I need help with an induction proof(s). If you could be as simple and clear as p

Question

I need help with an induction proof(s). If you could be as simple and clear as possible, that would be great. A response in image form would be most helpful. Thanks.

The goal of this exercise is to prove that the Principal of Mathematical Induction (PMI), Strong Induction (SI), and the Well-Ordering Principle (WOP) are logically equivalent. Using the simplest methods of proof available to you, prove the following in order to show that the PMI, SI, and WOP are all logically equivalent:

(a) PMI implies SI,

(b) SI implies WOP,

(c) WOP implies PMI.

Explanation / Answer

THIS IS WELL DOCUMENTED TOPIC

BEING A VERY LENGTHY PROOF , INSTEAD OF REWRITING THE SAME ,

I AM GIVING BELOW THE EXTRACT FROM A WELL EXPLAINED TEXT AS DESIRED BY YOU.

SHOULD YOU NEED ANY CLARIFICATION PLEASE COME BACK ..

LINK .....

https://bspace.berkeley.edu/access/content/group/2fb5bd3e-8d09-40ee-a371-6cc033d854b9/ho1.pdf

QUOTE:

Principle of Induction. Let P(n) be a statement (proposition) depending on a naturalnumber n . Assume that(i). (basis statement) P(1) is true, and(ii). (inductive step) for all n 2 N , if P(n) is true, then P(n + 1) is also true.Then P(n) is true for all n 2 N .Principle of Strong Induction. Let P(n) be a statement (proposition) depending on anatural number n . Assume that(i). (basis statement) P(1) is true, and(ii0). (inductive step) for all n 2 N , if P(k) is true for all k = 1; 2; : : : ; n , thenP(n + 1) is also true.Then P(n) is true for all n 2 N .Well-Ordering of N . Every nonempty subset of N has a smallest element.A. Strong Induction implies Induction.Since n 1 , we haveP(k) is true for all k = 1; 2; : : : ; n implies P(n) is true :Therefore condition (ii) implies condition (ii0). This is because if (ii) is true and if P(k)is true for all k = 1; 2; : : : ; n , then P(n) is true, and therefore by (ii) P(n+ 1) is true.This veries (ii0).By a similar argument, the Principal of Strong Induction implies the Principle ofInduction, because hypothesis (ii0) is weaker.B. Induction implies Strong Induction.Let Q(n) be the statement P(k) is true for all k = 1; 2; : : : ; n ." Then condition(i) for P implies condition (i) for Q . Also, assume that condition (ii0) holds for P .Then condition (ii) holds for Q(n) , because the truth of Q(n) implies that P(n + 1)is true (by (ii0)), and then Q(n+ 1) is true because both Q(n) and P(n+ 1) are true.By induction, it then follows that Q(n) is true for all n , so P(n) is also true forall n .C. Well-Ordering of N implies Induction.This is Theorem 0.3.6.12D. Strong Induction implies Well-Ordering of N .Suppose that the Principle of Strong Induction is true.In order to prove that N is well ordered, we prove the following statement P(n)by strong induction on n : If S is a subset of N and n 2 S , then S has a smallestelement.The basis step is true, because if 1 2 S then 1 is the smallest element of S ,because there are no smaller elements of N .For the inductive step, suppose that P(k) is true for all k = 1; 2; : : : ; n . To showthat P(n + 1) is true, let S be a subset of N that contains n + 1 . If n + 1 is thesmallest element of S , then we're done. Otherwise, S has a smaller element k , andP(k) is true by the inductive assumption, so again S has a smallest element.Therefore, by strong induction, P(n) is true for all n 2 N . This implies thewell-ordering of N , because if S is a nonempty subset of N , then pick n 2 S . Sincen 2 N , P(n) is true, and therefore S has a smallest element.(Note that parts C and D make part A redundant.)(Also note that part D is not really relevant, since well-ordering of N is an axiom,but induction is a theorem. However, it may be useful if one has an axiom systemin which induction is an axiom instead of well-ordering on N . It is also useful forexplaining why induction may or may not be valid for sets other than N .)

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