TITLE: Proof of the principle of backwards induction
QUESTION [6 upvotes]: I have difficulty in neatly writing down a proof for the following from Terence Tao's Analysis I, where $m{+\!+}$ is the successor function:

Let $n$ be a natural number, and let $P(m)$ be a property pertaining to the natural numbers such that whenever $P(m{+\!+})$ is true, then $P(m)$ is true. Suppose that $P(n)$ is also true. Prove that $P(m)$ is true for all natural numbers $m\leq n$; this is know as the principle of backwards induction. (Hint: apply induction to the variable $n$.)

First of all, I am unsure about what the base case should look like. For the induction step, I understand that if we suppose inductively that $P(n)$ is true, that then for a natural number $a$ s.t. $a{+\!+}=n$ it holds that $P(a)$ is true, and then for a natural number $b$ s.t. $b{+\!+}=a$ it holds that $P(b)$ is true etc. Hence for all natural numbers $m\leq n$, $P(m)$ is true.
Could anyone please tell me what the base case should look like, and whether there is a neater way of writing down the induction step?

REPLY [2 votes]: Same answer as @coffeemath but made more clear. The statement to be proved is this:
$\forall n\in \textbf{N}, \Bigg (P(n) \implies (\forall m \le n, P(m)) \Bigg )$
which we can rewrite as, taking $R(n) := (\forall m \le n, P(m))$
$\forall n\in \textbf{N}, \Bigg (P(n) \implies R(n) \Bigg )$
Define $Q(n):=(P(n) \implies R(n) )$. We want to show $\forall n\in \textbf{N}, Q(n)$ and we shall use the principle of mathematical induction on $Q(n)$.
Step 1: $Q(0):=(P(0) \implies R(0) )\equiv (P(0) \implies P(0) )$. Thus, Q(0) is true.
Step 2: Assume $Q(n)$ is true.
Step 3: Since $P(n{+\!+})\implies P(n)$ and $P(n) \implies R(n)$, we have $P(n{+\!+}) \implies R(n)$. (syllogism). Now $ (P(n{+\!+}) \implies R(n)) \equiv (P(n{+\!+}) \implies R(n)) \land P(n{+\!+})) \equiv (P(n{+\!+}) \implies R(n{+\!+})) \equiv Q(n{+\!+})$.
Hence $Q(n)$ is true $\forall n$. QED.