For an assertion $P(n)$ where $n\ge2$, the following information is given, There is no formal Induction assertion given. But there is an assertion $P(n)$ for $n\ge2$.
(a)We know that $P(2)$ is true.
(b) we know for any $a\ge2$, truth of $P(a)$ implies truth of $P(2a)$.
(c) we know for any $b\ge3$, truth of $P(b)$ implies truth of $P(b-1)$.
We have to use induction to prove that $P(n)$ is true for all $n\ge2$.
My Solution: I use the Well-ordering property (WOP) of the natural numbers.
T = { k>= 2; p(k) is True}
F= {l>= 2, p(l) is False}
so T U F = N(all natural Numbers), and T∩F = ∅
Case 1: (F = ∅ Then T = N so P(n) is always True)
Case 2: (F ≠ ∅ Then ∅ ≠ F ⊆ N s.t there is an element m = min F (WOP))
now, m ≠ 2 because 2 ∈ T (by Hypothesis)
I cant figure out how to use p(a) and p(b) to prove the statement p(n).
 A: Using (a), (b) and induction you can derive that $P(2^n)$ is true for any $n\in\mathbb{N}.$
Now suppose you want to show the validity of $P(n)$ for some random positive integer $n\ge 2.$ Notice that $2^n\gt n,$ and $P(2^n)$ is true. Then use (c) to verify the result for the (decreasing) sequence $P(2^n-1), P(2^n-2),\cdots, P(n+1), P(n).$
This method is known as (Cauchy's) forward backward induction. See here for an example.
A: I think it is easiest to prove using strong induction.
Induction hypothesis: Suppose $P(2), P(3), ..., P(n-1)$ are true.
From this we need to deduce that $P(n)$ is true.
If $n$ is even, then $n=2a$ for some $a<n$. By the induction hypothesis, $P(a)$ is true. From statement (b) it follows that $P(2a)=P(n)$ is true.
If $n$ is odd, then $n+1=2a$ for some $a<n$. As before it follows that $P(2a)=P(n+1)$ is true. From statement (c) with $b=n+1$ it follows that $P(b-1)=P(n)$ is true.
In either case, $P(n)$ is true. Therefore by induction $P(n)$ is true for all values of $n\ge2$.
