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.

Axiom of Choice
Care is required to avoid accidentally using the Axiom of Countable Choice for Finite Sets, which is independent of Zermelo-Fraenkel Set Theory.

Before beginning the descent, construct:
 * $\set {x : \map P x \land x < m}$

for each $m \leq n_\alpha$.

If it would be empty, define it to be any nonempty set instead.

Applying the Principle of Finite Choice, choose an element from each one.

Now, perform the descent, using the choice function to choose the next element at each step.

Because at each step, there are smaller solutions, it will never be the arbitrary set.

Therefore, the sequence is defined as required.