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So I understand that induction leads from the fifth axiom of natural numbers, where we say that we show the statement must be true for P(n + 1) if it is true for P(n) for some arbitrary n, then for all P(n) (starting at our base case), it must be true for its successor.

But can we use P(n - 1) in our proof?

For example, assume that P(n) says that $F(n)^2 + F(n + 1)^2 = F(2n + 1)$, are we allowed to also assume P(n - 1), namely that $F(n - 1)^2 + F(n)^2 = F(2n)$?

This question arose because I looked at the proof here: here and it seemed like they assumed exactly that.

My question is: why is this logically valid?

bob
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user129393192
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  • Generally induction needs starting point and chain step. You can prove sentence for chosen $k_0$ and then, having induction step after it, gives proof for every $n \geqslant k_0$. Chain step itself can be any: from $n$ to $n+1$, from $n-1$ to $n$ and any other between neighbors. – zkutch Oct 08 '23 at 18:33
  • I understand, but if we assume $P(n)$ and want to show $P(n + 1)$ in our chain step, can we also validly assume $P(n - 1)$ in our proof? This is because the proof I linked does exactly that. @zkutch – user129393192 Oct 08 '23 at 18:35
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    See complete (strong) induction (which is discussed in many prior answers) – Bill Dubuque Oct 08 '23 at 18:44
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    The dupe & answer below answer the Q. But it's worth pointing out you can prove the validity of "two base case" induction directly from weak induction (without going via strong). Namely - suppose you've shown $P(1)$ and $P(2)$ hold, and for all $n$, that $P(n)$ and $P(n + 1)$ together imply $P(n + 2)$. Then you can show by induction that for all $n$, $Q(n) = \text{"$P(n)$ and $P(n + 1)$"}$ holds. Do you see why? This is a formal justification. But I think it should also be kind of intuitive! You start with $P(1)$ and $P(2)$, and that gives you $P(3)$, and that gives you $P(4)$ and so on... – Izaak van Dongen Oct 08 '23 at 19:15
  • Your $F(2n)$ should be $F(2n-1)$. – J.G. Oct 08 '23 at 21:56

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Here $F(0)=0$, $F(1)=1$, etc.

If you assume $P(n)$, then you can't assume $P(n-1)$. But you can assume $P(i)$ holds for all $i\leqslant n$ and then prove $P(n+1)$. If you do that, you'd need to make sure you handle all appropriate base cases. The logic is as follows:

We prove it by contradiction. Assume there exists $m$ such that $P(m)$ is not true. Then let $n$ be the minimum $m\in\mathbb{N}$ for which $P(m)$ does not hold.

Case $1$: $n=0$. We know $F(0)=0$, $F(1)=1$, $F(2\cdot 0+1)=1$, and $0^2+1^2=1$, a contradiction. So $n=0$ is impossible, because $P(0)$ does hold.

Case $2$: $n=1$. We know $F(1)=1$, $F(2)=1$, and $F(2\cdot 1+1)=F(3)=2$. Then $1^2+1^2=2$, a contradiction. So $n=1$ is impossible, because $P(1)$ does hold.

Case $3$: $n>2$. In this case, $P(n-1)$ and $P(n-2)$ both hold, and you go as in the link.

That's the logic behind the linked argument, and the reason it refers to two base cases, $n=0$ and $n=1$.

In general, you can show that if $P(0)$ holds and $P(n)$ implies $P(n+1)$ for any $n$, then $P(n)$ holds for all $n$, This is called weak induction. But you can also argue that if $P(0)$ holds and if $P(k)$ holds for all $k\leqslant n$, then $P(n+1)$ also holds, then $P(n)$ holds for all $n$. This is called strong induction. In such a situation, if your inductive step to prove $P(n)$ involves $P(n-1)$, $\ldots$, and $P(n-k)$, then you will likely have to perform $k$ base cases.

Either way, the underlying logic is the same. If $A$ is the set of $n$ for which $P(n)$ fails, then either $A$ is empty ($P(n)$ holds for all $n$), or it has a minimum, $n$. But if we know that $P(0)$ holds, $n$ cannot be equal to zero. And if we know that $P(m)$ must be true whenever $P(i)$ is true for all $i<m$, then $P(n)$ must be true as well, by minimality, and we arrive at a contradiction.