Computable Execution Traces

Abstract

This paper gives an execution trace set based account of computability by imposing restrictions on sets of arbitrary sequences of objects, based on a supplied stock of unary tests and binary operations. This account of finite control computability provides a highly general, top down perspective on computability. We prove equivalence with the Turing machine model, under appropriate assumptions, and show how finite control computability can be used to provide a unified account of computability across multiple levels of abstraction.

A Technical Appendix

1.1 A.1 Preliminaries

Fix potentially infinite sets $\mathbb {G}$ and $\mathbb {S}$, and a distinguished element $\mathtt {B}\in \mathbb {G}$. $\varGamma ^*$ is the set of finite strings over any $\varGamma \subseteq \mathbb {G}$, with $\epsilon$ the empty string and $\varGamma ^+ = \varGamma ^* {\setminus }\{\epsilon \}$. We use $\mathtt {a}, \mathtt {b}, \dots$ for elements of $\varGamma$ and $\mathtt {u}, \mathtt {v}, \dots$ for elements of $\varGamma ^*$. Sequences of these symbols indicate concatenation.

Definition 6 (Turing machine)

A Turing machine is a tuple $\mathfrak {M}= \langle S, \varGamma , \delta , s_0 \rangle$, where $S \subseteq \mathbb {S}$ is a finite set; $\varGamma \subseteq \mathbb {G}$ is a finite set with $\mathtt {B}\in \varGamma$; $\delta : S \times \varGamma \rightarrow S \times \varGamma \times \{\mathsf {L}, \mathsf {R}\}$ is a partial function; $s_0 \in S$ is the start state.

A Turing machine tape is a pair of strings over $\varGamma$, written with | indicating the position of an imagined read/write head. A blank tape is denoted . The set of all possible Turing machine tapes is . A configuration of $\mathfrak {M}$ is a pair where $s \in S$ and is a machine tape. We also sometimes write for configurations. $\mathfrak {M}$ gives rise to a partial transition function on its configurations satisfying

is undefined if $\delta (s, \mathtt {a})$ is. generates a set of sequences of configurations labelled $\textsc {CRun}_\mathfrak {M}$. If is undefined then $\vert \mathsf {\sigma } \vert = \alpha + 1$.

Each $\mathsf {\sigma }\in \textsc {Run}_\mathfrak {M}$ corresponds to sequences $\mathsf {\sigma }^C \in \textsc {CRun}_\mathfrak {M}$ and $\mathsf {\sigma }^S \in \textsc {SRun}_\mathfrak {M}$ satisfying $\mathsf {\sigma }^C[\alpha ] = \langle \mathsf {\sigma }^S[\alpha ], \mathsf {\tau }[\alpha ] \rangle$. Write if there is $\mathsf {\sigma }\in \textsc {Run}_\mathfrak {M}$ with and .

1.2 A.2 Proof of Theorem 2

Proof. Let $\boxed {\textsc {a}} / {\mathrel {\leftrightarrows }_{\textsc {a}}} = \{ C_0, C_1, \dots C_n \}$, with $C_0$ the starting state. We will construct a Turing machine $\mathfrak {M}$ by considering the tests and operations performed at each $C_i$. Since $\mathfrak {q}$ and $\mathfrak {p}$ are Turing computable, $\textsc {a}$ is a trace set over $\mathcal {T}$.

Take $0 \le i \le n$. Let $\mathfrak {F}_i \subseteq \mathfrak {q}\times \mathfrak {p}$ be the construction set for $C_i$ and $\mathfrak {G}_i \subseteq \mathfrak {q}$ the halting set. Since $\mathfrak {q}$ and $\mathfrak {p}$ are both finite, so are $\mathfrak {F}_i$ and $\mathfrak {G}_i$. Suppose

$$\begin{aligned} \mathfrak {F}_i&= \{ \langle \textsc {q}_{i, 0}, \textsc {p}_{i, 0} \rangle , \langle \textsc {q}_{i, 1}, \textsc {p}_{i, 1} \rangle , \dots \langle \textsc {q}_{i, k_i-1}, \textsc {p}_{i, k_i - 1} \rangle \} \\ \mathfrak {G}_i&= \{ \textsc {q}_{i, k_i}, \textsc {q}_{i, k_i+1}, \dots \textsc {q}_{i, \ell _i - 1} \} \end{aligned}$$

for some $0 \le k_i \le \ell _i$. Each test and operation is Turing computable, so $\textsc {q}_{i, j}$ is associated with some Turing machine $\mathfrak {N}_{i, j}$ and $\textsc {p}_{i, j}$ some machine $\mathfrak {O}_{i, j}$. Since $\textsc {a}$ is fully deterministic, we may assume $\textsc {q}_{i, j} \cap \textsc {q}_{i,m} = \emptyset$ when $j \ne m$.² By consistency, each pair $\langle \textsc {q}_{i, j}, \textsc {p}_{i, j} \rangle$ leads to a new equivalence class $C_{c{(i, j)}}$.

Define a new machine $\mathfrak {M}= \langle S, \varGamma , \delta , s_0 \rangle$ as follows:

$\mathfrak {M}=$ “On tape contents with $\mathtt {u}\in \varGamma ^*$, $\mathtt {v}\in \varGamma ^+$:

• 1. Assign $\mathsf {x}\leftarrow 0$ and .

• 2. Replace the tape contents with $\mathsf {y}$.

• 3. Run all $\mathfrak {N}_{\mathsf {x}, j}$ with $j < \ell _\mathsf {x}$ on copies of $\mathsf {y}$ in parallel until some $\mathfrak {N}_{\mathsf {x}, \mathsf {z}}$ halts with output .

• 4. If $\mathsf {z}< k_\mathsf {x}$:

  • a. Run $\mathfrak {O}_{\mathsf {x}, \mathsf {z}}$ on $\mathsf {y}$.

  • b. Assign $\mathsf {x}\leftarrow c{(\mathsf {x}, \mathsf {z})}$ and $\mathsf {y}$ to the tape content generated in stage 4a and go to stage 2.

• 5. If $\mathsf {z}\ge k_\mathsf {x}$: halt.”

There are two meta-variables: $\mathsf {x}$ for the currently considered equivalence class of $\boxed {\textsc {a}} / {\mathrel {\leftrightarrows }_{\textsc {a}}}$ and $\mathsf {y}$ for the currently considered tape contents. $\mathfrak {M}$’s operation proper starts by carrying out all the tests required in $C_\mathsf {x}$ (Stage 3). By Lemma 1, some test $\textsc {q}_{\mathsf {x}, \mathsf {z}}$ will succeed (i.e. output ). If this $\textsc {q}_{\mathsf {x}, \mathsf {z}}$ belongs to $\mathfrak {F}_\mathsf {x}$ (i.e. $\mathsf {z}< k_\mathsf {x}$) $\mathfrak {M}$ runs the operation $\textsc {p}_{\mathsf {x}, \mathsf {z}}$ corresponding to $\textsc {q}_{\mathsf {x}, \mathsf {z}}$, stores the result in $\mathsf {y}$ and updates $\mathsf {x}$ to the new equivalence class from consistency and the process is repeated (Stage 4). Otherwise, the sequence halts and so $\mathfrak {M}$ halts (Stage 5). Since $\textsc {a}$ is finite control, $\mathsf {x}$ takes finitely many values and can be encoded in the states of $\mathfrak {M}$. In implementing $\mathfrak {M}$, we require for each $i \le n$ a unique state $s_i \in S$ corresponding to the end of Stage 2 with $\mathsf {x}= i$. This preserves $\mathsf {x}$’s value through Stage 2.

Let $S_\textsc {a}= \{ s_0, \dots s_n\}$. Intuitively, we will define an expansion mapping $(f, h)$ associating each $C_i \in \boxed {\textsc {a}} / {\mathrel {\leftrightarrows }_{\textsc {a}}}$ with $s_i$, satisfying $h(C_i) = \{ \mathsf {\tau }\in \boxed {\textsc {Run}_\mathfrak {M}} \mid \mathsf {\tau }^S[-1] = s_i \} \in \boxed {\textsc {Run}_\mathfrak {M}} / {\mathrel {\leftrightarrows }_{\textsc {Run}_\mathfrak {M}}}$ for all $i \le n$. Take $f: \textsc {f}\rightarrow \textsc {Run}_\mathfrak {M}$ so that $f(\mathsf {\sigma })[0] = \mathsf {\sigma }[0]$ for all $\mathsf {\sigma }\in \textsc {f}$. Since $\textsc {f}$ and $\textsc {Run}_\mathfrak {M}$ are both fully deterministic, starting with the same domain, this is a well-defined bijection. If we can show $\mathsf {\sigma }\mathrel {\ll }f(\mathsf {\sigma })$ and $\vert \mathsf {\sigma } \vert < \omega$ implies $\vert f(\mathsf {\sigma }) \vert < \omega$ then we can extend $f$ to an expansion mapping $(f, h)$ using Lemma 3. To that end, take $\mathsf {\sigma }\in \textsc {f}$ and let $\mathsf {\tau }= f(\mathsf {\sigma })$. $\mathsf {\tau }\in \textsc {Run}_\mathfrak {M}$ so is associated with sequences $\mathsf {\tau }^S \in \textsc {SRun}_\mathfrak {M}$ and $\mathsf {\tau }^C \in \textsc {CRun}_\mathfrak {M}$. Define $g_\mathsf {\sigma }: \vert \mathsf {\sigma } \vert \rightarrow \vert \mathsf {\tau } \vert$ so that:

• 1. $g_\mathsf {\sigma }(0)$ is the least ordinal $\beta$ for which $\mathsf {\tau }^S[\beta ] \in S_\textsc {a}$;

• 2. for all $\alpha$ with $\alpha + 1 < \vert \mathsf {\sigma } \vert$, $g_\mathsf {\sigma }(\alpha +1)$ is the least ordinal $\beta$ such that $g_\mathsf {\sigma }(\alpha ) < \beta$ and $\mathsf {\tau }^S[\beta ] \in S_\textsc {a}$;

Clearly, $g_\mathsf {\sigma }$ is strictly increasing. We can show by induction that (I) $\mathsf {\sigma }[\alpha ] = \mathsf {\tau }[g_\mathsf {\sigma }(\alpha )]$ and (II) $\mathsf {\sigma }[0, \alpha +1) \in C_i$ iff $\mathsf {\tau }^S[g_\mathsf {\sigma }(\alpha )] = s_i$ for all $\alpha < \vert \mathsf {\sigma } \vert$ .

Thus for each $\mathsf {\sigma }\in \textsc {a}$ we can define a strictly increasing $g_\mathsf {\sigma }: \vert \mathsf {\sigma } \vert \rightarrow \vert f(\mathsf {\sigma }) \vert$ such that $\mathsf {\sigma }[\alpha ] = f(\mathsf {\sigma })[g_\mathsf {\sigma }(\alpha )]$ for all $\alpha \ < \vert \mathsf {\sigma } \vert$, so $\mathsf {\sigma }\mathrel {\ll }f(\mathsf {\sigma })$ for all $\mathsf {\sigma }\in \textsc {a}$. Next we show that $\vert \mathsf {\sigma } \vert < \omega$ implies $\vert f(\mathsf {\sigma }) \vert < \omega$. Suppose $\mathsf {\sigma }\in \textsc {a}$ with $\vert \mathsf {\sigma } \vert < \omega$. Then $\mathsf {\sigma }\in C_i \cap \textsc {a}$ for some i and so by halting there is $\textsc {q}_{i,j}$ with $k_i \le j < \ell _i$ such that $\mathsf {\sigma }[-1] \in \textsc {q}_{i,j}$. Thus . Let $\mathsf {\tau }= f(\mathsf {\sigma })$. As established by Claims (I) and (II) above, $\mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)] = \mathsf {\sigma }[-1]$ and $\mathsf {\tau }^S[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)] = s_i$. So in $\mathsf {\tau }^C[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)]$ $\mathfrak {M}$ has just finished Stage 2 with $\mathsf {x}= i$. Since , when $\mathfrak {M}$ moves to Stage 3 it sets $\mathsf {z}\leftarrow j$. Thus $\mathsf {z}\ge k_\mathsf {x}$ and so $\mathfrak {M}$ next moves to Stage 5, halting. Thus, $\mathsf {\tau }$ is finite and $f$ meets the conditions of Lemma 3 and can be extended to an expansion mapping $(f, h)$ from $\textsc {a}$ to $\textsc {Run}_\mathfrak {M}$.

Finally, we turn to verifying that $f(C_i) \in \boxed {\textsc {Run}_\mathfrak {M}} / {\mathrel {\leftrightarrows }_{\textsc {Run}_\mathfrak {M}}}$ for all $C_i \in \boxed {\textsc {a}} / {\mathrel {\leftrightarrows }_{\textsc {a}}}$. It is a quick consequence of Claim (II) that if $\mathsf {\sigma }\in C_i$ then $h(\mathsf {\sigma })^S[-1] = s_i$. Conversely, suppose $\mathsf {\tau }^S[-1] = s_i$ for some $\mathsf {\tau }\in \boxed {\textsc {Run}_\mathfrak {M}}$. Since $f$ is surjective onto $\textsc {Run}_\mathfrak {M}$ there is $\mathsf {\sigma }\in \textsc {a}$ with $f(\mathsf {\sigma })[0, \vert \mathsf {\tau } \vert ) = \mathsf {\tau }$ and $f(\mathsf {\sigma })^S[\vert \mathsf {\tau } \vert - 1] = s_i$. Let $\alpha = \max \{ \beta \mid g_\mathsf {\sigma }(\beta ) < \vert \mathsf {\tau } \vert - 1 \}$. Using Claim (II), we can see that $\mathsf {\sigma }[0, \alpha + 2) \in C_i$. From this we see $f(C_i) = \{ \mathsf {\tau }\in \boxed {\textsc {Run}_\mathfrak {M}} \mid \mathsf {\tau }= f(\mathsf {\sigma }) \text { for some } \mathsf {\sigma }\in C_i \} = \{ \mathsf {\tau }\in \boxed {\textsc {Run}_\mathfrak {M}} \mid \mathsf {\tau }^S[-1] = s_i \}$. Hence $f(C_i) \in \boxed {\textsc {Run}_\mathfrak {M}} / {\mathrel {\leftrightarrows }_{\textsc {Run}_\mathfrak {M}}}$ by definition of $\mathrel {\leftrightarrows }_{\textsc {Run}_\mathfrak {M}}$.

1.3 A.3 Proof of Theorem 3

Definition 7

Let $\mathfrak {M}= \langle S, \varGamma , \delta , s_0 \rangle$ be a Turing machine and $T \subseteq S$. Write if either or for some $u \notin T$ with . Write if is undefined or for $t \notin T$ with .

Proof (Theorem 3). Part 1 is immediate from Lemma 6. Suppose $\boxed {\textsc {a}} / {\mathrel {\leftrightarrows }_{\textsc {a}}} = \{ C_0, C_1, \dots C_n\}$ and $\textsc {Run}_\mathfrak {M}$ carries out $\textsc {a}$ via $(f, h)$. For each $0\le i \le n$ there is an $s_i \in S$ with $h(C_i) = \{ \mathsf {\tau }\in \boxed {\textsc {Run}_\mathfrak {M}} \mid \mathsf {\tau }^S[-1] = s_i \}$. Take $S_\textsc {a}= \{s_0, \dots s_n\}$.

Let $0 \le i \le n$. Take the minimal set $\{D_{i,0}, D_{i,1}, \dots D_{i,k_i}\} \subseteq \boxed {\textsc {a}} / {\mathrel {\leftrightarrows }_{\textsc {a}}}$ where $\{ \mathsf {\sigma }\in \boxed {\textsc {a}} \mid \mathsf {\sigma }[0, -1) \in C_i \} \subseteq \bigcup _{0 \le j \le k_i} D_{i,j}$. Let $0 \le j \le k_i$ and suppose $h(D_{i,j}) = \{ \mathsf {\tau }\in \boxed {\textsc {Run}_\mathfrak {M}} \mid \mathsf {\tau }^S[-1] = t_{i,j} \}$ (so $t_{i,j} = s_\ell$ for some $\ell$). Define $\mathfrak {N}_{i,j}$ and $\mathfrak {O}_{i, j}$ as:

We have that

Let . and . Clearly, $\textsc {q}_{i, j}$ and $\textsc {p}_{i,j}$ are Turing computable. Repeating the above for each $0 \le j \le k_i$, take $\mathfrak {F}_i := \{ \langle \textsc {q}_{i,j}, \textsc {p}_{i,j} \rangle \mid 0 \le j \le k_i \}$. The halting set $\mathfrak {G}_i$ for $C_i$ is defined similarly. Repeating the same process for each $C_i$ provides $\mathfrak {q}$ and $\mathfrak {p}$.

In Stage (II), we verify that construction holds using $\mathfrak {q}$ and $\mathfrak {p}$. To do this, let $0 \le i \le n$. We start with composition. Let $\mathsf {\sigma }\in \{ \mathsf {\tau }\in \boxed {\textsc {a}} \mid \mathsf {\tau }[0, -1) \in C_i \}$. Then $\mathsf {\sigma }[0, -1) \in C_i$ and $\mathsf {\sigma }\in D_{i,j}$ for some j. Take $\mathsf {\tau }= h{(\mathsf {\sigma })}$. We have $h(\mathsf {\sigma }[0, -1)) = \mathsf {\tau }[0, g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2) + 1) \in h{(C_i)}$ and $\mathsf {\tau }\in h{(D_{i,j})}$. Thus we know $\mathsf {\tau }^S[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)] = s_i$ and $\mathsf {\tau }^S[-1] = \mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)] = t_{i,j}$. We claim that

$$ \langle \mathsf {\tau }^S[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)], \mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)] \rangle \underset{S_\textsc {a}}{\overset{\mathfrak {M}}{\rightsquigarrow }} \langle \mathsf {\tau }^S[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)], \mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)] \rangle . $$

If not, then there must be some $g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)< \beta < g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)$ with

$$ \langle \mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)], \mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)] \rangle \underset{S_\textsc {a}}{\overset{\mathfrak {M}}{\rightsquigarrow }} \langle \mathsf {\tau }^S[\beta ], \mathsf {\tau }[\beta ] \rangle $$

and $\mathsf {\tau }^S[\beta ] \in S_\textsc {a}$. But then $\mathsf {\tau }[0, \beta + 1) \in h(C_k)$ for some k, impossible by Corollary 1. Since $\mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)] = \mathsf {\sigma }[-2]$ and $\mathsf {\tau }[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)] = \mathsf {\sigma }[-1]$, we conclude that $\langle s_i, \mathsf {\sigma }[-2] \rangle \underset{S_\textsc {a}}{\overset{\mathfrak {M}}{\rightsquigarrow }} \langle t_{i,j}, \mathsf {\sigma }[-1] \rangle$. From this we have , so $\mathsf {\sigma }[-2] \in \textsc {q}_{i,j}$, and that $\mathfrak {O}_{i,j}(\mathsf {\sigma }[-2]) = \mathsf {\sigma }[-1]$, so . We have $\mathsf {\sigma }\in C_i \circ \textsc {q}_{i,j} \circ \textsc {p}_{i,j}$.

In the other direction, suppose for some j that $\mathsf {\sigma }\in C_i \circ \textsc {q}_{i, j} \circ \textsc {p}_{i,j}$, so as above $\mathsf {\sigma }[0, -1) \in C_i$, and $\mathfrak {O}_{i,j}(\mathsf {\sigma }[-2]) = \mathsf {\sigma }[-1]$. We wish to show that $\mathsf {\sigma }\in \boxed {\textsc {a}}$. Since $\mathsf {\sigma }[0, -1) \in C_i \subseteq \boxed {\textsc {a}}$, there is some $\mathsf {\tau }\in \textsc {a}$ with $\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert - 1) = \mathsf {\sigma }[0, -1)$. If we can establish that $\vert \mathsf {\sigma } \vert \le \vert \mathsf {\tau } \vert$ and $\mathsf {\tau }[\vert \mathsf {\sigma } \vert - 1] = \mathsf {\sigma }[-1]$ we will have that $\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert ) = \mathsf {\sigma }$, so $\mathsf {\sigma }\in \boxed {\textsc {a}}$. Since $\mathfrak {O}_{i,j}(\mathsf {\sigma }[-2]) = \mathsf {\sigma }[-1]$, we have $\langle s_i, \mathsf {\sigma }[-2] \rangle \underset{S_\textsc {a}}{\overset{\mathfrak {M}}{\rightsquigarrow }} \langle t_{i,j}, \mathsf {\sigma }[-1] \rangle$. By expansion, $\mathsf {\sigma }[-2] = \mathsf {\tau }[\vert \mathsf {\sigma } \vert - 2] = f(\mathsf {\tau })[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)]$. Since $\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert - 1) = \mathsf {\sigma }[0, -1) \in C_i$ we know $f(\mathsf {\tau })^S[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)] = h(\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert - 1))^S[-1] = s_i$. Hence we have $f(\mathsf {\tau })^C[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)] = \langle s_i, \mathsf {\sigma }[-2] \rangle$ so $f(\mathsf {\tau })^C[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)] \underset{S_\textsc {a}}{\overset{\mathfrak {M}}{\rightsquigarrow }} \langle t_{i,j}, \mathsf {\sigma }[-1] \rangle$. It follows there must be $\beta > g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 2)$ with $f(\mathsf {\tau })^C[\beta ] = \langle t_{i,j}, \mathsf {\sigma }[-1] \rangle$. We then have $f{(\mathsf {\tau })}[0, \beta + 1) \in h(D_{i,j})$ so by Lemma 5 $f{(\mathsf {\tau })}[0, \beta + 1) = h{(\mathsf {\tau }[0, \alpha + 1))}$ for some $\alpha < \vert \mathsf {\tau } \vert$, where $g_\mathsf {\tau }(\alpha ) = \beta$. Since $g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 2) < \beta$, $\vert \mathsf {\sigma } \vert - 2 < \alpha$. Thus $\vert \mathsf {\sigma } \vert \le \vert \mathsf {\tau } \vert$ and so $g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 1)$ is defined. $\beta \ge g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 1)$ otherwise $f{(\mathsf {\tau })}[0, \beta + 1) \notin h(D_{i,j})$ by Corollary 1. $\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert ) \in C_k$ for some k, so $f(\mathsf {\tau })^S[g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 1)] \in S_\textsc {a}$. Thus $\beta \not > g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 1)$ or else , contradicting $f(\mathsf {\tau })^C[\beta ] = \langle t_{i,j}, \mathsf {\sigma }[-1] \rangle$. Thus $g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 1) = \beta$. We have $\mathsf {\tau }[\vert \mathsf {\sigma } \vert - 1] = f(\mathsf {\tau })[g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 1)] = f(\mathsf {\tau })[\beta ] = \mathsf {\sigma }[-1]$. Since $\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert -1) = \mathsf {\sigma }[0, -1)$ we conclude $\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert ) = \mathsf {\sigma }$. Thus $\mathsf {\sigma }\in \{ \mathsf {\upsilon }\mid \mathsf {\upsilon }[0, -1) \in C_i \}$. This establishes composition.

Consistency is easy to verify, since if $\mathsf {\sigma }\in C_i \circ \textsc {q}_{i,j} \circ \textsc {p}_{i,j}$ then following the reasoning above where $\mathsf {\tau }\in \textsc {a}$ has $\mathsf {\tau }[0, \vert \mathsf {\sigma } \vert ) = \mathsf {\sigma }$, $h(\mathsf {\sigma }) = f(\mathsf {\tau })[0, g_\mathsf {\tau }(\vert \mathsf {\sigma } \vert - 1) + 1) \in h(D_{i,j})$ and since $h$ is injective $\mathsf {\sigma }\in D_{i,j}$. Thus, construction holds with $\mathfrak {q}$ and $\mathfrak {p}$. Halting follows similar reasoning to construction, except that we must establish

$$\begin{aligned} f{(\mathsf {\sigma })}^C[g_\mathsf {\sigma }(\vert \mathsf {\sigma } \vert - 1)] \underset{S_\textsc {a}}{\overset{\mathfrak {M}}{\rightsquigarrow }} \ \downarrow . \end{aligned}$$

The details are omitted for reasons of space.