Method of Infinite Descent/Proof 2

Proof
Suppose that $\map P {n_\alpha}$ holds.

Then from the descent step, $\exists n_\beta \in \N_{n_\alpha}: \map P {n_\beta}$.

The descent step then tell us we can deduce a smaller positive solution, $n_\gamma$, such that $\map P {n_\gamma}$ is true and $n_\gamma \in \N_{n_\beta}$.

And again, the descent step tells us we can deduce a still smaller positive solution, $n_\delta$, such that $\map P {n_\delta}$ is true and $n_\delta \in \N_{n_\gamma}$.

Now, consider the unending sequence: $n_\alpha > n_\beta > n_\gamma > n_\delta > \cdots > 0$.

The set $S = \set {n_\alpha, n_\beta, n_\gamma, n_\delta, \ldots}$ is not bounded below, as for any $\forall x \in S: \exists y \in S: y < x$.

By the Well-Ordering Principle, any non-empty subset of $\N$ must have a least element.

As $S$ is not bounded below, it has no least element, so must be empty.