TITLE: I was asked to prove the Principle of Cauchy Induction
QUESTION [5 upvotes]: Question: Carefully prove the Principle of Cauchy Induction: Suppose $P(n)$ is a property that a
natural number $n$ may or may not have. Suppose that

$P(2)$ holds,
For every $n ≥ 2$, if $P(n)$ holds, then $P(2n)$ holds, and
For every $n ≥ 3$, if $P(n)$ holds, then $P(n − 1)$ holds.

Then $P(n)$ holds for every natural number $n ≥ 2$.

REPLY [2 votes]: By standard induction you can prove that $P(2^k)$ holds for any $k$.
Base case:  Holds for $P(2)$ and inductive case: If $P(2^k)$ holds so does $P(2^k)$.
For every natural number $m\ge 3$ there is a natural $k$ so that $2^k \le m < 2^{k+1}$ and $k \ge 2$.
As it holds for $2^{k+1}$ and $2^{k+1}> 3$ then it holds for $2^{k+1}-1$ and inductively of $2^{k+1} -(2^k - m)$
If that last line is breezy consider.
Let $Q_k(m)$ be the statement "$P(2^{k+1}-m)$ if $2^{k+1}-m)\ge 3".
Base case: If $m=0$ then $P(2^{k+1})$ is true as we proved above so $Q_k(0)$ is true.
Base case:  If $Q_k(m)$ is true then $P(2^{k+1}-m)$ is true and $P(2^{k+1}-m-1)= Q_k(m+1)$ is true.
And that's it.  It is true for all $n=2^k$ and $n=2^{k+1}$ and all $n: 2^k \le n \le 2^{k+1}$.  So it is true for all natural $n \ge 2$.