Your question can be understood in different ways. András has answered one way, here is another way.

It is a consequence of the $\mathbb{N}$-elimination rule that, to define any function $f:\mathbb{N}\to A$, it is enough to define it for $0$, and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$. So we can say: "to define $f(x)$, we can suppose that $x$ is either $0$ or $s(a)$".

The classical recursion principle is also a consequence of the $\mathbb{N}$-elimination rule: to prove a property for all $x:\mathbb{N}$, it is enough to prove it for $0$ and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$.

Using this principle you can prove the property: $\Pi_{x:\mathbb{N}} (x=0) + (\Sigma_{a:\mathbb{N}} (x=s(a))$.

Your question can be understood in different ways. András has answered one way, here is another way.

It is a consequence of the $\mathbb{N}$-elimination rule that, to define any function $f:\mathbb{N}\to A$, it is enough to define it for $0$, and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$. So we can say: "to define $f(x)$, we can suppose that $x$ is either $0$ or $s(a)$".

The classical recursion principle is also a consequence of the $\mathbb{N}$-elimination rule: to prove a property for all $x:\mathbb{N}$, it is enough to prove it for $0$ and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$.

Using this principle you can prove the property: $\Pi_{x:\mathbb{N}} (x=0) + (\Sigma_{a:\mathbb{N}} (x=s(a)))$.

Your question can be understood in different ways. Andras has answered one way, here is another way.

It is a consequence of the $\mathbb{N}$-elimination rule that, to define any function $f:\mathbb{N}\to A$, it is enough to define it for $0$, and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$. So we can say: "to define $f(x)$, we can suppose that $x$ is either $0$ or $s(a)$".

The classical recursion principle is also a consequence of the $\mathbb{N}$-elimination rule: to prove a property for all $x:\mathbb{N}$, it is enough to prove it for $0$ and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$.

Using this principle you can prove the property  : $\Pi_{x:\mathbb{N}} (x=0) + (\Sigma_{a:\mathbb{N}} (x=s(a))$.

Your question can be understood in different ways. András has answered one way, here is another way.

It is a consequence of the $\mathbb{N}$-elimination rule that, to define any function $f:\mathbb{N}\to A$, it is enough to define it for $0$, and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$. So we can say: "to define $f(x)$, we can suppose that $x$ is either $0$ or $s(a)$".

The classical recursion principle is also a consequence of the $\mathbb{N}$-elimination rule: to prove a property for all $x:\mathbb{N}$, it is enough to prove it for $0$ and for $s(a)$, where $a$ is an arbitrary inhabitant of $\mathbb{N}$.

Using this principle you can prove the property: $\Pi_{x:\mathbb{N}} (x=0) + (\Sigma_{a:\mathbb{N}} (x=s(a))$.