--- a/Paper.thy Fri Jan 11 08:47:02 2013 +0000
+++ b/Paper.thy Fri Jan 11 13:18:33 2013 +0000
@@ -120,7 +120,7 @@
\end{quote}
\noindent
-In this paper we took on this daunting prospect and provide a
+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 interactive
@@ -213,7 +213,7 @@
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, written @{term hd}, of the right-list is
+ 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:
@@ -296,7 +296,7 @@
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 consider such a representation, but
+ \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
@@ -323,10 +323,10 @@
renaming can be quite cumbersome to reason about. Therefore we made
the choice of representing a state by a natural number and the states
of a Turing machine will always consist of the initial segment
- of natural numbers starting from @{text 0} up to number of states
+ of natural numbers starting from @{text 0} up to the number of states
of the machine. In doing so we can compose
two Turing machine by ``shifting'' the states of one by an appropriate
- amount to a higher segment and adjust some ``next states''.
+ amount to a higher segment and adjust some ``next states'' in the other.
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
@@ -354,22 +354,6 @@
The 0-state is special in that it will be used as the \emph{halting state}.
There are no instructions for the 0-state, but it will always
perform a @{term Nop}-operation and remain in the 0-state.
- A \emph{well-formed} Turing machine program @{term p} is one that 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 0-state.
Given a program @{term p}, a state
and the cell being read by the head, we need to fetch
@@ -397,7 +381,24 @@
(@{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 0th state with the @{term Nop}-action.
+ to the 0th state with the @{term Nop}-action. However, with introducing the
+ notion of \emph{well-formed} Turing machine programs we will exclude such
+ cases. 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 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
@@ -415,11 +416,11 @@
\end{center}
\noindent
- It is impossible in Isabelle/HOL to lift this function in order to realise
+ It is impossible in Isabelle/HOL to lift this 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 for which this does not hold. We can however define an evaluation
- function so that it performs exactly @{text n}:
+ programs which 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}
@@ -441,11 +442,13 @@
\begin{center}
\begin{tabular}{l}
@{term "halt 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))"}
+ \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 1 stands for the starting state and 0 for the halting state, or
+ in our settin it should better be called the final state.
Later on we need to consider specific Turing machines that
start with a tape in standard form and halt the computation
in standard form. By this we mean
@@ -455,15 +458,19 @@
\end{center}
\noindent
- the Turing machine starts with a tape containg @{text n} @{term Oc}s
+ 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 at the first @{term Oc}.
- The intuitive meaning of this definition is to start Turing machine with a
+ 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 tape).
+ 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 them. Given our setup, this is relatively straightforward,
+ if slightly fiddly.
shift and change of a p
@@ -480,8 +487,6 @@
measure for the copying TM, which we however omit.
-
-
halting problem
*}