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I would like some guidance in the right direction for my attempt.

Proof.

($\Rightarrow$): First assume strong induction to be true and we prove standard induction. Consider the set $A \subseteq \mathbb{N}$ defined by the conditions, $1 \in A$ and $n \in A \Rightarrow n+1 \in A$. We prove that $A = \mathbb{N}$. Define another set $B \subseteq \mathbb{N}$ such that $B$ contains all $m \in \mathbb{N}$ such that $m \notin A$. Clearly $1 \notin B$ since $1 \in A$. Now assume that $1,...,n \notin B \Rightarrow 1,...,n \in A$. In particular, notice that $n \in A \Rightarrow n+1 \in A$ (by definition of how $A$ is defined.). So by strong induction, we must have $A = \mathbb{N}$ as desired.

($\Leftarrow$): Now assume standard induction and we prove strong induction. Consider the set $C \subseteq \mathbb{N}$ defined by the conditions, $1 \in C$ and $1,...,k \in C \Rightarrow k+1 \in C$. Since $1 \in C$ and $k \in C \Rightarrow k+1 \in C$, we get that $C = \mathbb{N}$ (by use of standard induction). QED.

Please point out any circular reasoning, confusing use of notation or simply if some of my arguments don't make any sense.

Thanks in advance!

RobPratt
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Lucas
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  • Somehow I feel that your second step should be more complicated than your first – Henry Mar 13 '22 at 01:46
  • Previous related discussion: https://math.stackexchange.com/questions/335479/is-strong-induction-not-actually-stronger-than-normal-induction also https://math.stackexchange.com/questions/4286534/equivalence-of-mathematical-induction-and-strong-induction – Gerry Myerson Mar 13 '22 at 02:00
  • And https://math.stackexchange.com/questions/1184541/what-exactly-is-the-difference-between-weak-and-strong-induction – Gerry Myerson Mar 13 '22 at 02:08
  • Any thoughts on these links, Lucas, or on the answer that was posted yesterday? – Gerry Myerson Mar 14 '22 at 12:27
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    Hello Gerry. Sorry for the late response. The links are very helpful. I had polished off my proof with the feedback in mind. – Lucas Mar 14 '22 at 23:29
  • If you want to be sure I see a comment intended for me, you have to put @Gerry in it so I get notified. – Gerry Myerson Mar 15 '22 at 22:33

1 Answers1

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Your $\Leftarrow$ direction looks fine to me.

For the $\Rightarrow$ direction, I would be more explicit about justifying the assumption $1, \ldots, n \notin B$. In particular, you could mention something to the effect of "by way of contradiction, assume $B$ is non-empty. By the well-ordering of $\mathbb{N}$, this implies $B$ has a least element $m$ such that $m \leq b$ for all $b \in B$. Therefore since $1, \ldots, m-1$ are not in $B$, then they must be in $A$ by definition of $B$." Your argment then proceeds more or less as you've written it, modulo some fiddly bits.

Andrew L
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