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Could anyone please explain me what is Loaded Induction Hypothesis and its difference from a standard one? I imagine that it must be a stronger hypothesis in some way.

Charles Bronson
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    I haven't heard about a "loaded induction hypothesis". Do you have a definition, or some more context? – Arthur Aug 02 '17 at 14:35
  • No, this is the problem. I found an example here goo.gl/GPxER6, and don't quite get it. – Charles Bronson Aug 02 '17 at 14:38
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    Yes: he says: "then $\psi, \text {M-INT}_G(\psi) \in \Delta$. (Note that this is stronger than what is actaully needed..." Because the result needed is $\psi \in \Delta$. – Mauro ALLEGRANZA Aug 02 '17 at 15:00
  • So, does it mean that in this case we use a stronger assumption as IH than the one we want to prove? – Charles Bronson Aug 02 '17 at 15:04
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    (continuing the quote) "...such a loaded induction hypothesis makes the proof easier." – hardmath Aug 02 '17 at 15:12
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    That happens some times. The typical example is that proving $\frac11+\frac14+\frac19+\cdots+\frac1{n^2}<2$ will require an induction hypothesis which is stronger, like ${}<2-\frac1n$. – Arthur Aug 02 '17 at 15:23
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    I am not familiar with this term. But sometimes instead of trying to show that $P(n)\implies P(n+1),$ for every $n\in \mathbb N,$ we show that $[;\forall m \in \mathbb N; (m<n\implies P(m);]\implies P(n).$ – DanielWainfleet Aug 02 '17 at 17:19
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    @DanielWainfleet That's usually called "strong induction" (if I read your expression right). But yes, I thought about that too. – Arthur Aug 03 '17 at 08:23

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