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Material Implication implication applies to two independent variables p and q, whereas Induction applies to Set of Related Elements.

See Truth Table of Implication(p->q) says that if, q is True, then, it doesn't really matter what p is, Implication is going to be True.

That is why we ONLY EVER prove k+1th part, the equivalent of q in logical equation, as if we are saying, we don't really care whether it holds for kth step as long as it holds for k+1th, actually meaning for any arbitrary member, which here is known as k+1th member because of how Induction is Defined.

AND we prove it for Base Case(0), because this memeber is different from other "arbitrary members" of the Set and is equivalent to p.

So while implication sys, p->q,

Induction says (Base Case(0)) ∧ (Arbitray Case(k+1th))->(True for Whole Set). And this Imlication Holds preisely because of the definition of Natural Numbers as The Relation between Base Case(0) and any Arbitrary Case(k+1th) is EXACTLY SAME as the Relation Between Base Case(0) and any Member Before or After K+1th Member.

Or Considering Base Case(0) as Base Case(0) and ALSO the kth Case, then if it applies to 0th/nth case AND it applies to next 0+1/k+1 member then it applies to all bacuse Relation between any two adjacent members is EXACTLY SAME(By Definition/Construction) as Relation between any two other Adjacent Pairs.

Is My understanding Correct?

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    Yes and no. It is a sort of infinite chian of Modus Ponens: P0 and "if P0, then P1", from which P1. And from it and "if P1, then P2"...
    – Mauro ALLEGRANZA
    Commented Nov 16 at 7:49
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    If one can prove Base Case → Arbitrary Case directly then one does not need induction at all. The whole point of induction is that one can bypass an indefinitely long chain of implications (which we cannot follow all the way) with a single implication of generic form (which we can). And it needs elements of arithmetic in addition to logic to do it. So it is an 'extension' that introduces qualitatively new tools. Some mathematicians, like Poincare, even considered induction the hallmark of arithmetic that distinguishes it from logic.
    – Conifold
    Commented Nov 16 at 8:11
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    Logic calculus alone cannot define Natural numbers with its critical successor function thus cannot conservatively extend material implication to 1st order mathematical induction schema in any infinite fashion as you freely imagined. Conversely it could be viewed as a restriction of ZFC's much more powerful well ordering principle applied to the natural number set N only, therefore, without such restriction, you can go beyond and extend mathematical induction to transfinite induction like Gentzen did, a kind of levitation...
    – Double Knot
    Commented Nov 16 at 8:37
  • @conifold see I was looking at it like this, since the relation between base case and any arbitrary case(k+1) is exactly same between base case and any case before or after k+1th member is same hence proving just for base case and k+1th case is enough. OR assuming my base case as base case and kth case, if I prove it for next case k+1th, then it applies to all because relation between base case and next case is exactly same as any other pair of adjacent members. Either way, kth member somehow doesn't seem that important other than the fact that that's how you write math Statements generally.
    – Ashish Shukla
    Commented Nov 16 at 8:49
  • @DoubleKnot Make that an answer
    – Rushi
    Commented Nov 18 at 3:17

1 Answer 1

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Is My understanding Correct?

Yes. You've defined recursive induction using the successor function. That's perfectly legitimate.

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  • It's not what they did.
    – tkruse
    Commented Nov 18 at 8:37
  • @tkruse That's exactly what they did, but ooookay, we'll just leave it at that then.
    – Miss Understands
    Commented Nov 18 at 18:55

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