Theorem (Finite induction)
Let be P$(x)$ a property. So we suppose that
- P$(0)$,
- P$(n)\rightarrow$P$\big(S(n)\big)$, for all $n<k$
So P(n) is true for all $n<k$.
Unfortunately I can't formally prove the theorem. So could someone help me, please?
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Antonio Maria Di Mauro
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3Consider ${n\in\Bbb N:\mathbf{P}(n)\text{ or }k\le n}$. – Brian M. Scott Jun 24 '20 at 21:04
2 Answers
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Let $\pi:=\{h\in k:\mathbf{P}(h)\}$. So if it was $k\setminus \pi\neq\varnothing$ then there exist $m:=\min{k\setminus \pi}$ so we analise the case where $m=0$ and the case where $m\neq 0$. So for the hypothesis 1 it is impossible that $m=0$. Finally if $m\neq 0$ then $m=S(l)$ for some $l\in k$ and then $l\in\pi$ so that $\mathbf{P}(l)$ is true and so $\mathbf{P}(m)$ is true and this would be impossible.
Antonio Maria Di Mauro
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1Almost: either $m=S(\ell)$ for some $\ell\in k$ (and you do need to mention that $\ell$ is in $k$, not just in $\Bbb N$, or $m=0$. The $m=0$ case is trivial but does have to be mentioned. – Brian M. Scott Jun 24 '20 at 21:32
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You still haven’t dealt with the possibility that $m=0$. It’s not true that $m=S(\ell)$ for some $\ell\in k$: what’s true is that either $m=S(\ell)$ for some $\ell\in k$, or $m=0$. (Of course both turn out to be impossible, but for different reasons.) – Brian M. Scott Jun 24 '20 at 21:45
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@BrianM.Scott If for hypothesis $\mathbf{P}(0)$ is true then it seems to me that it is impossible that $m=0$, is this uncorrect? – Antonio Maria Di Mauro Jun 24 '20 at 21:50
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1But you have to say so. All you know about $m$ is that it is $\min k\setminus\pi$. That means that $m\in k$, but it does not immediately imply that $m\ne 0$; it says only that either $m=0$, or $m=S(\ell)$ for some $\ell\in k$. The first of these is impossible by clause (1) of the theorem, and the second is impossible by clause (2) — and both cases have to be ruled out. – Brian M. Scott Jun 24 '20 at 22:05
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@BrianM.Scott Okay, I edited the ansewer. Now is it correct? – Antonio Maria Di Mauro Jun 24 '20 at 22:10
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@BrianM.Scott Okay, thanks too much for your assistance! – Antonio Maria Di Mauro Jun 24 '20 at 22:13
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@BrianM.Scott Hi professor Scott, could I ask your assistance here? Excuse me for the bother. – Antonio Maria Di Mauro Jun 25 '20 at 21:16
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Hint:
Consider the set $S=\lbrace n\in \bigl[\mkern-4mu\bigl|0,k\bigr|\mkern-4.25mu\bigr]\mid \ P(n)\text{ false }\rbrace$ and suppose $S\ne\varnothing$. As any nonempty subset of $\mathbf N$, it has a least element. Can you deduce a contradiction?
Bernard
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Let $\pi:={n\in\Bbb{N}:\mathbf{P}(n)}$. So if it was $n\setminus \pi\neq\varnothing$ then there exist $m:=\min{n\setminus \pi}$ and so if $l\in\Bbb{N}$ is such that $m=S(l)$ then $l\in\pi$ and so $\mathbf{P}(l)$ is true and so $\mathbf{P}(m)$ is true and this would be impossible. Do you what say this? – Antonio Maria Di Mauro Jun 24 '20 at 21:20
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It's fine (I'd prefer a Π rather than a π). Τ To be quite complete, you have to mention that $m>0$. – Bernard Jun 24 '20 at 21:24
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@AntonioMariaDiMauro: No, this doesn’t work: $\mathbf{P}(\ell)$ does not imply $\mathbf{P}(m)$ if $\ell\ge k$. Replace $\mathbf{P}(n)$ with the property $\mathbf{Q}(n)$ that is ‘$\mathbf{P}(n)$ or $k\le n$ and prove that $\mathbf{Q}(n)\to\mathbf{Q}(S(n))$ for all $n\in\Bbb N$. – Brian M. Scott Jun 24 '20 at 21:24
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@BrianM.Scott Excuse me. I had do a terrible mistake: I wanted write $\pi:={n\in k:\mathbf{P}(n)}$. – Antonio Maria Di Mauro Jun 24 '20 at 21:27
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