Why is complete strong induction a valid proof method?

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I recently learned about complete strong induction. I am familiar with both strong induction and ordinary induction and make sense. But what particularly confuses me is why we do not to explicit the base cases for complete induction. They seem crucial for modus ponens to work and thus, actually show the stand alone proposition $p(n)$ to be true.



The claim for complete induction seems to be the following:




if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




These are my thoughts:



In induction we actually do two things (to show $ P(n) $ for all $ n in mathbb N$) :



  1. show P(0)

  2. show implication $ P(n-1) implies P(n) $

or for strong induction



  1. show P(0)

  2. show implication $ forall m, m < n, P(m) implies P(n) $

However, in complete induction we only show:



  1. $ forall m, m < n, P(m) implies P(n) $

now that I’ve thought of it more carefully, what bothers me is that in the inductive step we actually only show an implication is true, not that $P(n)$ is true. Intuitively, $P(n)$ ends up true because of Modus Ponens (MP), which forcibly requires checking some set of base cases, say $P(0)$.



What concerns me is the following: ff we show $ forall m, m < n, P(m) implies P(n) $ then we shown the implication is true, and not necessarily anything else. If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False. So sure, the implication is (vacuously) true, but that doesn’t necessarily say $P(0)$ is true stand alone (which is what induction ultimately cares about!).



My assumption is that the claim the wikipedia article does is that (somehow) whatever proof for $ forall m, m < n, P(m) implies P(n) $ that we have must also be a stand alone proof for $P(0)$. I guess I could abstractly believe that’s true (mostly on faith), but it seems rather odd to me. I’ve never seen False implies $P(n)$ implies $P(n)$. It’s nearly like the truth table for implication is built so that if you only know the antecedent is False, then you can’t decide if the consequent is true. Which makes sense. A false starting point should intuitively get you no where or get you everywhere (by principle of explosion).



So my question is, whats going on? Is it just that the proof for $ forall m, m < n, P(m) implies P(n) $ can also be plugged in for a proof for $P(0)$ and then $P(0)$ is true? Or am I missing something?



I have a feeling that this being so abstract makes it hard to be believable and a concrete example of how $ forall m, m < n, P(m) implies P(n) $ automagically makes $P(0)$ (the base cases) true would be really valuable.




Another useful source:



https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction










share|cite|improve this question



















  • 3




    " I am familiar with both strong induction and ordinary induction" Complete strong induction is strong induction.
    – fleablood
    10 hours ago










  • @fleablood yea, of course because of $P(m), m<n$, but I'd never seen before that the base cases were "automatically true" given a correct inductive step. As far as I've seen it must be show they are true, because of the Modus Ponens (MP) argument I sketched in my question.
    – Charlie Parker
    10 hours ago







  • 1




    Your statement "If n=0 then $P(m),m<nimplies P(n)$ is false" is incorrect (when interpreted properly); in fact, $P(m),m<0implies P(0)$ vacuously true (since there can't be a counterexample).
    – Noah Schweber
    9 hours ago











  • the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
    – Charlie Parker
    9 hours ago






  • 1




    I'm no expert but it looks like we're carefully discriminating between a) p is true if there's a preceding example, and b) p is true if there's no preceding counterexample.
    – Robert Frost
    18 mins ago














up vote
4
down vote

favorite
1












I recently learned about complete strong induction. I am familiar with both strong induction and ordinary induction and make sense. But what particularly confuses me is why we do not to explicit the base cases for complete induction. They seem crucial for modus ponens to work and thus, actually show the stand alone proposition $p(n)$ to be true.



The claim for complete induction seems to be the following:




if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




These are my thoughts:



In induction we actually do two things (to show $ P(n) $ for all $ n in mathbb N$) :



  1. show P(0)

  2. show implication $ P(n-1) implies P(n) $

or for strong induction



  1. show P(0)

  2. show implication $ forall m, m < n, P(m) implies P(n) $

However, in complete induction we only show:



  1. $ forall m, m < n, P(m) implies P(n) $

now that I’ve thought of it more carefully, what bothers me is that in the inductive step we actually only show an implication is true, not that $P(n)$ is true. Intuitively, $P(n)$ ends up true because of Modus Ponens (MP), which forcibly requires checking some set of base cases, say $P(0)$.



What concerns me is the following: ff we show $ forall m, m < n, P(m) implies P(n) $ then we shown the implication is true, and not necessarily anything else. If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False. So sure, the implication is (vacuously) true, but that doesn’t necessarily say $P(0)$ is true stand alone (which is what induction ultimately cares about!).



My assumption is that the claim the wikipedia article does is that (somehow) whatever proof for $ forall m, m < n, P(m) implies P(n) $ that we have must also be a stand alone proof for $P(0)$. I guess I could abstractly believe that’s true (mostly on faith), but it seems rather odd to me. I’ve never seen False implies $P(n)$ implies $P(n)$. It’s nearly like the truth table for implication is built so that if you only know the antecedent is False, then you can’t decide if the consequent is true. Which makes sense. A false starting point should intuitively get you no where or get you everywhere (by principle of explosion).



So my question is, whats going on? Is it just that the proof for $ forall m, m < n, P(m) implies P(n) $ can also be plugged in for a proof for $P(0)$ and then $P(0)$ is true? Or am I missing something?



I have a feeling that this being so abstract makes it hard to be believable and a concrete example of how $ forall m, m < n, P(m) implies P(n) $ automagically makes $P(0)$ (the base cases) true would be really valuable.




Another useful source:



https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction










share|cite|improve this question



















  • 3




    " I am familiar with both strong induction and ordinary induction" Complete strong induction is strong induction.
    – fleablood
    10 hours ago










  • @fleablood yea, of course because of $P(m), m<n$, but I'd never seen before that the base cases were "automatically true" given a correct inductive step. As far as I've seen it must be show they are true, because of the Modus Ponens (MP) argument I sketched in my question.
    – Charlie Parker
    10 hours ago







  • 1




    Your statement "If n=0 then $P(m),m<nimplies P(n)$ is false" is incorrect (when interpreted properly); in fact, $P(m),m<0implies P(0)$ vacuously true (since there can't be a counterexample).
    – Noah Schweber
    9 hours ago











  • the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
    – Charlie Parker
    9 hours ago






  • 1




    I'm no expert but it looks like we're carefully discriminating between a) p is true if there's a preceding example, and b) p is true if there's no preceding counterexample.
    – Robert Frost
    18 mins ago












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





I recently learned about complete strong induction. I am familiar with both strong induction and ordinary induction and make sense. But what particularly confuses me is why we do not to explicit the base cases for complete induction. They seem crucial for modus ponens to work and thus, actually show the stand alone proposition $p(n)$ to be true.



The claim for complete induction seems to be the following:




if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




These are my thoughts:



In induction we actually do two things (to show $ P(n) $ for all $ n in mathbb N$) :



  1. show P(0)

  2. show implication $ P(n-1) implies P(n) $

or for strong induction



  1. show P(0)

  2. show implication $ forall m, m < n, P(m) implies P(n) $

However, in complete induction we only show:



  1. $ forall m, m < n, P(m) implies P(n) $

now that I’ve thought of it more carefully, what bothers me is that in the inductive step we actually only show an implication is true, not that $P(n)$ is true. Intuitively, $P(n)$ ends up true because of Modus Ponens (MP), which forcibly requires checking some set of base cases, say $P(0)$.



What concerns me is the following: ff we show $ forall m, m < n, P(m) implies P(n) $ then we shown the implication is true, and not necessarily anything else. If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False. So sure, the implication is (vacuously) true, but that doesn’t necessarily say $P(0)$ is true stand alone (which is what induction ultimately cares about!).



My assumption is that the claim the wikipedia article does is that (somehow) whatever proof for $ forall m, m < n, P(m) implies P(n) $ that we have must also be a stand alone proof for $P(0)$. I guess I could abstractly believe that’s true (mostly on faith), but it seems rather odd to me. I’ve never seen False implies $P(n)$ implies $P(n)$. It’s nearly like the truth table for implication is built so that if you only know the antecedent is False, then you can’t decide if the consequent is true. Which makes sense. A false starting point should intuitively get you no where or get you everywhere (by principle of explosion).



So my question is, whats going on? Is it just that the proof for $ forall m, m < n, P(m) implies P(n) $ can also be plugged in for a proof for $P(0)$ and then $P(0)$ is true? Or am I missing something?



I have a feeling that this being so abstract makes it hard to be believable and a concrete example of how $ forall m, m < n, P(m) implies P(n) $ automagically makes $P(0)$ (the base cases) true would be really valuable.




Another useful source:



https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction










share|cite|improve this question















I recently learned about complete strong induction. I am familiar with both strong induction and ordinary induction and make sense. But what particularly confuses me is why we do not to explicit the base cases for complete induction. They seem crucial for modus ponens to work and thus, actually show the stand alone proposition $p(n)$ to be true.



The claim for complete induction seems to be the following:




if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




These are my thoughts:



In induction we actually do two things (to show $ P(n) $ for all $ n in mathbb N$) :



  1. show P(0)

  2. show implication $ P(n-1) implies P(n) $

or for strong induction



  1. show P(0)

  2. show implication $ forall m, m < n, P(m) implies P(n) $

However, in complete induction we only show:



  1. $ forall m, m < n, P(m) implies P(n) $

now that I’ve thought of it more carefully, what bothers me is that in the inductive step we actually only show an implication is true, not that $P(n)$ is true. Intuitively, $P(n)$ ends up true because of Modus Ponens (MP), which forcibly requires checking some set of base cases, say $P(0)$.



What concerns me is the following: ff we show $ forall m, m < n, P(m) implies P(n) $ then we shown the implication is true, and not necessarily anything else. If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False. So sure, the implication is (vacuously) true, but that doesn’t necessarily say $P(0)$ is true stand alone (which is what induction ultimately cares about!).



My assumption is that the claim the wikipedia article does is that (somehow) whatever proof for $ forall m, m < n, P(m) implies P(n) $ that we have must also be a stand alone proof for $P(0)$. I guess I could abstractly believe that’s true (mostly on faith), but it seems rather odd to me. I’ve never seen False implies $P(n)$ implies $P(n)$. It’s nearly like the truth table for implication is built so that if you only know the antecedent is False, then you can’t decide if the consequent is true. Which makes sense. A false starting point should intuitively get you no where or get you everywhere (by principle of explosion).



So my question is, whats going on? Is it just that the proof for $ forall m, m < n, P(m) implies P(n) $ can also be plugged in for a proof for $P(0)$ and then $P(0)$ is true? Or am I missing something?



I have a feeling that this being so abstract makes it hard to be believable and a concrete example of how $ forall m, m < n, P(m) implies P(n) $ automagically makes $P(0)$ (the base cases) true would be really valuable.




Another useful source:



https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction







proof-verification logic induction proof-explanation






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edited 9 hours ago

























asked 10 hours ago









Charlie Parker

1,0491128




1,0491128







  • 3




    " I am familiar with both strong induction and ordinary induction" Complete strong induction is strong induction.
    – fleablood
    10 hours ago










  • @fleablood yea, of course because of $P(m), m<n$, but I'd never seen before that the base cases were "automatically true" given a correct inductive step. As far as I've seen it must be show they are true, because of the Modus Ponens (MP) argument I sketched in my question.
    – Charlie Parker
    10 hours ago







  • 1




    Your statement "If n=0 then $P(m),m<nimplies P(n)$ is false" is incorrect (when interpreted properly); in fact, $P(m),m<0implies P(0)$ vacuously true (since there can't be a counterexample).
    – Noah Schweber
    9 hours ago











  • the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
    – Charlie Parker
    9 hours ago






  • 1




    I'm no expert but it looks like we're carefully discriminating between a) p is true if there's a preceding example, and b) p is true if there's no preceding counterexample.
    – Robert Frost
    18 mins ago












  • 3




    " I am familiar with both strong induction and ordinary induction" Complete strong induction is strong induction.
    – fleablood
    10 hours ago










  • @fleablood yea, of course because of $P(m), m<n$, but I'd never seen before that the base cases were "automatically true" given a correct inductive step. As far as I've seen it must be show they are true, because of the Modus Ponens (MP) argument I sketched in my question.
    – Charlie Parker
    10 hours ago







  • 1




    Your statement "If n=0 then $P(m),m<nimplies P(n)$ is false" is incorrect (when interpreted properly); in fact, $P(m),m<0implies P(0)$ vacuously true (since there can't be a counterexample).
    – Noah Schweber
    9 hours ago











  • the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
    – Charlie Parker
    9 hours ago






  • 1




    I'm no expert but it looks like we're carefully discriminating between a) p is true if there's a preceding example, and b) p is true if there's no preceding counterexample.
    – Robert Frost
    18 mins ago







3




3




" I am familiar with both strong induction and ordinary induction" Complete strong induction is strong induction.
– fleablood
10 hours ago




" I am familiar with both strong induction and ordinary induction" Complete strong induction is strong induction.
– fleablood
10 hours ago












@fleablood yea, of course because of $P(m), m<n$, but I'd never seen before that the base cases were "automatically true" given a correct inductive step. As far as I've seen it must be show they are true, because of the Modus Ponens (MP) argument I sketched in my question.
– Charlie Parker
10 hours ago





@fleablood yea, of course because of $P(m), m<n$, but I'd never seen before that the base cases were "automatically true" given a correct inductive step. As far as I've seen it must be show they are true, because of the Modus Ponens (MP) argument I sketched in my question.
– Charlie Parker
10 hours ago





1




1




Your statement "If n=0 then $P(m),m<nimplies P(n)$ is false" is incorrect (when interpreted properly); in fact, $P(m),m<0implies P(0)$ vacuously true (since there can't be a counterexample).
– Noah Schweber
9 hours ago





Your statement "If n=0 then $P(m),m<nimplies P(n)$ is false" is incorrect (when interpreted properly); in fact, $P(m),m<0implies P(0)$ vacuously true (since there can't be a counterexample).
– Noah Schweber
9 hours ago













the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
– Charlie Parker
9 hours ago




the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
– Charlie Parker
9 hours ago




1




1




I'm no expert but it looks like we're carefully discriminating between a) p is true if there's a preceding example, and b) p is true if there's no preceding counterexample.
– Robert Frost
18 mins ago




I'm no expert but it looks like we're carefully discriminating between a) p is true if there's a preceding example, and b) p is true if there's no preceding counterexample.
– Robert Frost
18 mins ago










7 Answers
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if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




Right, that's exactly correct: if there is nothing smaller than $0$ (as is the case for the natural numbers) then it is vacuously true that:



$$P(m) text holds for any m<0 tag1$$



So, if you have shown that:



$$textfor any n: text if P(m) text holds for any m<n, text then P(n) tag2$$



then in particular you have shown that:



$$text if P(m) text holds for any m<0, text then P(0) tag2'$$



and so we get



$$P(0)$$



by Modus Ponens on $(1)$ and $(2')$



So, there is indeed no need to prove an explicit base case.



That said, think about how in practice you would actually prove $(2)$. Probably, you would be able to show $P(n)$ based on the assumption that there actually are $m<n$ for which we can show that if they all have property $P(m)$, then $P(n)$. But for the edge case of $n=0$, there are no such $m<n$ ... so ... you need to show $P(0)$ by itself!



In other words, in practice, you often do have to treat the base cases as special cases that you prove as base cases after all.






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  • yea I think I came to this realization. $P(m)$ holds for $m<0$ holds true vacuously on its own. I think in fact is more clear if we write that out as an explicit implication $q(0) := forall m (m<0 to P(m) )$. Then $q(0)$ is true because $m<0$ is false.
    – Charlie Parker
    1 hour ago







  • 1




    @CharlieParker Right! Glad you realized this yourself :)
    – Bram28
    1 hour ago










  • Actually, I wish I did...someone pointed out that my interpretation of $forall m (m < n land P(m) )$ was incorrect and that the correct thing was to consider $forall m (m < n to P(m) )$. Then I realized the rest by myself...I have no idea why I thought the logical AND was the interpretation. Probably because of my horrible use of comma's. Thanks to everyone! :)
    – Charlie Parker
    1 hour ago

















up vote
9
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A valid proof by complete induction includes a uniform proof for all $k$ of the inferences listed below. As such it include a (vacuous) proof of the base case $color#c00,P(0).,$ Examine the schema carefully to see it.



$qquadqquadqquad beginalign
Rightarrow,color#c00 P(0)\
P(0)Rightarrow, P(1)\
P(0),P(1)Rightarrow, P(2)\
vdotsqquad \
P(0),P(1),ldots,P(k-1),Rightarrow,P(k)\
endalign$



Remark $ $ Its contrapositive form (infinite descent) is clearer: if given a counterexample $lnot P(n)$ we can prove there exists a smaller counterexample $lnot P(k), k < n,,$ then no counterexample exists, else iterating the proof would yield an infinite descending chain of counterexamples, contra $,Bbb N,$ is well-ordered. Or, reformulated, if there is a counterexample then, by well order, we can choose a minimal one ("minimal criminal"), contra the proof yields a smaller one.






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  • what if the inductive step (that only proves the implcation $forall m (m<n to P(m)) to P(n)$ assume implicitly $n not = 0$ by saying assume $m < n$ or pick an element from the non-empty set? I think in those cases then you wouldn't actually have the base case for free: "In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element." quoting wikipedia: en.wikipedia.org/wiki/…
    – Charlie Parker
    1 hour ago











  • @CharlieParker I recommend that you focus on some specific examples in your question. Then it will be much easier to be precise.
    – Bill Dubuque
    55 mins ago


















up vote
5
down vote














if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




It's not clear exactly how one ought to interpret "$P(m), m<n implies P(n)$",
but I think we have agreed that the following formula
(shown in the answer by José Carlos Santos)
represents the induction step according to the article:
$$((forall min0,1,2,ldots,n-1) P(m))implies P(n).$$



You seem to be looking at this and saying that for the case $n = 0,$
it is equivalent to $$bot implies P(n),$$
using $bot$ as a symbol for something that is always false.
This implication is vacuously true.
But in fact, a statement of the form
$$ (forall min emptyset) P(m) $$
is also vacuously true. That is, it is true because there is no value of $m$ that can make it false.
This may be a little more obvious if you write it this way:
$$ (forall m)(m in emptyset implies P(m)). $$



So what the induction step of complete induction actually says in the case $n = 0$ is that
$$topimplies P(0),$$
where $top$ is always true. If you prove that $topimplies P(0),$ you have proved $P(0).$



You might question whether the Wikipedia article does a good job of explaining itself, and I would sympathize. It's only an encyclopedia article, however;
a self-evident justification for everything may be too much too ask.
(Useful inline citations for details such as this are not too much to ask, however, and there is a notice at the top of that section of the article requesting them.)






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  • This seems to be the core of the OP's misunderstanding; other answers seem to skip over why P(0) is true and take it for granted.
    – Yakk
    8 hours ago










  • Yea, I've realized that was key, the correct thing to consider $forall m (m < n to P(m) )$. After thats cleared up it becomes much easier.
    – Charlie Parker
    1 hour ago

















up vote
4
down vote













You did not describe strong induction correctly; there's a quantifier missing. The second step should be:



$$bigl((forall min0,1,2,ldots,n-1):P(m)bigr)implies P(n).tag1$$



So, you prove that if $P(0)$, $P(1)$, …, $P(n-1)$, then $P(n)$ holds too. Why should this work? Suppose that you have proved $P(0)$ and also that $(1)$ holds. Then:



  • Since $P(0)$ holds, $P(1)$ holds, by $(1)$.

  • Since $P(0)$ and $P(1)$ hold, $P(2)$ holds, by $(1)$.

  • Since $P(0)$, $P(1)$, and $P(2)$ hold, $P(3)$ holds, by $(1)$.

And so on…






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  • doesn't change anything about my confusion because I was always assuming $m<n$ meant $forall m, m<n$, since $n$ was a free variable. I've fixed it I guess.
    – Charlie Parker
    9 hours ago







  • 3




    the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
    – Charlie Parker
    9 hours ago







  • 1




    Jose, if you see the Wikipedia and Quora link I gave it claims complete strong induction does NOT explicitly need the base cases proven. Which is what I'm puzzled about.
    – Charlie Parker
    9 hours ago







  • 1




    "the whole point is that complete strong induction does NOT need to show P(0)" Where in either of you sources was that stated?. "Thats my question. Why don't we need to show P(0)?" Who says you don't?
    – fleablood
    9 hours ago






  • 2




    @fleablood In one form of strong induction, the base case is a special case of the induction hypothesis. See Bill Dubuque’s answer. Or am I misunderstanding what the argument is about?
    – spaceisdarkgreen
    9 hours ago

















up vote
3
down vote













You write:




If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False.




This is where you are wrong. As you noticed, $ forall m, m < 0, P(0)$ is (vacuously) true. But that does not mean that the above statement is false, indeed




$forall m, m < 0, P(m) implies P(0)quad $ is equivalent to $quad P(0)$.




(If you doubt this: $mathrmtruerightarrow x iff neg mathrmtrue lor x iff mathrmfalse lor x iff x$.)



So in complete induction you actually have to show $P(0)$, there is just no reason to list it separately from the implications that you have to show.



Stated differently: In complete induction, for each $n$ you show $P(n)$, but you are allowed to use all $P(m)$ for $m < n$ in the proof of $P(n)$. For $n=0$ this does not allow you anything new as there is no $m<0$.






share|cite|improve this answer



























    up vote
    0
    down vote













    I think I finally understand my confusions after reading the wikipedia article more carefully and getting my notation right. First recall what the inductive step (that we have to proof) is in induction:



    $$ varphi(n) := forall m (m < n to P(m)) to P(n) $$



    what complete strong induction claims to my understanding is that the proof of the inductive step includes the base case automatically because the argument also holds for the base cases, $P(0)$ for example. So now define:



    $$ q(n) := forall m (m < n to P(m)) $$
    $$ p(n) := P(n) $$
    so:



    $$ varphi(n) = q(n) to p(n) $$



    if we assume we prove the inductive step and that argument holds for every $n$ including the base case then we have:



    $$ varphi(0) = q(0) to p(0) $$



    is true as a whole. However, if we carefully examine what $q(0)$ is we notice that its a tautology, i.e.



    $$ q(0) = forall m (m < 0 to P(m))$$



    because $m < 0$ is False because $m in mathbb N = 0,1,2,3,dots$ (i.e. $0<0$,$1<0,2<0cdots$ is always false), so $(m < 0 to P(m))$ is True for all values of m in consideration. So now we know $varphi(0) = q(0) to p(0)$ is True and $q(0)$ is True as a stand alone logical sentence (this is not usually true). So by modus pones (MP) we have:



    $$ q(0)$$
    $$ q(0) to p(0)$$



    and it follows by MP:



    $$ p(0) $$



    which eventually results in the cascading of logical implications usual to induction.



    Note however that the inductive step, depending on the content of the proof might or might NOT prove the base case automatically. For example, wikipedia made a good job of outlining why we need to be careful:




    In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element.




    the second case that talks about sets is important to note because we also have structural induction and the argument might depend on selecting an element from a set, but that only holds if the set is none empty to start with.



    So, if your unsure, proof the base cases, but you can do complete induction if your sure your proof does include $m=0$ as well as $m>0$.






    share|cite|improve this answer



























      up vote
      -3
      down vote














      "But what particularly confuses me is why we do not to explicit the base cases for complete induction."




      Short answer. You do explicitly prove the base cases. You misunderstood the description. You absolutely DO have to prove the base cases.




      "I am familiar with both strong induction and ordinary induction"




      Complete Strong induction is just another name for strong induction. These are two methods of induction. Not three. There is nothing new here that you do not already know.




      However, in complete induction we only show:



      1) ∀m,m<n,P(m)⟹P(n)



      No. That is your mistake. That is NOT what complete strong induction says. Complete strong induction (which is exactly strong induction) says you must show a base case. It simply DOES. Your were wrong in thinking it said you didn't.)



      Weak induction:



      1) Base step: Prove $P(0)$. (This will prove $P(m)$ is true for $m=0$.)



      2) Induction step: Assume $P(m)$. Prove P(m+1).



      (Complete) Strong induction



      1) Base step: Prove $P(0)$. (This will prove $P(m)$ and $P(k)$ are true for $m = 0$ and $0 le k le m = 0$.)



      2) Induction step: Assume $P(m)$ and $P(k)$ for $0 le k le m$. Prove $P(m+1)$.



      Notice that both say "assume" something. They do not mean "this is true and we don't need to prove it"; they both mean "we will assume for now; for actual cases we will prove the base case $m=0$ and once we prove this induction step will we use it repeatedly to get it true for any specific case"



      ==== looooong repetitvie answer below =====



      (from https://en.wikipedia.org/wiki/Mathematical_induction#Complete_(strong)_induction)




      one proves the statement P(m + 1) under the assumption that P(n) holds for all natural n less than m + 1; (emphasis mine)




      In other words you assume $P(0), P(1), P(2),...... P(m)$. Why do you assume them? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(0), P(1), P(2)...... P(m)$.



      ... and further....




      by contrast, the basic form (weak induction) only assumes P(m). (empahsis mine)




      THe assumption in both these description refer to a base case that must be done later.



      In weak induction you assume $P(m)$. Why do you assume $P(m)$? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(m)$.



      (frome https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction)




      Of course you do (need to prove base cases).




      Well, nothing I can really add to that.



      Okay, it's confusingly written in both your sources but you are citing the description of the inductive step part of the proof only. BOTH strong and weak induction MUST have a base case and neither of your cites claim they don't.



      Compare the descriptions of the weak and strong induction steps (via my paraphrasing).




      Weak induction: Assuming $P(m)$ is true, we show that $P(m)implies P(m+1)$.




      Well, how do we know $P(m)$ is true? Well, via the base case step. Assuming nothing we showed $P(0)$ was true. This doesn't say "We don't need to prove $P(0)$". Quite the opposite. It assumes we either have or will prove in our Base Case. But that was a different part of the proof. In describing the induction step we are assuming the base case has or will be done else where.




      Complete Strong Induction: Assuming for all $0le k le m$ that $P(m)$
      is true we show that $forall k le m: P(m)implies P(m+1)$.




      Well, how do we know $forall k le m: P(m)$ is true? Well, .... cut and paste argument above....



      The only difference in Complete Strong Induction (which is another term for "Strong Induction" which you claim to understand) is in the induction step assumption. The base case(s) step in both are exactly the same and absolutely essential. And NO-ONE is claiming they are not.






      share|cite|improve this answer






















      • why was this down voted?
        – Charlie Parker
        3 hours ago










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      7 Answers
      7






      active

      oldest

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      7 Answers
      7






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      up vote
      7
      down vote



      accepted











      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      Right, that's exactly correct: if there is nothing smaller than $0$ (as is the case for the natural numbers) then it is vacuously true that:



      $$P(m) text holds for any m<0 tag1$$



      So, if you have shown that:



      $$textfor any n: text if P(m) text holds for any m<n, text then P(n) tag2$$



      then in particular you have shown that:



      $$text if P(m) text holds for any m<0, text then P(0) tag2'$$



      and so we get



      $$P(0)$$



      by Modus Ponens on $(1)$ and $(2')$



      So, there is indeed no need to prove an explicit base case.



      That said, think about how in practice you would actually prove $(2)$. Probably, you would be able to show $P(n)$ based on the assumption that there actually are $m<n$ for which we can show that if they all have property $P(m)$, then $P(n)$. But for the edge case of $n=0$, there are no such $m<n$ ... so ... you need to show $P(0)$ by itself!



      In other words, in practice, you often do have to treat the base cases as special cases that you prove as base cases after all.






      share|cite|improve this answer




















      • yea I think I came to this realization. $P(m)$ holds for $m<0$ holds true vacuously on its own. I think in fact is more clear if we write that out as an explicit implication $q(0) := forall m (m<0 to P(m) )$. Then $q(0)$ is true because $m<0$ is false.
        – Charlie Parker
        1 hour ago







      • 1




        @CharlieParker Right! Glad you realized this yourself :)
        – Bram28
        1 hour ago










      • Actually, I wish I did...someone pointed out that my interpretation of $forall m (m < n land P(m) )$ was incorrect and that the correct thing was to consider $forall m (m < n to P(m) )$. Then I realized the rest by myself...I have no idea why I thought the logical AND was the interpretation. Probably because of my horrible use of comma's. Thanks to everyone! :)
        – Charlie Parker
        1 hour ago














      up vote
      7
      down vote



      accepted











      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      Right, that's exactly correct: if there is nothing smaller than $0$ (as is the case for the natural numbers) then it is vacuously true that:



      $$P(m) text holds for any m<0 tag1$$



      So, if you have shown that:



      $$textfor any n: text if P(m) text holds for any m<n, text then P(n) tag2$$



      then in particular you have shown that:



      $$text if P(m) text holds for any m<0, text then P(0) tag2'$$



      and so we get



      $$P(0)$$



      by Modus Ponens on $(1)$ and $(2')$



      So, there is indeed no need to prove an explicit base case.



      That said, think about how in practice you would actually prove $(2)$. Probably, you would be able to show $P(n)$ based on the assumption that there actually are $m<n$ for which we can show that if they all have property $P(m)$, then $P(n)$. But for the edge case of $n=0$, there are no such $m<n$ ... so ... you need to show $P(0)$ by itself!



      In other words, in practice, you often do have to treat the base cases as special cases that you prove as base cases after all.






      share|cite|improve this answer




















      • yea I think I came to this realization. $P(m)$ holds for $m<0$ holds true vacuously on its own. I think in fact is more clear if we write that out as an explicit implication $q(0) := forall m (m<0 to P(m) )$. Then $q(0)$ is true because $m<0$ is false.
        – Charlie Parker
        1 hour ago







      • 1




        @CharlieParker Right! Glad you realized this yourself :)
        – Bram28
        1 hour ago










      • Actually, I wish I did...someone pointed out that my interpretation of $forall m (m < n land P(m) )$ was incorrect and that the correct thing was to consider $forall m (m < n to P(m) )$. Then I realized the rest by myself...I have no idea why I thought the logical AND was the interpretation. Probably because of my horrible use of comma's. Thanks to everyone! :)
        – Charlie Parker
        1 hour ago












      up vote
      7
      down vote



      accepted







      up vote
      7
      down vote



      accepted







      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      Right, that's exactly correct: if there is nothing smaller than $0$ (as is the case for the natural numbers) then it is vacuously true that:



      $$P(m) text holds for any m<0 tag1$$



      So, if you have shown that:



      $$textfor any n: text if P(m) text holds for any m<n, text then P(n) tag2$$



      then in particular you have shown that:



      $$text if P(m) text holds for any m<0, text then P(0) tag2'$$



      and so we get



      $$P(0)$$



      by Modus Ponens on $(1)$ and $(2')$



      So, there is indeed no need to prove an explicit base case.



      That said, think about how in practice you would actually prove $(2)$. Probably, you would be able to show $P(n)$ based on the assumption that there actually are $m<n$ for which we can show that if they all have property $P(m)$, then $P(n)$. But for the edge case of $n=0$, there are no such $m<n$ ... so ... you need to show $P(0)$ by itself!



      In other words, in practice, you often do have to treat the base cases as special cases that you prove as base cases after all.






      share|cite|improve this answer













      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      Right, that's exactly correct: if there is nothing smaller than $0$ (as is the case for the natural numbers) then it is vacuously true that:



      $$P(m) text holds for any m<0 tag1$$



      So, if you have shown that:



      $$textfor any n: text if P(m) text holds for any m<n, text then P(n) tag2$$



      then in particular you have shown that:



      $$text if P(m) text holds for any m<0, text then P(0) tag2'$$



      and so we get



      $$P(0)$$



      by Modus Ponens on $(1)$ and $(2')$



      So, there is indeed no need to prove an explicit base case.



      That said, think about how in practice you would actually prove $(2)$. Probably, you would be able to show $P(n)$ based on the assumption that there actually are $m<n$ for which we can show that if they all have property $P(m)$, then $P(n)$. But for the edge case of $n=0$, there are no such $m<n$ ... so ... you need to show $P(0)$ by itself!



      In other words, in practice, you often do have to treat the base cases as special cases that you prove as base cases after all.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 8 hours ago









      Bram28

      56k33983




      56k33983











      • yea I think I came to this realization. $P(m)$ holds for $m<0$ holds true vacuously on its own. I think in fact is more clear if we write that out as an explicit implication $q(0) := forall m (m<0 to P(m) )$. Then $q(0)$ is true because $m<0$ is false.
        – Charlie Parker
        1 hour ago







      • 1




        @CharlieParker Right! Glad you realized this yourself :)
        – Bram28
        1 hour ago










      • Actually, I wish I did...someone pointed out that my interpretation of $forall m (m < n land P(m) )$ was incorrect and that the correct thing was to consider $forall m (m < n to P(m) )$. Then I realized the rest by myself...I have no idea why I thought the logical AND was the interpretation. Probably because of my horrible use of comma's. Thanks to everyone! :)
        – Charlie Parker
        1 hour ago
















      • yea I think I came to this realization. $P(m)$ holds for $m<0$ holds true vacuously on its own. I think in fact is more clear if we write that out as an explicit implication $q(0) := forall m (m<0 to P(m) )$. Then $q(0)$ is true because $m<0$ is false.
        – Charlie Parker
        1 hour ago







      • 1




        @CharlieParker Right! Glad you realized this yourself :)
        – Bram28
        1 hour ago










      • Actually, I wish I did...someone pointed out that my interpretation of $forall m (m < n land P(m) )$ was incorrect and that the correct thing was to consider $forall m (m < n to P(m) )$. Then I realized the rest by myself...I have no idea why I thought the logical AND was the interpretation. Probably because of my horrible use of comma's. Thanks to everyone! :)
        – Charlie Parker
        1 hour ago















      yea I think I came to this realization. $P(m)$ holds for $m<0$ holds true vacuously on its own. I think in fact is more clear if we write that out as an explicit implication $q(0) := forall m (m<0 to P(m) )$. Then $q(0)$ is true because $m<0$ is false.
      – Charlie Parker
      1 hour ago





      yea I think I came to this realization. $P(m)$ holds for $m<0$ holds true vacuously on its own. I think in fact is more clear if we write that out as an explicit implication $q(0) := forall m (m<0 to P(m) )$. Then $q(0)$ is true because $m<0$ is false.
      – Charlie Parker
      1 hour ago





      1




      1




      @CharlieParker Right! Glad you realized this yourself :)
      – Bram28
      1 hour ago




      @CharlieParker Right! Glad you realized this yourself :)
      – Bram28
      1 hour ago












      Actually, I wish I did...someone pointed out that my interpretation of $forall m (m < n land P(m) )$ was incorrect and that the correct thing was to consider $forall m (m < n to P(m) )$. Then I realized the rest by myself...I have no idea why I thought the logical AND was the interpretation. Probably because of my horrible use of comma's. Thanks to everyone! :)
      – Charlie Parker
      1 hour ago




      Actually, I wish I did...someone pointed out that my interpretation of $forall m (m < n land P(m) )$ was incorrect and that the correct thing was to consider $forall m (m < n to P(m) )$. Then I realized the rest by myself...I have no idea why I thought the logical AND was the interpretation. Probably because of my horrible use of comma's. Thanks to everyone! :)
      – Charlie Parker
      1 hour ago










      up vote
      9
      down vote













      A valid proof by complete induction includes a uniform proof for all $k$ of the inferences listed below. As such it include a (vacuous) proof of the base case $color#c00,P(0).,$ Examine the schema carefully to see it.



      $qquadqquadqquad beginalign
      Rightarrow,color#c00 P(0)\
      P(0)Rightarrow, P(1)\
      P(0),P(1)Rightarrow, P(2)\
      vdotsqquad \
      P(0),P(1),ldots,P(k-1),Rightarrow,P(k)\
      endalign$



      Remark $ $ Its contrapositive form (infinite descent) is clearer: if given a counterexample $lnot P(n)$ we can prove there exists a smaller counterexample $lnot P(k), k < n,,$ then no counterexample exists, else iterating the proof would yield an infinite descending chain of counterexamples, contra $,Bbb N,$ is well-ordered. Or, reformulated, if there is a counterexample then, by well order, we can choose a minimal one ("minimal criminal"), contra the proof yields a smaller one.






      share|cite|improve this answer






















      • what if the inductive step (that only proves the implcation $forall m (m<n to P(m)) to P(n)$ assume implicitly $n not = 0$ by saying assume $m < n$ or pick an element from the non-empty set? I think in those cases then you wouldn't actually have the base case for free: "In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element." quoting wikipedia: en.wikipedia.org/wiki/…
        – Charlie Parker
        1 hour ago











      • @CharlieParker I recommend that you focus on some specific examples in your question. Then it will be much easier to be precise.
        – Bill Dubuque
        55 mins ago















      up vote
      9
      down vote













      A valid proof by complete induction includes a uniform proof for all $k$ of the inferences listed below. As such it include a (vacuous) proof of the base case $color#c00,P(0).,$ Examine the schema carefully to see it.



      $qquadqquadqquad beginalign
      Rightarrow,color#c00 P(0)\
      P(0)Rightarrow, P(1)\
      P(0),P(1)Rightarrow, P(2)\
      vdotsqquad \
      P(0),P(1),ldots,P(k-1),Rightarrow,P(k)\
      endalign$



      Remark $ $ Its contrapositive form (infinite descent) is clearer: if given a counterexample $lnot P(n)$ we can prove there exists a smaller counterexample $lnot P(k), k < n,,$ then no counterexample exists, else iterating the proof would yield an infinite descending chain of counterexamples, contra $,Bbb N,$ is well-ordered. Or, reformulated, if there is a counterexample then, by well order, we can choose a minimal one ("minimal criminal"), contra the proof yields a smaller one.






      share|cite|improve this answer






















      • what if the inductive step (that only proves the implcation $forall m (m<n to P(m)) to P(n)$ assume implicitly $n not = 0$ by saying assume $m < n$ or pick an element from the non-empty set? I think in those cases then you wouldn't actually have the base case for free: "In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element." quoting wikipedia: en.wikipedia.org/wiki/…
        – Charlie Parker
        1 hour ago











      • @CharlieParker I recommend that you focus on some specific examples in your question. Then it will be much easier to be precise.
        – Bill Dubuque
        55 mins ago













      up vote
      9
      down vote










      up vote
      9
      down vote









      A valid proof by complete induction includes a uniform proof for all $k$ of the inferences listed below. As such it include a (vacuous) proof of the base case $color#c00,P(0).,$ Examine the schema carefully to see it.



      $qquadqquadqquad beginalign
      Rightarrow,color#c00 P(0)\
      P(0)Rightarrow, P(1)\
      P(0),P(1)Rightarrow, P(2)\
      vdotsqquad \
      P(0),P(1),ldots,P(k-1),Rightarrow,P(k)\
      endalign$



      Remark $ $ Its contrapositive form (infinite descent) is clearer: if given a counterexample $lnot P(n)$ we can prove there exists a smaller counterexample $lnot P(k), k < n,,$ then no counterexample exists, else iterating the proof would yield an infinite descending chain of counterexamples, contra $,Bbb N,$ is well-ordered. Or, reformulated, if there is a counterexample then, by well order, we can choose a minimal one ("minimal criminal"), contra the proof yields a smaller one.






      share|cite|improve this answer














      A valid proof by complete induction includes a uniform proof for all $k$ of the inferences listed below. As such it include a (vacuous) proof of the base case $color#c00,P(0).,$ Examine the schema carefully to see it.



      $qquadqquadqquad beginalign
      Rightarrow,color#c00 P(0)\
      P(0)Rightarrow, P(1)\
      P(0),P(1)Rightarrow, P(2)\
      vdotsqquad \
      P(0),P(1),ldots,P(k-1),Rightarrow,P(k)\
      endalign$



      Remark $ $ Its contrapositive form (infinite descent) is clearer: if given a counterexample $lnot P(n)$ we can prove there exists a smaller counterexample $lnot P(k), k < n,,$ then no counterexample exists, else iterating the proof would yield an infinite descending chain of counterexamples, contra $,Bbb N,$ is well-ordered. Or, reformulated, if there is a counterexample then, by well order, we can choose a minimal one ("minimal criminal"), contra the proof yields a smaller one.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 9 hours ago

























      answered 9 hours ago









      Bill Dubuque

      203k29188612




      203k29188612











      • what if the inductive step (that only proves the implcation $forall m (m<n to P(m)) to P(n)$ assume implicitly $n not = 0$ by saying assume $m < n$ or pick an element from the non-empty set? I think in those cases then you wouldn't actually have the base case for free: "In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element." quoting wikipedia: en.wikipedia.org/wiki/…
        – Charlie Parker
        1 hour ago











      • @CharlieParker I recommend that you focus on some specific examples in your question. Then it will be much easier to be precise.
        – Bill Dubuque
        55 mins ago

















      • what if the inductive step (that only proves the implcation $forall m (m<n to P(m)) to P(n)$ assume implicitly $n not = 0$ by saying assume $m < n$ or pick an element from the non-empty set? I think in those cases then you wouldn't actually have the base case for free: "In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element." quoting wikipedia: en.wikipedia.org/wiki/…
        – Charlie Parker
        1 hour ago











      • @CharlieParker I recommend that you focus on some specific examples in your question. Then it will be much easier to be precise.
        – Bill Dubuque
        55 mins ago
















      what if the inductive step (that only proves the implcation $forall m (m<n to P(m)) to P(n)$ assume implicitly $n not = 0$ by saying assume $m < n$ or pick an element from the non-empty set? I think in those cases then you wouldn't actually have the base case for free: "In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element." quoting wikipedia: en.wikipedia.org/wiki/…
      – Charlie Parker
      1 hour ago





      what if the inductive step (that only proves the implcation $forall m (m<n to P(m)) to P(n)$ assume implicitly $n not = 0$ by saying assume $m < n$ or pick an element from the non-empty set? I think in those cases then you wouldn't actually have the base case for free: "In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element." quoting wikipedia: en.wikipedia.org/wiki/…
      – Charlie Parker
      1 hour ago













      @CharlieParker I recommend that you focus on some specific examples in your question. Then it will be much easier to be precise.
      – Bill Dubuque
      55 mins ago





      @CharlieParker I recommend that you focus on some specific examples in your question. Then it will be much easier to be precise.
      – Bill Dubuque
      55 mins ago











      up vote
      5
      down vote














      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      It's not clear exactly how one ought to interpret "$P(m), m<n implies P(n)$",
      but I think we have agreed that the following formula
      (shown in the answer by José Carlos Santos)
      represents the induction step according to the article:
      $$((forall min0,1,2,ldots,n-1) P(m))implies P(n).$$



      You seem to be looking at this and saying that for the case $n = 0,$
      it is equivalent to $$bot implies P(n),$$
      using $bot$ as a symbol for something that is always false.
      This implication is vacuously true.
      But in fact, a statement of the form
      $$ (forall min emptyset) P(m) $$
      is also vacuously true. That is, it is true because there is no value of $m$ that can make it false.
      This may be a little more obvious if you write it this way:
      $$ (forall m)(m in emptyset implies P(m)). $$



      So what the induction step of complete induction actually says in the case $n = 0$ is that
      $$topimplies P(0),$$
      where $top$ is always true. If you prove that $topimplies P(0),$ you have proved $P(0).$



      You might question whether the Wikipedia article does a good job of explaining itself, and I would sympathize. It's only an encyclopedia article, however;
      a self-evident justification for everything may be too much too ask.
      (Useful inline citations for details such as this are not too much to ask, however, and there is a notice at the top of that section of the article requesting them.)






      share|cite|improve this answer




















      • This seems to be the core of the OP's misunderstanding; other answers seem to skip over why P(0) is true and take it for granted.
        – Yakk
        8 hours ago










      • Yea, I've realized that was key, the correct thing to consider $forall m (m < n to P(m) )$. After thats cleared up it becomes much easier.
        – Charlie Parker
        1 hour ago














      up vote
      5
      down vote














      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      It's not clear exactly how one ought to interpret "$P(m), m<n implies P(n)$",
      but I think we have agreed that the following formula
      (shown in the answer by José Carlos Santos)
      represents the induction step according to the article:
      $$((forall min0,1,2,ldots,n-1) P(m))implies P(n).$$



      You seem to be looking at this and saying that for the case $n = 0,$
      it is equivalent to $$bot implies P(n),$$
      using $bot$ as a symbol for something that is always false.
      This implication is vacuously true.
      But in fact, a statement of the form
      $$ (forall min emptyset) P(m) $$
      is also vacuously true. That is, it is true because there is no value of $m$ that can make it false.
      This may be a little more obvious if you write it this way:
      $$ (forall m)(m in emptyset implies P(m)). $$



      So what the induction step of complete induction actually says in the case $n = 0$ is that
      $$topimplies P(0),$$
      where $top$ is always true. If you prove that $topimplies P(0),$ you have proved $P(0).$



      You might question whether the Wikipedia article does a good job of explaining itself, and I would sympathize. It's only an encyclopedia article, however;
      a self-evident justification for everything may be too much too ask.
      (Useful inline citations for details such as this are not too much to ask, however, and there is a notice at the top of that section of the article requesting them.)






      share|cite|improve this answer




















      • This seems to be the core of the OP's misunderstanding; other answers seem to skip over why P(0) is true and take it for granted.
        – Yakk
        8 hours ago










      • Yea, I've realized that was key, the correct thing to consider $forall m (m < n to P(m) )$. After thats cleared up it becomes much easier.
        – Charlie Parker
        1 hour ago












      up vote
      5
      down vote










      up vote
      5
      down vote










      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      It's not clear exactly how one ought to interpret "$P(m), m<n implies P(n)$",
      but I think we have agreed that the following formula
      (shown in the answer by José Carlos Santos)
      represents the induction step according to the article:
      $$((forall min0,1,2,ldots,n-1) P(m))implies P(n).$$



      You seem to be looking at this and saying that for the case $n = 0,$
      it is equivalent to $$bot implies P(n),$$
      using $bot$ as a symbol for something that is always false.
      This implication is vacuously true.
      But in fact, a statement of the form
      $$ (forall min emptyset) P(m) $$
      is also vacuously true. That is, it is true because there is no value of $m$ that can make it false.
      This may be a little more obvious if you write it this way:
      $$ (forall m)(m in emptyset implies P(m)). $$



      So what the induction step of complete induction actually says in the case $n = 0$ is that
      $$topimplies P(0),$$
      where $top$ is always true. If you prove that $topimplies P(0),$ you have proved $P(0).$



      You might question whether the Wikipedia article does a good job of explaining itself, and I would sympathize. It's only an encyclopedia article, however;
      a self-evident justification for everything may be too much too ask.
      (Useful inline citations for details such as this are not too much to ask, however, and there is a notice at the top of that section of the article requesting them.)






      share|cite|improve this answer













      if we show $ P(m), m<n implies P(n) $ then somehow that must mean that $P(0)$ is true (or $P(b)$ is true for some set of base cases $b in BaseCases$)




      It's not clear exactly how one ought to interpret "$P(m), m<n implies P(n)$",
      but I think we have agreed that the following formula
      (shown in the answer by José Carlos Santos)
      represents the induction step according to the article:
      $$((forall min0,1,2,ldots,n-1) P(m))implies P(n).$$



      You seem to be looking at this and saying that for the case $n = 0,$
      it is equivalent to $$bot implies P(n),$$
      using $bot$ as a symbol for something that is always false.
      This implication is vacuously true.
      But in fact, a statement of the form
      $$ (forall min emptyset) P(m) $$
      is also vacuously true. That is, it is true because there is no value of $m$ that can make it false.
      This may be a little more obvious if you write it this way:
      $$ (forall m)(m in emptyset implies P(m)). $$



      So what the induction step of complete induction actually says in the case $n = 0$ is that
      $$topimplies P(0),$$
      where $top$ is always true. If you prove that $topimplies P(0),$ you have proved $P(0).$



      You might question whether the Wikipedia article does a good job of explaining itself, and I would sympathize. It's only an encyclopedia article, however;
      a self-evident justification for everything may be too much too ask.
      (Useful inline citations for details such as this are not too much to ask, however, and there is a notice at the top of that section of the article requesting them.)







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 9 hours ago









      David K

      50k340111




      50k340111











      • This seems to be the core of the OP's misunderstanding; other answers seem to skip over why P(0) is true and take it for granted.
        – Yakk
        8 hours ago










      • Yea, I've realized that was key, the correct thing to consider $forall m (m < n to P(m) )$. After thats cleared up it becomes much easier.
        – Charlie Parker
        1 hour ago
















      • This seems to be the core of the OP's misunderstanding; other answers seem to skip over why P(0) is true and take it for granted.
        – Yakk
        8 hours ago










      • Yea, I've realized that was key, the correct thing to consider $forall m (m < n to P(m) )$. After thats cleared up it becomes much easier.
        – Charlie Parker
        1 hour ago















      This seems to be the core of the OP's misunderstanding; other answers seem to skip over why P(0) is true and take it for granted.
      – Yakk
      8 hours ago




      This seems to be the core of the OP's misunderstanding; other answers seem to skip over why P(0) is true and take it for granted.
      – Yakk
      8 hours ago












      Yea, I've realized that was key, the correct thing to consider $forall m (m < n to P(m) )$. After thats cleared up it becomes much easier.
      – Charlie Parker
      1 hour ago




      Yea, I've realized that was key, the correct thing to consider $forall m (m < n to P(m) )$. After thats cleared up it becomes much easier.
      – Charlie Parker
      1 hour ago










      up vote
      4
      down vote













      You did not describe strong induction correctly; there's a quantifier missing. The second step should be:



      $$bigl((forall min0,1,2,ldots,n-1):P(m)bigr)implies P(n).tag1$$



      So, you prove that if $P(0)$, $P(1)$, …, $P(n-1)$, then $P(n)$ holds too. Why should this work? Suppose that you have proved $P(0)$ and also that $(1)$ holds. Then:



      • Since $P(0)$ holds, $P(1)$ holds, by $(1)$.

      • Since $P(0)$ and $P(1)$ hold, $P(2)$ holds, by $(1)$.

      • Since $P(0)$, $P(1)$, and $P(2)$ hold, $P(3)$ holds, by $(1)$.

      And so on…






      share|cite|improve this answer




















      • doesn't change anything about my confusion because I was always assuming $m<n$ meant $forall m, m<n$, since $n$ was a free variable. I've fixed it I guess.
        – Charlie Parker
        9 hours ago







      • 3




        the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
        – Charlie Parker
        9 hours ago







      • 1




        Jose, if you see the Wikipedia and Quora link I gave it claims complete strong induction does NOT explicitly need the base cases proven. Which is what I'm puzzled about.
        – Charlie Parker
        9 hours ago







      • 1




        "the whole point is that complete strong induction does NOT need to show P(0)" Where in either of you sources was that stated?. "Thats my question. Why don't we need to show P(0)?" Who says you don't?
        – fleablood
        9 hours ago






      • 2




        @fleablood In one form of strong induction, the base case is a special case of the induction hypothesis. See Bill Dubuque’s answer. Or am I misunderstanding what the argument is about?
        – spaceisdarkgreen
        9 hours ago














      up vote
      4
      down vote













      You did not describe strong induction correctly; there's a quantifier missing. The second step should be:



      $$bigl((forall min0,1,2,ldots,n-1):P(m)bigr)implies P(n).tag1$$



      So, you prove that if $P(0)$, $P(1)$, …, $P(n-1)$, then $P(n)$ holds too. Why should this work? Suppose that you have proved $P(0)$ and also that $(1)$ holds. Then:



      • Since $P(0)$ holds, $P(1)$ holds, by $(1)$.

      • Since $P(0)$ and $P(1)$ hold, $P(2)$ holds, by $(1)$.

      • Since $P(0)$, $P(1)$, and $P(2)$ hold, $P(3)$ holds, by $(1)$.

      And so on…






      share|cite|improve this answer




















      • doesn't change anything about my confusion because I was always assuming $m<n$ meant $forall m, m<n$, since $n$ was a free variable. I've fixed it I guess.
        – Charlie Parker
        9 hours ago







      • 3




        the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
        – Charlie Parker
        9 hours ago







      • 1




        Jose, if you see the Wikipedia and Quora link I gave it claims complete strong induction does NOT explicitly need the base cases proven. Which is what I'm puzzled about.
        – Charlie Parker
        9 hours ago







      • 1




        "the whole point is that complete strong induction does NOT need to show P(0)" Where in either of you sources was that stated?. "Thats my question. Why don't we need to show P(0)?" Who says you don't?
        – fleablood
        9 hours ago






      • 2




        @fleablood In one form of strong induction, the base case is a special case of the induction hypothesis. See Bill Dubuque’s answer. Or am I misunderstanding what the argument is about?
        – spaceisdarkgreen
        9 hours ago












      up vote
      4
      down vote










      up vote
      4
      down vote









      You did not describe strong induction correctly; there's a quantifier missing. The second step should be:



      $$bigl((forall min0,1,2,ldots,n-1):P(m)bigr)implies P(n).tag1$$



      So, you prove that if $P(0)$, $P(1)$, …, $P(n-1)$, then $P(n)$ holds too. Why should this work? Suppose that you have proved $P(0)$ and also that $(1)$ holds. Then:



      • Since $P(0)$ holds, $P(1)$ holds, by $(1)$.

      • Since $P(0)$ and $P(1)$ hold, $P(2)$ holds, by $(1)$.

      • Since $P(0)$, $P(1)$, and $P(2)$ hold, $P(3)$ holds, by $(1)$.

      And so on…






      share|cite|improve this answer












      You did not describe strong induction correctly; there's a quantifier missing. The second step should be:



      $$bigl((forall min0,1,2,ldots,n-1):P(m)bigr)implies P(n).tag1$$



      So, you prove that if $P(0)$, $P(1)$, …, $P(n-1)$, then $P(n)$ holds too. Why should this work? Suppose that you have proved $P(0)$ and also that $(1)$ holds. Then:



      • Since $P(0)$ holds, $P(1)$ holds, by $(1)$.

      • Since $P(0)$ and $P(1)$ hold, $P(2)$ holds, by $(1)$.

      • Since $P(0)$, $P(1)$, and $P(2)$ hold, $P(3)$ holds, by $(1)$.

      And so on…







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 10 hours ago









      José Carlos Santos

      128k17102189




      128k17102189











      • doesn't change anything about my confusion because I was always assuming $m<n$ meant $forall m, m<n$, since $n$ was a free variable. I've fixed it I guess.
        – Charlie Parker
        9 hours ago







      • 3




        the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
        – Charlie Parker
        9 hours ago







      • 1




        Jose, if you see the Wikipedia and Quora link I gave it claims complete strong induction does NOT explicitly need the base cases proven. Which is what I'm puzzled about.
        – Charlie Parker
        9 hours ago







      • 1




        "the whole point is that complete strong induction does NOT need to show P(0)" Where in either of you sources was that stated?. "Thats my question. Why don't we need to show P(0)?" Who says you don't?
        – fleablood
        9 hours ago






      • 2




        @fleablood In one form of strong induction, the base case is a special case of the induction hypothesis. See Bill Dubuque’s answer. Or am I misunderstanding what the argument is about?
        – spaceisdarkgreen
        9 hours ago
















      • doesn't change anything about my confusion because I was always assuming $m<n$ meant $forall m, m<n$, since $n$ was a free variable. I've fixed it I guess.
        – Charlie Parker
        9 hours ago







      • 3




        the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
        – Charlie Parker
        9 hours ago







      • 1




        Jose, if you see the Wikipedia and Quora link I gave it claims complete strong induction does NOT explicitly need the base cases proven. Which is what I'm puzzled about.
        – Charlie Parker
        9 hours ago







      • 1




        "the whole point is that complete strong induction does NOT need to show P(0)" Where in either of you sources was that stated?. "Thats my question. Why don't we need to show P(0)?" Who says you don't?
        – fleablood
        9 hours ago






      • 2




        @fleablood In one form of strong induction, the base case is a special case of the induction hypothesis. See Bill Dubuque’s answer. Or am I misunderstanding what the argument is about?
        – spaceisdarkgreen
        9 hours ago















      doesn't change anything about my confusion because I was always assuming $m<n$ meant $forall m, m<n$, since $n$ was a free variable. I've fixed it I guess.
      – Charlie Parker
      9 hours ago





      doesn't change anything about my confusion because I was always assuming $m<n$ meant $forall m, m<n$, since $n$ was a free variable. I've fixed it I guess.
      – Charlie Parker
      9 hours ago





      3




      3




      the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
      – Charlie Parker
      9 hours ago





      the whole point is that complete strong induction does NOT need to show $P(0)$. Thats my question. Why don't we need to show $P(0)$? I understand modus ponens (MP). If we have P(0) then we can get the next proposition if we have $ P(0) implies P(1) $ shown true (by the inductive step). My question is if we have the inductive step, why do we have the base case automatically?
      – Charlie Parker
      9 hours ago





      1




      1




      Jose, if you see the Wikipedia and Quora link I gave it claims complete strong induction does NOT explicitly need the base cases proven. Which is what I'm puzzled about.
      – Charlie Parker
      9 hours ago





      Jose, if you see the Wikipedia and Quora link I gave it claims complete strong induction does NOT explicitly need the base cases proven. Which is what I'm puzzled about.
      – Charlie Parker
      9 hours ago





      1




      1




      "the whole point is that complete strong induction does NOT need to show P(0)" Where in either of you sources was that stated?. "Thats my question. Why don't we need to show P(0)?" Who says you don't?
      – fleablood
      9 hours ago




      "the whole point is that complete strong induction does NOT need to show P(0)" Where in either of you sources was that stated?. "Thats my question. Why don't we need to show P(0)?" Who says you don't?
      – fleablood
      9 hours ago




      2




      2




      @fleablood In one form of strong induction, the base case is a special case of the induction hypothesis. See Bill Dubuque’s answer. Or am I misunderstanding what the argument is about?
      – spaceisdarkgreen
      9 hours ago




      @fleablood In one form of strong induction, the base case is a special case of the induction hypothesis. See Bill Dubuque’s answer. Or am I misunderstanding what the argument is about?
      – spaceisdarkgreen
      9 hours ago










      up vote
      3
      down vote













      You write:




      If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False.




      This is where you are wrong. As you noticed, $ forall m, m < 0, P(0)$ is (vacuously) true. But that does not mean that the above statement is false, indeed




      $forall m, m < 0, P(m) implies P(0)quad $ is equivalent to $quad P(0)$.




      (If you doubt this: $mathrmtruerightarrow x iff neg mathrmtrue lor x iff mathrmfalse lor x iff x$.)



      So in complete induction you actually have to show $P(0)$, there is just no reason to list it separately from the implications that you have to show.



      Stated differently: In complete induction, for each $n$ you show $P(n)$, but you are allowed to use all $P(m)$ for $m < n$ in the proof of $P(n)$. For $n=0$ this does not allow you anything new as there is no $m<0$.






      share|cite|improve this answer
























        up vote
        3
        down vote













        You write:




        If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False.




        This is where you are wrong. As you noticed, $ forall m, m < 0, P(0)$ is (vacuously) true. But that does not mean that the above statement is false, indeed




        $forall m, m < 0, P(m) implies P(0)quad $ is equivalent to $quad P(0)$.




        (If you doubt this: $mathrmtruerightarrow x iff neg mathrmtrue lor x iff mathrmfalse lor x iff x$.)



        So in complete induction you actually have to show $P(0)$, there is just no reason to list it separately from the implications that you have to show.



        Stated differently: In complete induction, for each $n$ you show $P(n)$, but you are allowed to use all $P(m)$ for $m < n$ in the proof of $P(n)$. For $n=0$ this does not allow you anything new as there is no $m<0$.






        share|cite|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          You write:




          If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False.




          This is where you are wrong. As you noticed, $ forall m, m < 0, P(0)$ is (vacuously) true. But that does not mean that the above statement is false, indeed




          $forall m, m < 0, P(m) implies P(0)quad $ is equivalent to $quad P(0)$.




          (If you doubt this: $mathrmtruerightarrow x iff neg mathrmtrue lor x iff mathrmfalse lor x iff x$.)



          So in complete induction you actually have to show $P(0)$, there is just no reason to list it separately from the implications that you have to show.



          Stated differently: In complete induction, for each $n$ you show $P(n)$, but you are allowed to use all $P(m)$ for $m < n$ in the proof of $P(n)$. For $n=0$ this does not allow you anything new as there is no $m<0$.






          share|cite|improve this answer












          You write:




          If $n=0$ then $ forall m, m < n, P(m) implies P(n) $ is False.




          This is where you are wrong. As you noticed, $ forall m, m < 0, P(0)$ is (vacuously) true. But that does not mean that the above statement is false, indeed




          $forall m, m < 0, P(m) implies P(0)quad $ is equivalent to $quad P(0)$.




          (If you doubt this: $mathrmtruerightarrow x iff neg mathrmtrue lor x iff mathrmfalse lor x iff x$.)



          So in complete induction you actually have to show $P(0)$, there is just no reason to list it separately from the implications that you have to show.



          Stated differently: In complete induction, for each $n$ you show $P(n)$, but you are allowed to use all $P(m)$ for $m < n$ in the proof of $P(n)$. For $n=0$ this does not allow you anything new as there is no $m<0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Carsten S

          6,75411334




          6,75411334




















              up vote
              0
              down vote













              I think I finally understand my confusions after reading the wikipedia article more carefully and getting my notation right. First recall what the inductive step (that we have to proof) is in induction:



              $$ varphi(n) := forall m (m < n to P(m)) to P(n) $$



              what complete strong induction claims to my understanding is that the proof of the inductive step includes the base case automatically because the argument also holds for the base cases, $P(0)$ for example. So now define:



              $$ q(n) := forall m (m < n to P(m)) $$
              $$ p(n) := P(n) $$
              so:



              $$ varphi(n) = q(n) to p(n) $$



              if we assume we prove the inductive step and that argument holds for every $n$ including the base case then we have:



              $$ varphi(0) = q(0) to p(0) $$



              is true as a whole. However, if we carefully examine what $q(0)$ is we notice that its a tautology, i.e.



              $$ q(0) = forall m (m < 0 to P(m))$$



              because $m < 0$ is False because $m in mathbb N = 0,1,2,3,dots$ (i.e. $0<0$,$1<0,2<0cdots$ is always false), so $(m < 0 to P(m))$ is True for all values of m in consideration. So now we know $varphi(0) = q(0) to p(0)$ is True and $q(0)$ is True as a stand alone logical sentence (this is not usually true). So by modus pones (MP) we have:



              $$ q(0)$$
              $$ q(0) to p(0)$$



              and it follows by MP:



              $$ p(0) $$



              which eventually results in the cascading of logical implications usual to induction.



              Note however that the inductive step, depending on the content of the proof might or might NOT prove the base case automatically. For example, wikipedia made a good job of outlining why we need to be careful:




              In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element.




              the second case that talks about sets is important to note because we also have structural induction and the argument might depend on selecting an element from a set, but that only holds if the set is none empty to start with.



              So, if your unsure, proof the base cases, but you can do complete induction if your sure your proof does include $m=0$ as well as $m>0$.






              share|cite|improve this answer
























                up vote
                0
                down vote













                I think I finally understand my confusions after reading the wikipedia article more carefully and getting my notation right. First recall what the inductive step (that we have to proof) is in induction:



                $$ varphi(n) := forall m (m < n to P(m)) to P(n) $$



                what complete strong induction claims to my understanding is that the proof of the inductive step includes the base case automatically because the argument also holds for the base cases, $P(0)$ for example. So now define:



                $$ q(n) := forall m (m < n to P(m)) $$
                $$ p(n) := P(n) $$
                so:



                $$ varphi(n) = q(n) to p(n) $$



                if we assume we prove the inductive step and that argument holds for every $n$ including the base case then we have:



                $$ varphi(0) = q(0) to p(0) $$



                is true as a whole. However, if we carefully examine what $q(0)$ is we notice that its a tautology, i.e.



                $$ q(0) = forall m (m < 0 to P(m))$$



                because $m < 0$ is False because $m in mathbb N = 0,1,2,3,dots$ (i.e. $0<0$,$1<0,2<0cdots$ is always false), so $(m < 0 to P(m))$ is True for all values of m in consideration. So now we know $varphi(0) = q(0) to p(0)$ is True and $q(0)$ is True as a stand alone logical sentence (this is not usually true). So by modus pones (MP) we have:



                $$ q(0)$$
                $$ q(0) to p(0)$$



                and it follows by MP:



                $$ p(0) $$



                which eventually results in the cascading of logical implications usual to induction.



                Note however that the inductive step, depending on the content of the proof might or might NOT prove the base case automatically. For example, wikipedia made a good job of outlining why we need to be careful:




                In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element.




                the second case that talks about sets is important to note because we also have structural induction and the argument might depend on selecting an element from a set, but that only holds if the set is none empty to start with.



                So, if your unsure, proof the base cases, but you can do complete induction if your sure your proof does include $m=0$ as well as $m>0$.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  I think I finally understand my confusions after reading the wikipedia article more carefully and getting my notation right. First recall what the inductive step (that we have to proof) is in induction:



                  $$ varphi(n) := forall m (m < n to P(m)) to P(n) $$



                  what complete strong induction claims to my understanding is that the proof of the inductive step includes the base case automatically because the argument also holds for the base cases, $P(0)$ for example. So now define:



                  $$ q(n) := forall m (m < n to P(m)) $$
                  $$ p(n) := P(n) $$
                  so:



                  $$ varphi(n) = q(n) to p(n) $$



                  if we assume we prove the inductive step and that argument holds for every $n$ including the base case then we have:



                  $$ varphi(0) = q(0) to p(0) $$



                  is true as a whole. However, if we carefully examine what $q(0)$ is we notice that its a tautology, i.e.



                  $$ q(0) = forall m (m < 0 to P(m))$$



                  because $m < 0$ is False because $m in mathbb N = 0,1,2,3,dots$ (i.e. $0<0$,$1<0,2<0cdots$ is always false), so $(m < 0 to P(m))$ is True for all values of m in consideration. So now we know $varphi(0) = q(0) to p(0)$ is True and $q(0)$ is True as a stand alone logical sentence (this is not usually true). So by modus pones (MP) we have:



                  $$ q(0)$$
                  $$ q(0) to p(0)$$



                  and it follows by MP:



                  $$ p(0) $$



                  which eventually results in the cascading of logical implications usual to induction.



                  Note however that the inductive step, depending on the content of the proof might or might NOT prove the base case automatically. For example, wikipedia made a good job of outlining why we need to be careful:




                  In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element.




                  the second case that talks about sets is important to note because we also have structural induction and the argument might depend on selecting an element from a set, but that only holds if the set is none empty to start with.



                  So, if your unsure, proof the base cases, but you can do complete induction if your sure your proof does include $m=0$ as well as $m>0$.






                  share|cite|improve this answer












                  I think I finally understand my confusions after reading the wikipedia article more carefully and getting my notation right. First recall what the inductive step (that we have to proof) is in induction:



                  $$ varphi(n) := forall m (m < n to P(m)) to P(n) $$



                  what complete strong induction claims to my understanding is that the proof of the inductive step includes the base case automatically because the argument also holds for the base cases, $P(0)$ for example. So now define:



                  $$ q(n) := forall m (m < n to P(m)) $$
                  $$ p(n) := P(n) $$
                  so:



                  $$ varphi(n) = q(n) to p(n) $$



                  if we assume we prove the inductive step and that argument holds for every $n$ including the base case then we have:



                  $$ varphi(0) = q(0) to p(0) $$



                  is true as a whole. However, if we carefully examine what $q(0)$ is we notice that its a tautology, i.e.



                  $$ q(0) = forall m (m < 0 to P(m))$$



                  because $m < 0$ is False because $m in mathbb N = 0,1,2,3,dots$ (i.e. $0<0$,$1<0,2<0cdots$ is always false), so $(m < 0 to P(m))$ is True for all values of m in consideration. So now we know $varphi(0) = q(0) to p(0)$ is True and $q(0)$ is True as a stand alone logical sentence (this is not usually true). So by modus pones (MP) we have:



                  $$ q(0)$$
                  $$ q(0) to p(0)$$



                  and it follows by MP:



                  $$ p(0) $$



                  which eventually results in the cascading of logical implications usual to induction.



                  Note however that the inductive step, depending on the content of the proof might or might NOT prove the base case automatically. For example, wikipedia made a good job of outlining why we need to be careful:




                  In this method it is, however, vital to ensure that the proof of P(m) does not implicitly assume that m > 0, e.g. by saying "choose an arbitrary n < m" or assuming that a set of m elements has an element.




                  the second case that talks about sets is important to note because we also have structural induction and the argument might depend on selecting an element from a set, but that only holds if the set is none empty to start with.



                  So, if your unsure, proof the base cases, but you can do complete induction if your sure your proof does include $m=0$ as well as $m>0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Charlie Parker

                  1,0491128




                  1,0491128




















                      up vote
                      -3
                      down vote














                      "But what particularly confuses me is why we do not to explicit the base cases for complete induction."




                      Short answer. You do explicitly prove the base cases. You misunderstood the description. You absolutely DO have to prove the base cases.




                      "I am familiar with both strong induction and ordinary induction"




                      Complete Strong induction is just another name for strong induction. These are two methods of induction. Not three. There is nothing new here that you do not already know.




                      However, in complete induction we only show:



                      1) ∀m,m<n,P(m)⟹P(n)



                      No. That is your mistake. That is NOT what complete strong induction says. Complete strong induction (which is exactly strong induction) says you must show a base case. It simply DOES. Your were wrong in thinking it said you didn't.)



                      Weak induction:



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ is true for $m=0$.)



                      2) Induction step: Assume $P(m)$. Prove P(m+1).



                      (Complete) Strong induction



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ and $P(k)$ are true for $m = 0$ and $0 le k le m = 0$.)



                      2) Induction step: Assume $P(m)$ and $P(k)$ for $0 le k le m$. Prove $P(m+1)$.



                      Notice that both say "assume" something. They do not mean "this is true and we don't need to prove it"; they both mean "we will assume for now; for actual cases we will prove the base case $m=0$ and once we prove this induction step will we use it repeatedly to get it true for any specific case"



                      ==== looooong repetitvie answer below =====



                      (from https://en.wikipedia.org/wiki/Mathematical_induction#Complete_(strong)_induction)




                      one proves the statement P(m + 1) under the assumption that P(n) holds for all natural n less than m + 1; (emphasis mine)




                      In other words you assume $P(0), P(1), P(2),...... P(m)$. Why do you assume them? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(0), P(1), P(2)...... P(m)$.



                      ... and further....




                      by contrast, the basic form (weak induction) only assumes P(m). (empahsis mine)




                      THe assumption in both these description refer to a base case that must be done later.



                      In weak induction you assume $P(m)$. Why do you assume $P(m)$? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(m)$.



                      (frome https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction)




                      Of course you do (need to prove base cases).




                      Well, nothing I can really add to that.



                      Okay, it's confusingly written in both your sources but you are citing the description of the inductive step part of the proof only. BOTH strong and weak induction MUST have a base case and neither of your cites claim they don't.



                      Compare the descriptions of the weak and strong induction steps (via my paraphrasing).




                      Weak induction: Assuming $P(m)$ is true, we show that $P(m)implies P(m+1)$.




                      Well, how do we know $P(m)$ is true? Well, via the base case step. Assuming nothing we showed $P(0)$ was true. This doesn't say "We don't need to prove $P(0)$". Quite the opposite. It assumes we either have or will prove in our Base Case. But that was a different part of the proof. In describing the induction step we are assuming the base case has or will be done else where.




                      Complete Strong Induction: Assuming for all $0le k le m$ that $P(m)$
                      is true we show that $forall k le m: P(m)implies P(m+1)$.




                      Well, how do we know $forall k le m: P(m)$ is true? Well, .... cut and paste argument above....



                      The only difference in Complete Strong Induction (which is another term for "Strong Induction" which you claim to understand) is in the induction step assumption. The base case(s) step in both are exactly the same and absolutely essential. And NO-ONE is claiming they are not.






                      share|cite|improve this answer






















                      • why was this down voted?
                        – Charlie Parker
                        3 hours ago














                      up vote
                      -3
                      down vote














                      "But what particularly confuses me is why we do not to explicit the base cases for complete induction."




                      Short answer. You do explicitly prove the base cases. You misunderstood the description. You absolutely DO have to prove the base cases.




                      "I am familiar with both strong induction and ordinary induction"




                      Complete Strong induction is just another name for strong induction. These are two methods of induction. Not three. There is nothing new here that you do not already know.




                      However, in complete induction we only show:



                      1) ∀m,m<n,P(m)⟹P(n)



                      No. That is your mistake. That is NOT what complete strong induction says. Complete strong induction (which is exactly strong induction) says you must show a base case. It simply DOES. Your were wrong in thinking it said you didn't.)



                      Weak induction:



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ is true for $m=0$.)



                      2) Induction step: Assume $P(m)$. Prove P(m+1).



                      (Complete) Strong induction



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ and $P(k)$ are true for $m = 0$ and $0 le k le m = 0$.)



                      2) Induction step: Assume $P(m)$ and $P(k)$ for $0 le k le m$. Prove $P(m+1)$.



                      Notice that both say "assume" something. They do not mean "this is true and we don't need to prove it"; they both mean "we will assume for now; for actual cases we will prove the base case $m=0$ and once we prove this induction step will we use it repeatedly to get it true for any specific case"



                      ==== looooong repetitvie answer below =====



                      (from https://en.wikipedia.org/wiki/Mathematical_induction#Complete_(strong)_induction)




                      one proves the statement P(m + 1) under the assumption that P(n) holds for all natural n less than m + 1; (emphasis mine)




                      In other words you assume $P(0), P(1), P(2),...... P(m)$. Why do you assume them? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(0), P(1), P(2)...... P(m)$.



                      ... and further....




                      by contrast, the basic form (weak induction) only assumes P(m). (empahsis mine)




                      THe assumption in both these description refer to a base case that must be done later.



                      In weak induction you assume $P(m)$. Why do you assume $P(m)$? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(m)$.



                      (frome https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction)




                      Of course you do (need to prove base cases).




                      Well, nothing I can really add to that.



                      Okay, it's confusingly written in both your sources but you are citing the description of the inductive step part of the proof only. BOTH strong and weak induction MUST have a base case and neither of your cites claim they don't.



                      Compare the descriptions of the weak and strong induction steps (via my paraphrasing).




                      Weak induction: Assuming $P(m)$ is true, we show that $P(m)implies P(m+1)$.




                      Well, how do we know $P(m)$ is true? Well, via the base case step. Assuming nothing we showed $P(0)$ was true. This doesn't say "We don't need to prove $P(0)$". Quite the opposite. It assumes we either have or will prove in our Base Case. But that was a different part of the proof. In describing the induction step we are assuming the base case has or will be done else where.




                      Complete Strong Induction: Assuming for all $0le k le m$ that $P(m)$
                      is true we show that $forall k le m: P(m)implies P(m+1)$.




                      Well, how do we know $forall k le m: P(m)$ is true? Well, .... cut and paste argument above....



                      The only difference in Complete Strong Induction (which is another term for "Strong Induction" which you claim to understand) is in the induction step assumption. The base case(s) step in both are exactly the same and absolutely essential. And NO-ONE is claiming they are not.






                      share|cite|improve this answer






















                      • why was this down voted?
                        – Charlie Parker
                        3 hours ago












                      up vote
                      -3
                      down vote










                      up vote
                      -3
                      down vote










                      "But what particularly confuses me is why we do not to explicit the base cases for complete induction."




                      Short answer. You do explicitly prove the base cases. You misunderstood the description. You absolutely DO have to prove the base cases.




                      "I am familiar with both strong induction and ordinary induction"




                      Complete Strong induction is just another name for strong induction. These are two methods of induction. Not three. There is nothing new here that you do not already know.




                      However, in complete induction we only show:



                      1) ∀m,m<n,P(m)⟹P(n)



                      No. That is your mistake. That is NOT what complete strong induction says. Complete strong induction (which is exactly strong induction) says you must show a base case. It simply DOES. Your were wrong in thinking it said you didn't.)



                      Weak induction:



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ is true for $m=0$.)



                      2) Induction step: Assume $P(m)$. Prove P(m+1).



                      (Complete) Strong induction



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ and $P(k)$ are true for $m = 0$ and $0 le k le m = 0$.)



                      2) Induction step: Assume $P(m)$ and $P(k)$ for $0 le k le m$. Prove $P(m+1)$.



                      Notice that both say "assume" something. They do not mean "this is true and we don't need to prove it"; they both mean "we will assume for now; for actual cases we will prove the base case $m=0$ and once we prove this induction step will we use it repeatedly to get it true for any specific case"



                      ==== looooong repetitvie answer below =====



                      (from https://en.wikipedia.org/wiki/Mathematical_induction#Complete_(strong)_induction)




                      one proves the statement P(m + 1) under the assumption that P(n) holds for all natural n less than m + 1; (emphasis mine)




                      In other words you assume $P(0), P(1), P(2),...... P(m)$. Why do you assume them? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(0), P(1), P(2)...... P(m)$.



                      ... and further....




                      by contrast, the basic form (weak induction) only assumes P(m). (empahsis mine)




                      THe assumption in both these description refer to a base case that must be done later.



                      In weak induction you assume $P(m)$. Why do you assume $P(m)$? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(m)$.



                      (frome https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction)




                      Of course you do (need to prove base cases).




                      Well, nothing I can really add to that.



                      Okay, it's confusingly written in both your sources but you are citing the description of the inductive step part of the proof only. BOTH strong and weak induction MUST have a base case and neither of your cites claim they don't.



                      Compare the descriptions of the weak and strong induction steps (via my paraphrasing).




                      Weak induction: Assuming $P(m)$ is true, we show that $P(m)implies P(m+1)$.




                      Well, how do we know $P(m)$ is true? Well, via the base case step. Assuming nothing we showed $P(0)$ was true. This doesn't say "We don't need to prove $P(0)$". Quite the opposite. It assumes we either have or will prove in our Base Case. But that was a different part of the proof. In describing the induction step we are assuming the base case has or will be done else where.




                      Complete Strong Induction: Assuming for all $0le k le m$ that $P(m)$
                      is true we show that $forall k le m: P(m)implies P(m+1)$.




                      Well, how do we know $forall k le m: P(m)$ is true? Well, .... cut and paste argument above....



                      The only difference in Complete Strong Induction (which is another term for "Strong Induction" which you claim to understand) is in the induction step assumption. The base case(s) step in both are exactly the same and absolutely essential. And NO-ONE is claiming they are not.






                      share|cite|improve this answer















                      "But what particularly confuses me is why we do not to explicit the base cases for complete induction."




                      Short answer. You do explicitly prove the base cases. You misunderstood the description. You absolutely DO have to prove the base cases.




                      "I am familiar with both strong induction and ordinary induction"




                      Complete Strong induction is just another name for strong induction. These are two methods of induction. Not three. There is nothing new here that you do not already know.




                      However, in complete induction we only show:



                      1) ∀m,m<n,P(m)⟹P(n)



                      No. That is your mistake. That is NOT what complete strong induction says. Complete strong induction (which is exactly strong induction) says you must show a base case. It simply DOES. Your were wrong in thinking it said you didn't.)



                      Weak induction:



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ is true for $m=0$.)



                      2) Induction step: Assume $P(m)$. Prove P(m+1).



                      (Complete) Strong induction



                      1) Base step: Prove $P(0)$. (This will prove $P(m)$ and $P(k)$ are true for $m = 0$ and $0 le k le m = 0$.)



                      2) Induction step: Assume $P(m)$ and $P(k)$ for $0 le k le m$. Prove $P(m+1)$.



                      Notice that both say "assume" something. They do not mean "this is true and we don't need to prove it"; they both mean "we will assume for now; for actual cases we will prove the base case $m=0$ and once we prove this induction step will we use it repeatedly to get it true for any specific case"



                      ==== looooong repetitvie answer below =====



                      (from https://en.wikipedia.org/wiki/Mathematical_induction#Complete_(strong)_induction)




                      one proves the statement P(m + 1) under the assumption that P(n) holds for all natural n less than m + 1; (emphasis mine)




                      In other words you assume $P(0), P(1), P(2),...... P(m)$. Why do you assume them? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(0), P(1), P(2)...... P(m)$.



                      ... and further....




                      by contrast, the basic form (weak induction) only assumes P(m). (empahsis mine)




                      THe assumption in both these description refer to a base case that must be done later.



                      In weak induction you assume $P(m)$. Why do you assume $P(m)$? Because you did a base case with $m =0$. And once you prove this step you will repeat it to get up $m$ and thus you will know $P(m)$.



                      (frome https://www.quora.com/Why-dont-you-need-to-prove-base-cases-for-complete-strong-induction)




                      Of course you do (need to prove base cases).




                      Well, nothing I can really add to that.



                      Okay, it's confusingly written in both your sources but you are citing the description of the inductive step part of the proof only. BOTH strong and weak induction MUST have a base case and neither of your cites claim they don't.



                      Compare the descriptions of the weak and strong induction steps (via my paraphrasing).




                      Weak induction: Assuming $P(m)$ is true, we show that $P(m)implies P(m+1)$.




                      Well, how do we know $P(m)$ is true? Well, via the base case step. Assuming nothing we showed $P(0)$ was true. This doesn't say "We don't need to prove $P(0)$". Quite the opposite. It assumes we either have or will prove in our Base Case. But that was a different part of the proof. In describing the induction step we are assuming the base case has or will be done else where.




                      Complete Strong Induction: Assuming for all $0le k le m$ that $P(m)$
                      is true we show that $forall k le m: P(m)implies P(m+1)$.




                      Well, how do we know $forall k le m: P(m)$ is true? Well, .... cut and paste argument above....



                      The only difference in Complete Strong Induction (which is another term for "Strong Induction" which you claim to understand) is in the induction step assumption. The base case(s) step in both are exactly the same and absolutely essential. And NO-ONE is claiming they are not.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 8 hours ago

























                      answered 9 hours ago









                      fleablood

                      1




                      1











                      • why was this down voted?
                        – Charlie Parker
                        3 hours ago
















                      • why was this down voted?
                        – Charlie Parker
                        3 hours ago















                      why was this down voted?
                      – Charlie Parker
                      3 hours ago




                      why was this down voted?
                      – Charlie Parker
                      3 hours ago

















                       

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