diff -r a5f86aed785b -r a977245dfc8a doc-src/TutorialI/ToyList/document/ToyList.tex
--- a/doc-src/TutorialI/ToyList/document/ToyList.tex	Tue Aug 29 15:43:29 2000 +0200
+++ b/doc-src/TutorialI/ToyList/document/ToyList.tex	Tue Aug 29 16:05:13 2000 +0200
@@ -132,24 +132,24 @@
 The name and the simplification attribute are optional.
 \end{itemize}
 Isabelle's response is to print
-\begin{isabellepar}%
+\begin{isabelle}
 proof(prove):~step~0\isanewline
 \isanewline
 goal~(theorem~rev\_rev):\isanewline
 rev~(rev~xs)~=~xs\isanewline
 ~1.~rev~(rev~xs)~=~xs
-\end{isabellepar}%
+\end{isabelle}
 The first three lines tell us that we are 0 steps into the proof of
 theorem \isa{rev_rev}; for compactness reasons we rarely show these
 initial lines in this tutorial. The remaining lines display the current
 proof state.
 Until we have finished a proof, the proof state always looks like this:
-\begin{isabellepar}%
+\begin{isabelle}
 $G$\isanewline
 ~1.~$G\sb{1}$\isanewline
 ~~\vdots~~\isanewline
 ~$n$.~$G\sb{n}$
-\end{isabellepar}%
+\end{isabelle}
 where $G$
 is the overall goal that we are trying to prove, and the numbered lines
 contain the subgoals $G\sb{1}$, \dots, $G\sb{n}$ that we need to prove to
@@ -169,15 +169,15 @@
 By default, induction acts on the first subgoal. The new proof state contains
 two subgoals, namely the base case (\isa{Nil}) and the induction step
 (\isa{Cons}):
-\begin{isabellepar}%
+\begin{isabelle}
 ~1.~rev~(rev~[])~=~[]\isanewline
 ~2.~{\isasymAnd}a~list.~rev(rev~list)~=~list~{\isasymLongrightarrow}~rev(rev(a~\#~list))~=~a~\#~list%
-\end{isabellepar}%
+\end{isabelle}
 
 The induction step is an example of the general format of a subgoal:
-\begin{isabellepar}%
+\begin{isabelle}
 ~$i$.~{\indexboldpos{\isasymAnd}{$IsaAnd}}$x\sb{1}$~\dots~$x\sb{n}$.~{\it assumptions}~{\isasymLongrightarrow}~{\it conclusion}
-\end{isabellepar}%
+\end{isabelle}
 The prefix of bound variables \isasymAnd$x\sb{1}$~\dots~$x\sb{n}$ can be
 ignored most of the time, or simply treated as a list of variables local to
 this subgoal. Their deeper significance is explained in \S\ref{sec:PCproofs}.
@@ -200,15 +200,15 @@
 ``simplify'' the subgoals.  In our case, subgoal~1 is solved completely (thanks
 to the equation \isa{rev [] = []}) and disappears; the simplified version
 of subgoal~2 becomes the new subgoal~1:
-\begin{isabellepar}%
+\begin{isabelle}
 ~1.~\dots~rev(rev~list)~=~list~{\isasymLongrightarrow}~rev(rev~list~@~a~\#~[])~=~a~\#~list
-\end{isabellepar}%
+\end{isabelle}
 In order to simplify this subgoal further, a lemma suggests itself.%
 \end{isamarkuptxt}%
 %
+\isamarkupsubsubsection{First lemma: \texttt{rev(xs \at~ys) = (rev ys) \at~(rev xs)}}
+%
 \begin{isamarkuptext}%
-\subsubsection*{First lemma: \texttt{rev(xs \at~ys) = (rev ys) \at~(rev xs)}}
-
 After abandoning the above proof attempt\indexbold{abandon
 proof}\indexbold{proof!abandon} (at the shell level type
 \isacommand{oops}\indexbold{*oops}) we start a new proof:%
@@ -232,18 +232,18 @@
 \end{isamarkuptxt}%
 \isacommand{apply}{\isacharparenleft}auto{\isacharparenright}%
 \begin{isamarkuptxt}%
-\begin{isabellepar}%
+\begin{isabelle}
 ~1.~rev~ys~=~rev~ys~@~[]\isanewline
 ~2. \dots
-\end{isabellepar}%
+\end{isabelle}
 Again, we need to abandon this proof attempt and prove another simple lemma first.
 In the future the step of abandoning an incomplete proof before embarking on
 the proof of a lemma usually remains implicit.%
 \end{isamarkuptxt}%
 %
+\isamarkupsubsubsection{Second lemma: \texttt{xs \at~[] = xs}}
+%
 \begin{isamarkuptext}%
-\subsubsection*{Second lemma: \texttt{xs \at~[] = xs}}
-
 This time the canonical proof procedure%
 \end{isamarkuptext}%
 \isacommand{lemma}\ app{\isacharunderscore}Nil\isadigit{2}\ {\isacharbrackleft}simp{\isacharbrackright}{\isacharcolon}\ {\isachardoublequote}xs\ {\isacharat}\ {\isacharbrackleft}{\isacharbrackright}\ {\isacharequal}\ xs{\isachardoublequote}\isanewline
@@ -252,10 +252,10 @@
 \begin{isamarkuptxt}%
 \noindent
 leads to the desired message \isa{No subgoals!}:
-\begin{isabellepar}%
+\begin{isabelle}
 xs~@~[]~=~xs\isanewline
 No~subgoals!
-\end{isabellepar}%
+\end{isabelle}
 
 We still need to confirm that the proof is now finished:%
 \end{isamarkuptxt}%
@@ -279,34 +279,31 @@
 \noindent
 we find that this time \isa{auto} solves the base case, but the
 induction step merely simplifies to
-\begin{isabellepar}
+\begin{isabelle}
 ~1.~{\isasymAnd}a~list.\isanewline
 ~~~~~~~rev~(list~@~ys)~=~rev~ys~@~rev~list~{\isasymLongrightarrow}\isanewline
 ~~~~~~~(rev~ys~@~rev~list)~@~a~\#~[]~=~rev~ys~@~rev~list~@~a~\#~[]
-\end{isabellepar}%
+\end{isabelle}
 Now we need to remember that \isa{\at} associates to the right, and that
 \isa{\#} and \isa{\at} have the same priority (namely the \isa{65}
 in their \isacommand{infixr} annotation). Thus the conclusion really is
-\begin{isabellepar}%
+\begin{isabelle}
 ~~~~~(rev~ys~@~rev~list)~@~(a~\#~[])~=~rev~ys~@~(rev~list~@~(a~\#~[]))%
-\end{isabellepar}%
-and the missing lemma is associativity of \isa{\at}.
-
-\subsubsection*{Third lemma: \texttt{(xs \at~ys) \at~zs = xs \at~(ys \at~zs)}}
-
-Abandoning the previous proof, the canonical proof procedure%
+\end{isabelle}
+and the missing lemma is associativity of \isa{\at}.%
 \end{isamarkuptxt}%
 %
-\begin{comment}
-\isacommand{oops}%
-\end{comment}
+\isamarkupsubsubsection{Third lemma: \texttt{(xs \at~ys) \at~zs = xs \at~(ys \at~zs)}}
+%
+\begin{isamarkuptext}%
+Abandoning the previous proof, the canonical proof procedure%
+\end{isamarkuptext}%
 \isacommand{lemma}\ app{\isacharunderscore}assoc\ {\isacharbrackleft}simp{\isacharbrackright}{\isacharcolon}\ {\isachardoublequote}{\isacharparenleft}xs\ {\isacharat}\ ys{\isacharparenright}\ {\isacharat}\ zs\ {\isacharequal}\ xs\ {\isacharat}\ {\isacharparenleft}ys\ {\isacharat}\ zs{\isacharparenright}{\isachardoublequote}\isanewline
 \isacommand{apply}{\isacharparenleft}induct{\isacharunderscore}tac\ xs{\isacharparenright}\isanewline
 \isacommand{by}{\isacharparenleft}auto{\isacharparenright}%
 \begin{isamarkuptext}%
 \noindent
 succeeds without further ado.
-
 Now we can go back and prove the first lemma%
 \end{isamarkuptext}%
 \isacommand{lemma}\ rev{\isacharunderscore}app\ {\isacharbrackleft}simp{\isacharbrackright}{\isacharcolon}\ {\isachardoublequote}rev{\isacharparenleft}xs\ {\isacharat}\ ys{\isacharparenright}\ {\isacharequal}\ {\isacharparenleft}rev\ ys{\isacharparenright}\ {\isacharat}\ {\isacharparenleft}rev\ xs{\isacharparenright}{\isachardoublequote}\isanewline