# Disparity between Induction and Well-ordering Principles

Over classical logic, the induction and well-ordering schemas are equivalent. These schemas state the following, given any linear ordering $$(W,<)$$ and property $$Q$$ on $$W$$:

Induction: $$∀k{∈}W\ ( \ ∀i{∈}W_{.

Well-ordering: $$∃k{∈}W\ ( \ Q(k) \ ) ⇒ ∃k{∈}W\ ( \ Q(k) ∧ ∀i{∈}W_{.

When applied to the ordering on $$\mathbb{N}$$, these yield the (so-called) "strong induction" schema and the "well-ordering principle". It sometimes seems as though the latter gives a quicker proof, but on the other hand proofs based on just induction feel more direct. Is there any substance to this feeling? Can non-classical logics illuminate the disparity between these two principles, and explain why they feel different even in ordinary mathematics?

• I think your well-ordering principle even on $2$ would in fact imply LEM: given proposition $p$ define $Q(k) := (k = 0 \wedge p) \vee (k = 1)$. Then if 0 is the minimal witness for $Q$ then $p$, and if 1 is the minimal witness for $Q$ then $\lnot p$. (And in Coq, for example, the standard library defines an "accessibility" predicate inductively as the smallest subset satisfying the induction principle given above, and defines "well-ordered" as meaning every element is accessible.) – Daniel Schepler May 15 at 18:22
• (Might you also need to add a vector of parameters $\vec t$ as input to $Q$ and prefix both with $\forall \vec t := \forall t_1 \cdots \forall t_n$, as is common in formulating for example the replacement axiom schema in ZFC?) – Daniel Schepler May 15 at 18:35
• @DanielSchepler: I'll respond to your second comment first. You're of course right that $Q$ in the schemas would need parameters (besides the one shown) in some variants of FOL (e.g. Hilbert-style). As for your first comment, indeed well-ordering on {0,1} immediately gives LEM, and that is pretty much a special case of the explanation I gave using 3VL. The reason for looking at 3VL is that it satisfies the rule ( $¬¬Q ⊢ Q$ ) but not LEM. In any case, thanks for mentioning how "well-ordered" is defined in Coq! – user21820 May 16 at 4:45

## 1 Answer

The equivalence between induction and well-ordering breaks down in the absence of LEM (law of excluded middle). To isolate the reliance on LEM, consider the corresponding Fitch-style rules over Kleene's 3-valued logic 3VL for any given linear order $$(W,<)$$:

Induction rule: $$∀k{∈}W\ ( \ ∀i{∈}W_{.

Well-ordering rule: $$∃k{∈}W\ ( \ Q(k) \ ) ⊢ ∃k{∈}W\ ( \ Q(k) ∧ ∀i{∈}W_{.

Note for induction on $$(\mathbb{N},<)$$, the induction rule is equivalent over 3VL to the basic induction rule (i.e. "$$Q(0) ∧ ∀k{∈}\mathbb{N}\ ( \ Q(k)⊢Q(k+1) \ ) ⊢ ∀k{∈}\mathbb{N}\ ( \ Q(k) \ )$$" for each property $$Q$$ on $$\mathbb{N}$$). But these are very different from the well-ordering rule even for $$\mathbb{N}$$.

The induction rule is sound over 3VL if $$(W,<)$$ is truly a well-order, and we can easily observe this fact by transfinite induction in the (classical) meta-system. But the well-ordering rule is not sound over 3VL even if $$(W,<)$$ is a well-order, because it may be that for some $$k,m∈W$$ we have $$k and $$Q(k) ≡ \text{null}$$ but $$Q(m) ≡ \text{true}$$.

So there is a logically significant disparity between these two principles. Intuitively, well-ordering generates more information than induction.

Furthermore, induction on $$\mathbb{N}$$ is intuitionistically sound in the sense that every instance is witnessed by a program as per the BHK interpretation. In contrast, well-ordering on $$\mathbb{N}$$ is not intuitionistically sound, because if $$Q(k,x)$$ says "there is a program that has length $$k$$ and outputs string $$x$$", then any program witnessing "$$∀x{∈}\mathbb{N}\ ( \ ∃k{∈}\mathbb{N}\ ( \ Q(k,x) \ ) ⇒ ∃k{∈}\mathbb{N}\ ( \ Q(k,x) ∧ ∀i{∈}\mathbb{N}_{" can be used to compute Kolmogorov complexity, which is impossible.

This disparity actually shows up in a wide variety of mathematical problems. For example, any proof that Kolmogorov complexity is well-defined requires LEM as shown above, and is trivial via well-ordering on $$\mathbb{N}$$. Another example is the proof that every positive integer $$n > 1$$ is not a factor of $$2^n-1$$:

Take any positive integer $$n > 1$$ such that $$n \mid 2^n-1$$. Let $$p$$ be the smallest prime factor of $$n$$, which exists by well-ordering on $$\mathbb{N}$$ since $$n$$ has a prime factor, and let $$k$$ be the positive integer such that $$p·k = n$$. Then $$p \mid 2^n-1$$. Note that $$p \nmid 2$$ since $$2 \nmid 2^n-1$$, and so $$2^{p-1} ≡ 1 \pmod{p}$$ by Fermat's little theorem. Thus $$1 ≡ 2^n ≡ (2^p)^k ≡ 2^k \pmod{p}$$. Let $$c$$ be the minimum positive integer such that $$2^c ≡ 1 \pmod{p}$$, again by well-ordering on $$\mathbb{N}$$. Then $$c > 1$$ and $$c \mid k , p-1$$ (otherwise by the division lemma and Euclid's lemma we can obtain a contradiction). Now let $$q$$ be the smallest prime factor of $$c$$, yet again by well-ordering on $$\mathbb{N}$$. Thus $$q \mid c \mid k,p-1$$, and hence $$q$$ is a prime factor of $$n$$ that is smaller than $$p$$, contradicting minimality of $$p$$.

Classically, every proof using well-ordering can be mechanically translated into a proof using only induction, but as the above examples illustrate, well-ordering sometimes seems to 'generate' more information, and this extra information is actually coming from LEM. It feels especially unnatural to use induction instead of well-ordering in the number theory example above, because the intrinsic structure of the problem does not follow the structure of the natural ordering on $$\mathbb{N}$$.