Reverse induction? Typically, induction is done by proving that $P(1)$ and $P(n) \implies P(n+1)$ for all $n$. Is it possible to instead prove $P(1)$ and $P(n+1) \implies P(n)$? Provided that $n$ is arbitrarily large, it seems that this should prove the statement for all $n$.
 A: There is no flat out pure reverse induction, but there is a form of "forward-backward" induction where you prove it works for some sequence that goes to infinity,  then prove that if it works for $n$ it works for $n-1$.
The first place people tend to see this is in proving the generalized arithmetic/geometric mean inequality:
$$\frac {x_1 +x_2 + \dots +x_n}n\geq (x_1x_2x_3 \dots x_n)^{\frac 1 n}$$
First you prove it works for arbitrary powers of 2,  i.e for $n=2^k$.   Then you prove that if it works for any $n$,  it works for $n-1$.  This gets you all natural numbers because you can get to any natural number by first reaching a power of two above it, then going backward.
This generalizes to saying that forward/backward induction works if you can show for some sequence $a_n$ where $a_n\to \infty$ $p(a_n)$  holds and that if $n\geq (\text{starting number }+1)$  and $p(n)$ holds then $p(n-1)$ holds, then $p$ holds for all numbers from your start upward.
A: Proving $P(1)$ and $\forall n[P(n+1)\to P(n)]$ won't buy anything; indeed things break at the first step: since you don't know whether $P(n)$ holds, you can't infer $P(n-1)$.
You can do reverse induction when things are bounded above. Say you want to show that $P(n)$ holds for every $n\in \{1,\dots,N\}$, where $N$ is fixed. Then, if you can prove $P(N)$ and $P(n)\to P(n-1)$ for every $n\in \{2,\dots,N\}$, you can infer $\forall n\in \{1,\dots,N\} P(n)$.
Note that this is just regular induction with a fake mustache: let $Q(n)$ be the assertion $P(N-n)$. Then "reverse induction" is just regular induction with $Q$.
A: Reverse induction is meaningless or useless to "prove the statement for all $n$". The scope of induction is to "conquer" step by step all the numerable values for $\mathbb{N}$, in such way to show that $P(n)$ holds for all $n\ge n_0$. The base case is the first step and to proceed we need to go forward not backward.
A: In a sense this is true.  If we have $P(1)$ and also $\forall n [P(n+1) \Rightarrow P(n)]$, then we can conclude that $P(k)$ holds for all integers $\le 1$.  That is, for $k = 1,0,-1,-2,-3,\dots$.
Example.  Let $P(n)$ be: "$n-2 < 0$".  We know $1-2 < 0$ and from $n+1 < 0$ we can deduce $n = (n+1)-1 < 0-1 = -1 < 0$ so $n < 0$.  The conclusion would be $k-2 < 0$ for all integers $k \le 1$.
