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In the induction step, we want to prove that $\forall n\in \mathbb{N}\,\big( P(n) \implies P(n+1)\big)\tag1,$ so we first assume $P(n)$ "for some natural $n.$" What sort of quantifier are we applying to $n$ here?

Initially, it seems like we are just applying the existential quantifier, which is often translated to "for some." But this doesn't seem to make much sense in the context of an induction hypothesis, because the "some" $n$ for which $P(n)$ is true could very well not include the base case.

It also couldn't be that "for some" here means "for all," since that begs the question.

Instead of just "some $n,$" I have also seen $P(n)$ assumed for "some arbitrary $n,$" and this seems to make more sense. It feels like we are not taking any particular $n,$ nor every $n,$ but just something that represents $n$ in an abstract way and using that to prove the implication. But I would be interested to hear a proper explanation of this.

M. A.
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Vpie649
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  • We do not assume $P(n)$ just for "some" $n$, but for arbitrary given $n$, and then conclude $P(n+1)$. So it is as you said. – Dietrich Burde Sep 29 '18 at 18:38
  • And since $n$ is arbitrary, it can be any $n$, so it needs to apply to all $n$; we are still assuming the universal quantifier. – amWhy Sep 29 '18 at 18:41
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    "any" rather than "some". – Michal Adamaszek Sep 29 '18 at 18:53
  • The "logic" is : assume $P(n)$ for $n$ whatever and prove - on this assumption - that $P(n+1)$ holds. Then, apply Conditional proof to conclude that $P(n) \to P(n+1)$ holds (without assumptions). – Mauro ALLEGRANZA Sep 30 '18 at 08:54
  • Finally, apply Universal generalization to conclude with $\forall n (P(n) \to P(n+1))$. Universal generalization formalizes exactly the intuitive argument that if some "facts" holds for an object whatever, it must hold for every object. – Mauro ALLEGRANZA Sep 30 '18 at 08:56
  • It is not necessarily the case that an expression in a subproof can be interpreted as a fully quantified statement. It is just the way some FOLs work. People think that deductive logics always take fully meaningful statements and infer new fully meaningful statements from them, but not many (formal) logics actually work like that. – DanielV Mar 04 '23 at 10:45

1 Answers1

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Your reasoning is totally correct, and here's the answer to your question: the induction hypothesis actually involves no quantification, which is introduced only at the end of the induction step, where Universal Introduction is finally (implicitly) invoked to deduce statement $(1).$

Some clear ways of phrasing the induction hypothesis:

  1. Let $k$ be an arbitrary natural number; suppose that $P(k)$
  2. Take an arbitrary natural number $k;$ suppose that $P(k)$
  3. Suppose that $P(k)$ for an arbitrary natural number $k.$

Other ways:

  1. Suppose that $P(k)$ for some arbitrary natural number $k$
  2. Suppose that $P(k)$ for some natural number $k$
  3. Suppose that $P(k)$ for any natural number $k$
  4. Take any natural number $k;$ suppose that $P(k)$
  5. Let $k$ be any natural number; suppose that $P(k).$

I'd just steer away from #5 and #6, for being potentially ambiguous or even misleading.

Fuller explanation here.

ryang
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