Forward-Backward Induction

Theorem
Let $P$ be a propositional function on the natural numbers $\N$.

Suppose that:
 * $(1): \quad \forall n \in \N: \map P {2^n}$ holds.
 * $(2): \quad \map P n \implies \map P {n - 1}$.

Then $\map P n$ holds for all $\forall n \in \N$.

The proof technique based on this result is called forward-backward induction.

Proof
$\exists k \in \N$ such that $\map P k$ is false.

From Power of Real Number greater than One is Unbounded Above‎:
 * $\set {2^n: n \in \N}$ is unbounded above.

Therefore we can find:
 * $M = 2^N > k$

Now let us create the set:
 * $S = \set {n \in \N: n < M, \map P n \text { is false} }$

As $k < M$ and $\map P k$ is false:
 * $S \ne \O$

As $\forall x \in S: x < M$ it follows that $S$ is bounded above.

So from Set of Integers Bounded Above by Integer has Greatest Element, $S$ has a greatest element, which we call $m$.

We have that $\map P n$ holds for $m < n \le M$.

Hence $\map P {m + 1}$ holds.

But $\map P {m + 1} \implies \map P m$.

This contradicts our assertion that $\map P n$ is false.

Hence by Proof by Contradiction there can be no such $k \in \N$ such that $\map P k$ is false.

Hence the result.