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I came across an exercise which asks us to prove by induction a statement that has 3 variables.

The exercise provides a hint that we should induct on one variable, and fix the other two variables.

Question: Why is it valid and sufficient to induct on one variable? It feels insufficient to induct on one variable when the statement to be proved has three free variables.

The specific example is Exercise 2.2.1 from Terence Tao's Analysis I:

For any natural numbers a,b,c, we have (a+b)+c = a + (b+c).

The idea is to prove this using only the Peano axioms provided in the chapter. Someone else has provided a solution here: link.

In addition, the author provides worked examples where he inducts on one variable and proves a statement with more than one free variable.

Wider reading. I have reads the replies to this question, but the answers aren't clear to me. In fact the last answer counter-intuitively suggests "fix" creates a variable.

Penelope
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  • Can you be more specific about what you find unclear about the answers to the question you linked? In particular I think augurar's answer quite clearly says what is going on: you are proving the statement $\forall a\forall b (a+b)+c=a+(b+c)$ by induction on $c$. – Eric Wofsey Sep 05 '23 at 02:37

2 Answers2

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Suppose you fix a variable $x$, and can prove some statement $Q(x)$ without using any assumptions about $x$. Then since $x$ was arbitrary, that proves the statement $\forall x . Q(x)$.

You want to prove something of the form $\forall x,y . \varphi(x,y)$. Let $Q(x)$ be the statement $\forall y . \varphi(x,y)$. The hint is proposing that you fix $x$, and prove $Q(x)$ (without using any information about $x$) using induction on $y$; from that, you can conclude $\forall x . Q(x)$, and thus conclude $\forall x,y . \varphi(x,y)$. So fixing $x$ and inducting over $y$ is a valid way to prove a statement quantified over all $x,y$.

Now generalize that to three variables, and you will see why the suggest strategy is valid.

D.W.
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I don't know exactly what axioms you're using, but consider the following:

Suppose $+$ is specified as following the schematic rules:

  • $0 + y = y$
  • $(\mathsf{suc}\ x) + y = \mathsf{suc}(x + y)$

Now, suppose you want to prove $∀y. 0+y = y$. This is possible by:

  • Using $∀$ introduction to reduce our goal to proving $0 + y = y$ for some abstract $y$
  • Applying the first rule above

You could also do induction on $y$, applying the first rule above in each case. But the induction isn't helping at all. It just results in having to apply the second step above twice.

If you pick the right variable for the associativity proof, you should find that you are in a similar situation. Doing induction one one of the variables yields cases that are easy to prove by other rules.

There is no obligation to always do induction on every natural number involved in a statement. Induction is a tool for breaking down a proof into simpler and/or more concrete components when there is no general way to proceed. But often you will have rules or previous theorems that apply without considering how the problem breaks down by individual cases on each number.

Dan Doel
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