(*<*)
theory Paper
imports "../thys/uncomputable"
begin

(*
hide_const (open) s 
*)

hide_const (open) Divides.adjust

abbreviation
  "update2 p a \<equiv> update a p"

consts DUMMY::'a

notation (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
  tm_wf0 ("wf") and
  (*is_even ("iseven") and*)
  tcopy_begin ("copy\<^bsub>begin\<^esub>") and
  tcopy_loop ("copy\<^bsub>loop\<^esub>") and
  tcopy_end ("copy\<^bsub>end\<^esub>") and
  step0 ("step") and
  steps0 ("steps") and
  exponent ("_\<^bsup>_\<^esup>") and
(*  abc_lm_v ("lookup") and
  abc_lm_s ("set") and*)
  haltP ("stdhalt") and 
  tcopy ("copy") and 
  tape_of ("\<langle>_\<rangle>") and 
  tm_comp ("_ \<oplus> _") and 
  DUMMY  ("\<^raw:\mbox{$\_\!\_\,$}>")

declare [[show_question_marks = false]]

(* THEOREMS *)
notation (Rule output)
  "==>"  ("\<^raw:\mbox{}\inferrule{\mbox{>_\<^raw:}}>\<^raw:{\mbox{>_\<^raw:}}>")

syntax (Rule output)
  "_bigimpl" :: "asms \<Rightarrow> prop \<Rightarrow> prop"
  ("\<^raw:\mbox{}\inferrule{>_\<^raw:}>\<^raw:{\mbox{>_\<^raw:}}>")

  "_asms" :: "prop \<Rightarrow> asms \<Rightarrow> asms" 
  ("\<^raw:\mbox{>_\<^raw:}\\>/ _")

  "_asm" :: "prop \<Rightarrow> asms" ("\<^raw:\mbox{>_\<^raw:}>")

notation (Axiom output)
  "Trueprop"  ("\<^raw:\mbox{}\inferrule{\mbox{}}{\mbox{>_\<^raw:}}>")

notation (IfThen output)
  "==>"  ("\<^raw:{\normalsize{}>If\<^raw:\,}> _/ \<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
syntax (IfThen output)
  "_bigimpl" :: "asms \<Rightarrow> prop \<Rightarrow> prop"
  ("\<^raw:{\normalsize{}>If\<^raw:\,}> _ /\<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
  "_asms" :: "prop \<Rightarrow> asms \<Rightarrow> asms" ("\<^raw:\mbox{>_\<^raw:}> /\<^raw:{\normalsize \,>and\<^raw:\,}>/ _")
  "_asm" :: "prop \<Rightarrow> asms" ("\<^raw:\mbox{>_\<^raw:}>")

notation (IfThenNoBox output)
  "==>"  ("\<^raw:{\normalsize{}>If\<^raw:\,}> _/ \<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
syntax (IfThenNoBox output)
  "_bigimpl" :: "asms \<Rightarrow> prop \<Rightarrow> prop"
  ("\<^raw:{\normalsize{}>If\<^raw:\,}> _ /\<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
  "_asms" :: "prop \<Rightarrow> asms \<Rightarrow> asms" ("_ /\<^raw:{\normalsize \,>and\<^raw:\,}>/ _")
  "_asm" :: "prop \<Rightarrow> asms" ("_")

lemma nats2tape:
  shows "<([]::nat list)> = []"
  and "<[n]> = Oc \<up> (n + 1)"
  and "ns \<noteq> [] \<Longrightarrow> <n#ns> = Oc \<up> (n + 1) @ [Bk] @ <ns>"
apply(auto simp add: tape_of_nl_abv tape_of_nat_list.simps)
apply(case_tac ns)
apply(auto)
done

(*>*)

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. 


\noindent
Suppose you want to mechanise a proof for whether a predicate @{term
P}, say, is decidable or not. Decidability of @{text P} usually
amounts to showing whether \mbox{@{term "P \<or> \<not>P"}} holds. But this
does \emph{not} work in Isabelle/HOL and other HOL theorem provers,
since they are based on classical logic where the law of excluded
middle ensures that \mbox{@{term "P \<or> \<not>P"}} is always provable no
matter whether @{text P} is constructed by computable means. We hit on
this limitation previously when we mechanised the correctness proofs
of two algorithms \cite{UrbanCheneyBerghofer11,WuZhangUrban12}, but
were unable to formalise arguments about decidability.

%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 based
on classical logic is to formalise a theory of computability. Norrish
provided such a formalisation for the HOL. He choose
the $\lambda$-calculus as the starting point for his formalisation of
computability theory, because of its ``simplicity'' \cite[Page
297]{Norrish11}.  Part of his formalisation is a clever infrastructure
for reducing $\lambda$-terms. He also established the computational
equivalence between the $\lambda$-calculus and recursive functions.
Nevertheless he concluded that it would be appealing
 to have formalisations for more operational models of
computations, such as Turing machines or register machines.  One
reason is that many proofs in the literature use them.  He noted
however 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}

\noindent
In this paper we take on this daunting prospect and provide a
formalisation of Turing machines, as well as abacus machines (a kind
of register machines) and recursive functions. To see the difficulties
involved with this work, one has to understand that Turing machine
programs can be completely \emph{unstructured}, behaving similar to
Basic programs involving the infamous goto \cite{Dijkstra68}. This
precludes in the general case a compositional Hoare-style reasoning
about Turing programs.  We provide such Hoare-rules for when it
\emph{is} possible to reason 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
reasoning about concrete Turing machine programs is usually
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 of
Turing machines in the Matita theorem prover, including a universal
Turing machine. However, they do \emph{not} formalise the undecidability of the 
halting problem since their main focus is complexity, rather than
computability theory. They also report that the informal proofs from which they
started are not ``sufficiently accurate to be directly usable as a
guideline for formalization'' \cite[Page 2]{AspertiRicciotti12}. For
our formalisation we follow mainly the proofs from the textbook
\cite{Boolos87} and found that the description there is quite
detailed. Some details are left out however: for example, constructing
the \emph{copy Turing program} is left as an excerise to the reader; also 
\cite{Boolos87} only shows how the universal Turing machine is constructed for Turing
machines computing unary functions. We had to figure out a way to
generalise this result to $n$-ary functions. Similarly, when compiling
recursive functions to abacus machines, the textbook again only shows
how it can be done for 2- and 3-ary functions, but in the
formalisation we need arbitrary functions. But the general ideas for
how 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 Asperti
and Ricciotti is that their universal Turing machine uses a different
alphabet than the machines it simulates. They write \cite[Page
23]{AspertiRicciotti12}:

\begin{quote}\it
``In particular, the fact that the universal machine operates with a
different alphabet with respect to the machines it simulates is
annoying.'' 
\end{quote}

\noindent
In this paper we follow the approach by Boolos et al \cite{Boolos87},
which goes back to Post \cite{Post36}, where all Turing machines
operate on tapes that contain only \emph{blank} or \emph{occupied} cells. 
Traditionally the content of a cell can be any
character from a finite alphabet. Although computationally equivalent,
the more restrictive notion of Turing machines in \cite{Boolos87} makes
the reasoning more uniform. In addition some proofs \emph{about} Turing
machines are simpler.  The reason is that one often needs to encode
Turing machines---consequently if the Turing machines are simpler, then the coding
functions are simpler too. Unfortunately, the restrictiveness also makes
it harder to design programs for these Turing machines. In order
to construct a universal Turing machine we therefore do not follow 
\cite{AspertiRicciotti12}, instead follow the proof in
\cite{Boolos87} by translating abacus machines to Turing machines and in
turn recursive functions to abacus machines. The universal Turing
machine can then be constructed as a recursive function.

\smallskip
\noindent
{\bf Contributions:} We formalised in Isabelle/HOL Turing machines following the
description of Boolos et al \cite{Boolos87} where tapes only have blank or
occupied cells. We mechanise the undecidability of the halting problem and
prove the correctness of concrete Turing machines that are needed
in this proof; such correctness proofs are left out in the informal literature.  
For reasoning about Turing machine programs we derive Hoare-rules.
We also construct the universal Turing machine from \cite{Boolos87} by
translating recursive functions to abacus machines and abacus machines to
Turing machines. Since we have set up in Isabelle/HOL a very general computability
model and undecidability result, we are able to formalise other
results: we describe a proof of the computational equivalence
of single-sided Turing machines, which is not given in \cite{Boolos87},
but needed for example for formalising the undecidability proof of 
Wang's tiling problem \cite{Robinson71}.
%We are not aware of any other
%formalisation 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 use
  the notation @{term "Bk \<up> n"} (similarly @{term "Oc \<up> n"}) for lists
  composed of @{term n} elements of @{term Bk}s.  We also 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
  scannes. This can be pictured as follows:
  %
  \begin{center}
  \begin{tikzpicture}[scale=0.9]
  \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[fill]     (-1.65,0.1) rectangle (-1.35,0.4);
  \draw (-0.25,0.8) -- (-0.25,-0.8);
  \draw[<->] (-1.25,-0.7) -- (0.75,-0.7);
  \node [anchor=base] at (-0.85,-0.5) {\small left list};
  \node [anchor=base] at (0.40,-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} 
  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} and \cite{AspertiRicciotti12} 
  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 follow the choice made in \cite{AspertiRicciotti12} 
  by representing a state with a natural number and the states in a Turing
  machine program by the initial segment of natural numbers starting from @{text 0}.
  In doing so we can compose two Turing machine programs by
  shifting the states of one by an appropriate amount to a higher
  segment and adjusting some ``next states'' in the other. 

  An \emph{instruction} 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,-0.64) 
  {$\underbrace{\hspace{21mm}}_{\text{\begin{tabular}{@ {}l@ {}}1st state\\[-2mm] 
  = starting state\end{tabular}}}$};
  
  \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 such a
  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 directly done for our programs-as-lists, but is
  slightly 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{equation}\label{fetch}
  \mbox{\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{equation}

  \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). We often need to restrict Turing machine programs 
  to be well-formed: a program @{term p} is \emph{well-formed} if it 
  satisfies the following three properties:

  \begin{center}
  @{thm tm_wf.simps[where p="p" and off="0::nat", simplified, THEN eq_reflection]}
  \end{center}

  \noindent
  The first states 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.

  We need to be able to sequentially compose Turing machine programs. Given our
  concrete representation, 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 "Suc(length p div 2)"} to the @{text
  0}-state, thus redirecting 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"} as

  \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.

  A \emph{configuration} @{term c} of a Turing machine is a state
  together with a tape. This is written as @{text "(s, (l, r))"}.  We
  say a configuration \emph{is final} if @{term "s = (0::nat)"} and we
  say a predicate @{text P} \emph{holds for} a configuration if @{text
  "P"} holds for the tape @{text "(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 step}

  \begin{center}
  \begin{tabular}{l@ {\hspace{1mm}}c@ {\hspace{1mm}}l}
  @{text "step (s, (l, r)) p"} & @{text "\<equiv>"} & @{text "let (a, s') = fetch p s (read r)"}\\
  & & @{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 to realise
  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 that 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} (shown in
  \eqref{fetch}) with the default value for the @{text 0}-state. In
  case a Turing program takes according to the usual textbook
  definition \cite{Boolos87} less than @{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 (the
  final state) and remains there performing @{text Nop}-actions until
  @{text n} is reached. 
  
  \begin{figure}[t]
  \begin{center}
  \begin{tabular}{@ {}c@ {\hspace{3mm}}c@ {\hspace{3mm}}c}
  \begin{tabular}[t]{@ {}l@ {}}
  @{thm (lhs) tcopy_begin_def} @{text "\<equiv>"}\\
  \hspace{2mm}@{text "["}@{text "(W\<^bsub>Bk\<^esub>, 0), (R, 2), (R, 3),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(R, 2), (W1, 3), (L, 4),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(L, 4), (L, 0)"}@{text "]"} 
  \end{tabular}
  &
  \begin{tabular}[t]{@ {}l@ {}}
  @{thm (lhs) tcopy_loop_def} @{text "\<equiv>"}\\
  \hspace{2mm}@{text "["}@{text "(R, 0), (R, 2), (R, 3),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(W\<^bsub>Bk\<^esub>, 2), (R, 3), (R, 4),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(W\<^bsub>Oc\<^esub>, 5), (R, 4), (L, 6),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(L, 5), (L, 6), (L, 1)"}@{text "]"} 
  \end{tabular}
  &
  \begin{tabular}[t]{@ {}l@ {}}
  @{thm (lhs) tcopy_end_def} @{text "\<equiv>"}\\
  \hspace{2mm}@{text "["}@{text "(L, 0), (R, 2), (W\<^bsub>Oc\<^esub>, 3),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(L, 4), (R, 2), (R, 2),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(L, 5), (W\<^bsub>Bk\<^esub>, 4), (R, 0),"}\\
  \hspace{2mm}\phantom{@{text "["}}@{text "(L, 5)"}@{text "]"} 
  \end{tabular}
  \end{tabular}\\[2mm]

  \begin{tikzpicture}[scale=0.7]
  \node [anchor=base] at (2.2,0.1) {\small$\Rightarrow$};
  \node [anchor=base] at (5.6,0.1) {\small$\Rightarrow$};
  \node [anchor=base] at (10.5,0.1) {\small$\Rightarrow$};
  \node [anchor=base] at (2.2,-0.6) {\small$\overbrace{@{term "tcopy_init"}}^{}$};
  \node [anchor=base] at (5.6,-0.6) {\small$\overbrace{@{term "tcopy_loop"}}^{}$};
  \node [anchor=base] at (10.5,-0.6) {\small$\overbrace{@{term "tcopy_end"}}^{}$};


  \begin{scope}[shift={(0.5,0)}]
  \draw[very thick] (-0.25,0)   -- ( 1.25,0);
  \draw[very thick] (-0.25,0.5) -- ( 1.25,0.5);
  \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] ( 1.25,0)   -- ( 1.25,0.5);
  \draw[rounded corners=1mm] (-0.35,-0.1) rectangle (0.35,0.6);
  \draw[fill]     (-0.15,0.1) rectangle (0.15,0.4);
  \draw[fill]     ( 0.35,0.1) rectangle (0.65,0.4);
  \draw[fill]     ( 0.85,0.1) rectangle (1.15,0.4);
  \end{scope}

  \begin{scope}[shift={(2.9,0)}]
  \draw[very thick] (-0.25,0)   -- ( 2.25,0);
  \draw[very thick] (-0.25,0.5) -- ( 2.25,0.5);
  \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] ( 1.25,0)   -- ( 1.25,0.5);
  \draw[very thick] ( 1.75,0)   -- ( 1.75,0.5);
  \draw[very thick] ( 2.25,0)   -- ( 2.25,0.5);
  \draw[rounded corners=1mm] (0.15,-0.1) rectangle (0.85,0.6);
  \draw[fill]     (-0.15,0.1) rectangle (0.15,0.4);
  \draw[fill]     ( 0.35,0.1) rectangle (0.65,0.4);
  \draw[fill]     ( 0.85,0.1) rectangle (1.15,0.4);
  \draw[fill]     ( 1.85,0.1) rectangle (2.15,0.4);
  \end{scope}

  \begin{scope}[shift={(6.8,0)}]
  \draw[very thick] (-0.75,0)   -- ( 3.25,0);
  \draw[very thick] (-0.75,0.5) -- ( 3.25,0.5);
  \draw[very thick] (-0.75,0)   -- (-0.75,0.5);
  \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] ( 1.25,0)   -- ( 1.25,0.5);
  \draw[very thick] ( 1.75,0)   -- ( 1.75,0.5);
  \draw[very thick] ( 2.25,0)   -- ( 2.25,0.5);
  \draw[very thick] ( 2.75,0)   -- ( 2.75,0.5);
  \draw[very thick] ( 3.25,0)   -- ( 3.25,0.5);
  \draw[rounded corners=1mm] (-0.35,-0.1) rectangle (0.35,0.6);
  \draw[fill]     (-0.15,0.1) rectangle (0.15,0.4);
  \draw[fill]     ( 2.35,0.1) rectangle (2.65,0.4);
  \draw[fill]     ( 2.85,0.1) rectangle (3.15,0.4);
  \draw[fill]     ( 1.85,0.1) rectangle (2.15,0.4);
  \end{scope}

  \begin{scope}[shift={(11.7,0)}]
  \draw[very thick] (-0.75,0)   -- ( 3.25,0);
  \draw[very thick] (-0.75,0.5) -- ( 3.25,0.5);
  \draw[very thick] (-0.75,0)   -- (-0.75,0.5);
  \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] ( 1.25,0)   -- ( 1.25,0.5);
  \draw[very thick] ( 1.75,0)   -- ( 1.75,0.5);
  \draw[very thick] ( 2.25,0)   -- ( 2.25,0.5);
  \draw[very thick] ( 2.75,0)   -- ( 2.75,0.5);
  \draw[very thick] ( 3.25,0)   -- ( 3.25,0.5);
  \draw[rounded corners=1mm] (-0.35,-0.1) rectangle (0.35,0.6);
  \draw[fill]     (-0.15,0.1) rectangle (0.15,0.4);
  \draw[fill]     ( 2.35,0.1) rectangle (2.65,0.4);
  \draw[fill]     ( 2.85,0.1) rectangle (3.15,0.4);
  \draw[fill]     ( 1.85,0.1) rectangle (2.15,0.4);
  \draw[fill]     ( 0.35,0.1) rectangle (0.65,0.4);
  \draw[fill]     ( 0.85,0.1) rectangle (1.15,0.4);
  \end{scope}
  \end{tikzpicture}\\[-8mm]\mbox{} 
  \end{center}
  \caption{The components of the \emph{copy Turing machine} (above). If started 
  with the tape @{term "([], <[2::nat]>)"} the first machine appends @{term "[Bk, Oc]"} at 
  the end of the right tape; the second then ``moves'' all @{term Oc}s except the 
  first from the beginning of the tape to the end; the third ``refills'' the original 
  block of @{term "Oc"}s. The resulting tape is @{term "([Bk], <[2::nat, 2::nat]>)"}.}
  \label{copy}
  \end{figure}


  We often need to restrict tapes to be in \emph{standard form}, which means 
  the left list of the tape is either empty or only contains @{text "Bk"}s, and 
  the right list contains some ``clusters'' of @{text "Oc"}s separted by single 
  blanks. To make this formal we define the following function
  
  \begin{center}
  \begin{tabular}{@ {}l@ {\hspace{1mm}}c@ {\hspace{1mm}}l@ {}}
  @{thm (lhs) nats2tape(1)} & @{text "\<equiv>"} & @{thm (rhs) nats2tape(1)}\\
  @{thm (lhs) nats2tape(2)} & @{text "\<equiv>"} & @{thm (rhs) nats2tape(2)}\\
  @{thm (lhs) nats2tape(3)} & @{text "\<equiv>"} & @{thm (rhs) nats2tape(3)}
  \end{tabular}
  \end{center}

  \noindent
  A standard tape is then of the form @{text "(Bk\<^isup>l,\<langle>[n\<^isub>1,...,n\<^isub>m]\<rangle>)"} for some @{text l} 
  and @{text "n\<^isub>i"}. Note that the head in a standard tape ``points'' to the 
  leftmost @{term "Oc"} on the tape. Note also that @{text 0} is represented by
  a single filled cell, @{text 1} by two filled cells and so on.

  Before we can prove the undecidability of the halting problem for our
  Turing machines, we need to analyse two concrete Turing machine
  programs and establish that they are correct---that means they are
  ``doing what they are supposed to be doing''.  Such correctness proofs are usually left
  out in the informal literature, for example \cite{Boolos87}.  One program 
  we need to prove correct is the @{term dither} program shown in \eqref{dither} 
  and the other program is @{term "tcopy"} defined as

  \begin{equation}
  \mbox{\begin{tabular}{@ {}l@ {\hspace{1mm}}c@ {\hspace{1mm}}l@ {}}
  @{thm (lhs) tcopy_def} & @{text "\<equiv>"} & @{thm (rhs) tcopy_def}
  \end{tabular}}\label{tcopy}
  \end{equation}

  \noindent
  whose three components are given in Figure~\ref{copy}. To the prove
  correctness of these Turing machine programs, we introduce the
  notion of total correctness defined in terms of
  \emph{Hoare-triples}, written @{term "{P} p {Q}"}. They are very similar
  to the notion of \emph{realisability} in \cite{AspertiRicciotti12}.  
  They implement the 
  idea that a program @{term p} started in state @{term "1::nat"} with
  a tape satisfying @{term P} will after some @{text n} steps halt (have
  transitioned into the halting state) with a tape satisfying @{term
  Q}. We also have \emph{Hoare-pairs} of the form @{term "{P} p \<up>"}
  realising the case that a program @{term p} started with a tape
  satisfying @{term P} will loop (never transition into the halting
  state). Both notion are formally defined as

  \begin{center}
  \begin{tabular}{@ {}c@ {\hspace{4mm}}c@ {}}
  \begin{tabular}[t]{@ {}l@ {}}
  \colorbox{mygrey}{@{thm (lhs) Hoare_halt_def}} @{text "\<equiv>"}\\[1mm]
  \hspace{5mm}@{text "\<forall>"} @{term "(l, r)"}.\\
  \hspace{7mm}if @{term "P (l, r)"} holds then\\
  \hspace{7mm}@{text "\<exists>"} @{term n}. such that\\
  \hspace{7mm}@{text "is_final (steps (1, (l, r)) p n)"} \hspace{1mm}@{text "\<and>"}\\
  \hspace{7mm}@{text "Q holds_for (steps (1, (l, r)) p n)"}
  \end{tabular} &
  \begin{tabular}[t]{@ {}l@ {}}
  \colorbox{mygrey}{@{thm (lhs) Hoare_unhalt_def}} @{text "\<equiv>"}\\[1mm]
  \hspace{5mm}@{text "\<forall>"} @{term "(l, r)"}.\\ 
  \hspace{7mm}if @{term "P (l, r)"} holds then\\
  \hspace{7mm}@{text "\<forall>"} @{term n}. @{text "\<not> is_final (steps (1, (l, r)) p n)"}
  \end{tabular}
  \end{tabular}
  \end{center}
  
  \noindent
  Like Asperti and Ricciotti with their notion of realisability, we
  have set up our Hoare-style reasoning so that we can deal explicitly
  with total correctness and non-terminantion, rather than have
  notions for partial correctness and termination. Although the latter
  would allow us to reason more uniformly (only using Hoare-triples),
  we prefer our definitions because we can derive (below) simple
  Hoare-rules for sequentially composed Turing programs.  In this way
  we can reason about the correctness of @{term "tcopy_init"}, for
  example, completely separately from @{term "tcopy_loop"} and @{term
  "tcopy_end"}.

  It is realatively straightforward to prove that the Turing program 
  @{term "dither"} shown in \eqref{dither} is correct. This program
  should be the ``identity'' when started with a standard tape representing 
  @{text "1"} but loop when started with @{text 0} instead.


  \begin{center}
  \begin{tabular}{l@ {\hspace{3mm}}lcl}
  & \multicolumn{1}{c}{start tape}\\[1mm]
  \raisebox{2mm}{halting case:} &
  \begin{tikzpicture}[scale=0.8]
  \draw[very thick] (-2,0)   -- ( 0.75,0);
  \draw[very thick] (-2,0.5) -- ( 0.75,0.5);
  \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[rounded corners=1mm] (-0.35,-0.1) rectangle (0.35,0.6);
  \draw[fill]     (-0.15,0.1) rectangle (0.15,0.4);
  \draw[fill]     ( 0.35,0.1) rectangle (0.65,0.4);
  \node [anchor=base] at (-1.7,0.2) {\ldots};
  \end{tikzpicture} 
  & \raisebox{2mm}{$\;\;\large\Rightarrow\;\;$} &
  \begin{tikzpicture}[scale=0.8]
  \draw[very thick] (-2,0)   -- ( 0.75,0);
  \draw[very thick] (-2,0.5) -- ( 0.75,0.5);
  \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[rounded corners=1mm] (-0.35,-0.1) rectangle (0.35,0.6);
  \draw[fill]     (-0.15,0.1) rectangle (0.15,0.4);
  \draw[fill]     ( 0.35,0.1) rectangle (0.65,0.4);
  \node [anchor=base] at (-1.7,0.2) {\ldots};
  \end{tikzpicture}\\

  \raisebox{2mm}{non-halting case:} &
  \begin{tikzpicture}[scale=0.8]
  \draw[very thick] (-2,0)   -- ( 0.25,0);
  \draw[very thick] (-2,0.5) -- ( 0.25,0.5);
  \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] (-1.25,0)   -- (-1.25,0.5);
  \draw[rounded corners=1mm] (-0.35,-0.1) rectangle (0.35,0.6);
  \draw[fill]     (-0.15,0.1) rectangle (0.15,0.4);
  \node [anchor=base] at (-1.7,0.2) {\ldots};
  \end{tikzpicture} 
  & \raisebox{2mm}{$\;\;\large\Rightarrow\;\;$} &
  \raisebox{2mm}{loops}
  \end{tabular}
  \end{center}

  \noindent
  We can prove the following Hoare-statements:
 
  \begin{center}
  \begin{tabular}{l}
  @{thm dither_halts}\\
  @{thm dither_loops}
  \end{tabular}
  \end{center}

  \noindent
  The first is by a simple calculation. The second is by induction on the
  number of steps we can perform starting from the input tape.

  The purpose of the @{term tcopy} program defined in \eqref{tcopy} is
  to produce the standard tape @{term "(DUMMY, <[n, n::nat]>)"} when
  started with @{term "(DUMMY, <[n::nat]>)"}, that is making a copy of
  a value on the tape.  Reasoning about this program is substantially
  harder than about @{term dither}. To ease the burden, we can prove
  the following Hoare-rules for sequentially composed programs.

  \begin{center}
  \begin{tabular}{@ {}p{3cm}@ {\hspace{9mm}}p{3cm}@ {}}
  @{thm[mode=Rule] 
  Hoare_plus_halt[where ?P1.0="P\<^isub>1" and ?P2.0="P\<^isub>2" and ?Q1.0="Q\<^isub>1" and ?Q2.0="Q\<^isub>2" and ?A="p\<^isub>1" and B="p\<^isub>2"]}
  &
  @{thm[mode=Rule] 
  Hoare_plus_unhalt[where ?P1.0="P\<^isub>1" and ?P2.0="P\<^isub>2" and ?Q1.0="Q\<^isub>1" and ?A="p\<^isub>1" and B="p\<^isub>2"]}
  \end{tabular}
  \end{center}
   
  \noindent
  The first corresponds to the usual Hoare-rule for composition of imperative
  programs, where @{term "Q\<^isub>1 \<mapsto> P\<^isub>2"} means @{term "Q\<^isub>1"} implies @{term "P\<^isub>2"} for 
  all tapes @{term "(l, r)"}. The second rule covers the case where rughly the 
  first program terminates generating a tape for which the second program
  loops. These are two cases we need in our proof for undecidability.
  
  The Hoare-rules above allow us to prove the correctness of @{term tcopy}
  by considering the correctness of @{term "tcopy_begin"}, @{term "tcopy_loop"} 
  and @{term "tcopy_end"} in isolation. We will show some details for the 
  the program @{term "tcopy_begin"}. 


  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

  Magnus: invariants -- Section 5.4.5 on page 75.
*}


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 program 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)


  
  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).

*}


(*
Questions:

Can this be done: Ackerman function is not primitive 
recursive (Nora Szasz)

Tape is represented as two lists (finite - usually infinite tape)?

*)


(*<*)
end
(*>*)