Confusion on a proof from Terence Tao's Analysis 1 I recently started learning mathematical induction and I saw this question on here about the proof of backwards induction.
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 $n$.)

I've seen quite a few answers on this yet its still confusing me mainly because I don't quite grasp the assumption about whenever $P(m{+\!+})$ is true, then $P(m)$ is true i know that for induction you must first prove the base case showing that from what you assume is true actually leads to a truth then you carry out the induction however assuming that that $\mathbb{N}=\{1,2,3,\dots\}$ for the base case $n=1$ how can I show that the assumption $P(m{+\!+})$ $\implies$ $P(m)$ is true as if what exactly is $P(m)$  I saw on other posts that we just say that only $1\leq 1$ and by assumption $P(1)$ was true but doing this we didn't show that $P(m{+\!+}) \implies P(m)$ is true. I'm not quite sure exactly how to deal with the base case here to show that all the assumptions are true this has confused the other users on s.e. too some help would be greatly appreciated.
Thanks in advance.
 A: The base case is straightforward (and it does not change if we "start from" $1$ instead of $0$. According to Tao, page 15, natural numbers start from $0$).
If $P(1)$ holds, then $P(k)$ holds for every $k \le 1$ (there is only $1$).
The fact $P(m++) \to P(m)$ is not used here; as you can see, it is not necessary. Also, $1$ (the "first" number) is not a successor of any number: thus, you cannot find an $m$ such that $1=m++$.
If instead we admit also $0$, from $P(1)$ and $P(m++) \to P(m)$, we have immediately that $P(0)$ also holds.
In both cases: $P(k)$ holds for every $k \le 1$.
Assume now the Induction Hypothesis, i.e. that:

if $P(n)$ holds, then $P(k)$ holds for every $k \le n$.

We want to prove the same for $n++$.
Assume that $P(n++)$ holds. Using the fact that $P(m++) \to P(m)$ holds, for every $m$, we have that (Modus Ponens) also $P(n)$ holds.
Thus, using IH, we have that $P(k)$ holds for every $k \le n$. But we have assumed that $P(n++)$ holds.
Thus: $P(k)$ holds for every $k \le n++$.
A: We wish to prove by induction that $\varphi(n)$ for all $n\ge0$, where $\varphi(n)$ means$$P(n)\to\forall m\le n(P(m)),$$so $\varphi(0)$ is vacuously true. If $\varphi(k)$ then $P(k++)\to P(k)$ and $P(k)\to\forall m\le k(P(m))$, and so$$P(k+1)\to\forall m\le k++(P(m)),$$which is just $\varphi(k++)$.
A: The intuitive idea:
$$\forall m: P(m++)\implies P(m)$$
means than all the implication chains
$$P(n)\implies\cdots\implies P(2)\implies P(1)$$
are true.
Now, what happens if (for some $n$) $P(n)$ is true?
