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What is a simple way of proving strong induction implies weak induction and vice versa using simple predicate logic and quantifiers?

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closed as off-topic by Carl Mummert, RedMushroom, G. Sassatelli, hardmath, wythagoras Mar 10 at 15:48

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One direction is easy, right? An instance of weak induction can be "embedded" in the strong induction scheme, simply ignoring the extra strength of the induction hypothesis. You seem familiar with terminology of first-order logic. As your Question hints, the strong induction hypothesis can be recast as a weak induction hypothesis using a quantified predicate. – hardmath Mar 10 at 14:38

Weak induction: $$\Phi(0), \forall n\colon \Phi(n)\to\Phi(n+1)\vdash\forall n\colon \Phi(n).$$ Strong induction: $$\forall n\colon (\forall m<n\colon \Psi(m))\to\Psi(n)\vdash\forall n\colon \Psi(n)$$ (All vars implied to be in $\mathbb N_0$ for brevity).

To show that weak induction implies strong induction, one lets $\Phi(n)\equiv \forall m<n\colon \Psi(m)$ for given $\Psi$: We have $\Phi(0)$ because $\forall m<0\colon\Psi(m)$ is vacuously true. Using $m<n+1\to m<n\lor m=n$, we find that $$\forall m<n\colon \Psi(m)\implies\Psi(n)\land\forall m<n\colon \Psi(m)\implies \forall m<n+1\colon \Psi(m)$$ and thus $\forall n\colon \Phi(n)\to\Phi(n+1)$. By weak induction, $\forall n\colon \Phi(n)$. Especially, for arbitrary $n$ we find $\forall m<n\colon \Phi(m)$, i.e. $\forall n\colon\Psi(n)$.

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