(*<*)theory Paperimports "../thys/uncomputable"begin(*hide_const (open) s *)hide_const (open) Divides.adjustabbreviation "update2 p a \<equiv> update a p"consts DUMMY::'anotation (latex output) Cons ("_::_" [78,77] 73) and set ("") and W0 ("W\<^bsub>\<^raw:\hspace{-2pt}>Bk\<^esub>") and W1 ("W\<^bsub>\<^raw:\hspace{-2pt}>Oc\<^esub>") and update2 ("update") and(* abc_lm_v ("lookup") and abc_lm_s ("set") and*) haltP ("stdhalt") and tcopy ("copy") and tape_of_nat_list ("\<ulcorner>_\<urcorner>") and tm_comp ("_ \<oplus> _") and DUMMY ("\<^raw:\mbox{$\_$}>")declare [[show_question_marks = false]](*>*)section {* Introduction *}text {*%\noindent%We formalised in earlier work the correctness proofs for two%algorithms in Isabelle/HOL---one about type-checking in%LF~\cite{UrbanCheneyBerghofer11} and another about deciding requests%in access control~\cite{WuZhangUrban12}. The formalisations%uncovered a gap in the informal correctness proof of the former and%made us realise that important details were left out in the informal%model for the latter. However, in both cases we were unable to%formalise in Isabelle/HOL computability arguments about the%algorithms. \noindentSuppose you want to mechanise a proof about whether a predicate @{term P}, say, isdecidable or not. Decidability of @{text P} usually amounts to showingwhether \mbox{@{term "P \<or> \<not>P"}} holds. But this does \emph{not} workin Isabelle/HOL and other HOL theorem provers, since they are based on classical logicwhere the law of excluded middle ensures that \mbox{@{term "P \<or> \<not>P"}}is always provable no matter whether @{text P} is constructed bycomputable means. %The same problem would arise if we had formulated%the algorithms as recursive functions, because internally in%Isabelle/HOL, like in all HOL-based theorem provers, functions are%represented as inductively defined predicates too.The only satisfying way out of this problem in a theorem prover basedon classical logic is to formalise a theory of computability. Norrishprovided such a formalisation for the HOL4. He choosethe $\lambda$-calculus as the starting point for his formalisation ofcomputability theory, because of its ``simplicity'' \cite[Page297]{Norrish11}. Part of his formalisation is a clever infrastructurefor reducing $\lambda$-terms. He also established the computationalequivalence between the $\lambda$-calculus and recursive functions.Nevertheless he concluded that it would be appealing to have formalisations for more operational models ofcomputations, such as Turing machines or register machines. Onereason is that many proofs in the literature use them. He notedhowever that \cite[Page 310]{Norrish11}:\begin{quote}\it``If register machines are unappealing because of their general fiddliness,\\ Turing machines are an even more daunting prospect.''\end{quote}\noindentIn this paper we take on this daunting prospect and provide aformalisation of Turing machines, as well as abacus machines (a kindof register machines) and recursive functions. To see the difficultiesinvolved with this work, one has to understand that Turing machineprograms can be completely \emph{unstructured}, behaving similar to Basic's infamous goto \cite{Dijkstra68}. This precludes in thegeneral case a compositional Hoare-style reasoning about Turingprograms. We provide such Hoare-rules for when it is possible toreason in a compositional manner (which is fortunately quite often), but also tackle the more complicated case when we translate abacus programs into Turing programs. This aspect of reasoning about computability theory is usually completely left out in the informal literature, e.g.~\cite{Boolos87}.%To see the difficulties%involved with this work, one has to understand that interactive%theorem provers, like Isabelle/HOL, are at their best when the%data-structures at hand are ``structurally'' defined, like lists,%natural numbers, regular expressions, etc. Such data-structures come%with convenient reasoning infrastructures (for example induction%principles, recursion combinators and so on). But this is \emph{not}%the case with Turing machines (and also not with register machines):%underlying their definitions are sets of states together with %transition functions, all of which are not structurally defined. This%means we have to implement our own reasoning infrastructure in order%to prove properties about them. This leads to annoyingly fiddly%formalisations. We noticed first the difference between both,%structural and non-structural, ``worlds'' when formalising the%Myhill-Nerode theorem, where regular expressions fared much better%than automata \cite{WuZhangUrban11}. However, with Turing machines%there seems to be no alternative if one wants to formalise the great%many proofs from the literature that use them. We will analyse one%example---undecidability of Wang's tiling problem---in Section~\ref{Wang}. The%standard proof of this property uses the notion of universal%Turing machines.We are not the first who formalised Turing machines: we are aware of the preliminary work by Asperti and Ricciotti\cite{AspertiRicciotti12}. They describe a complete formalisation ofTuring machines in the Matita theorem prover, including a universalTuring machine. They report that the informal proofs from which theystarted are \emph{not} ``sufficiently accurate to be directly usable as aguideline for formalization'' \cite[Page 2]{AspertiRicciotti12}. Forour formalisation we followed mainly the proofs from the textbook\cite{Boolos87} and found that the description there is quitedetailed. Some details are left out however: for example, it is onlyshown how the universal Turing machine is constructed for Turingmachines computing unary functions. We had to figure out a way togeneralise this result to $n$-ary functions. Similarly, when compilingrecursive functions to abacus machines, the textbook again only showshow it can be done for 2- and 3-ary functions, but in theformalisation we need arbitrary functions. But the general ideas forhow to do this are clear enough in \cite{Boolos87}. %However, one%aspect that is completely left out from the informal description in%\cite{Boolos87}, and similar ones we are aware of, is arguments why certain Turing%machines are correct. We will introduce Hoare-style proof rules%which help us with such correctness arguments of Turing machines.The main difference between our formalisation and the one by Aspertiand Ricciotti is that their universal Turing machine uses a differentalphabet than the machines it simulates. They write \cite[Page23]{AspertiRicciotti12}:\begin{quote}\it``In particular, the fact that the universal machine operates with adifferent alphabet with respect to the machines it simulates isannoying.'' \end{quote}\noindentIn this paper we follow the approach by Boolos et al \cite{Boolos87},which goes back to Post \cite{Post36}, where all Turing machinesoperate on tapes that contain only \emph{blank} or \emph{occupied} cells. Traditionally the content of a cell can be anycharacter from a finite alphabet. Although computationally equivalent,the more restrictive notion of Turing machines in \cite{Boolos87} makesthe reasoning more uniform. In addition some proofs \emph{about} Turingmachines are simpler. The reason is that one often needs to encodeTuring machines---consequently if the Turing machines are simpler, then the codingfunctions are simpler too. Unfortunately, the restrictiveness also makesit harder to design programs for these Turing machines. In orderto construct a universal Turing machine we therefore do not follow \cite{AspertiRicciotti12}, instead follow the proof in\cite{Boolos87} by relating abacus machines to Turing machines and inturn recursive functions to abacus machines. The universal Turingmachine can then be constructed as a recursive function.\smallskip\noindent{\bf Contributions:} We formalised in Isabelle/HOL Turing machines following thedescription of Boolos et al \cite{Boolos87} where tapes only have blank oroccupied cells. We mechanise the undecidability of the halting problem andprove the correctness of concrete Turing machines that are neededin this proof; such correctness proofs are left out in the informal literature. We construct the universal Turing machine from \cite{Boolos87} byrelating recursive functions to abacus machines and abacus machines toTuring machines. Since we have set up in Isabelle/HOL a very general computabilitymodel and undecidability result, we are able to formalise theundecidability of Wang's tiling problem. We are not aware of any otherformalisation of a substantial undecidability problem.*}section {* Turing Machines *}text {* \noindent Turing machines can be thought of as having a \emph{head}, ``gliding'' over a potentially infinite tape. Boolos et al~\cite{Boolos87} only consider tapes with cells being either blank or occupied, which we represent by a datatype having two constructors, namely @{text Bk} and @{text Oc}. One way to represent such tapes is to use a pair of lists, written @{term "(l, r)"}, where @{term l} stands for the tape on the left-hand side of the head and @{term r} for the tape on the right-hand side. We have the convention that the head, abbreviated @{term hd}, of the right-list is the cell on which the head of the Turing machine currently operates. This can be pictured as follows: % \begin{center} \begin{tikzpicture} \draw[very thick] (-3.0,0) -- ( 3.0,0); \draw[very thick] (-3.0,0.5) -- ( 3.0,0.5); \draw[very thick] (-0.25,0) -- (-0.25,0.5); \draw[very thick] ( 0.25,0) -- ( 0.25,0.5); \draw[very thick] (-0.75,0) -- (-0.75,0.5); \draw[very thick] ( 0.75,0) -- ( 0.75,0.5); \draw[very thick] (-1.25,0) -- (-1.25,0.5); \draw[very thick] ( 1.25,0) -- ( 1.25,0.5); \draw[very thick] (-1.75,0) -- (-1.75,0.5); \draw[very thick] ( 1.75,0) -- ( 1.75,0.5); \draw[rounded corners=1mm] (-0.35,-0.1) rectangle (0.35,0.6); \draw[fill] (1.35,0.1) rectangle (1.65,0.4); \draw[fill] (0.85,0.1) rectangle (1.15,0.4); \draw[fill] (-0.35,0.1) rectangle (-0.65,0.4); \draw (-0.25,0.8) -- (-0.25,-0.8); \draw[<->] (-1.25,-0.7) -- (0.75,-0.7); \node [anchor=base] at (-0.8,-0.5) {\small left list}; \node [anchor=base] at (0.35,-0.5) {\small right list}; \node [anchor=base] at (0.1,0.7) {\small head}; \node [anchor=base] at (-2.2,0.2) {\ldots}; \node [anchor=base] at ( 2.3,0.2) {\ldots}; \end{tikzpicture} \end{center} \noindent Note that by using lists each side of the tape is only finite. The potential infinity is achieved by adding an appropriate blank or occupied cell whenever the head goes over the ``edge'' of the tape. To make this formal we define five possible \emph{actions}, @{text a} the Turing machine can perform: \begin{center} \begin{tabular}[t]{@ {}rcl@ {\hspace{2mm}}l} @{text "a"} & $::=$ & @{term "W0"} & (write blank, @{term Bk})\\ & $\mid$ & @{term "W1"} & (write occupied, @{term Oc})\\ \end{tabular} \begin{tabular}[t]{rcl@ {\hspace{2mm}}l} & $\mid$ & @{term L} & (move left)\\ & $\mid$ & @{term R} & (move right)\\ \end{tabular} \begin{tabular}[t]{rcl@ {\hspace{2mm}}l@ {}} & $\mid$ & @{term Nop} & (do-nothing operation)\\ \end{tabular} \end{center} \noindent We slightly deviate from the presentation in \cite{Boolos87} by using the @{term Nop} operation; however its use will become important when we formalise halting computations and also universal Turing machines. Given a tape and an action, we can define the following tape updating function: \begin{center} \begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}l} @{thm (lhs) update.simps(1)} & @{text "\<equiv>"} & @{thm (rhs) update.simps(1)}\\ @{thm (lhs) update.simps(2)} & @{text "\<equiv>"} & @{thm (rhs) update.simps(2)}\\ @{thm (lhs) update.simps(3)} & @{text "\<equiv>"} & @{thm (rhs) update.simps(3)}\\ @{thm (lhs) update.simps(4)} & @{text "\<equiv>"} & @{thm (rhs) update.simps(4)}\\ @{thm (lhs) update.simps(5)} & @{text "\<equiv>"} & @{thm (rhs) update.simps(5)}\\ \end{tabular} \end{center} \noindent The first two clauses replace the head of the right-list with a new @{term Bk} or @{term Oc}, respectively. To see that these two clauses make sense in case where @{text r} is the empty list, one has to know that the tail function, @{term tl}, is defined such that @{term "tl [] == []"} holds. The third clause implements the move of the head one step to the left: we need to test if the left-list @{term l} is empty; if yes, then we just prepend a blank cell to the right-list; otherwise we have to remove the head from the left-list and prepend it to the right-list. Similarly in the fourth clause for a right move action. The @{term Nop} operation leaves the the tape unchanged. %Note that our treatment of the tape is rather ``unsymmetric''---we %have the convention that the head of the right-list is where the %head is currently positioned. Asperti and Ricciotti %\cite{AspertiRicciotti12} also considered such a representation, but %dismiss it as it complicates their definition for \emph{tape %equality}. The reason is that moving the head one step to %the left and then back to the right might change the tape (in case %of going over the ``edge''). Therefore they distinguish four types %of tapes: one where the tape is empty; another where the head %is on the left edge, respectively right edge, and in the middle %of the tape. The reading, writing and moving of the tape is then %defined in terms of these four cases. In this way they can keep the %tape in a ``normalised'' form, and thus making a left-move followed %by a right-move being the identity on tapes. Since we are not using %the notion of tape equality, we can get away with the unsymmetric %definition above, and by using the @{term update} function %cover uniformly all cases including corner cases. Next we need to define the \emph{states} of a Turing machine. %Given %how little is usually said about how to represent them in informal %presentations, it might be surprising that in a theorem prover we %have to select carefully a representation. If we use the naive %representation where a Turing machine consists of a finite set of %states, then we will have difficulties composing two Turing %machines: we would need to combine two finite sets of states, %possibly renaming states apart whenever both machines share %states.\footnote{The usual disjoint union operation in Isabelle/HOL %cannot be used as it does not preserve types.} This renaming can be %quite cumbersome to reason about. We followed the choice made by \cite{AspertiRicciotti12} representing a state by a natural number and the states of a Turing machine by the initial segment of natural numbers starting from @{text 0}. In doing so we can compose two Turing machine by shifting the states of one by an appropriate amount to a higher segment and adjusting some ``next states'' in the other. {\it composition here?} An \emph{instruction} @{term i} of a Turing machine is a pair consisting of an action and a natural number (the next state). A \emph{program} @{term p} of a Turing machine is then a list of such pairs. Using as an example the following Turing machine program, which consists of four instructions \begin{equation} \begin{tikzpicture} \node [anchor=base] at (0,0) {@{thm dither_def}}; \node [anchor=west] at (-1.5,-0.42) {$\underbrace{\hspace{21mm}}_{\text{1st state}}$}; \node [anchor=west] at ( 1.1,-0.42) {$\underbrace{\hspace{17mm}}_{\text{2nd state}}$}; \node [anchor=west] at (-1.5,0.65) {$\overbrace{\hspace{10mm}}^{\text{@{term Bk}-case}}$}; \node [anchor=west] at (-0.1,0.65) {$\overbrace{\hspace{6mm}}^{\text{@{term Oc}-case}}$}; \end{tikzpicture} \label{dither} \end{equation} \noindent the reader can see we have organised our Turing machine programs so that segments of two belong to a state. The first component of the segment determines what action should be taken and which next state should be transitioned to in case the head reads a @{term Bk}; similarly the second component determines what should be done in case of reading @{term Oc}. We have the convention that the first state is always the \emph{starting state} of the Turing machine. The @{text 0}-state is special in that it will be used as the ``halting state''. There are no instructions for the @{text 0}-state, but it will always perform a @{term Nop}-operation and remain in the @{text 0}-state. Unlike Asperti and Riccioti \cite{AspertiRicciotti12}, we have chosen a very concrete representation for programs, because when constructing a universal Turing machine, we need to define a coding function for programs. This can be easily done for our programs-as-lists, but is more difficult for the functions used by Asperti and Ricciotti. Given a program @{term p}, a state and the cell being read by the head, we need to fetch the corresponding instruction from the program. For this we define the function @{term fetch} \begin{center} \begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}l} \multicolumn{3}{l}{@{thm fetch.simps(1)[where b=DUMMY]}}\\ @{thm (lhs) fetch.simps(2)} & @{text "\<equiv>"} & @{text "case nth_of p (2 * s) of"}\\ \multicolumn{3}{@ {\hspace{4cm}}l}{@{text "None \<Rightarrow> (Nop, 0) | Some i \<Rightarrow> i"}}\\ @{thm (lhs) fetch.simps(3)} & @{text "\<equiv>"} & @{text "case nth_of p (2 * s + 1) of"}\\ \multicolumn{3}{@ {\hspace{4cm}}l}{@{text "None \<Rightarrow> (Nop, 0) | Some i \<Rightarrow> i"}} \end{tabular} \end{center} \noindent In this definition the function @{term nth_of} returns the @{text n}th element from a list, provided it exists (@{term Some}-case), or if it does not, it returns the default action @{term Nop} and the default state @{text 0} (@{term None}-case). In doing so we slightly deviate from the description in \cite{Boolos87}: if their Turing machines transition to a non-existing state, then the computation is halted. We will transition in such cases to the @{text 0}-state.\footnote{\it However, with introducing the notion of \emph{well-formed} Turing machine programs we will later exclude such cases and make the @{text 0}-state the only ``halting state''. A program @{term p} is said to be well-formed if it satisfies the following three properties: \begin{center} \begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}l} @{term "t_correct p"} & @{text "\<equiv>"} & @{term "2 <= length p"}\\ & @{text "\<and>"} & @{term "iseven (length p)"}\\ & @{text "\<and>"} & @{term "\<forall> (a, s) \<in> set p. s <= length p div 2"} \end{tabular} \end{center} \noindent The first says that @{text p} must have at least an instruction for the starting state; the second that @{text p} has a @{term Bk} and @{term Oc} instruction for every state, and the third that every next-state is one of the states mentioned in the program or being the @{text 0}-state. } A \emph{configuration} @{term c} of a Turing machine is a state together with a tape. This is written as @{text "(s, (l, r))"}. If we have a configuration and a program, we can calculate what the next configuration is by fetching the appropriate action and next state from the program, and by updating the state and tape accordingly. This single step of execution is defined as the function @{term tstep} \begin{center} \begin{tabular}{l} @{text "step (s, (l, r)) p"} @{text "\<equiv>"}\\ \hspace{10mm}@{text "let (a, s) = fetch p s (read r)"}\\ \hspace{10mm}@{text "in (s', update (l, r) a)"} \end{tabular} \end{center} \noindent where @{term "read r"} returns the head of the list @{text r}, or if @{text r} is empty it returns @{term Bk}. It is impossible in Isabelle/HOL to lift the @{term step}-function realising a general evaluation function for Turing machines. The reason is that functions in HOL-based provers need to be terminating, and clearly there are Turing machine programs that are not. We can however define an evaluation function so that it performs exactly @{text n} steps: \begin{center} \begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}l} @{thm (lhs) steps.simps(1)} & @{text "\<equiv>"} & @{thm (rhs) steps.simps(1)}\\ @{thm (lhs) steps.simps(2)} & @{text "\<equiv>"} & @{thm (rhs) steps.simps(2)}\\ \end{tabular} \end{center} \noindent Recall our definition of @{term fetch} with the default value for the @{text 0}-state. In case a Turing program takes in \cite{Boolos87} less then @{text n} steps before it halts, then in our setting the @{term steps}-evaluation does not actually halt, but rather transitions to the @{text 0}-state and remains there performing @{text Nop}-actions until @{text n} is reached. Given some input tape @{text "(l\<^isub>i,r\<^isub>i)"}, we can define when a program @{term p} generates a specific output tape @{text "(l\<^isub>o,r\<^isub>o)"} \begin{center} \begin{tabular}{l} @{term "runs p (l\<^isub>i, r\<^isub>i) (l\<^isub>o,r\<^isub>o)"} @{text "\<equiv>"}\\ \hspace{6mm}@{text "\<exists>n. nsteps (1, (l\<^isub>i,r\<^isub>i)) p n = (0, (l\<^isub>o,r\<^isub>o))"} \end{tabular} \end{center} \noindent where @{text 1} stands for the starting state and @{text 0} for our final state. A program @{text p} with input tape @{term "(l\<^isub>i, r\<^isub>i)"} \emph{halts} iff \begin{center} @{term "halts p (l\<^isub>i, r\<^isub>i) \<equiv> \<exists>l\<^isub>o r\<^isub>o. runs p (l\<^isub>i, r\<^isub>i) (l\<^isub>o,r\<^isub>o)"} \end{center} \noindent Later on we need to consider specific Turing machines that start with a tape in standard form and halt the computation in standard form. To define a tape in standard form, it is useful to have an operation %@{ term "tape_of_nat_list DUMMY"} that translates lists of natural numbers into tapes. \begin{center} \begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}l} %@ { thm (lhs) tape_of_nat_list_def2(1)} & @{text "\<equiv>"} & @ { thm (rhs) tape_of_nat_list_def2(1)}\\ %@ { thm (lhs) tape_of_nat_list_def2(2)} & @{text "\<equiv>"} & @ { thm (rhs) tape_of_nat_list_def2(2)}\\ %@ { thm (lhs) tape_of_nat_list_def2(3)} & @{text "\<equiv>"} & @ { thm (rhs) tape_of_nat_list_def2(3)}\\ \end{tabular} \end{center} By this we mean \begin{center} %@ {thm haltP_def2[where p="p" and n="n", THEN eq_reflection]} \end{center} \noindent This means the Turing machine starts with a tape containg @{text n} @{term Oc}s and the head pointing to the first one; the Turing machine halts with a tape consisting of some @{term Bk}s, followed by a ``cluster'' of @{term Oc}s and after that by some @{term Bk}s. The head in the output is pointing again at the first @{term Oc}. The intuitive meaning of this definition is to start the Turing machine with a tape corresponding to a value @{term n} and producing a new tape corresponding to the value @{term l} (the number of @{term Oc}s clustered on the output tape). Before we can prove the undecidability of the halting problem for Turing machines, we have to define how to compose two Turing machines. Given our setup, this is relatively straightforward, if slightly fiddly. We use the following two auxiliary functions: \begin{center} \begin{tabular}{@ {}l@ {\hspace{1mm}}c@ {\hspace{1mm}}l@ {}} @{thm (lhs) shift.simps} @{text "\<equiv>"} @{thm (rhs) shift.simps}\\ @{thm (lhs) adjust.simps} @{text "\<equiv>"} @{thm (rhs) adjust.simps}\\ \end{tabular} \end{center} \noindent The first adds @{text n} to all states, exept the @{text 0}-state, thus moving all ``regular'' states to the segment starting at @{text n}; the second adds @{term "length p div 2 + 1"} to the @{text 0}-state, thus ridirecting all references to the ``halting state'' to the first state after the program @{text p}. With these two functions in place, we can define the \emph{sequential composition} of two Turing machine programs @{text "p\<^isub>1"} and @{text "p\<^isub>2"} \begin{center} @{thm tm_comp.simps[where ?p1.0="p\<^isub>1" and ?p2.0="p\<^isub>2", THEN eq_reflection]} \end{center} \noindent This means @{text "p\<^isub>1"} is executed first. Whenever it originally transitioned to the @{text 0}-state, it will in the composed program transition to the starting state of @{text "p\<^isub>2"} instead. All the states of @{text "p\<^isub>2"} have been shifted in order to make sure that the states of the composed program @{text "p\<^isub>1 \<oplus> p\<^isub>2"} still only ``occupy'' an initial segment of the natural numbers. \begin{center} \begin{tabular}{@ {}l@ {\hspace{1mm}}c@ {\hspace{1mm}}p{6.9cm}@ {}} @{thm (lhs) tcopy_def} & @{text "\<equiv>"} & @{thm (rhs) tcopy_def} \end{tabular} \end{center} assertion holds for all tapes Hoare rule for composition For showing the undecidability of the halting problem, we need to consider two specific Turing machines. copying TM and dithering TM correctness of the copying TM measure for the copying TM, which we however omit. halting problem*}section {* Abacus Machines *}text {* \noindent Boolos et al \cite{Boolos87} use abacus machines as a stepping stone for making it less laborious to write programs for Turing machines. Abacus machines operate over an unlimited number of registers $R_0$, $R_1$, \ldots each being able to hold an arbitrary large natural number. We use natural numbers to refer to registers, but also to refer to \emph{opcodes} of abacus machines. Obcodes are given by the datatype \begin{center} \begin{tabular}{rcll} @{text "o"} & $::=$ & @{term "Inc R\<iota>"} & increment register $R$ by one\\ & $\mid$ & @{term "Dec R\<iota> o\<iota>"} & if content of $R$ is non-zero,\\ & & & then decrement it by one\\ & & & otherwise jump to opcode $o$\\ & $\mid$ & @{term "Goto o\<iota>"} & jump to opcode $o$ \end{tabular} \end{center} \noindent A \emph{program} of an abacus machine is a list of such obcodes. For example the program clearing the register $R$ (setting it to 0) can be defined as follows: \begin{center} %@ {thm clear.simps[where n="R\<iota>" and e="o\<iota>", THEN eq_reflection]} \end{center} \noindent The second opcode @{term "Goto 0"} in this programm means we jump back to the first opcode, namely @{text "Dec R o"}. The \emph{memory} $m$ of an abacus machine holding the values of the registers is represented as a list of natural numbers. We have a lookup function for this memory, written @{term "abc_lm_v m R\<iota>"}, which looks up the content of register $R$; if $R$ is not in this list, then we return 0. Similarly we have a setting function, written @{term "abc_lm_s m R\<iota> n"}, which sets the value of $R$ to $n$, and if $R$ was not yet in $m$ it pads it approriately with 0s. Abacus machine halts when it jumps out of range.*}section {* Recursive Functions *}section {* Wang Tiles\label{Wang} *}text {* Used in texture mapings - graphics*}section {* Related Work *}text {* The most closely related work is by Norrish \cite{Norrish11}, and Asperti and Ricciotti \cite{AspertiRicciotti12}. Norrish bases his approach on lambda-terms. For this he introduced a clever rewriting technology based on combinators and de-Bruijn indices for rewriting modulo $\beta$-equivalence (to keep it manageable)*}(*Questions:Can this be done: Ackerman function is not primitive recursive (Nora Szasz)Tape is represented as two lists (finite - usually infinite tape)?*)(*<*)end(*>*)