theory Solutions

imports First_Steps "Recipes/Timing"

begin

chapter {* Solutions to Most Exercises\label{ch:solutions} *}

text {* \solution{fun:revsum} *}

ML %grayML{*fun rev_sum 

  ((p as Const (@{const_name plus}, _)) $ t $ u) = p $ u $ rev_sum t

  | rev_sum t = t *}

text {* 

  An alternative solution using the function @{ML_ind mk_binop in HOLogic} is:

 *}

ML %grayML{*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 *}

text {* \solution{fun:makesum} *}

ML %grayML{*fun make_sum t1 t2 =

  HOLogic.mk_nat (HOLogic.dest_nat t1 + HOLogic.dest_nat t2) *}

text {* \solution{fun:killqnt} *}

ML %linenosgray{*val quantifiers = [@{const_name All}, @{const_name Ex}]

fun kill_trivial_quantifiers trm =

let

  fun aux t =

    case t of

      Const (s1, T1) $ Abs (x, T2, t2) =>

        if member (op =) quantifiers s1 andalso not (loose_bvar1 (t2, 0)) 

        then incr_boundvars ~1 (aux t2)

        else Const (s1, T1) $ Abs (x, T2, aux t2)

    | t1 $ t2 => aux t1 $ aux t2

    | Abs (s, T, t') => Abs (s, T, aux t')

    | _ => t

in 

  aux trm

end*}

text {*

  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 

  as follows:

  @{ML_response_fake [display,gray]

  "@{prop \"\<forall>x y z. P x = P z\"}

  |> kill_trivial_quantifiers

  |> pretty_term @{context} 

  |> pwriteln"

  "\<forall>x z. P x = P z"}

*}

text {* \solution{fun:makelist} *}

ML %grayML{*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"}*}

text {* \solution{ex:debruijn} *}

ML %grayML{*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))

fun lhs 1 n = HOLogic.mk_imp (HOLogic.mk_eq (P 1, P n), rhs n)

  | lhs m n = HOLogic.mk_conj (HOLogic.mk_imp 

                 (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))*}

text {* \solution{ex:scancmts} *}

ML %grayML{*val any = Scan.one (Symbol.not_eof)

val scan_cmt =

let

  val begin_cmt = Scan.this_string "(*" 

  val end_cmt = Scan.this_string "*)"

in

  begin_cmt |-- Scan.repeat (Scan.unless end_cmt any) --| end_cmt 

  >> (enclose "(**" "**)" o implode)

end

val parser = Scan.repeat (scan_cmt || any)

val scan_all =

      Scan.finite Symbol.stopper parser >> implode #> fst *}

text {*

  By using @{text "#> fst"} in the last line, the function 

  @{ML scan_all} retruns a string, instead of the pair a parser would

  normally return. For example:

  @{ML_response [display,gray]

"let

  val input1 = (Symbol.explode \"foo bar\")

  val input2 = (Symbol.explode \"foo (*test*) bar (*test*)\")

in

  (scan_all input1, scan_all input2)

end"

"(\"foo bar\", \"foo (**test**) bar (**test**)\")"}

*}

text {* \solution{ex:contextfree} *}

ML %grayML{*datatype expr = 

   Number of int

 | Mult of expr * expr 

 | Add of expr * expr

fun parse_basic xs =

  (Parse.nat >> Number 

   || Parse.$$$ "(" |-- parse_expr --| Parse.$$$ ")") xs

and parse_factor xs =

  (parse_basic --| Parse.$$$ "*" -- parse_factor >> Mult 

   || parse_basic) xs

and parse_expr xs =

  (parse_factor --| Parse.$$$ "+" -- parse_expr >> Add 

   || parse_factor) xs*}

text {* \solution{ex:dyckhoff} *}

text {* 

  rules correspond to the theorems:

  \begin{center}

  \begin{tabular}{cc}

  \begin{tabular}{rl}

  $\wedge_R$ & @{thm [source] conjI}\\

  $\vee_{R_1}$ & @{thm [source] disjI1}\\

  $\vee_{R_2}$ & @{thm [source] disjI2}\\

  $\longrightarrow_R$ & @{thm [source] impI}\\

  $=_R$ & @{thm [source] iffI}\\

  \end{tabular}

  &

  \begin{tabular}{rl}

  $False$ & @{thm [source] FalseE}\\

  $\wedge_L$ & @{thm [source] conjE}\\

  $\vee_L$ & @{thm [source] disjE}\\

  $=_L$ & @{thm [source] iffE}

  \end{tabular}

  \end{tabular}

  \end{center}

  For the other rules we need to prove the following lemmas.

*}

lemma impE1:

  shows "\<lbrakk>A \<longrightarrow> B; A; B \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R"

  by iprover

lemma impE2:

  shows "\<lbrakk>(C \<and> D) \<longrightarrow> B; C \<longrightarrow> (D \<longrightarrow>B) \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R"

  and   "\<lbrakk>(C \<or> D) \<longrightarrow> B; \<lbrakk>C \<longrightarrow> B; D \<longrightarrow> B\<rbrakk> \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R"

  and   "\<lbrakk>(C \<longrightarrow> D) \<longrightarrow> B; D \<longrightarrow> B \<Longrightarrow> C \<longrightarrow> D; B \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R"

  and   "\<lbrakk>(C = D) \<longrightarrow> B; (C \<longrightarrow> D) \<longrightarrow> ((D \<longrightarrow> C) \<longrightarrow> B) \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R"

  by iprover+

text {*

  Now the tactic which applies a single rule can be implemented

  as follows.

*}

let

  val intros = @{thms conjI disjI1 disjI2 impI iffI}

  val elims = @{thms FalseE conjE disjE iffE impE2}

in

end*}

text {*

  In Line 11 we apply the rule @{thm [source] impE1} in concjunction 

  need to be explored. You can use the tactic as follows.

*}

lemma

  shows "((((P \<longrightarrow> Q) \<longrightarrow> P) \<longrightarrow> P) \<longrightarrow> Q) \<longrightarrow> Q"

done

text {*

  We can use the tactic to prove or disprove automatically the

  de Bruijn formulae from Exercise \ref{ex:debruijn}.

*}

ML %grayML{*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

end*}

text {* 

  You can use this function to prove de Bruijn formulae.

*}

ML %grayML{*de_bruijn_prove @{context} 3 *}

text {* \solution{ex:addsimproc} *}

ML %grayML{*fun dest_sum term =

  case term of 

        (snd (HOLogic.dest_number t1), snd (HOLogic.dest_number t2))

  | _ => raise TERM ("dest_sum", [term])

fun get_sum_thm ctxt t (n1, n2) =  

let 

  val sum = HOLogic.mk_number @{typ "nat"} (n1 + n2)

  val goal = Logic.mk_equals (t, sum)

in

  Goal.prove ctxt [] [] goal (K (Arith_Data.arith_tac ctxt 1))

end

fun add_sp_aux ctxt t =

let 

in

  SOME (get_sum_thm ctxt t' (dest_sum t'))

  handle TERM _ => NONE

end*}

text {* The setup for the simproc is *}

simproc_setup %gray add_sp ("t1 + t2") = {* K add_sp_aux *}

text {* and a test case is the lemma *}

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 {* 

  where the simproc produces the goal state

  \begin{minipage}{\textwidth}

  @{subgoals [display]}

  \end{minipage}\bigskip

*}(*<*)oops(*>*)

text {* \solution{ex:addconversion} *}

text {*

  The following code assumes the function @{ML dest_sum} from the previous

  exercise.

*}

ML %grayML{*fun add_simple_conv ctxt ctrm =

let

  val trm = Thm.term_of ctrm

in

  case trm of

        get_sum_thm ctxt trm (dest_sum trm)

    | _ => Conv.all_conv ctrm

end

val add_conv = Conv.bottom_conv add_simple_conv

fun add_tac ctxt = CONVERSION (add_conv ctxt)*}

text {*

  A test case for this conversion is as follows

*}

lemma "P (Suc (99 + 1)) ((0 + 0)::nat) (Suc (3 + 3 + 3)) ((4 + 1)::nat)"

  apply(tactic {* add_tac @{context} 1 *})?

txt {* 

  where it produces the goal state

  \begin{minipage}{\textwidth}

  @{subgoals [display]}

  \end{minipage}\bigskip

*}(*<*)oops(*>*)

text {* \solution{ex:compare} *}

text {* 

  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 

  simplified. We have to be careful to set up the goal so that

  other parts of the simplifier do not interfere. For this we construct an

  unprovable goal which, after simplification, we are going to ``prove'' with

  the help of ``\isacommand{sorry}'', that is the method @{ML Skip_Proof.cheat_tac}

  For constructing test cases, we first define a function that returns a 

  complete binary tree whose leaves are numbers and the nodes are 

  additions.

*}

ML %grayML{*fun term_tree n =

let

  val count = Unsynchronized.ref 0; 

  fun term_tree_aux n =

    case n of

      0 => (count := !count + 1; HOLogic.mk_number @{typ nat} (!count))

    | _ => Const (@{const_name "plus"}, @{typ "nat\<Rightarrow>nat\<Rightarrow>nat"}) 

             $ (term_tree_aux (n - 1)) $ (term_tree_aux (n - 1))

in

  term_tree_aux n

end*}

text {*

  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 

  produced by @{ML term_tree} filled in.

*}

ML %grayML{*fun goal n = HOLogic.mk_Trueprop (@{term "P::nat\<Rightarrow> bool"} $ (term_tree n))*}

text {*

  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,

  respectively. The idea is to first apply the conversion (respectively simproc) and 

  then prove the remaining goal using @{ML "cheat_tac" in Skip_Proof}.

*}

ML Skip_Proof.cheat_tac

ML %grayML{*local

in

    (put_simpset HOL_basic_ss ctxt addsimprocs [@{simproc add_sp}]))

end*}

text {*

  This is all we need to let the conversion run against the simproc:

*}

ML %grayML{*val _ = Goal.prove @{context} [] [] (goal 8) (fn {context, ...} => c_tac context)

val _ = Goal.prove @{context} [] [] (goal 8) (fn {context, ...} => s_tac context)*}

text {*

  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

  conversions, is quite efficient for tasks it is designed for. It usually does not

  make sense to implement general-purpose rewriting using

  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.

*}

end