--- a/ProgTutorial/Solutions.thy	Tue May 14 16:59:53 2019 +0200
+++ b/ProgTutorial/Solutions.thy	Tue May 14 17:10:47 2019 +0200
@@ -2,34 +2,34 @@
 imports First_Steps "Recipes/Timing"
 begin
 
-chapter {* Solutions to Most Exercises\label{ch:solutions} *}
+chapter \<open>Solutions to Most Exercises\label{ch:solutions}\<close>
 
-text {* \solution{fun:revsum} *}
+text \<open>\solution{fun:revsum}\<close>
 
-ML %grayML{*fun rev_sum 
+ML %grayML\<open>fun rev_sum 
   ((p as Const (@{const_name plus}, _)) $ t $ u) = p $ u $ rev_sum t
-  | rev_sum t = t *}
+  | rev_sum t = t\<close>
 
-text {* 
+text \<open>
   An alternative solution using the function @{ML_ind mk_binop in HOLogic} is:
- *}
+\<close>
 
-ML %grayML{*fun rev_sum t =
+ML %grayML\<open>fun rev_sum t =
 let
   fun dest_sum (Const (@{const_name plus}, _) $ u $ u') = u' :: dest_sum u
     | dest_sum u = [u]
 in
   foldl1 (HOLogic.mk_binop @{const_name plus}) (dest_sum t)
-end *}
+end\<close>
 
-text {* \solution{fun:makesum} *}
+text \<open>\solution{fun:makesum}\<close>
 
-ML %grayML{*fun make_sum t1 t2 =
-  HOLogic.mk_nat (HOLogic.dest_nat t1 + HOLogic.dest_nat t2) *}
+ML %grayML\<open>fun make_sum t1 t2 =
+  HOLogic.mk_nat (HOLogic.dest_nat t1 + HOLogic.dest_nat t2)\<close>
 
-text {* \solution{fun:killqnt} *}
+text \<open>\solution{fun:killqnt}\<close>
 
-ML %linenosgray{*val quantifiers = [@{const_name All}, @{const_name Ex}]
+ML %linenosgray\<open>val quantifiers = [@{const_name All}, @{const_name Ex}]
 
 fun kill_trivial_quantifiers trm =
 let
@@ -44,15 +44,15 @@
     | _ => t
 in 
   aux trm
-end*}
+end\<close>
 
-text {*
+text \<open>
   In line 7 we traverse the term, by first checking whether a term is an
   application of a constant with an abstraction. If the constant stands for
   a listed quantifier (see Line 1) and the bound variable does not occur
   as a loose bound variable in the body, then we delete the quantifier. 
   For this we have to increase all other dangling de Bruijn indices by
-  @{text "-1"} to account for the deleted quantifier. An example is 
+  \<open>-1\<close> to account for the deleted quantifier. An example is 
   as follows:
 
   @{ML_response_fake [display,gray]
@@ -61,21 +61,21 @@
   |> pretty_term @{context} 
   |> pwriteln"
   "\<forall>x z. P x = P z"}
-*}
+\<close>
 
 
 
-text {* \solution{fun:makelist} *}
+text \<open>\solution{fun:makelist}\<close>
 
-ML %grayML{*fun mk_rev_upto i = 
+ML %grayML\<open>fun mk_rev_upto i = 
   1 upto i
   |> map (HOLogic.mk_number @{typ int})
   |> HOLogic.mk_list @{typ int}
-  |> curry (op $) @{term "rev :: int list \<Rightarrow> int list"}*}
+  |> curry (op $) @{term "rev :: int list \<Rightarrow> int list"}\<close>
 
-text {* \solution{ex:debruijn} *}
+text \<open>\solution{ex:debruijn}\<close>
 
-ML %grayML{*fun P n = @{term "P::nat \<Rightarrow> bool"} $ (HOLogic.mk_number @{typ "nat"} n) 
+ML %grayML\<open>fun P n = @{term "P::nat \<Rightarrow> bool"} $ (HOLogic.mk_number @{typ "nat"} n) 
 
 fun rhs 1 = P 1
   | rhs n = HOLogic.mk_conj (P n, rhs (n - 1))
@@ -85,11 +85,11 @@
                  (HOLogic.mk_eq (P (m - 1), P m), rhs n), lhs (m - 1) n)
 
 fun de_bruijn n =
-  HOLogic.mk_Trueprop (HOLogic.mk_imp (lhs n n, rhs n))*}
+  HOLogic.mk_Trueprop (HOLogic.mk_imp (lhs n n, rhs n))\<close>
 
-text {* \solution{ex:scancmts} *}
+text \<open>\solution{ex:scancmts}\<close>
 
-ML %grayML{*val any = Scan.one (Symbol.not_eof)
+ML %grayML\<open>val any = Scan.one (Symbol.not_eof)
 
 val scan_cmt =
 let
@@ -103,10 +103,10 @@
 val parser = Scan.repeat (scan_cmt || any)
 
 val scan_all =
-      Scan.finite Symbol.stopper parser >> implode #> fst *}
+      Scan.finite Symbol.stopper parser >> implode #> fst\<close>
 
-text {*
-  By using @{text "#> fst"} in the last line, the function 
+text \<open>
+  By using \<open>#> fst\<close> in the last line, the function 
   @{ML scan_all} retruns a string, instead of the pair a parser would
   normally return. For example:
 
@@ -118,11 +118,11 @@
   (scan_all input1, scan_all input2)
 end"
 "(\"foo bar\", \"foo (**test**) bar (**test**)\")"}
-*}
+\<close>
 
-text {* \solution{ex:contextfree} *}
+text \<open>\solution{ex:contextfree}\<close>
 
-ML %grayML{*datatype expr = 
+ML %grayML\<open>datatype expr = 
    Number of int
  | Mult of expr * expr 
  | Add of expr * expr
@@ -135,12 +135,12 @@
    || parse_basic) xs
 and parse_expr xs =
   (parse_factor --| Parse.$$$ "+" -- parse_expr >> Add 
-   || parse_factor) xs*}
+   || parse_factor) xs\<close>
 
 
-text {* \solution{ex:dyckhoff} *}
+text \<open>\solution{ex:dyckhoff}\<close>
 
-text {* 
+text \<open>
   The axiom rule can be implemented with the function @{ML assume_tac}. The other
   rules correspond to the theorems:
 
@@ -164,7 +164,7 @@
   \end{center}
 
   For the other rules we need to prove the following lemmas.
-*}
+\<close>
 
 lemma impE1:
   shows "\<lbrakk>A \<longrightarrow> B; A; B \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R"
@@ -177,12 +177,12 @@
   and   "\<lbrakk>(C = D) \<longrightarrow> B; (C \<longrightarrow> D) \<longrightarrow> ((D \<longrightarrow> C) \<longrightarrow> B) \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R"
   by iprover+
 
-text {*
+text \<open>
   Now the tactic which applies a single rule can be implemented
   as follows.
-*}
+\<close>
 
-ML %linenosgray{*fun apply_tac ctxt =
+ML %linenosgray\<open>fun apply_tac ctxt =
 let
   val intros = @{thms conjI disjI1 disjI2 impI iffI}
   val elims = @{thms FalseE conjE disjE iffE impE2}
@@ -191,41 +191,41 @@
   ORELSE' resolve_tac ctxt intros
   ORELSE' eresolve_tac ctxt elims
   ORELSE' (eresolve_tac ctxt [@{thm impE1}] THEN' assume_tac ctxt)
-end*}
+end\<close>
 
-text {*
+text \<open>
   In Line 11 we apply the rule @{thm [source] impE1} in concjunction 
   with @{ML assume_tac} in order to reduce the number of possibilities that
   need to be explored. You can use the tactic as follows.
-*}
+\<close>
 
 lemma
   shows "((((P \<longrightarrow> Q) \<longrightarrow> P) \<longrightarrow> P) \<longrightarrow> Q) \<longrightarrow> Q"
-apply(tactic {* (DEPTH_SOLVE o apply_tac @{context}) 1 *})
+apply(tactic \<open>(DEPTH_SOLVE o apply_tac @{context}) 1\<close>)
 done
 
-text {*
+text \<open>
   We can use the tactic to prove or disprove automatically the
   de Bruijn formulae from Exercise \ref{ex:debruijn}.
-*}
+\<close>
 
-ML %grayML{*fun de_bruijn_prove ctxt n =
+ML %grayML\<open>fun de_bruijn_prove ctxt n =
 let 
   val goal = HOLogic.mk_Trueprop (HOLogic.mk_imp (lhs n n, rhs n))
 in
   Goal.prove ctxt ["P"] [] goal
    (fn _ => (DEPTH_SOLVE o apply_tac ctxt) 1)
-end*}
+end\<close>
 
-text {* 
+text \<open>
   You can use this function to prove de Bruijn formulae.
-*}
+\<close>
 
-ML %grayML{*de_bruijn_prove @{context} 3 *}
+ML %grayML\<open>de_bruijn_prove @{context} 3\<close>
 
-text {* \solution{ex:addsimproc} *}
+text \<open>\solution{ex:addsimproc}\<close>
 
-ML %grayML{*fun dest_sum term =
+ML %grayML\<open>fun dest_sum term =
   case term of 
     (@{term "(+):: nat \<Rightarrow> nat \<Rightarrow> nat"} $ t1 $ t2) =>
         (snd (HOLogic.dest_number t1), snd (HOLogic.dest_number t2))
@@ -245,32 +245,32 @@
 in
   SOME (get_sum_thm ctxt t' (dest_sum t'))
   handle TERM _ => NONE
-end*}
+end\<close>
 
-text {* The setup for the simproc is *}
+text \<open>The setup for the simproc is\<close>
 
-simproc_setup %gray add_sp ("t1 + t2") = {* K add_sp_aux *}
+simproc_setup %gray add_sp ("t1 + t2") = \<open>K add_sp_aux\<close>
  
-text {* and a test case is the lemma *}
+text \<open>and a test case is the lemma\<close>
 
 lemma "P (Suc (99 + 1)) ((0 + 0)::nat) (Suc (3 + 3 + 3)) ((4 + 1)::nat)"
-  apply(tactic {* simp_tac (put_simpset HOL_basic_ss @{context} addsimprocs [@{simproc add_sp}]) 1 *})
-txt {* 
+  apply(tactic \<open>simp_tac (put_simpset HOL_basic_ss @{context} addsimprocs [@{simproc add_sp}]) 1\<close>)
+txt \<open>
   where the simproc produces the goal state
   
   \begin{minipage}{\textwidth}
   @{subgoals [display]}
   \end{minipage}\bigskip
-*}(*<*)oops(*>*)
+\<close>(*<*)oops(*>*)
 
-text {* \solution{ex:addconversion} *}
+text \<open>\solution{ex:addconversion}\<close>
 
-text {*
+text \<open>
   The following code assumes the function @{ML dest_sum} from the previous
   exercise.
-*}
+\<close>
 
-ML %grayML{*fun add_simple_conv ctxt ctrm =
+ML %grayML\<open>fun add_simple_conv ctxt ctrm =
 let
   val trm = Thm.term_of ctrm
 in
@@ -282,25 +282,25 @@
 
 val add_conv = Conv.bottom_conv add_simple_conv
 
-fun add_tac ctxt = CONVERSION (add_conv ctxt)*}
+fun add_tac ctxt = CONVERSION (add_conv ctxt)\<close>
 
-text {*
+text \<open>
   A test case for this conversion is as follows
-*}
+\<close>
 
 lemma "P (Suc (99 + 1)) ((0 + 0)::nat) (Suc (3 + 3 + 3)) ((4 + 1)::nat)"
-  apply(tactic {* add_tac @{context} 1 *})?
-txt {* 
+  apply(tactic \<open>add_tac @{context} 1\<close>)?
+txt \<open>
   where it produces the goal state
   
   \begin{minipage}{\textwidth}
   @{subgoals [display]}
   \end{minipage}\bigskip
-*}(*<*)oops(*>*)
+\<close>(*<*)oops(*>*)
 
-text {* \solution{ex:compare} *}
+text \<open>\solution{ex:compare}\<close>
 
-text {* 
+text \<open>
   We use the timing function @{ML timing_wrapper} from Recipe~\ref{rec:timing}.
   To measure any difference between the simproc and conversion, we will create 
   mechanically terms involving additions and then set up a goal to be 
@@ -312,9 +312,9 @@
   For constructing test cases, we first define a function that returns a 
   complete binary tree whose leaves are numbers and the nodes are 
   additions.
-*}
+\<close>
 
-ML %grayML{*fun term_tree n =
+ML %grayML\<open>fun term_tree n =
 let
   val count = Unsynchronized.ref 0; 
 
@@ -325,47 +325,47 @@
              $ (term_tree_aux (n - 1)) $ (term_tree_aux (n - 1))
 in
   term_tree_aux n
-end*}
+end\<close>
 
-text {*
+text \<open>
   This function generates for example:
 
   @{ML_response_fake [display,gray] 
   "pwriteln (pretty_term @{context} (term_tree 2))" 
   "(1 + 2) + (3 + 4)"} 
 
-  The next function constructs a goal of the form @{text "P \<dots>"} with a term 
+  The next function constructs a goal of the form \<open>P \<dots>\<close> with a term 
   produced by @{ML term_tree} filled in.
-*}
+\<close>
 
-ML %grayML{*fun goal n = HOLogic.mk_Trueprop (@{term "P::nat\<Rightarrow> bool"} $ (term_tree n))*}
+ML %grayML\<open>fun goal n = HOLogic.mk_Trueprop (@{term "P::nat\<Rightarrow> bool"} $ (term_tree n))\<close>
 
-text {*
+text \<open>
   Note that the goal needs to be wrapped in a @{term "Trueprop"}. Next we define
-  two tactics, @{text "c_tac"} and @{text "s_tac"}, for the conversion and simproc,
+  two tactics, \<open>c_tac\<close> and \<open>s_tac\<close>, for the conversion and simproc,
   respectively. The idea is to first apply the conversion (respectively simproc) and 
   then prove the remaining goal using @{ML "cheat_tac" in Skip_Proof}.
-*}
+\<close>
 
 ML Skip_Proof.cheat_tac
 
-ML %grayML{*local
+ML %grayML\<open>local
   fun mk_tac ctxt tac = 
         timing_wrapper (EVERY1 [tac, Skip_Proof.cheat_tac ctxt])
 in
   fun c_tac ctxt = mk_tac ctxt (add_tac ctxt) 
   fun s_tac ctxt = mk_tac ctxt (simp_tac 
     (put_simpset HOL_basic_ss ctxt addsimprocs [@{simproc add_sp}]))
-end*}
+end\<close>
 
-text {*
+text \<open>
   This is all we need to let the conversion run against the simproc:
-*}
+\<close>
 
-ML %grayML{*val _ = Goal.prove @{context} [] [] (goal 8) (fn {context, ...} => c_tac context)
-val _ = Goal.prove @{context} [] [] (goal 8) (fn {context, ...} => s_tac context)*}
+ML %grayML\<open>val _ = Goal.prove @{context} [] [] (goal 8) (fn {context, ...} => c_tac context)
+val _ = Goal.prove @{context} [] [] (goal 8) (fn {context, ...} => s_tac context)\<close>
 
-text {*
+text \<open>
   If you do the exercise, you can see that both ways of simplifying additions
   perform relatively similar with perhaps some advantages for the
   simproc. That means the simplifier, even if much more complicated than
@@ -374,6 +374,6 @@
   conversions. Conversions only have clear advantages in special situations:
   for example if you need to have control over innermost or outermost
   rewriting, or when rewriting rules are lead to non-termination.
-*}
+\<close>
 
 end