# Strong Induction on Two Variables

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?

• 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, $S=n+m$. – lulu Jun 16 at 15:50

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.

• 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 $m,n$? – Shivashriganesh Mahato Jun 16 at 16:05
• @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). – saulspatz Jun 16 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 $$a \ne b \land a \le b$$.

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