The construction of the natural integers with the Peano lemma is at the root of a powerful way to prove theorems, the proof by recursion.

But let's study an example.

Let's prove that the natural integers are alternatively even and odd.

An even number n is either zero or the k-th follower of a given number k: n equals 2k.

And an odd number n is either 1 or the follower of the k-th follower of a given number k: n equals 2k plus 1.

We will formulate the rule of alternated parities of natural integers in a theorem.

The successive natural integers are alternatively even and odd.

That means that every natural integer is either even or odd.

If n is even, then its follower is odd.

And if n is odd, then its follower is even.

Let's start the proof by recursion with the initialization.

If n is equal to 0, 0 is even and its follower 1 is odd.

For n equals 1, 1 is odd and its follower 2 is the first follower of k equals 1, thus it is even.

For n equals 2, 2 is even and its follower 3 is the follower of the first follower of 1, thus it is odd.

n equals 3, 3 is odd and its follower 4 is the second follower of 2, thus it is even.

Now we shall make one step.

The recursion hypothesis is that the theorem is true for n that is non-zero or one.

Then n is either even or odd.

The question is, is the theorem also true for the follower of n? The proof depends on whether n is even or odd.

Let's first study the case where n is even non-zero.

Then n is the kth follower of a given number k.

And then the follower of n is the follower of the k-th follower of a number k.

Thus, the follower of n is odd.

Now let's make the step in the second case.

n is odd, non 1.

Then n is the follower of the k-th follower of a given number k.

Consequently, n is also the k-th follower of the follower k plus one of the number k, because we may switch the k-th follower and the follower.

Consequently, the follower of n is the k plus one follower, one more, of the number k plus one.

Consequently, the follower of n is even.

And now let's draw a conclusion.

The theorem is true for the first four natural integers 0, 1, 2 and 3.

And for the follower of any natural integer n for which it is true.

The theorem is completely proved.

What's coming now? Now we shall compare natural integers and we shall see N as an increasing sequence.