revise the computability theory chapter

Author: rzach

content/computability/computability-theory/ce-closed-cup-cap.tex

@@ -50,7 +50,7 @@
 \end{cases}
 \]
 Then $k$ enumerates $A \cup B$; the idea is that $k$ just alternates
-between the enumerations offered by $f$ and $g$. Enumerating $A \cap
+between the enumerations offered by $f$ and~$g$. Enumerating $A \cap
 B$ is tricker. If $A \cap B$ is empty, it is trivially computably
 enumerable. Otherwise, let $c$ be any element of $A \cap B$, and
 define $l$ by

content/computability/computability-theory/ce-sets.tex

@@ -28,11 +28,13 @@
 \[
 S = \{ f(0), f(1), f(2), \dots \},
 \]
-and so $f$ can be seen as ``enumerating'' the elements of $S$. Note
-that according to the definition, $f$ need not be an increasing
+and so $f$ can be seen as ``enumerating'' the elements of~$S$. Note
+that according to the definition, $f$~need not be an increasing
 function, i.e., the enumeration need not be in increasing order. In
-fact, $f$ need not even be injective, so that the constant function
-$f(x) = 0$ enumerates the set $\{ 0 \}$.
+fact, $f$ need not even be injective, i.e., repetitions in the
+enumeration $f(0)$, $f(1)$, $f(2)$, \dots{} of~$S$ are allowed. For
+instance, the constant function $f(x) = 0$ enumerates the set $\{ 0
+\}$.
 \end{explain}
 
 Any computable set is computably enumerable. To see this, suppose

content/computability/computability-theory/coding-computations.tex

@@ -22,7 +22,8 @@
   step of computation.
 \item Test whether a putative record of a computation is in fact the
   record of how a computable function with a given definition would be
-  computed for a given input.
+  computed for a given input (on which the function is
+  defined, i.e., the computation halts).
 \item Extract from such a description of the complete record of a
   computation the value of the function for a given input. For
   instance, the contents of the tape in the very last step of a
@@ -33,8 +34,8 @@
 computable function a numerical \emph{index} in such a way that the
 instructions can be recovered from the index in a computable way.
 Similarly, the complete record of a computation can be coded by a
-single number as well. The resulting arithmetical relation ``$s$
-codes the record of computation of the function with index~$e$ for
+single number as well. The resulting arithmetical relation ``$s$~codes
+the record of computation of the function with index~$e$ for
 input~$x$'' and the function ``output of computation sequence with
 code~$s$'' are then computable; in fact, they are primitive recursive.
 

content/computability/computability-theory/complement-ce.tex

@@ -9,8 +9,8 @@
 \olfileid{cmp}{thy}{cmp}
 \olsection{Computably Enumerable Sets not Closed under Complement}
 
-Suppose $A$ is computably enumerable. Is the complement of $A$,
-$\Complement{A} = \Nat \setminus A$, necessarily computably enumerable
+Suppose $A$ is computably enumerable. Is the complement of~$A$,
+$\Complement{A} = \Nat \setminus A$, always computably enumerable
 as well? The following theorem and corollary show that the answer is
 ``no.''
 
@@ -25,19 +25,19 @@
 $\Complement{A}$ is computable as well ($\Char{A} = 1 \tsub
 \Char{\Complement{A}}$), and so both are computably enumerable.
 
-In the other direction, suppose $A$ and $\Complement{A}$ are both
-computably enumerable. Let $A$ be the domain of $\cfind{d}$, and let
-$\Complement{A}$ be the domain of $\cfind{e}$. Define $h$ by
+In the other direction, suppose $A$ and~$\Complement{A}$ are both
+computably enumerable. Let $A$ be the domain of~$\cfind{d}$, and let
+$\Complement{A}$ be the domain of~$\cfind{e}$. Define $h$ by
 \[
 h(x) = \umin{s}{(T(d,x,s) \lor T(e,x,s))}.
 \]
-In other words, on input $x$, $h$ searches for either a halting
-computation of~$\cfind{d}$ or a halting computation
-of~$\cfind{e}$. Now, if $x \in A$, it will succeed in the first case,
-and if $x \in \Complement{A}$, it will succeed in the second case. So,
-$h$~is a total computable function. But now we have that for every
-$x$, $x \in A$ if and only if $T(e, x, h(x))$, i.e., if $\cfind{e}$ is
-the one that is defined. Since $T(e, x, h(x))$ is a computable relation,
+In other words, on input~$x$, $h$~searches for either a halting
+computation of~$\cfind{d}$ or a halting computation of~$\cfind{e}$.
+Now, if $x \in A$, it will succeed in the first case, and if $x \in
+\Complement{A}$, it will succeed in the second case. So, $h$~is a
+total computable function. But now we have that for every~$x$, $x \in
+A$ if and only if $T(e, x, h(x))$, i.e., if $\cfind{e}$ is the one
+that is defined. Since $T(e, x, h(x))$ is a computable relation,
 $A$~is computable.
 \end{proof}
 
@@ -56,7 +56,7 @@
 \begin{proof}
 We know that $K_0$ is computably enumerable, but not computable. If
 $\Complement{K_0}$ were computably enumerable, then $K_0$ would be
-computable by \olref{thm:ce-comp}.
+computable by \olref{thm:ce-comp}, contradicting \olref[ncp]{thm:K-0}.
 \end{proof}
 
 \end{document}

content/computability/computability-theory/complete-ce-sets.tex

@@ -10,7 +10,7 @@
 \olsection{Complete Computably Enumerable Sets}
 
 \begin{defn}
-A set $A$ is a \emph{complete computably enumerable set}
+A set $A$ is a \emph{complete !!{computably enumerable} set}
 (under many-one reducibility) if
 \begin{enumerate}
 \item $A$ is computably enumerable, and
@@ -19,7 +19,7 @@
 \end{defn}
 
 In other words, complete computably enumerable sets are the
-``hardest'' computably enumerable sets possible; they allow one to
+``hardest'' computably enumerable sets possible. They allow one to
 answer questions about \emph{any} computably enumerable set.
 
 \begin{thm}
@@ -34,12 +34,12 @@
 \]
 Let $f$ be the function $f(x) = \tuple{e, x}$. Then for every natural
 number $x$, $x \in B$ if and only if $f(x) \in K_0$. In other words, $f$
-reduces $B$ to $K_0$.
+reduces $B$ to~$K_0$.
 
 To see that $K_1$ is complete, note that in the proof of
 \olref[k1]{prop:k1} we reduced $K_0$ to it. So, by
 \olref[ppr]{prop:trans-red}, any computably enumerable set can be
-reduced to $K_1$ as well.
+reduced to~$K_1$ as well.
 
 $K$ can be reduced to $K_0$ in much the same way.
 \end{proof}

content/computability/computability-theory/computability-theory.tex

@@ -34,6 +34,8 @@
 
 \olimport{equiv-ce-defs}
 
+\olimport{non-comp-set}
+
 \olimport{ce-closed-cup-cap}
 
 \olimport{complement-ce}
@@ -56,10 +58,6 @@
 
 \olimport{def-functions-self-reference}
 
-\iftag{lambda}{
-  \olimport{minimization-lambda}
-}{}
-
 \OLEndChapterHook
 
 \end{document}

content/computability/computability-theory/computable-sets.tex

@@ -14,22 +14,22 @@
 
 \begin{defn}
   Let $S$ be a set of natural numbers. Then $S$ is \emph{computable}
-  iff its characteristic function is. In other words, $S$ is
-  computable iff the function
+  iff its characteristic function~$\Char{S}$ is. In other words,
+  $S$~is computable iff the function
 \[
 \Char{S}(x) =
 \begin{cases}
 1 & \text{if $x \in S$} \\
 0 & \text{otherwise}
 \end{cases}
 \]
-is computable. Similarly, a relation $\Atom{R}{x_0, \dots, x_{k-1}}$ is
+is computable. Similarly, a relation $R(x_0, \dots, x_{k-1})$ is
 computable if and only if its characteristic function is.
+
+Computable sets and relations are also called \emph{decidable}.
 \end{defn}
 
 \begin{explain}
-Computable sets are also called \emph{decidable}.
-
 Notice that we now have a number of notions of computability: for
 partial functions, for functions, and for sets. Do not get them
 confused!{} \iftag{TMs}{The Turing machine computing a partial function

content/computability/computability-theory/def-functions-self-reference.tex

@@ -36,7 +36,7 @@
 \[
 \fn{gcd}(u,v) \simeq
 \begin{cases}
-v & \text{if $0 = 0$} \\
+v & \text{if $u = 0$} \\
 \fn{gcd}(\fn{mod}(v, u), u) & \text{otherwise}
 \end{cases}
 \]
@@ -50,7 +50,7 @@
 fancier than the examples just discussed. Most programming languages
 support definitions of functions in terms of themselves, one way or
 another. Note that this is a little bit less dramatic than being able
-to define a function in terms of an {\em index} for an algorithm
+to define a function in terms of an \emph{index} for an algorithm
 computing the functions, which is what, in full generality, the
 fixed-point theorem lets you do.
 

content/computability/computability-theory/equiv-ce-defs.tex

@@ -20,10 +20,10 @@
 Let $S$ be a set of natural numbers. Then the following are
 equivalent:
 \begin{enumerate}
-\item $S$ is computably enumerable.
-\item $S$ is the range of a \emph{partial} computable function.
-\item $S$ is empty or the range of a primitive recursive function.
-\item $S$ is the \emph{domain} of a partial computable function.
+\item\ollabel{case:ce} $S$ is computably enumerable.
+\item\ollabel{case:ran-pc} $S$ is the range of a \emph{partial} computable function.
+\item\ollabel{case:ran-prim} $S$ is empty or the range of a primitive recursive function.
+\item\ollabel{case:ce-domain} $S$ is the \emph{domain} of a partial computable function.
 \end{enumerate}
 \end{thm}
 
@@ -32,7 +32,7 @@
 computably enumerable set to be enumerated by either a computable
 function, a partial computable function, or a primitive recursive
 function. The fourth clause tells us that if $S$ is computably
-enumerable, then for some index $e$,
+enumerable, then for some index~$e$,
 \[
 S = \Setabs{x}{\cfind{e}(x) \fdefined}.
 \]
@@ -44,66 +44,69 @@
 
 \begin{proof}
 Since every primitive recursive function is computable and every
-computable function is partial computable, (3) implies (1) and (1)
-implies~(2). (Note that if $S$ is empty, $S$ is the range of the
-partial computable function that is nowhere defined.) If we show that
-(2) implies (3), we will have shown the first three clauses
-equivalent.
+computable function is partial computable, \olref{case:ran-prim}
+implies \olref{case:ce} and \olref{case:ce}
+implies~\olref{case:ran-pc}. (Note that if $S$ is empty, $S$~is the
+range of the partial computable function that is nowhere defined.) If
+we show that \olref{case:ran-pc} implies \olref{case:ran-prim}, we
+will have shown the first three clauses equivalent.
 
 So, suppose $S$ is the range of the partial computable function
 $\cfind{e}$. If $S$ is empty, we are done. Otherwise, let $a$ be any
 element of~$S$. By Kleene's normal form theorem, we can write
 \[
-\cfind{e}(x) = U (\umin{s}{T(e, x, s)}).
+\cfind{e}(x) = U(\umin{s}{T(e, x, s)}).
 \]
 In particular, $\cfind{e}(x) \fdefined$ and $= y$ if and only if there
 is an $s$ such that $T(e, x, s)$ and $U(s) = y$. Define $f(z)$ by
 \[
 f(z) = \begin{cases}
-U((z)_1) & \text{if $T(e, (z)_0, (z)_1)$} \\
-a        & \text{otherwise.}
+  U((z)_1) & \text{if $T(e, (z)_0, (z)_1)$} \\
+  a        & \text{otherwise.}
 \end{cases}
 \]
 Then $f$ is primitive recursive, because $T$ and $U$
 are. \iftag{TMs}{Expressed in terms of Turing machines, if $z$ codes a
   pair $\tuple{(z)_0, (z)_1}$ such that $(z)_1$ is a halting
-  computation of machine $e$ on input $(z)_0$, then $f$ returns the
-  output of the computation; otherwise, it returns $a$.}
+  computation of machine~$M_e$ on input $(z)_0$, then $f$ returns the
+  output of the computation; otherwise, it returns~$a$.}
 
-We need to show that $S$ is the range of $f$, i.e., for any natural
-number $y$, $y \in S$ if and only if it is in the range of~$f$. In the
+We need to show that $S$ is the range of~$f$, i.e., for any natural
+number~$y$, $y \in S$ if and only if it is in the range of~$f$. In the
 forwards direction, suppose $y \in S$. Then $y$ is in the range of
-$\cfind{e}$, so for some $x$ and $s$, $T(e,x,s)$ and $U(s) = y$; but then
-$y = f(\tuple{x,s})$. Conversely, suppose $y$ is in the range of
-$f$. Then either $y = a$, or for some $z$, $T(e,(z)_0,(z)_1)$ and
-$U((z)_1) = y$. Since, in the latter case, $\cfind{e}(x) \fdefined = y$,
-either way, $y$ is in $S$.
+$\cfind{e}$, so for some $x$ and~$s$, $T(e,x,s)$ holds and $U(s) = y$.
+But then $y = f(\tuple{x,s})$. Conversely, suppose $y$ is in the range
+of~$f$. Then either $y = a$, or for some~$z$, $T(e,(z)_0,(z)_1)$ and
+$U((z)_1) = y$. Since, in the latter case, $\cfind{e}(x) \fdefined =
+y$, either way, $y$ is in~$S$.
 
 (The notation $\cfind{e}(x) \fdefined = y$ means ``$\cfind{e}(x)$ is
 defined and equal to $y$.'' We could just as well use $\cfind{e}(x) =
 y$, but the extra arrow is sometimes helpful in reminding us that we
 are dealing with a partial function.)
 
 To finish up the proof of \olref{thm:ce-equiv}, it suffices to show
-that (1) and (4) are equivalent. First, let us show that (1) implies
-(4). Suppose $S$ is the range of a computable function~$f$, i.e.,
+that \olref{case:ce} and~\olref{case:ce-domain} are equivalent. First,
+let us show that \olref{case:ce} implies~\olref{case:ce-domain}.
+Suppose $S$ is the range of a computable function~$f$, i.e.,
 \[
-S = \Setabs{y}{\text{for some $x$,} f(x) = y}.
+S = \Setabs{y}{\text{for some $x$, } f(x) = y}.
 \]
 Let
 \[
-g(y) = \umin{x}{f(x) = y}.
+g(y) = \umin{x}{(f(x) = y)}.
 \]
 Then $g$ is a partial computable function, and $g(y)$ is defined if
-and only if for some $x$, $f(x) = y$. In other words, the domain of
+and only if for some~$x$, $f(x) = y$. In other words, the domain of
 $g$~is the range of~$f$. \iftag{TMs}{Expressed in terms of Turing
-  machines: given a Turing machine $F$ that enumerates the elements of
-  $S$, let $G$ be the Turing machine that semi-decides $S$ by
-  searching through the outputs of~$F$ to see if a given element is in
-  the set.}{}
-
-Finally, to show (4) implies (1), suppose that $S$~is the domain of the
-partial computable function~$\cfind{e}$, i.e.,
+machines: given a Turing machine~$F$ that enumerates the elements
+of~$S$, let $G$ be the Turing machine that semi-decides $S$ by
+searching through the outputs of~$F$ to see if a given element is in
+the set, halts if it is and keeps searching forever if it isn't.}{}
+
+Finally, to show \olref{case:ce-domain} implies~\olref{case:ce},
+suppose that $S$~is the domain of the partial computable
+function~$\cfind{e}$, i.e.,
 \[
 S = \Setabs{x}{\cfind{e}(x) \fdefined}.
 \]
@@ -123,34 +126,37 @@
 to halting computations.}{}
 \end{proof}
 
-The fourth clause of \olref{thm:ce-equiv} provides us with a
-convenient way of enumerating the computably enumerable sets: for each
-$e$, let $W_e$ denote the domain of $\cfind{e}$. Then if $A$ is any
-computably enumerable set, $A = W_e$, for some $e$.
+Clause~\olref{case:ce-domain} of \olref{thm:ce-equiv} provides us with
+a convenient way of enumerating the computably enumerable sets: for
+each~$e$, let $W_e$ denote the domain of $\cfind{e}$, i.e.,
+\[
+W_e = \Setabs{x}{\cfind{e}(x) \fdefined}.
+\] 
+Then if $A$ is any computably enumerable set, $A = W_e$, for some~$e$.
 
 The following provides yet another characterization of the computably
 enumerable sets.
 
 \begin{thm}
 \ollabel{thm:exists-char}
 A set $S$ is computably enumerable if and only if there is a
-computable relation $\Atom{R}{x,y}$ such that
+computable relation $R(x,y)$ such that
 \[
-S = \Setabs{ x }{ \lexists[y][\Atom{R}{x,y}] }.
+S = \Setabs{ x }{ \lexists[y][R(x,y)] }.
 \]
 \end{thm}
 
 \begin{proof}
 In the forward direction, suppose $S$ is computably
-enumerable. Then for some $e$, $S = W_e$. For this value of $e$
+enumerable. Then for some $e$, $S = W_e$. For this value of~$e$
 we can write $S$ as
 \[
 S = \Setabs{ x }{ \lexists[y][T(e, x, y)] }.
 \]
 In the reverse direction, suppose $S = \Setabs{ x }{
-  \lexists[y][\Atom{R}{x, y}] }$. Define $f$~by
+  \lexists[y][R(x, y)] }$. Define $f$~by
 \[
-f(x) \simeq \umin{y}{Atom{R}{x,y}}.
+f(x) \simeq \umin{y}{R(x, y)}.
 \]
 Then $f$ is partial computable, and $S$ is the domain of~$f$.
 \end{proof}

content/computability/computability-theory/fixed-point-thm.tex

@@ -184,7 +184,8 @@
 string $x$ repeatedly, $y$ times. Then the ``program''
 \begin{quote}
 \begin{verbatim}
-getinput(y); print(diag('getinput(y); print(diag(x), y)'), y)
+getinput(y);
+print(diag('getinput(y); print(diag(x), y)'), y)
 \end{verbatim}
 \end{quote}
 prints itself out $y$ times, on input $y$. Replacing the