^n_P(p_0,p_s) : P(n) }$$

4. computation rules

   $$\frac{ x : \mathbb{N} \vdash P(x) : Type; \;\; \vdash p_0 : P(0); \;\; x : \mathbb{N}, p : P(x) \vdash p_s(x,p) : P(s x) }{ rec^0_P(p_0,p_s) = p_0 : P(0) }$$

   and

   $$\frac{ x : \mathbb{N} \vdash P(x) : Type; \;\; \vdash p_0 : P(0); \;\; x : \mathbb{N}, p : P(x) \vdash p_s(x,p) : P(s x); \;\; \vdash n : \mathbb{N} }{ rec^{s n}_P(p_0,p_s) = p_s(n,rec^n_P(p_0,p_s)) : P(s n) }$$

See for instance (Pfenning, section 2).

In Coq-syntax the natural numbers are the inductive type defined by

Inductive nat : Type :=
 | zero : nat
 | succ : nat -> nat.

In the categorical semantics this is interpreted as the initial algebra for the endofunctor $F$ that sends an object to its coproduct with the terminal object

$$F(X) = * \coprod X \,,$$

or in different equivalent notation, which is very suggestive here:

$$F(X) = 1 + X \,.$$

That initial algebra is (as explained there) precisely a natural number object $\mathbb{N}$. The two components of the morphism $F(\mathbb{N}) \to \mathbb{N}$ that defines the algebra structure are the 0-element $zero : * \to \mathbb{N}$ and the successor endomorphism $successor : \mathbb{N} \to \mathbb{N}$

$$(zero,successor) : * \coprod \mathbb{N} \to \mathbb{N} \,.$$

In the following we write of course for short $0 : * \to \mathbb{N}$ and $s : \mathbb{N} \to \mathbb{N}$.

Properties

Induction

Example

We spell out in detail how the fact that $\mathbb{N}$ satisfied def. is the classical induction principle.

That principle says informally that if a proposition $P$ depending on the natural numbers is true at $n = 0$ and such that if it is true for some $n$ then it is true for $n+1$, then it is true for all natural numbers.

Here is how this is formalized in type theory and then interpreted in some suitable ambient category $\mathcal{C}$.

First of all, that $P$ is a proposition depending on the natural numbers means that it is a dependent type

$$n \in \mathbb{N} \vdash P(n) : Type \,.$$

The categorical interpretation of this is by a display map

$$\array{ P \\ \downarrow \\ \mathbb{N} }$$

in the given category $\mathcal{C}$.

Next, the fact that $P$ holds at 0 means that there is a (proof-)term

$$\vdash p_0 \in P(0) \,.$$

In the categorical semantics the substitution of $n$ for 0 that gives $P(0)$ is given by the pullback of the above fibration

$$\array{ 0^* P &\to& P \\ \downarrow && \downarrow \\ * &\stackrel{0}{\to}& \mathbb{N} }$$

and the term $p_0$ is interpreted as a section of the resultig fibration over the terminal object

$$\array{ * &\stackrel{p_0}{\to}& 0^* P &\to& P \\ &\searrow& \downarrow && \downarrow \\ && * &\stackrel{0}{\to}& \mathbb{N} } \,.$$

But by the defining universal property of the pullback, this is equivalently just a commuting diagram

$$\array{ * &\stackrel{p_0}{\to}& P \\ \downarrow && \downarrow \\ * &\stackrel{0}{\to}& \mathbb{N} } \,.$$

Next the induction step. Formally it says that for all $n \in \mathbb{N}$ there is an implication $p_s(n) : P(n) \to P(n+1)$

$$n \in \mathbb{N} \vdash p_s(n) : P(n) \to P(n+1)$$

The categorical semantics of the substitution of $n+1$ for $n$ is now given by the pullback

$$\array{ P((-)+1) \coloneqq & s^*P &\to& P \\ & \downarrow && \downarrow \\ & \mathbb{N} &\stackrel{s}{\to}& \mathbb{N} }$$

and the interpretation of the implication term $p_s(n)$ is as a morphism $P \to s^* P$ in $\mathcal{C}_{/\mathbb{N}}$

$$\array{ P & \stackrel{p_s}{\to} & s^*P &\to& P \\ &\searrow & \downarrow && \downarrow \\ && \mathbb{N} &\stackrel{s}{\to}& \mathbb{N} } \,.$$

Again by the universal property of the pullback this is the same as a commuting diagram

$$\array{ P &\stackrel{p_s}{\to}& P \\ \downarrow && \downarrow \\ \mathbb{N} &\stackrel{s}{\to}& \mathbb{N} } \,.$$

In summary this shows that the fact that $P$ is a proposition depending on natural numbers which holds at 0 and which holds at $n+1$ if it holds at $n$ is interpreted precisely as an $F$-algebra homomorphism

$$\array{ P \\ \downarrow \\ \mathbb{N} } \,.$$

The induction principle is supposed to deduce from this that $P$ holds for every $n$, hence that there is a proof $p_n : P(n)$ for all $n$:

$$n \in \mathbb{N} \vdash p(n) : P(n) \,.$$

The categorical interpretation of this is as a morphism $p : \mathbb{N} \to P$ in $\mathcal{C}_{/\mathbb{N}}$. The existence of this is indeed exactly what the interpretation of the elimination rule, def. , gives, or (equivalently by prop. ) exactly what the initiality of the $F$-algebra $\mathbb{N}$ gives.

Recursion

Example

We spell out how the fact that $\mathbb{N}$ satisfies def. is the classical recursion principle.

So let $A$ be an $F$-algebra object, hence equipped with a morphism

$$a_0 : * \to A$$

and a morphism

$$h : A \to A \,.$$

By initiality of the $F$-algebra $\mathbb{N}$, there is then a (unique) morphism

$$f : \mathbb{N} \to A$$

such that the diagram

$$\array{ * &\stackrel{0}{\to}& \mathbb{N} &\stackrel{(-)+1}{\to}& \mathbb{N} \\ \downarrow && \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{f}} \\ * &\stackrel{a_0}{\to}& A &\stackrel{h}{\to}& A }$$

commutes. This means precisely that $f$ is the function defined recursively by

1. $f(0) = a_0$;

2. $f(n+1) = h(f(n))$.

Related concepts

• natural numbers, natural numbers object

References

• Frank Pfenning, Lecture notes on natural numbers (2009) (pdf)