2.13 Natural numbers

We use the encode-decode method to characterize the path space of the natural numbers, which are also a positive type. In this case, rather than fixing one endpoint, we characterize the two-sided path space all at once. Thus, the codes for identities are a type family

$\mathsf{code}:\mathbb{N}\to\mathbb{N}\to\mathcal{U},$

defined by double recursion over $\mathbb{N}$ as follows:

	$\displaystyle\mathsf{code}(0,0)$	$\displaystyle:\!\!\equiv\mathbf{1}$
	$\displaystyle\mathsf{code}(\mathsf{succ}(m),0)$	$\displaystyle:\!\!\equiv\mathbf{0}$
	$\displaystyle\mathsf{code}(0,\mathsf{succ}(n))$	$\displaystyle:\!\!\equiv\mathbf{0}$
	$\displaystyle\mathsf{code}(\mathsf{succ}(m),\mathsf{succ}(n))$	$\displaystyle:\!\!\equiv\mathsf{code}(m,n).$

We also define by recursion a dependent function $r:\mathchoice{\prod_{n:\mathbb{N}}\,}{\mathchoice{{\textstyle\prod_{(n:\mathbb% {N})}}}{\prod_{(n:\mathbb{N})}}{\prod_{(n:\mathbb{N})}}{\prod_{(n:\mathbb{N})}% }}{\mathchoice{{\textstyle\prod_{(n:\mathbb{N})}}}{\prod_{(n:\mathbb{N})}}{% \prod_{(n:\mathbb{N})}}{\prod_{(n:\mathbb{N})}}}{\mathchoice{{\textstyle\prod_% {(n:\mathbb{N})}}}{\prod_{(n:\mathbb{N})}}{\prod_{(n:\mathbb{N})}}{\prod_{(n:% \mathbb{N})}}}\mathsf{code}(n,n)$ , with

	$\displaystyle r(0)$	$\displaystyle:\!\!\equiv\star$
	$\displaystyle r(\mathsf{succ}(n))$	$\displaystyle:\!\!\equiv r(n).$

Theorem 2.13.1.

For all $m,n:\mathbb{N}$ we have $(m=n)\simeq\mathsf{code}(m,n)$ .

Proof.

We define

$\mathsf{encode}:\mathchoice{\prod_{m,n:\mathbb{N}}\,}{\mathchoice{{\textstyle% \prod_{(m,n:\mathbb{N})}}}{\prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}% {\prod_{(m,n:\mathbb{N})}}}{\mathchoice{{\textstyle\prod_{(m,n:\mathbb{N})}}}{% \prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}}% {\mathchoice{{\textstyle\prod_{(m,n:\mathbb{N})}}}{\prod_{(m,n:\mathbb{N})}}{% \prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}}(m=n)\to\mathsf{code}(m,n)$

by transporting, $\mathsf{encode}(m,n,p):\!\!\equiv\mathsf{transport}^{\mathsf{code}(m,{\mathord% {\hskip 1.0pt\text{--}\hskip 1.0pt}})}(p,r(m))$ . And we define

$\mathsf{decode}:\mathchoice{\prod_{m,n:\mathbb{N}}\,}{\mathchoice{{\textstyle% \prod_{(m,n:\mathbb{N})}}}{\prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}% {\prod_{(m,n:\mathbb{N})}}}{\mathchoice{{\textstyle\prod_{(m,n:\mathbb{N})}}}{% \prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}}% {\mathchoice{{\textstyle\prod_{(m,n:\mathbb{N})}}}{\prod_{(m,n:\mathbb{N})}}{% \prod_{(m,n:\mathbb{N})}}{\prod_{(m,n:\mathbb{N})}}}\mathsf{code}(m,n)\to(m=n)$

by double induction on $m, n$ . When $m$ and $n$ are both $0$ , we need a function $\mathbf{1}\to(0=0)$ , which we define to send everything to $\mathsf{refl}_{0}$ . When $m$ is a successor and $n$ is $0$ or vice versa, the domain $\mathsf{code}(m,n)$ is $\mathbf{0}$ , so the eliminator for $\mathbf{0}$ suffices. And when both are successors, we can define $\mathsf{decode}(\mathsf{succ}(m),\mathsf{succ}(n))$ to be the composite

$\mathsf{code}(\mathsf{succ}(m),\mathsf{succ}(n))\equiv\mathsf{code}(m,n)% \xrightarrow{\mathsf{decode}(m,n)}(m=n)\xrightarrow{\mathsf{ap}_{\mathsf{succ}% }}(\mathsf{succ}(m)=\mathsf{succ}(n)).$

Next we show that $\mathsf{encode}(m,n)$ and $\mathsf{decode}(m,n)$ are quasi-inverses for all $m, n$ .

On one hand, if we start with $p:m=n$ , then by induction on $p$ it suffices to show

$\mathsf{decode}(n,n,\mathsf{encode}(n,n,\mathsf{refl}_{n}))=\mathsf{refl}_{n}.$

But $\mathsf{encode}(n,n,\mathsf{refl}_{n})\equiv r(n)$ , so it suffices to show that $\mathsf{decode}(n,n,r(n))=\mathsf{refl}_{n}$ . We can prove this by induction on $n$ . If $n\equiv 0$ , then $\mathsf{decode}(0,0,r(0))=\mathsf{refl}_{0}$ by definition of $\mathsf{decode}$ . And in the case of a successor, by the inductive hypothesis we have $\mathsf{decode}(n,n,r(n))=\mathsf{refl}_{n}$ , so it suffices to observe that $\mathsf{ap}_{\mathsf{succ}}(\mathsf{refl}_{n})\equiv\mathsf{refl}_{\mathsf{% succ}(n)}$ .

On the other hand, if we start with $c:\mathsf{code}(m,n)$ , then we proceed by double induction on $m$ and $n$ . If both are $0$ , then $\mathsf{decode}(0,0,c)\equiv\mathsf{refl}_{0}$ , while $\mathsf{encode}(0,0,\mathsf{refl}_{0})\equiv r(0)\equiv\star$ . Thus, it suffices to recall from §2.8 (http://planetmath.org/28theunittype) that every inhabitant of $\mathbf{1}$ is equal to $\star$ . If $m$ is $0$ but $n$ is a successor, or vice versa, then $c:\mathbf{0}$ , so we are done. And in the case of two successors, we have

$\displaystyle\mathsf{encode}(\mathsf{succ}(m),\mathsf{succ}(n),\mathsf{decode}% (\mathsf{succ}(m),\mathsf{succ}(n),c))\\ \displaystyle\begin{aligned} &\displaystyle=\mathsf{encode}(\mathsf{succ}(m),% \mathsf{succ}(n),\mathsf{ap}_{\mathsf{succ}}(\mathsf{decode}(m,n,c)))\\ &\displaystyle=\mathsf{transport}^{\mathsf{code}(\mathsf{succ}(m),{\mathord{% \hskip 1.0pt\text{--}\hskip 1.0pt}})}(\mathsf{ap}_{\mathsf{succ}}(\mathsf{% decode}(m,n,c)),r(\mathsf{succ}(m)))\\ &\displaystyle=\mathsf{transport}^{\mathsf{code}(\mathsf{succ}(m),\mathsf{succ% }({\mathord{\hskip 1.0pt\text{--}\hskip 1.0pt}}))}(\mathsf{decode}(m,n,c),r(% \mathsf{succ}(m)))\\ &\displaystyle=\mathsf{transport}^{\mathsf{code}(m,{\mathord{\hskip 1.0pt\text% {--}\hskip 1.0pt}})}(\mathsf{decode}(m,n,c),r(m))\\ &\displaystyle=\mathsf{encode}(m,n,\mathsf{decode}(m,n,c))\\ &\displaystyle=c\end{aligned}$

using the inductive hypothesis. ∎

In particular, we have

$\mathsf{encode}(\mathsf{succ}(m),0):(\mathsf{succ}(m)=0)\to\mathbf{0}$

(2.13.2)

which shows that “ $0$ is not the successor of any natural number”. We also have the composite

$(\mathsf{succ}(m)=\mathsf{succ}(n))\xrightarrow{\mathsf{encode}}\mathsf{code}(% \mathsf{succ}(m),\mathsf{succ}(n))\equiv\mathsf{code}(m,n)\xrightarrow{\mathsf% {decode}}(m=n)$

(2.13.3)

which shows that the function $\mathsf{succ}$ is injective.

We will study more general positive types in http://planetmath.org/node/87578Chapter 5,http://planetmath.org/node/87579Chapter 6. In http://planetmath.org/node/87582Chapter 8, we will see that the same technique used here to characterize the identity types of coproducts and $\mathbb{N}$ can also be used to calculate homotopy groups of spheres.

Title	2.13 Natural numbers
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