I know that for proving some $P(m,n)$ we can apply induction as follows: prove some base case $P(m_0, n_0)$, an inductive step given $P(m-1,n)$ to show $P(m,n)$, and an inductive step given $P(m,n-1)$ to show $P(m,n)$. However in a proof I am working on I need a stronger hypothesis. In particular, in order to show $P(m,n)$ given $P(m-1,n)$ I also need to assume that $P(m-1,n-k)$ holds for $1\leq k\leq n$ too, a strong induction of sorts. Is this logically sound, and if not what is the issue with it? If it is sound, how much stronger can we get?

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    $\begingroup$ As long as you prove enough "base cases", this is fine. If you like, think of it as doing the usual sort of induction on the sum, $S=n+m$. $\endgroup$
    – lulu
    Jun 16, 2021 at 15:50

2 Answers 2


One way to do this is to prove $P(1,n)$ for all values of $n\geq1$. Then make the induction hypothesis that for some $m>1$, $P(m,n)$ is true for all values of $n\geq1$, and prove that $P(m+1,n)$ is true for all values of $n\geq1$. This proves that that $P(m,n)$ is true for $m\geq1$ and all values of $m$. In proving that $P(m+1,n)$ is true for all $m,n\geq1$, you are, of course, free to use any sort of induction you choose.

  • $\begingroup$ Ah so to clarify, by proving $P(1,n)$ for all $n$, then given $P(m-1,n-k)$ for all $1\leq k\leq n-1$ showing $P(m,n)$, we've successfully proven $P(m,n)$ for all $m,n$? $\endgroup$ Jun 16, 2021 at 16:05
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    $\begingroup$ @ShivashriganeshMahato That isn't what I said, and I think you need to state it more carefully. Suppose we've proved $P(1,n)$ fot all $n$. How do you conclude that $P(2,1)$ is true? There is no $1\leq k\leq 1$ so your proof must include a proof that $Pm,1)$ is true for all $m$. After that, I think your argument works. (I still think that the approach I outlined is conceptually simpler). $\endgroup$
    – saulspatz
    Jun 16, 2021 at 16:17

Given $P(m, n)$ you can treat $P(m-1, 0), P(m-1, 1), \cdots P(m-1, n)$ as accessible to your induction step, but you can also go further and take all pairs of natural numbers that are smaller lexicographically, which includes statements of the form $P(m-1, \cdot)$.

I think it would help to explicitly pick a set, explicitly pick a partial order on that set, prove that that partial order is a well-order, and then proceed from there.

You can perform induction on any set equipped with a well-order.

Let's define $\le$ on pairs of numbers as follows. This is the lexicographic order on $\mathbb{N} \times \mathbb{N}$.

$$ (a, b) \le (c, d) \;\;\text{iff}\;\; a \le c \lor (a = c \land b \le d) $$

This order is kind of interesting. There are no infinite descending chains, but if we start with $(1, 0)$ in our chain, we can make an arbitrarily long descending chain by picking the next element to be $(0, n)$ for some $n$ in $\mathbb{N}$. The next next element to be $(0, n-1)$ and so on.

In order to convince ourselves that there are no infinite descending chains, we notice that if we start on the row $(n, \cdot)$ then we must reach the row $(n-1, \cdot)$ in finitely many steps. And if we are on the row $(0, \cdot)$ we must reach $(0, 0)$ in finitely many steps. The sum of finitely many finite numbers is finite.

The strong induction schema looks like this. Here, $\implies$ is a low-precedence version of $\to$. $\vec{0}$ is the minimum element of $\le$. Here, $a < b$ is an abbreviation for $a \ne b \land a \le b$.

$$ \bigg((\forall u < x \mathop. \varphi(u)) \to \varphi(x)\bigg) \land \varphi\!\left( \vec{0} \right) \implies (\forall u \mathop. \varphi(u)) $$


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