# Formal induction over two variables

## Induction.

According to the Peano Axioms in this article https://en.wikipedia.org/wiki/Peano_axioms, one axioms states that if $$\varphi$$ is a unary predicate such that $$\varphi(0)$$ is true and $$\forall n \in \mathbf{N}[\varphi(n) \implies \varphi(s(n))],$$ then $$\forall n \in \mathbf{N}[P(n)]$$ is true.

This does not allow $$n$$-ary predicates for $$n \geq 2$$, however I have seen that induction can be used for those as well. What is the formal reason for that?

## Example.

Consider a formula of the form $$\forall n (\forall m [(m \in \mathbf{N} \wedge n \in \mathbf{N}) \implies P(m,n)])$$ (I hope this is a correct translation of "For all $$n \in \mathbf{N}$$ and for all $$m \in \mathbf{N}$$ the statement $$P(m,n)$$ is true"). One can prove this statement in the following way. Let $$n \in \mathbf{N}$$ and $$m \in \mathbf{N}$$. Then (...) and thus $$P(n,m)$$. Now, I have read that I can prove such a statement by induction the following way. Let $$n \in \mathbf{N}$$, that is, fix an element $$n$$. Prove $$P(0,n)$$ and for all $$k \in \mathbf{N}$$ with $$P(k,n)$$ show that $$P(s(k),n)$$ is true as well. Now, if I could apply induction to this, one could deduce $$\forall m \in \mathbf{N} [P(m,n)] (*).$$ Since $$n$$ was arbitrary, it holds that $$\forall n (\forall m[m \in \mathbf{N} \wedge n \in \mathbf{N} \implies P(m,n)]).$$

## Questions.

The argument of my example, if correct, needs that induction can be applied to predicates of the form $$P(m,n)$$. $$(1)$$ Why can this be done/Why is this formally possible?

$$(2)$$ I have also read the following question Complete Induction on two variables, in which the answer defines $$Q(x)=\forall y P(x,y)$$. However, is $$Q(x)$$ then a valid unary predicate, that is, if $$P(x,y)$$ is a predicate, then $$\forall y P(x,y)$$ is a unary predicate?

$$(3)$$ I have also read the following question Double induction. Here the answer says that the formula $$\forall m P(m,n)$$ is not a sentence and can thus not be proven. However, in $$(*)$$ I have a similar expression. Did I do something wrong in my argument above?

• Re: (2), there's a bit of a terminology confusion here: $\varphi$ isn't a unary predicate, it's a formula. And indeed if $P(x,y)$ is a formula then $\forall yP(x,y)$ is a formula as well: formulas are closed under the basic syntactic operations, including 'quantifying out' a variable. Jan 29, 2022 at 23:26
• @NoahSchweber Do you mean the $\varphi$ that is also denoted as $\varphi$ in my post? If that is the case, then I am confused, since I just copied what was written on the wikipedia page. Before the Arithmetic section, point 9 literally says "unary predicate". If you don't mean that $\varphi$, then it would be nice to know which one you mean exactly. Thanks! Jan 29, 2022 at 23:31
• I do mean that $\varphi$. And wikipedia is not infallible: the term "unary predicate" is somewhat informal (and when formal, often refers to something narrower than intended here). Jan 29, 2022 at 23:34
• @NoahSchweber I see, thanks. It was more likely to me that I got something wrong though, which is why I wanted to make sure. So am I correct that the $\varphi$ from wikipedia should be a formula, not a unary predicate? Jan 29, 2022 at 23:36
• @NoahSchweber: Though for PA one doesn't need induction for formulae; it is enough to have induction for 1-parameter sentences. I'm sure you know that, but probably the asker doesn't, and it might explain why wikipedia doesn't bother to say there can be free parameters in the induction scheme. Jan 30, 2022 at 8:48

The argument of my example, if correct, needs that induction can be applied to predicates of the form $$P(m,n)$$. $$(1)$$ Why can this be done/Why is this formally possible?

Note that in the example proof you gave, you fix the $$n$$, and then do induction over $$m$$. So in that sense the formula $$P(m,n)$$ really has only one variable: $$m$$.

$$(2)$$ I have also read the following question Complete Induction on two variables, in which the answer defines $$Q(x)=\forall y P(x,y)$$. However, is $$Q(x)$$ then a valid unary predicate, that is, if $$P(x,y)$$ is a predicate, then $$\forall y P(x,y)$$ is a unary predicate?

Here it is even more clear that $$Q(x) = \forall y P(x,y)$$ is a formula with one unbounded variable $$x$$. The predicate $$P(x,y)$$ is still a 2-place predicate, but the formula $$\forall y P(x,y)$$ has only one free variable. So yes, that fits the Peano Axiom scheme just fine.

$$(3)$$ I have also read the following question Double induction. Here the answer says that the formula $$\forall m P(m,n)$$ is not a sentence and can thus not be proven. However, in $$(*)$$ I have a similar expression. Did I do something wrong in my argument above?

Again, your argument started by saying the $$n$$ is some arbitrary number: it is the start of any universal proof, where we try to prove $$\phi(n)$$ and, since $$n$$ was assumed to be arbtrarily picked, we can then conclude $$\forall n \ \phi(n)$$. So, within that context of a universal proof, we can treat $$\forall m P(m,n)$$ as a sentence, even if we don't know what $$n$$ exactly is, and thus also don't know what $$\forall m P(m,n)$$ exactly says.

• Thanks a lot for your answer. I think I know what caused the confusion and it should now be clear! Jan 30, 2022 at 21:09
• @user1578232 Glad I could help! :) Jan 30, 2022 at 23:20

A simple reason that induction works on $$n$$-ary formulae/predicates is that you can use iterated induction on unary formulae to derive induction on $$n$$-ary predicates. There is also the method you state. Lets look at the former. Suppose you want to prove $$\forall n\in\mathbb{N} \forall m\in\mathbb{N} \phi(n,m)$$. We can do this by induction on $$n$$ in the formula $$\forall m\in\mathbb{N} \phi(n,m)$$. This is in fact a formula. Since it is a formula with one free variable, it is akin to unary predicate. It just might not be atomic. Now, via induction, our problem has been reduced to proving $$\forall m\in\mathbb{N}\phi(0,m)$$ and showing that $$\forall n\in\mathbb{N}(\forall m\in\mathbb{N}\phi(n,m)\rightarrow\forall m\in\mathbb{N}\phi(S(n),m))$$. Proving $$\forall m\in\mathbb{N}\phi(0,m)$$ can be done by induction on the formula $$\phi(0,m)$$, i.e., show $$\phi(0,0)$$ and $$\phi(0,m)$$ implies $$\phi(0,S(m))$$. Similarly, assuming the inductive hypothesis, one can prove $$\forall m\in\mathbb{N}\phi(S(n),m)$$ by induction on $$m$$.

So, formally, $$n$$-ary induction can be reduced to iterated unary induction. However, this is not always done in practice. You will see, as you mentioned, proofs of the form "take an arbitrary $$n\in\mathbb{N}$$, and prove $$\phi(n,m)$$ by induction on $$m$$. This is often more economical than iterated induction. The formal reason this work is due to the constructor for dependent functions ($$\Pi$$-types). Some times (mostly in the context of natural deduction) this is called $$\forall$$-introduction.

Aside - The reason induction works is due to the eliminator rules for the type $$\mathbb{N}$$. Using the eliminator rules for $$\mathbb{N}$$ and using the constructor for $$\Pi$$-types with the assumption $$(n : \mathbb{N})$$ both serve to do the same thing -- define a dependent function out of $$\mathbb{N}$$.