TITLE: Axiom of Double Induction?
QUESTION [6 upvotes]: 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:  

'Double Induction'
'Good Examples of Double Induction'
'A case of double induction?'
'Divisibility Proof with Induction - Stuck on Induction Step' (this answer, in particular…)

To other sources:  

'Proof method:  Multidimensional induction'
'Mathematical Induction' (PDF) (See §14.2.4, 'Appendix 2 — The Basic Schemes of Induction:  Induction for the Natural Numbers:  Double Induction (Weak Form)' on p. 15…)
'Mathematical induction:  variants and subtleties' (PDF) (See §3, which starts at the end of p. 2…)
'Different kinds of Mathematical Induction' (PDF) (See variant 11 at the end of p. 2…)
'Proof by Mathematical Induction'/'[The] Principle of Mathematical Induction' (PDF) (See the section on 'Double Induction' that starts at the end of p. 12…)

REPLY [13 votes]: 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 $\prec_2$.