Forward-Backward Induction

Theorem
Let $$P \left({n}\right)$$ be a propositional function depending on $$n \in \N$$.

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

Then $$P \left({n}\right)$$ holds for all $$\forall n \in \N$$.

Proof
Suppose $$\exists k \in \N$$ such that $$P \left({k}\right)$$ is false.

Since $$\left\{{2^n: n \in \N}\right\}$$ is unbounded above by Boundedness of Nth Powers‎, we can find $$M = 2^N > k$$.

Now let us create the set $$S = \left\{{n \in \N: n < M, P \left({n}\right) \mbox{ is false}}\right\}$$.

As $$k < M$$ and $$P \left({k}\right)$$ is false, $$S \ne \varnothing$$ and as $$\forall x \in S: x < M$$ it follows that $$S$$ is bounded above.

So from Integers Bounded Above has Maximal Element, $$S$$ has a maximal element.

However, $$P \left({n}\right)$$ holds for $$m < n \le M$$ and hence $$P \left({m+1}\right)$$ holds.

But $$P \left({m+1}\right) \implies P \left({m}\right)$$ and so $$P \left({m}\right)$$ holds after all.

So there can be no such $$m$$ and therefore $$S = \varnothing$$, hence there can be no such $$k \in \N$$ such that $$P \left({k}\right)$$ is false.

Hence the result.