Dependent choice as a termination principle

Abstract

We introduce a new formulation of the axiom of dependent choice, which can be viewed as an abstract termination principle that in particular generalises recursive path orderings, the latter being fundamental tools used to establish termination of rewrite systems. We consider several variants of our termination principle, and relate them to general termination theorems in the literature.

Direct proof of Corollary 2

${\mathrm {DC}}_{\rho }\rightarrow \mathrm {TP}_\rho [\rhd ]$: This uses the famous minimal bad-sequence construction, which in turn uses dependent choice in the following sequential variant:

$$\begin{aligned} \begin{aligned} {{\mathrm {DC\text{- }seq}}}_\rho \; :\; A([])\wedge \forall a^{\rho ^*}(A(a)\rightarrow \exists x^\rho A(a*x))\rightarrow \exists \alpha ^{\mathtt {Nat}\rightarrow \rho }\forall n A([{\alpha }]({n})) \end{aligned} \end{aligned}$$

For a proof that ${\mathrm {DC\text{- }seq}}_\rho$ is implied by ${\mathrm {DC}}_{\rho ^*}$ (which can in turn be coded as an instance of ${\mathrm {DC}}_\rho$) see [17, Section 6.1], and also [13, Lemma 5.11] for the two-way equivalence between ${\mathrm {DC}}_\rho$ and a sequential variant closely related to ${\mathrm {DC\text{- }seq}}_\rho$. Let us call a sequence $\alpha$bad if $\lnot { WF}[\succ ](\alpha )$ holds: in other words, $\alpha$ is an infinite $\succ$-descending chain. Suppose that the premise of $\mathrm{TP}[\rhd ]$ holds, and that for contradiction there exists at least one bad sequence. Using ${\mathrm {DC\text{- }seq}}_\rho$ together with $\mathrm{TI}[\rhd ]$, construct a minimal sequence $\alpha$ as follows:

Assuming we have already constructed $[{\alpha _0,\ldots ,\alpha _{n-1}}]$, choose $\alpha _n$ in such a way that $[{\alpha _0,\ldots ,\alpha _{n-1},\alpha _n}]$ extends to a bad sequence, but $[{\alpha _0,\ldots ,\alpha _{n-1},x}]$ does not for any $x\lhd \alpha _n$.

For the empty sequence in the first step this is guaranteed by the initial assumption that at least one bad sequence exists. It is easy to see that $\alpha$ must satisfy ${ MIN}(\alpha )$. However, $\alpha$ itself must also be bad: if on the contrary we would have $\alpha _n\nsucc \alpha _{n+1}$ for some n, then $[{\alpha _0,\ldots ,\alpha _{n+1}}]$ could not extend to a bad sequence, a contradiction.

$\mathrm{RBI}_\rho \rightarrow \mathrm{TP}_\rho [\rhd ]$: Define

$$\begin{aligned} \begin{aligned} S(a)&:\equiv (\forall n<|a|,\beta ^{\rho ^\mathtt {Nat}})([{a}]({n})\blacktriangleleft \beta \wedge a_n\rhd \beta _n\rightarrow { WF}(\beta ))\\ P(a)&:\equiv \forall \alpha (a\blacktriangleleft \alpha \rightarrow { WF}(\alpha )).\end{aligned} \end{aligned}$$

From the premise of $\mathrm{TP}$ we derive the three premises of $\mathrm{RBI}$ w.r.t P and S. Note that $S([{}])$ is trivially true, and if $\alpha \in S$ then this is completely equivalent to saying that ${ MIN}(\alpha )$ holds, and hence $\alpha _n\nsucc \alpha _{n+1}$ for some n and thus $P([{\alpha }]({n+2}))$ holds.

For the third premise, take some $a\in S$ and assume that $(\forall x)(S(a*x)\rightarrow P(a*x))$. We establish $(\forall x)P(a*x)$ via a side induction on $\rhd$, from which we trivially obtain P(a) since for any $\alpha$ with $a\blacktriangleleft \alpha$ we have $a*\alpha _{|a|}\blacktriangleleft \alpha$ and therefore ${ WF}(\alpha )$ follows from $P(a*\alpha _{|a|})$.

Suppose that $(\forall y\lhd x) P(a*y)$ holds. Then to prove $P(a*x)$ it suffices to prove $S(a*x)$. Since we already have $a\in S$, it suffices to check the last point of the sequence i.e.

$$\begin{aligned} (\forall \beta )(a\blacktriangleleft \beta \wedge x\rhd \beta _{|a|}\rightarrow { WF}(\beta )). \end{aligned}$$

But this follows from the side induction hypothesis, setting $y:=\beta _{|a|}$, which completes the side induction. Therefore, we can now apply bar induction to obtain $P([{}])$ which is just $(\forall \alpha ){ WF}(\alpha )$.