diff -r a8a442ba0dbf -r e93760534354 Myhill_2.thy --- a/Myhill_2.thy Thu May 12 05:55:05 2011 +0000 +++ b/Myhill_2.thy Wed May 18 19:54:43 2011 +0000 @@ -1,64 +1,15 @@ theory Myhill_2 - imports Myhill_1 - Prefix_subtract + imports Myhill_1 Prefix_subtract "~~/src/HOL/Library/List_Prefix" begin section {* Direction @{text "regular language \finite partition"} *} -subsection {* The scheme*} - -text {* - The following convenient notation @{text "x \A y"} means: - string @{text "x"} and @{text "y"} are equivalent with respect to - language @{text "A"}. - *} - definition str_eq :: "string \ lang \ string \ bool" ("_ \_ _") where "x \A y \ (x, y) \ (\A)" -text {* - The main lemma (@{text "rexp_imp_finite"}) is proved by a structural - induction over regular expressions. where base cases (cases for @{const - "NULL"}, @{const "EMPTY"}, @{const "CHAR"}) are quite straightforward to - proof. Real difficulty lies in inductive cases. By inductive hypothesis, - languages defined by sub-expressions induce finite partitiions. Under such - hypothsis, we need to prove that the language defined by the composite - regular expression gives rise to finite partion. The basic idea is to - attach a tag @{text "tag(x)"} to every string @{text "x"}. The tagging - fuction @{text "tag"} is carefully devised, which returns tags made of - equivalent classes of the partitions induced by subexpressoins, and - therefore has a finite range. Let @{text "Lang"} be the composite language, - it is proved that: - \begin{quote} - If strings with the same tag are equivalent with respect to @{text "Lang"}, expressed as: - \[ - @{text "tag(x) = tag(y) \ x \Lang y"} - \] - then the partition induced by @{text "Lang"} must be finite. - \end{quote} - There are two arguments for this. The first goes as the following: - \begin{enumerate} - \item First, the tagging function @{text "tag"} induces an equivalent relation @{text "(=tag=)"} - (defiintion of @{text "f_eq_rel"} and lemma @{text "equiv_f_eq_rel"}). - \item It is shown that: if the range of @{text "tag"} (denoted @{text "range(tag)"}) is finite, - the partition given rise by @{text "(=tag=)"} is finite (lemma @{text "finite_eq_f_rel"}). - Since tags are made from equivalent classes from component partitions, and the inductive - hypothesis ensures the finiteness of these partitions, it is not difficult to prove - the finiteness of @{text "range(tag)"}. - \item It is proved that if equivalent relation @{text "R1"} is more refined than @{text "R2"} - (expressed as @{text "R1 \ R2"}), - and the partition induced by @{text "R1"} is finite, then the partition induced by @{text "R2"} - is finite as well (lemma @{text "refined_partition_finite"}). - \item The injectivity assumption @{text "tag(x) = tag(y) \ x \Lang y"} implies that - @{text "(=tag=)"} is more refined than @{text "(\Lang)"}. - \item Combining the points above, we have: the partition induced by language @{text "Lang"} - is finite (lemma @{text "tag_finite_imageD"}). - \end{enumerate} -*} - definition tag_eq_rel :: "(string \ 'b) \ (string \ string) set" ("=_=") where @@ -69,7 +20,6 @@ shows "finite (UNIV // =tag=)" proof - let "?f" = "\X. tag ` X" and ?A = "(UNIV // =tag=)" - -- {* The finiteness of @{text "f"}-image is a consequence of @{text "rng_fnt"} *} have "finite (?f ` ?A)" proof - have "range ?f \ (Pow (range tag))" unfolding Pow_def by auto @@ -82,25 +32,23 @@ ultimately show "finite (?f ` ?A)" by (rule rev_finite_subset) qed moreover - -- {* The injectivity of @{text "f"}-image follows from the definition of @{text "(=tag=)"} *} have "inj_on ?f ?A" proof - { fix X Y assume X_in: "X \ ?A" and Y_in: "Y \ ?A" and tag_eq: "?f X = ?f Y" - then - obtain x y + then obtain x y where "x \ X" "y \ Y" "tag x = tag y" unfolding quotient_def Image_def image_def tag_eq_rel_def by (simp) (blast) with X_in Y_in have "X = Y" unfolding quotient_def tag_eq_rel_def by auto - } then show "inj_on ?f ?A" unfolding inj_on_def by auto + } + then show "inj_on ?f ?A" unfolding inj_on_def by auto qed - ultimately - show "finite (UNIV // =tag=)" by (rule finite_imageD) + ultimately show "finite (UNIV // =tag=)" by (rule finite_imageD) qed lemma refined_partition_finite: @@ -142,7 +90,7 @@ lemma tag_finite_imageD: assumes rng_fnt: "finite (range tag)" - and same_tag_eqvt: "\ m n. tag m = tag (n::string) \ m \A n" + and same_tag_eqvt: "\m n. tag m = tag n \ m \A n" shows "finite (UNIV // \A)" proof (rule_tac refined_partition_finite [of "=tag="]) show "finite (UNIV // =tag=)" by (rule finite_eq_tag_rel[OF rng_fnt]) @@ -161,48 +109,23 @@ qed -subsection {* The proof*} - -text {* - Each case is given in a separate section, as well as the final main lemma. Detailed explainations accompanied by - illustrations are given for non-trivial cases. - - For ever inductive case, there are two tasks, the easier one is to show the range finiteness of - of the tagging function based on the finiteness of component partitions, the - difficult one is to show that strings with the same tag are equivalent with respect to the - composite language. Suppose the composite language be @{text "Lang"}, tagging function be - @{text "tag"}, it amounts to show: - \[ - @{text "tag(x) = tag(y) \ x \Lang y"} - \] - expanding the definition of @{text "\Lang"}, it amounts to show: - \[ - @{text "tag(x) = tag(y) \ (\ z. x@z \ Lang \ y@z \ Lang)"} - \] - Because the assumed tag equlity @{text "tag(x) = tag(y)"} is symmetric, - it is suffcient to show just one direction: - \[ - @{text "\ x y z. \tag(x) = tag(y); x@z \ Lang\ \ y@z \ Lang"} - \] - This is the pattern followed by every inductive case. - *} +subsection {* The proof *} subsubsection {* The base case for @{const "NULL"} *} lemma quot_null_eq: - shows "(UNIV // \{}) = ({UNIV}::lang set)" - unfolding quotient_def Image_def str_eq_rel_def by auto + shows "UNIV // \{} = {UNIV}" +unfolding quotient_def Image_def str_eq_rel_def by auto lemma quot_null_finiteI [intro]: - shows "finite ((UNIV // \{})::lang set)" + shows "finite (UNIV // \{})" unfolding quot_null_eq by simp subsubsection {* The base case for @{const "EMPTY"} *} - lemma quot_empty_subset: - "UNIV // (\{[]}) \ {{[]}, UNIV - {[]}}" + shows "UNIV // \{[]} \ {{[]}, UNIV - {[]}}" proof fix x assume "x \ UNIV // \{[]}" @@ -221,7 +144,7 @@ qed lemma quot_empty_finiteI [intro]: - shows "finite (UNIV // (\{[]}))" + shows "finite (UNIV // \{[]})" by (rule finite_subset[OF quot_empty_subset]) (simp) @@ -237,23 +160,24 @@ show "x \ {{[]},{[c]}, UNIV - {[], [c]}}" proof - { assume "y = []" hence "x = {[]}" using h - by (auto simp:str_eq_rel_def) - } moreover { - assume "y = [c]" hence "x = {[c]}" using h - by (auto dest!:spec[where x = "[]"] simp:str_eq_rel_def) - } moreover { - assume "y \ []" and "y \ [c]" + by (auto simp:str_eq_rel_def) } + moreover + { assume "y = [c]" hence "x = {[c]}" using h + by (auto dest!:spec[where x = "[]"] simp:str_eq_rel_def) } + moreover + { assume "y \ []" and "y \ [c]" hence "\ z. (y @ z) \ [c]" by (case_tac y, auto) moreover have "\ p. (p \ [] \ p \ [c]) = (\ q. p @ q \ [c])" by (case_tac p, auto) ultimately have "x = UNIV - {[],[c]}" using h by (auto simp add:str_eq_rel_def) - } ultimately show ?thesis by blast + } + ultimately show ?thesis by blast qed qed lemma quot_char_finiteI [intro]: - shows "finite (UNIV // (\{[c]}))" + shows "finite (UNIV // \{[c]})" by (rule finite_subset[OF quot_char_subset]) (simp) @@ -265,7 +189,6 @@ "tag_str_ALT A B \ (\x. (\A `` {x}, \B `` {x}))" lemma quot_union_finiteI [intro]: - fixes L1 L2::"lang" assumes finite1: "finite (UNIV // \A)" and finite2: "finite (UNIV // \B)" shows "finite (UNIV // \(A \ B))" @@ -283,140 +206,79 @@ by auto qed + subsubsection {* The inductive case for @{text "SEQ"}*} -text {* - For case @{const "SEQ"}, the language @{text "L"} is @{text "L\<^isub>1 ;; L\<^isub>2"}. - Given @{text "x @ z \ L\<^isub>1 ;; L\<^isub>2"}, according to the defintion of @{text " L\<^isub>1 ;; L\<^isub>2"}, - string @{text "x @ z"} can be splitted with the prefix in @{text "L\<^isub>1"} and suffix in @{text "L\<^isub>2"}. - The split point can either be in @{text "x"} (as shown in Fig. \ref{seq_first_split}), - or in @{text "z"} (as shown in Fig. \ref{seq_snd_split}). Whichever way it goes, the structure - on @{text "x @ z"} cn be transfered faithfully onto @{text "y @ z"} - (as shown in Fig. \ref{seq_trans_first_split} and \ref{seq_trans_snd_split}) with the the help of the assumed - tag equality. The following tag function @{text "tag_str_SEQ"} is such designed to facilitate - such transfers and lemma @{text "tag_str_SEQ_injI"} formalizes the informal argument above. The details - of structure transfer will be given their. -\input{fig_seq} - - *} - definition tag_str_SEQ :: "lang \ lang \ string \ (lang \ lang set)" where "tag_str_SEQ L1 L2 \ - (\x. (\L1 `` {x}, {(\L2 `` {x - x'}) | x'. x' \ x \ x' \ L1}))" - -text {* The following is a techical lemma which helps to split the @{text "x @ z \ L\<^isub>1 ;; L\<^isub>2"} mentioned above.*} + (\x. (\L1 `` {x}, {(\L2 `` {x - xa}) | xa. xa \ x \ xa \ L1}))" -lemma append_seq_elim: - assumes "x @ y \ L\<^isub>1 ;; L\<^isub>2" - shows "(\ xa \ x. xa \ L\<^isub>1 \ (x - xa) @ y \ L\<^isub>2) \ - (\ ya \ y. (x @ ya) \ L\<^isub>1 \ (y - ya) \ L\<^isub>2)" -proof- - from assms obtain s\<^isub>1 s\<^isub>2 - where eq_xys: "x @ y = s\<^isub>1 @ s\<^isub>2" - and in_seq: "s\<^isub>1 \ L\<^isub>1 \ s\<^isub>2 \ L\<^isub>2" - by (auto simp:Seq_def) - from app_eq_dest [OF eq_xys] - have - "(x \ s\<^isub>1 \ (s\<^isub>1 - x) @ s\<^isub>2 = y) \ (s\<^isub>1 \ x \ (x - s\<^isub>1) @ y = s\<^isub>2)" - (is "?Split1 \ ?Split2") . - moreover have "?Split1 \ \ ya \ y. (x @ ya) \ L\<^isub>1 \ (y - ya) \ L\<^isub>2" - using in_seq by (rule_tac x = "s\<^isub>1 - x" in exI, auto elim:prefixE) - moreover have "?Split2 \ \ xa \ x. xa \ L\<^isub>1 \ (x - xa) @ y \ L\<^isub>2" - using in_seq by (rule_tac x = s\<^isub>1 in exI, auto) - ultimately show ?thesis by blast -qed - +lemma Seq_in_cases: + assumes "x @ z \ A ;; B" + shows "(\ x' \ x. x' \ A \ (x - x') @ z \ B) \ + (\ z' \ z. (x @ z') \ A \ (z - z') \ B)" +using assms +unfolding Seq_def prefix_def +by (auto simp add: append_eq_append_conv2) lemma tag_str_SEQ_injI: - fixes v w - assumes eq_tag: "tag_str_SEQ L\<^isub>1 L\<^isub>2 v = tag_str_SEQ L\<^isub>1 L\<^isub>2 w" - shows "v \(L\<^isub>1 ;; L\<^isub>2) w" -proof- - -- {* As explained before, a pattern for just one direction needs to be dealt with:*} + assumes eq_tag: "tag_str_SEQ A B x = tag_str_SEQ A B y" + shows "x \(A ;; B) y" +proof - { fix x y z - assume xz_in_seq: "x @ z \ L\<^isub>1 ;; L\<^isub>2" - and tag_xy: "tag_str_SEQ L\<^isub>1 L\<^isub>2 x = tag_str_SEQ L\<^isub>1 L\<^isub>2 y" - have"y @ z \ L\<^isub>1 ;; L\<^isub>2" - proof- - -- {* There are two ways to split @{text "x@z"}: *} - from append_seq_elim [OF xz_in_seq] - have "(\ xa \ x. xa \ L\<^isub>1 \ (x - xa) @ z \ L\<^isub>2) \ - (\ za \ z. (x @ za) \ L\<^isub>1 \ (z - za) \ L\<^isub>2)" . - -- {* It can be shown that @{text "?thesis"} holds in either case: *} - moreover { - -- {* The case for the first split:*} - fix xa - assume h1: "xa \ x" and h2: "xa \ L\<^isub>1" and h3: "(x - xa) @ z \ L\<^isub>2" - -- {* The following subgoal implements the structure transfer:*} - obtain ya - where "ya \ y" - and "ya \ L\<^isub>1" - and "(y - ya) @ z \ L\<^isub>2" + assume xz_in_seq: "x @ z \ A ;; B" + and tag_xy: "tag_str_SEQ A B x = tag_str_SEQ A B y" + have"y @ z \ A ;; B" + proof - + { (* first case with x' in A and (x - x') @ z in B *) + fix x' + assume h1: "x' \ x" and h2: "x' \ A" and h3: "(x - x') @ z \ B" + obtain y' + where "y' \ y" + and "y' \ A" + and "(y - y') @ z \ B" proof - - -- {* - \begin{minipage}{0.8\textwidth} - By expanding the definition of - @{thm [display] "tag_xy"} - and extracting the second compoent, we get: - \end{minipage} - *} - have "{\L\<^isub>2 `` {x - xa} |xa. xa \ x \ xa \ L\<^isub>1} = - {\L\<^isub>2 `` {y - ya} |ya. ya \ y \ ya \ L\<^isub>1}" (is "?Left = ?Right") + have "{\B `` {x - x'} |x'. x' \ x \ x' \ A} = + {\B `` {y - y'} |y'. y' \ y \ y' \ A}" (is "?Left = ?Right") using tag_xy unfolding tag_str_SEQ_def by simp - -- {* Since @{thm "h1"} and @{thm "h2"} hold, it is not difficult to show: *} - moreover have "\L\<^isub>2 `` {x - xa} \ ?Left" using h1 h2 by auto - -- {* - \begin{minipage}{0.7\textwidth} - Through tag equality, equivalent class @{term "\L\<^isub>2 `` {x - xa}"} also - belongs to the @{text "?Right"}: - \end{minipage} - *} - ultimately have "\L\<^isub>2 `` {x - xa} \ ?Right" by simp - -- {* From this, the counterpart of @{text "xa"} in @{text "y"} is obtained:*} - then obtain ya - where eq_xya: "\L\<^isub>2 `` {x - xa} = \L\<^isub>2 `` {y - ya}" - and pref_ya: "ya \ y" and ya_in: "ya \ L\<^isub>1" + moreover + have "\B `` {x - x'} \ ?Left" using h1 h2 by auto + ultimately + have "\B `` {x - x'} \ ?Right" by simp + then obtain y' + where eq_xy': "\B `` {x - x'} = \B `` {y - y'}" + and pref_y': "y' \ y" and y'_in: "y' \ A" by simp blast - -- {* It can be proved that @{text "ya"} has the desired property:*} - have "(y - ya)@z \ L\<^isub>2" - proof - - from eq_xya have "(x - xa) \L\<^isub>2 (y - ya)" - unfolding Image_def str_eq_rel_def str_eq_def by auto - with h3 show ?thesis unfolding str_eq_rel_def str_eq_def by simp - qed - -- {* Now, @{text "ya"} has all properties to be a qualified candidate:*} - with pref_ya ya_in + + have "(x - x') \B (y - y')" using eq_xy' + unfolding Image_def str_eq_rel_def str_eq_def by auto + with h3 have "(y - y') @ z \ B" + unfolding str_eq_rel_def str_eq_def by simp + with pref_y' y'_in show ?thesis using that by blast qed - -- {* From the properties of @{text "ya"}, @{text "y @ z \ L\<^isub>1 ;; L\<^isub>2"} is derived easily.*} - hence "y @ z \ L\<^isub>1 ;; L\<^isub>2" by (erule_tac prefixE, auto simp:Seq_def) - } moreover { - -- {* The other case is even more simpler: *} - fix za - assume h1: "za \ z" and h2: "(x @ za) \ L\<^isub>1" and h3: "z - za \ L\<^isub>2" - have "y @ za \ L\<^isub>1" - proof- - have "\L\<^isub>1 `` {x} = \L\<^isub>1 `` {y}" - using tag_xy unfolding tag_str_SEQ_def by simp - with h2 show ?thesis + then have "y @ z \ A ;; B" by (erule_tac prefixE) (auto simp: Seq_def) + } + moreover + { (* second case with x @ z' in A and z - z' in B *) + fix z' + assume h1: "z' \ z" and h2: "(x @ z') \ A" and h3: "z - z' \ B" + have "\A `` {x} = \A `` {y}" + using tag_xy unfolding tag_str_SEQ_def by simp + with h2 have "y @ z' \ A" unfolding Image_def str_eq_rel_def str_eq_def by auto - qed - with h1 h3 have "y @ z \ L\<^isub>1 ;; L\<^isub>2" - by (drule_tac A = L\<^isub>1 in seq_intro, auto elim:prefixE) + with h1 h3 have "y @ z \ A ;; B" + unfolding prefix_def Seq_def + by (auto) (metis append_assoc) } - ultimately show ?thesis by blast + ultimately show "y @ z \ A ;; B" + using Seq_in_cases [OF xz_in_seq] by blast qed - } - -- {* - \begin{minipage}{0.8\textwidth} - @{text "?thesis"} is proved by exploiting the symmetry of - @{thm [source] "eq_tag"}: - \end{minipage} - *} + } from this [OF _ eq_tag] and this [OF _ eq_tag [THEN sym]] - show ?thesis unfolding str_eq_def str_eq_rel_def by blast + show "x \(A ;; B) y" unfolding str_eq_def str_eq_rel_def by blast qed lemma quot_seq_finiteI [intro]: @@ -437,53 +299,13 @@ by auto qed + subsubsection {* The inductive case for @{const "STAR"} *} -text {* - This turned out to be the trickiest case. The essential goal is - to proved @{text "y @ z \ L\<^isub>1*"} under the assumptions that @{text "x @ z \ L\<^isub>1*"} - and that @{text "x"} and @{text "y"} have the same tag. The reasoning goes as the following: - \begin{enumerate} - \item Since @{text "x @ z \ L\<^isub>1*"} holds, a prefix @{text "xa"} of @{text "x"} can be found - such that @{text "xa \ L\<^isub>1*"} and @{text "(x - xa)@z \ L\<^isub>1*"}, as shown in Fig. \ref{first_split}. - Such a prefix always exists, @{text "xa = []"}, for example, is one. - \item There could be many but fintie many of such @{text "xa"}, from which we can find the longest - and name it @{text "xa_max"}, as shown in Fig. \ref{max_split}. - \item The next step is to split @{text "z"} into @{text "za"} and @{text "zb"} such that - @{text "(x - xa_max) @ za \ L\<^isub>1"} and @{text "zb \ L\<^isub>1*"} as shown in Fig. \ref{last_split}. - Such a split always exists because: - \begin{enumerate} - \item Because @{text "(x - x_max) @ z \ L\<^isub>1*"}, it can always be splitted into prefix @{text "a"} - and suffix @{text "b"}, such that @{text "a \ L\<^isub>1"} and @{text "b \ L\<^isub>1*"}, - as shown in Fig. \ref{ab_split}. - \item But the prefix @{text "a"} CANNOT be shorter than @{text "x - xa_max"} - (as shown in Fig. \ref{ab_split_wrong}), becasue otherwise, - @{text "ma_max@a"} would be in the same kind as @{text "xa_max"} but with - a larger size, conflicting with the fact that @{text "xa_max"} is the longest. - \end{enumerate} - \item \label{tansfer_step} - By the assumption that @{text "x"} and @{text "y"} have the same tag, the structure on @{text "x @ z"} - can be transferred to @{text "y @ z"} as shown in Fig. \ref{trans_split}. The detailed steps are: - \begin{enumerate} - \item A @{text "y"}-prefix @{text "ya"} corresponding to @{text "xa"} can be found, - which satisfies conditions: @{text "ya \ L\<^isub>1*"} and @{text "(y - ya)@za \ L\<^isub>1"}. - \item Since we already know @{text "zb \ L\<^isub>1*"}, we get @{text "(y - ya)@za@zb \ L\<^isub>1*"}, - and this is just @{text "(y - ya)@z \ L\<^isub>1*"}. - \item With fact @{text "ya \ L\<^isub>1*"}, we finally get @{text "y@z \ L\<^isub>1*"}. - \end{enumerate} - \end{enumerate} - - The formal proof of lemma @{text "tag_str_STAR_injI"} faithfully follows this informal argument - while the tagging function @{text "tag_str_STAR"} is defined to make the transfer in step - \ref{ansfer_step} feasible. - - \input{fig_star} -*} - definition tag_str_STAR :: "lang \ string \ lang set" where - "tag_str_STAR L1 \ (\x. {\L1 `` {x - x'} | x'. x' < x \ x' \ L1\})" + "tag_str_STAR L1 \ (\x. {\L1 `` {x - xa} | xa. xa < x \ xa \ L1\})" text {* A technical lemma. *} lemma finite_set_has_max: "\finite A; A \ {}\ \ @@ -513,7 +335,8 @@ qed -text {* The following is a technical lemma.which helps to show the range finiteness of tag function. *} +text {* The following is a technical lemma, which helps to show the range finiteness of tag function. *} + lemma finite_strict_prefix_set: "finite {xa. xa < (x::string)}" apply (induct x rule:rev_induct, simp) apply (subgoal_tac "{xa. xa < xs @ [x]} = {xa. xa < xs} \ {xs}") @@ -521,46 +344,26 @@ lemma tag_str_STAR_injI: - fixes v w assumes eq_tag: "tag_str_STAR L\<^isub>1 v = tag_str_STAR L\<^isub>1 w" - shows "(v::string) \(L\<^isub>1\) w" + shows "v \(L\<^isub>1\) w" proof- - -- {* As explained before, a pattern for just one direction needs to be dealt with:*} { fix x y z assume xz_in_star: "x @ z \ L\<^isub>1\" and tag_xy: "tag_str_STAR L\<^isub>1 x = tag_str_STAR L\<^isub>1 y" have "y @ z \ L\<^isub>1\" proof(cases "x = []") - -- {* - The degenerated case when @{text "x"} is a null string is easy to prove: - *} case True with tag_xy have "y = []" by (auto simp add: tag_str_STAR_def strict_prefix_def) thus ?thesis using xz_in_star True by simp next - -- {* The nontrival case: - *} case False - -- {* - \begin{minipage}{0.8\textwidth} - Since @{text "x @ z \ L\<^isub>1\"}, @{text "x"} can always be splitted - by a prefix @{text "xa"} together with its suffix @{text "x - xa"}, such - that both @{text "xa"} and @{text "(x - xa) @ z"} are in @{text "L\<^isub>1\"}, - and there could be many such splittings.Therefore, the following set @{text "?S"} - is nonempty, and finite as well: - \end{minipage} - *} let ?S = "{xa. xa < x \ xa \ L\<^isub>1\ \ (x - xa) @ z \ L\<^isub>1\}" have "finite ?S" by (rule_tac B = "{xa. xa < x}" in finite_subset, auto simp:finite_strict_prefix_set) moreover have "?S \ {}" using False xz_in_star by (simp, rule_tac x = "[]" in exI, auto simp:strict_prefix_def) - -- {* \begin{minipage}{0.7\textwidth} - Since @{text "?S"} is finite, we can always single out the longest and name it @{text "xa_max"}: - \end{minipage} - *} ultimately have "\ xa_max \ ?S. \ xa \ ?S. length xa \ length xa_max" using finite_set_has_max by blast then obtain xa_max @@ -570,12 +373,6 @@ and h4:"\ xa < x. xa \ L\<^isub>1\ \ (x - xa) @ z \ L\<^isub>1\ \ length xa \ length xa_max" by blast - -- {* - \begin{minipage}{0.8\textwidth} - By the equality of tags, the counterpart of @{text "xa_max"} among - @{text "y"}-prefixes, named @{text "ya"}, can be found: - \end{minipage} - *} obtain ya where h5: "ya < y" and h6: "ya \ L\<^isub>1\" and eq_xya: "(x - xa_max) \L\<^isub>1 (y - ya)" @@ -588,47 +385,25 @@ thus ?thesis using that apply (simp add:Image_def str_eq_rel_def str_eq_def) by blast qed - -- {* - \begin{minipage}{0.8\textwidth} - The @{text "?thesis"}, @{prop "y @ z \ L\<^isub>1\"}, is a simple consequence - of the following proposition: - \end{minipage} - *} have "(y - ya) @ z \ L\<^isub>1\" proof- - -- {* The idea is to split the suffix @{text "z"} into @{text "za"} and @{text "zb"}, - such that: *} obtain za zb where eq_zab: "z = za @ zb" and l_za: "(y - ya)@za \ L\<^isub>1" and ls_zb: "zb \ L\<^isub>1\" proof - - -- {* - \begin{minipage}{0.8\textwidth} - Since @{thm "h1"}, @{text "x"} can be splitted into - @{text "a"} and @{text "b"} such that: - \end{minipage} - *} from h1 have "(x - xa_max) @ z \ []" by (auto simp:strict_prefix_def elim:prefixE) from star_decom [OF h3 this] obtain a b where a_in: "a \ L\<^isub>1" and a_neq: "a \ []" and b_in: "b \ L\<^isub>1\" and ab_max: "(x - xa_max) @ z = a @ b" by blast - -- {* Now the candiates for @{text "za"} and @{text "zb"} are found:*} let ?za = "a - (x - xa_max)" and ?zb = "b" have pfx: "(x - xa_max) \ a" (is "?P1") and eq_z: "z = ?za @ ?zb" (is "?P2") proof - - -- {* - \begin{minipage}{0.8\textwidth} - Since @{text "(x - xa_max) @ z = a @ b"}, string @{text "(x - xa_max) @ z"} - can be splitted in two ways: - \end{minipage} - *} have "((x - xa_max) \ a \ (a - (x - xa_max)) @ b = z) \ (a < (x - xa_max) \ ((x - xa_max) - a) @ z = b)" - using app_eq_dest[OF ab_max] by (auto simp:strict_prefix_def) + using append_eq_dest[OF ab_max] by (auto simp:strict_prefix_def) moreover { - -- {* However, the undsired way can be refuted by absurdity: *} assume np: "a < (x - xa_max)" and b_eqs: "((x - xa_max) - a) @ z = b" have "False" @@ -639,24 +414,19 @@ moreover have "?xa_max' \ L\<^isub>1\" using a_in h2 by (simp add:star_intro3) moreover have "(x - ?xa_max') @ z \ L\<^isub>1\" - using b_eqs b_in np h1 by (simp add:diff_diff_appd) + using b_eqs b_in np h1 by (simp add:diff_diff_append) moreover have "\ (length ?xa_max' \ length xa_max)" using a_neq by simp ultimately show ?thesis using h4 by blast qed } - -- {* Now it can be shown that the splitting goes the way we desired. *} ultimately show ?P1 and ?P2 by auto qed hence "(x - xa_max)@?za \ L\<^isub>1" using a_in by (auto elim:prefixE) - -- {* Now candidates @{text "?za"} and @{text "?zb"} have all the requred properteis. *} with eq_xya have "(y - ya) @ ?za \ L\<^isub>1" by (auto simp:str_eq_def str_eq_rel_def) with eq_z and b_in show ?thesis using that by blast qed - -- {* - @{text "?thesis"} can easily be shown using properties of @{text "za"} and @{text "zb"}: - *} have "((y - ya) @ za) @ zb \ L\<^isub>1\" using l_za ls_zb by blast with eq_zab show ?thesis by simp qed @@ -664,123 +434,11 @@ by (drule_tac star_intro1) (auto simp:strict_prefix_def elim:prefixE) qed } - -- {* By instantiating the reasoning pattern just derived for both directions:*} from this [OF _ eq_tag] and this [OF _ eq_tag [THEN sym]] - -- {* The thesis is proved as a trival consequence: *} - show ?thesis unfolding str_eq_def str_eq_rel_def by blast -qed - -lemma -- {* The oringal version with less explicit details. *} - fixes v w - assumes eq_tag: "tag_str_STAR L\<^isub>1 v = tag_str_STAR L\<^isub>1 w" - shows "(v::string) \(L\<^isub>1\) w" -proof- - -- {* - \begin{minipage}{0.8\textwidth} - According to the definition of @{text "\Lang"}, - proving @{text "v \(L\<^isub>1\) w"} amounts to - showing: for any string @{text "u"}, - if @{text "v @ u \ (L\<^isub>1\)"} then @{text "w @ u \ (L\<^isub>1\)"} and vice versa. - The reasoning pattern for both directions are the same, as derived - in the following: - \end{minipage} - *} - { fix x y z - assume xz_in_star: "x @ z \ L\<^isub>1\" - and tag_xy: "tag_str_STAR L\<^isub>1 x = tag_str_STAR L\<^isub>1 y" - have "y @ z \ L\<^isub>1\" - proof(cases "x = []") - -- {* - The degenerated case when @{text "x"} is a null string is easy to prove: - *} - case True - with tag_xy have "y = []" - by (auto simp:tag_str_STAR_def strict_prefix_def) - thus ?thesis using xz_in_star True by simp - next - -- {* - \begin{minipage}{0.8\textwidth} - The case when @{text "x"} is not null, and - @{text "x @ z"} is in @{text "L\<^isub>1\"}, - \end{minipage} - *} - case False - obtain x_max - where h1: "x_max < x" - and h2: "x_max \ L\<^isub>1\" - and h3: "(x - x_max) @ z \ L\<^isub>1\" - and h4:"\ xa < x. xa \ L\<^isub>1\ \ (x - xa) @ z \ L\<^isub>1\ - \ length xa \ length x_max" - proof- - let ?S = "{xa. xa < x \ xa \ L\<^isub>1\ \ (x - xa) @ z \ L\<^isub>1\}" - have "finite ?S" - by (rule_tac B = "{xa. xa < x}" in finite_subset, - auto simp:finite_strict_prefix_set) - moreover have "?S \ {}" using False xz_in_star - by (simp, rule_tac x = "[]" in exI, auto simp:strict_prefix_def) - ultimately have "\ max \ ?S. \ a \ ?S. length a \ length max" - using finite_set_has_max by blast - thus ?thesis using that by blast - qed - obtain ya - where h5: "ya < y" and h6: "ya \ L\<^isub>1\" and h7: "(x - x_max) \L\<^isub>1 (y - ya)" - proof- - from tag_xy have "{\L\<^isub>1 `` {x - xa} |xa. xa < x \ xa \ L\<^isub>1\} = - {\L\<^isub>1 `` {y - xa} |xa. xa < y \ xa \ L\<^isub>1\}" (is "?left = ?right") - by (auto simp:tag_str_STAR_def) - moreover have "\L\<^isub>1 `` {x - x_max} \ ?left" using h1 h2 by auto - ultimately have "\L\<^isub>1 `` {x - x_max} \ ?right" by simp - with that show ?thesis apply - (simp add:Image_def str_eq_rel_def str_eq_def) by blast - qed - have "(y - ya) @ z \ L\<^isub>1\" - proof- - from h3 h1 obtain a b where a_in: "a \ L\<^isub>1" - and a_neq: "a \ []" and b_in: "b \ L\<^isub>1\" - and ab_max: "(x - x_max) @ z = a @ b" - by (drule_tac star_decom, auto simp:strict_prefix_def elim:prefixE) - have "(x - x_max) \ a \ (a - (x - x_max)) @ b = z" - proof - - have "((x - x_max) \ a \ (a - (x - x_max)) @ b = z) \ - (a < (x - x_max) \ ((x - x_max) - a) @ z = b)" - using app_eq_dest[OF ab_max] by (auto simp:strict_prefix_def) - moreover { - assume np: "a < (x - x_max)" and b_eqs: " ((x - x_max) - a) @ z = b" - have "False" - proof - - let ?x_max' = "x_max @ a" - have "?x_max' < x" - using np h1 by (clarsimp simp:strict_prefix_def diff_prefix) - moreover have "?x_max' \ L\<^isub>1\" - using a_in h2 by (simp add:star_intro3) - moreover have "(x - ?x_max') @ z \ L\<^isub>1\" - using b_eqs b_in np h1 by (simp add:diff_diff_appd) - moreover have "\ (length ?x_max' \ length x_max)" - using a_neq by simp - ultimately show ?thesis using h4 by blast - qed - } ultimately show ?thesis by blast - qed - then obtain za where z_decom: "z = za @ b" - and x_za: "(x - x_max) @ za \ L\<^isub>1" - using a_in by (auto elim:prefixE) - from x_za h7 have "(y - ya) @ za \ L\<^isub>1" - by (auto simp:str_eq_def str_eq_rel_def) - with b_in have "((y - ya) @ za) @ b \ L\<^isub>1\" by blast - with z_decom show ?thesis by auto - qed - with h5 h6 show ?thesis - by (drule_tac star_intro1) (auto simp:strict_prefix_def elim:prefixE) - qed - } - -- {* By instantiating the reasoning pattern just derived for both directions:*} - from this [OF _ eq_tag] and this [OF _ eq_tag [THEN sym]] - -- {* The thesis is proved as a trival consequence: *} show ?thesis unfolding str_eq_def str_eq_rel_def by blast qed lemma quot_star_finiteI [intro]: - fixes L1::"lang" assumes finite1: "finite (UNIV // \L1)" shows "finite (UNIV // \(L1\))" proof (rule_tac tag = "tag_str_STAR L1" in tag_finite_imageD) @@ -803,76 +461,9 @@ shows "finite (UNIV // \(L r))" by (induct r) (auto) + theorem Myhill_Nerode: shows "(\r::rexp. A = L r) \ finite (UNIV // \A)" -using Myhill_Nerode1 Myhill_Nerode2 by metis - -(* -section {* Closure properties *} - -abbreviation - reg :: "lang \ bool" -where - "reg A \ \r::rexp. A = L r" - - - -lemma closure_union[intro]: - assumes "reg A" "reg B" - shows "reg (A \ B)" -using assms -apply(auto) -apply(rule_tac x="ALT r ra" in exI) -apply(auto) -done - -lemma closure_seq[intro]: - assumes "reg A" "reg B" - shows "reg (A ;; B)" -using assms -apply(auto) -apply(rule_tac x="SEQ r ra" in exI) -apply(auto) -done - -lemma closure_star[intro]: - assumes "reg A" - shows "reg (A\)" -using assms -apply(auto) -apply(rule_tac x="STAR r" in exI) -apply(auto) -done - -lemma closure_complement[intro]: - assumes "reg A" - shows "reg (- A)" -using assms -unfolding Myhill_Nerode -unfolding str_eq_rel_def -by auto - -lemma closure_difference[intro]: - assumes "reg A" "reg B" - shows "reg (A - B)" -proof - - have "A - B = - ((- A) \ B)" by blast - moreover - have "reg (- ((- A) \ B))" - using assms by blast - ultimately show "reg (A - B)" by simp -qed - -lemma closure_intersection[intro]: - assumes "reg A" "reg B" - shows "reg (A \ B)" -proof - - have "A \ B = - ((- A) \ (- B))" by blast - moreover - have "reg (- ((- A) \ (- B)))" - using assms by blast - ultimately show "reg (A \ B)" by simp -qed -*) +using Myhill_Nerode1 Myhill_Nerode2 by auto end