Is this description of "ordinary induction" from Velleman's *How to Prove It* correct?

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The following is from How to Prove It: A Structured Approach, 2nd edition, by Daniel J. Velleman, page 289.

To see why strong induction works, it might help if we first review briefly why ordinary induction works. Recall that a proof by ordinary induction enables us to go through all the natural numbers in order and see that each of them has some property $P$. The base case gets the process started, and the induction step shows that the process can always be continued from one number to the next. But note that in this process, by the time we check that some natural number $n$ has the property $P$, we've already checked that all smaller numbers have the property. In other words, we already know that $∀k < nP(k)$. The idea behind strong induction is that we should be allowed to use this information in our proof of $P(n)$.

That characterization of "ordinary induction" is certainly not how I think of standard (not strong) complete induction. I think of it as showing that $P\left[1\right]$ holds (the initial case) and then showing that $P\left[n\right]\implies{P\left[n+1\right]}$, where $P\left[n\right]$ is called the induction hypothesis, and $P\left[n\right]\mapsto{P\left[n+1\right]}$ is the induction step.

Are these characterizations of complete induction mutually contradictory? Are they essentially the same?

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The phrase "recall that" in the quoted paragraph makes it clear that this paragraph is referring back to an earlier explanation of why ordinary induction works. It would be helpful to go back and look at that earlier explanation. Induction is introduced on p. 260, and the definition is the usual one: Prove $P(0)$ (the base case) and then prove $\forall n \in \mathbb{N}(P(n) \to P(n+1))$ (the induction step). (Some people include 0 in the natural numbers and some don't. In How To Prove It, 0 is included.) On p. 262 there is an explanation of why induction works. That explanation is that the base case proves that $P(0)$ is true. Plugging in $n=0$ in the induction step you get $P(0) \to P(1)$, and from $P(0)$ and $P(0) \to P(1)$ you get $P(1)$. Then plugging in $n=1$ in the induction step you get $P(1) \to P(2)$, and from $P(1)$ and $P(1) \to P(2)$ you get $P(2)$, etc. The explanation ends with the statement "Continuing in this way, you should be able to see that by repeatedly applying the induction step you can show that $P(n)$ must be true for every natural number $n$." When you carry out this process of starting with the base case and then repeatedly applying the induction step, you confirm first $P(0)$, then $P(1)$, then $P(2)$, and so on. It is this process that I was referring to in the sentence "Recall that a proof by ordinary induction enables us to go through all the natural numbers in order and see that each of them has some property $P$."

Notice that the quoted passage says that mathematical induction enables us to go through all the natural numbers in order, it doesn't say that a proof by induction consists of going through the natural numbers in order. A proof by induction consists of a base case and an induction step. But if you've proven the base case and the induction step, then that would enable you to go through the natural numbers in order and verify that they all have property $P$, as described above. And that's why induction works. For example, if you've done a proof by induction, why is, say, $P(10)$ true? The answer is that you have to start with $P(0)$ and apply the induction step 10 times to work your way from $P(0)$ up through $P(1)$, $P(2)$, and so on up to $P(10)$.