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What would the set-theoretical axiom of induction look like for double induction* when stated in the mathematical language of first- or second-order logic?

*References as to What Double Induction Is:

To questions on this StackExchange:

To other sources:

Community
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RandomDSdevel
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    @WillieWong: 'Double induction' is the use of mathematical induction to prove the truth of a logical predicate that depends on two variables instead of just one, hence the 'double' in its name. As I understand it, the technique can be implemented either by using a map from the bivariate predicate $\phi(x, y)$ in question to a univariate $\phi(z)$ as a shim allowing one to perform univariate mathematical induction over that or as recursive mathematical induction. I'm wondering what the axiom of induction might look like for a use of the latter implementation. – RandomDSdevel Nov 11 '15 at 21:22
  • Judging by the examples, it is not clear that "double induction" would have an axiom or axiom schema separate from an "axiom of induction" in number theory or from some principle of well-ordering/axiom of choice in set theory. – hardmath Nov 13 '15 at 23:04
  • @hardmath: Right, an 'axiom of double induction' would likely just be the axiom resulting from some kind of substitution of the univariate axiom of induction back into itself in one or more places. The reason I asked this question is because I can't seem to figure out what, exactly, that result would turn out to look like. – RandomDSdevel Nov 13 '15 at 23:13
  • One improvement to the Question would be to clarify whether the formal "axiom" is for number theory or set theory. It seems related to the topic of defining arithmetic functions by recursion. – hardmath Nov 13 '15 at 23:22
  • @hardmath: OK, then I'll go clarify that now. – RandomDSdevel Nov 14 '15 at 00:00
  • @hardmath: The edit should be live now. – RandomDSdevel Nov 14 '15 at 00:07
  • Do any of those of you who have seen this question think it needs any more refinement? – RandomDSdevel Nov 17 '15 at 23:35
  • Doesn't this (your second link) answer it? – Akiva Weinberger Nov 19 '15 at 00:26
  • @AkivaWeinberger: Perhaps, but I'm not familiar enough with the notation used by strong induction to be able to parse that. Besides, I was asking this question in the context of the form of mathematical induction described in the Wikipedia article I linked as being the more common form of the proof method (is that weak induction?) In any case, I'm most familiar with first- and second-order logic, so that's what I was expecting my answer to be in; I'll go clarify that in my original question and its bounty's description. – RandomDSdevel Nov 19 '15 at 01:01
  • Huh, turns out I can't edit my bounty description. Oh, well; I was able to edit my original question. Will that do? – RandomDSdevel Nov 19 '15 at 01:03
  • It would be of the form: $\forall m,n$, if *induction hypothesis* then $P(m,n)$. Right? – Akiva Weinberger Nov 19 '15 at 01:10
  • @AkivaWeinberger: Sort of, but you'd have to fill in the generic form of the 'induction hypothesis' such that it is specializable $\forall m, n, P(m, n)$ in order to truly answer this question since it's looking for a form of the axiom of induction that can be used with two variables. – RandomDSdevel Nov 22 '15 at 00:58
  • You may also be interested in the blog post http://math.blogoverflow.com/2015/03/10/when-can-we-do-induction/ – Tobias Kildetoft Nov 23 '15 at 08:57
  • @TobiasKildetoft: Thanks for the interesting reference! – RandomDSdevel Nov 24 '15 at 00:21

1 Answers1

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Here's a straight application of simple induction (not strong induction), twice:

We want to prove $P(m,n)$ by induction over $n$. Thus we need to prove $P(m,0)$ and $P(m,n)\to P(m,n+1)$. But in order to prove $P(m,0)$ we use induction over $m$, so we need to prove $P(0,0)$ and $P(m,0)\to P(m+1,0)$.

In symbols, this amounts to the following assertion:

$$P(0,0)\,\land\,[\forall m,P(m,0)\to P(m+1,0)]\ \land\ [\forall mn,P(m,n)\to P(m,n+1)]\implies\forall xy, P(x,y)$$

In the language of well-founded induction, this corresponds to the order $$(m,n)\prec_1(m',n')\iff (m=m'\land n<n')\lor(n=n'=0\land m<m'),$$ which is not a total order but is well-founded anyway, because there is a (unique) path from $(m,n)$ to the minimum element $(0,0)$ of length $m+n$, so there are no infinite descending sequences.

Alternatively, you could simplify the argument, by encompassing both inductions into one:

We want to prove $P(m,n)$ by induction over $m+n$. Thus we need to prove $P(0,0)$ and $P(m,n)\to P(m,n+1),P(m+1,n)$.

This can be expressed as:

$$P(0,0)\ \land\ [\forall mn,P(m,n)\to P(m+1,n)\land P(m,n+1)]\implies\forall xy, P(x,y)$$

As a partial order, this is talking about the product order on $\Bbb N^2$, that is $$(m,n)\preceq_2(m',n')\iff m\le m'\lor n\le n'.$$ Since this order is an extension of the first one, that is $(m,n)\preceq_1(m',n')$ implies $(m,n)\preceq_2(m',n')$, that means that the first induction theorem is the stronger one (applies to more $P$'s), but the well-foundedness of the second order implies that of the first. The argument is the same: any path from $(m,n)$ to $(0,0)$ must be of length at most $m+n$, so there are no infinite descending sequences.

Using the same partial order, we can even use two values which are less under the order in the induction:

If $P(m-1,n)$ and $P(m,n-1)$ together imply $P(m,n)$ (if one or the other is not defined then this should be provable with the remaining hypothesis), then $P(m,n)$ is true for all $m,n$.

This is a special case of strong induction over $\prec_2$ or simple induction over $z=m+n$ (where the induction hypothesis is $\forall mn,m+n=z\to P(m,n)$).

If we break off the base case and reindex so that it can be written as $P(m+1,n)\land P(m,n+1)\to P(m+1,n+1)$, this leaves the obligations $P(0,0)$, $P(m,0)\to P(m+1,0)$, $P(0,n)\to P(0,n+1)$, and if we simplify this to just $[\forall m,P(m,0)]\land[\forall n,P(0,n)]$, we get the same thing as variant 11 of 'Different kinds of mathematical induction':

If $P(0,n)$ and $P(n,0)$ are true for all $n$, and $P(m+1,n)\land P(m,n+1)\to P(m+1,n+1)$ for all $m,n$, then $P(m,n)$ is true for all $m,n$.

There is yet another alternative approach, which is a bit closer to some of your links:

We want to prove $P(m,n)$, which follows from $\forall n,P(m,n)$. This latter property is proven by induction on $m$, so we need to prove $\forall n,P(0,n)$ and $[\forall n,P(m,n)]\to[\forall n,P(m+1,n)]$. In each case, we have a secondary induction over $n$ to perform.

This translates as:

\begin{align}P(0,0)\ &\land\ [\forall n,P(0,n)\to P(0,n+1)]\ \land\\ (\forall m,[\forall n,P(m,n)]\to P(m+1,0))\ &\land \ \forall mn',[\forall n,P(m,n)]\land P(m+1,n')\to P(m+1,n'+1)\\&\implies\forall xy, P(x,y)\end{align}

In the language of well-orders, this is lexicographic order: $$(m,n)\prec_3(m',n')\iff m<m'\lor(m=m'\land n<n').$$

This last one is easier to state as a strong induction:

We want to prove $P(m,n)$, which follows from $\forall n,P(m,n)$. This is proven by strong induction on $m$, so we need to prove $\forall m'<m,\forall n,P(m',n)$ implies $\forall n,P(m,n)$. The latter for-all is proven by a strong induction over $n$, so assuming additionally that $\forall n'<n,P(m,n')$, we need to prove $P(m,n)$.

Expressed as a closed form rule, this is:

$$\forall m,[\forall m'<m,\forall n,P(m',n)]\to\forall n,[\forall n'<n,P(m,n')]\to P(m,n)\implies\forall xy,P(x,y)$$

which can be simplified to

$$\forall mn,[\forall m'n',m'<m\lor (m=m'\land n<n')\to P(m',n')]\to P(m,n)\implies\forall xy,P(x,y).$$

This is the most generalizable form of the induction principle, using strong instead of simple induction. In general it looks like:

$$\forall x,[\forall x'\prec x,P(x')]\to P(x)\implies\forall y,P(y)$$

where $\prec$ is a well-founded relation or a well-order over the domain. In this case we are using $\prec_3$ as a well-order of $\Bbb N^2$, and the previous cases used $\prec_1$ and $\prec_2$.

Mario Carneiro
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  • What's a 'closed form rule?' – RandomDSdevel Nov 27 '15 at 22:41
  • Also, what about the approach used on page 15 of the second document linked to by the second section of my question's resource appendix? What would that look like in formal mathematical logic? – RandomDSdevel Nov 27 '15 at 22:43
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    @RandomDSdevel 'closed form rule': basically the same thing as what you mean by 'axiom': one giant mathematical expression for the whole thing with no intervening english words such as in the preceding boxed paragraphs. – Mario Carneiro Nov 28 '15 at 03:58
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    @RandomDSdevel The approach on page 15 matches the first box of the section using $\prec_3$. The four parts of the antecedent correspond to the steps BC1, IS1, BC2, IS2; the hypotheses are written explicitly in my version but they wrote it out on several lines, that is IH1 -> IS1, IH0 -> BC2, IH0 & IH2 -> IS2. So it really is the same formula being applied. – Mario Carneiro Nov 28 '15 at 04:09
  • In reply to your first point: OK; that makes sense, and I thought it might be something like that… – RandomDSdevel Nov 30 '15 at 23:28
  • In reply to your second point: Ah, I see it now! One more question, though: what do the square brackets ('… [ … ] …') around the places where you say '$\forall n, P(m, n)$' mean? Are they Iverson brackets? If so, then I haven't encountered them before; do they basically allow to use the thing you're trying to prove as an assumption inside of that statement's own proof? If not, then what are those brackets doing there? – RandomDSdevel Nov 30 '15 at 23:50
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    @RandomDSdevel No Iverson brackets here, jut trying to distinguish all the scopes of the expression without making it look too hairy. They are the same as parentheses. I use the convention that $\forall$ has a looser scope than connectives, so $\forall x,P\to Q$ means $\forall x,(P\to Q)$ and the brackets are needed when that's not the case. – Mario Carneiro Dec 01 '15 at 03:52
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    @RandomDSdevel I edited the question to adhere more strictly to the precedence rules. The precedence order is: $\land,\lor$; $\to$ (right assoc); $\forall,\exists$; $\implies$ (which is the same as $\to$ except for the precedence). $\forall$ binds everything of equal or higher precedence to the right of it, so $A\land\forall x,B\lor C$ means $A\land (\forall x,(B\lor C))$. – Mario Carneiro Dec 01 '15 at 04:09
  • OK, thanks for clearing that up for me! – RandomDSdevel Dec 01 '15 at 21:08
  • Upon further inspection, I have some more questions about $\prec_3$. First, why does the part of it which corresponds to BC2 in the reference context I gave earlier universally quantified over $n$ when that variable is, in that step, fixed to equal 0? Second, shouldn't the first part of its equivalent of BC2 read '$P(m, 0)$' instead of '$[\forall n, P(m, n)]$?' Third, what is '$n^\prime$' for in $\prec_3$'s equivalent of IS2? Lastly, are $x$ & $y$ used in the final, proved statement instead of $m$ & $n$ to distinguish them as being arbitrary inputs instead of counters? – RandomDSdevel Dec 04 '15 at 22:23
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    @RandomDSdevel The extra variables $x,y$ and $n'$ are used so that they are not confused with different bound variables in adjacent scopes. The BC2 and IS2 parts each use an antecedent $\forall n,P(m,n)$, corresponding to IH0. This is because they are both part of the top level induction step, proving $[\forall n,P(m,n)]\to[\forall n,P(m+1,n)]$, so they both have access to the assumption $\forall n,P(m,n)$. – Mario Carneiro Dec 05 '15 at 05:47
  • On the extra variables and their use in scope disambiguation: OK, that makes sense! – RandomDSdevel Dec 05 '15 at 22:09
  • On BC2 and IS2 each using an antecedent equivalent to IH0: So that means that '$\forall n, P(m, n)$' is basically logically equivalent to and interchangeable with '$\forall n, P(0, n) \rightarrow P(0, n + 1)$?' If so, then wouldn't it make sense to just make that textual replacement in both places since that isn't defined anywhere? Then again, doing that wouldn't make sense in the statement's first occurrence, would it? I mean, shouldn't that part read something more like '$\forall m, P(m, 0) \rightarrow P(m + 1, 0)$' instead of '$\forall m, [\forall n, P(m, n)] \rightarrow P(m + 1)$?' – RandomDSdevel Dec 05 '15 at 22:22
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    @RandomDSdevel Think of it this way: we are proving this by $\prec_3$ induction, that is, by lexicographic order on $(m,n)$. Thus we prove $P(0,0)$, then $P(0,1)$, then $P(0,2)$, etc., then after that $P(1,0),P(1,1),P(1,2),$ etc., and continue in this fashion until we get to any desired number. Thus when you are at the stage of proving $P(2,0)$, you have already proved $P(0,n)$ and $P(1,n)$ for all $n$, not just $P(0,0)$ and $P(0,1)$ (which is what you would get with $\prec_1$ or $\prec_2$). Thus we are proving a step of the form $\forall m,[\forall n,P(m,n)]\to P(m+1,0)$. – Mario Carneiro Dec 07 '15 at 17:18
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    @RandomDSdevel Those two are not logically equivalent, because you are missing the base case $P(0,0)$. It is true that $P(m,0)\land\forall n,P(m,n)\to P(m,n+1)$ is logically equivalent to $\forall n,P(m,n)$; this is just the induction principle and its (trivial) converse. – Mario Carneiro Dec 07 '15 at 17:23
  • On your second point: Ah, OK! – RandomDSdevel Dec 07 '15 at 22:40
  • On your first point: Right. – RandomDSdevel Dec 07 '15 at 22:41
  • Just one last, quick question: I assume that all of the versions of induction that you give in your answer are usable even if you start with a base case other than $(0, 0)$ — say, $(1, 1)$ — if your proof requires you to do that, correct? – RandomDSdevel Dec 07 '15 at 22:47
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    @RandomDSdevel Yes. Formally, this is exploiting the order isomorphism from $\Bbb N$ (starting at $0$) to $\Bbb Z^{\ge k}$ given by $f(x)=x+k$. This will preserve any well-orders, so the same induction principles hold for these sets as well, with the base case being $f(0)=k$. The successor steps keep their shape because the successor function is preserved by $f$, that is $f(x+1)=f(x)+1$. – Mario Carneiro Dec 07 '15 at 23:26