src/HOL/Divides.thy
author haftmann
Thu Oct 29 22:13:09 2009 +0100 (2009-10-29)
changeset 33340 a165b97f3658
parent 33318 ddd97d9dfbfb
child 33361 1f18de40b43f
permissions -rw-r--r--
moved Nat_Transfer before Divides; distributed Nat_Transfer setup accordingly
     1 (*  Title:      HOL/Divides.thy
     2     Author:     Lawrence C Paulson, Cambridge University Computer Laboratory
     3     Copyright   1999  University of Cambridge
     4 *)
     5 
     6 header {* The division operators div and mod *}
     7 
     8 theory Divides
     9 imports Nat_Numeral Nat_Transfer
    10 uses "~~/src/Provers/Arith/cancel_div_mod.ML"
    11 begin
    12 
    13 subsection {* Syntactic division operations *}
    14 
    15 class div = dvd +
    16   fixes div :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "div" 70)
    17     and mod :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "mod" 70)
    18 
    19 
    20 subsection {* Abstract division in commutative semirings. *}
    21 
    22 class semiring_div = comm_semiring_1_cancel + no_zero_divisors + div +
    23   assumes mod_div_equality: "a div b * b + a mod b = a"
    24     and div_by_0 [simp]: "a div 0 = 0"
    25     and div_0 [simp]: "0 div a = 0"
    26     and div_mult_self1 [simp]: "b \<noteq> 0 \<Longrightarrow> (a + c * b) div b = c + a div b"
    27     and div_mult_mult1 [simp]: "c \<noteq> 0 \<Longrightarrow> (c * a) div (c * b) = a div b"
    28 begin
    29 
    30 text {* @{const div} and @{const mod} *}
    31 
    32 lemma mod_div_equality2: "b * (a div b) + a mod b = a"
    33   unfolding mult_commute [of b]
    34   by (rule mod_div_equality)
    35 
    36 lemma mod_div_equality': "a mod b + a div b * b = a"
    37   using mod_div_equality [of a b]
    38   by (simp only: add_ac)
    39 
    40 lemma div_mod_equality: "((a div b) * b + a mod b) + c = a + c"
    41   by (simp add: mod_div_equality)
    42 
    43 lemma div_mod_equality2: "(b * (a div b) + a mod b) + c = a + c"
    44   by (simp add: mod_div_equality2)
    45 
    46 lemma mod_by_0 [simp]: "a mod 0 = a"
    47   using mod_div_equality [of a zero] by simp
    48 
    49 lemma mod_0 [simp]: "0 mod a = 0"
    50   using mod_div_equality [of zero a] div_0 by simp
    51 
    52 lemma div_mult_self2 [simp]:
    53   assumes "b \<noteq> 0"
    54   shows "(a + b * c) div b = c + a div b"
    55   using assms div_mult_self1 [of b a c] by (simp add: mult_commute)
    56 
    57 lemma mod_mult_self1 [simp]: "(a + c * b) mod b = a mod b"
    58 proof (cases "b = 0")
    59   case True then show ?thesis by simp
    60 next
    61   case False
    62   have "a + c * b = (a + c * b) div b * b + (a + c * b) mod b"
    63     by (simp add: mod_div_equality)
    64   also from False div_mult_self1 [of b a c] have
    65     "\<dots> = (c + a div b) * b + (a + c * b) mod b"
    66       by (simp add: algebra_simps)
    67   finally have "a = a div b * b + (a + c * b) mod b"
    68     by (simp add: add_commute [of a] add_assoc left_distrib)
    69   then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"
    70     by (simp add: mod_div_equality)
    71   then show ?thesis by simp
    72 qed
    73 
    74 lemma mod_mult_self2 [simp]: "(a + b * c) mod b = a mod b"
    75   by (simp add: mult_commute [of b])
    76 
    77 lemma div_mult_self1_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> b * a div b = a"
    78   using div_mult_self2 [of b 0 a] by simp
    79 
    80 lemma div_mult_self2_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> a * b div b = a"
    81   using div_mult_self1 [of b 0 a] by simp
    82 
    83 lemma mod_mult_self1_is_0 [simp]: "b * a mod b = 0"
    84   using mod_mult_self2 [of 0 b a] by simp
    85 
    86 lemma mod_mult_self2_is_0 [simp]: "a * b mod b = 0"
    87   using mod_mult_self1 [of 0 a b] by simp
    88 
    89 lemma div_by_1 [simp]: "a div 1 = a"
    90   using div_mult_self2_is_id [of 1 a] zero_neq_one by simp
    91 
    92 lemma mod_by_1 [simp]: "a mod 1 = 0"
    93 proof -
    94   from mod_div_equality [of a one] div_by_1 have "a + a mod 1 = a" by simp
    95   then have "a + a mod 1 = a + 0" by simp
    96   then show ?thesis by (rule add_left_imp_eq)
    97 qed
    98 
    99 lemma mod_self [simp]: "a mod a = 0"
   100   using mod_mult_self2_is_0 [of 1] by simp
   101 
   102 lemma div_self [simp]: "a \<noteq> 0 \<Longrightarrow> a div a = 1"
   103   using div_mult_self2_is_id [of _ 1] by simp
   104 
   105 lemma div_add_self1 [simp]:
   106   assumes "b \<noteq> 0"
   107   shows "(b + a) div b = a div b + 1"
   108   using assms div_mult_self1 [of b a 1] by (simp add: add_commute)
   109 
   110 lemma div_add_self2 [simp]:
   111   assumes "b \<noteq> 0"
   112   shows "(a + b) div b = a div b + 1"
   113   using assms div_add_self1 [of b a] by (simp add: add_commute)
   114 
   115 lemma mod_add_self1 [simp]:
   116   "(b + a) mod b = a mod b"
   117   using mod_mult_self1 [of a 1 b] by (simp add: add_commute)
   118 
   119 lemma mod_add_self2 [simp]:
   120   "(a + b) mod b = a mod b"
   121   using mod_mult_self1 [of a 1 b] by simp
   122 
   123 lemma mod_div_decomp:
   124   fixes a b
   125   obtains q r where "q = a div b" and "r = a mod b"
   126     and "a = q * b + r"
   127 proof -
   128   from mod_div_equality have "a = a div b * b + a mod b" by simp
   129   moreover have "a div b = a div b" ..
   130   moreover have "a mod b = a mod b" ..
   131   note that ultimately show thesis by blast
   132 qed
   133 
   134 lemma dvd_eq_mod_eq_0 [code_unfold]: "a dvd b \<longleftrightarrow> b mod a = 0"
   135 proof
   136   assume "b mod a = 0"
   137   with mod_div_equality [of b a] have "b div a * a = b" by simp
   138   then have "b = a * (b div a)" unfolding mult_commute ..
   139   then have "\<exists>c. b = a * c" ..
   140   then show "a dvd b" unfolding dvd_def .
   141 next
   142   assume "a dvd b"
   143   then have "\<exists>c. b = a * c" unfolding dvd_def .
   144   then obtain c where "b = a * c" ..
   145   then have "b mod a = a * c mod a" by simp
   146   then have "b mod a = c * a mod a" by (simp add: mult_commute)
   147   then show "b mod a = 0" by simp
   148 qed
   149 
   150 lemma mod_div_trivial [simp]: "a mod b div b = 0"
   151 proof (cases "b = 0")
   152   assume "b = 0"
   153   thus ?thesis by simp
   154 next
   155   assume "b \<noteq> 0"
   156   hence "a div b + a mod b div b = (a mod b + a div b * b) div b"
   157     by (rule div_mult_self1 [symmetric])
   158   also have "\<dots> = a div b"
   159     by (simp only: mod_div_equality')
   160   also have "\<dots> = a div b + 0"
   161     by simp
   162   finally show ?thesis
   163     by (rule add_left_imp_eq)
   164 qed
   165 
   166 lemma mod_mod_trivial [simp]: "a mod b mod b = a mod b"
   167 proof -
   168   have "a mod b mod b = (a mod b + a div b * b) mod b"
   169     by (simp only: mod_mult_self1)
   170   also have "\<dots> = a mod b"
   171     by (simp only: mod_div_equality')
   172   finally show ?thesis .
   173 qed
   174 
   175 lemma dvd_imp_mod_0: "a dvd b \<Longrightarrow> b mod a = 0"
   176 by (rule dvd_eq_mod_eq_0[THEN iffD1])
   177 
   178 lemma dvd_div_mult_self: "a dvd b \<Longrightarrow> (b div a) * a = b"
   179 by (subst (2) mod_div_equality [of b a, symmetric]) (simp add:dvd_imp_mod_0)
   180 
   181 lemma dvd_mult_div_cancel: "a dvd b \<Longrightarrow> a * (b div a) = b"
   182 by (drule dvd_div_mult_self) (simp add: mult_commute)
   183 
   184 lemma dvd_div_mult: "a dvd b \<Longrightarrow> (b div a) * c = b * c div a"
   185 apply (cases "a = 0")
   186  apply simp
   187 apply (auto simp: dvd_def mult_assoc)
   188 done
   189 
   190 lemma div_dvd_div[simp]:
   191   "a dvd b \<Longrightarrow> a dvd c \<Longrightarrow> (b div a dvd c div a) = (b dvd c)"
   192 apply (cases "a = 0")
   193  apply simp
   194 apply (unfold dvd_def)
   195 apply auto
   196  apply(blast intro:mult_assoc[symmetric])
   197 apply(fastsimp simp add: mult_assoc)
   198 done
   199 
   200 lemma dvd_mod_imp_dvd: "[| k dvd m mod n;  k dvd n |] ==> k dvd m"
   201   apply (subgoal_tac "k dvd (m div n) *n + m mod n")
   202    apply (simp add: mod_div_equality)
   203   apply (simp only: dvd_add dvd_mult)
   204   done
   205 
   206 text {* Addition respects modular equivalence. *}
   207 
   208 lemma mod_add_left_eq: "(a + b) mod c = (a mod c + b) mod c"
   209 proof -
   210   have "(a + b) mod c = (a div c * c + a mod c + b) mod c"
   211     by (simp only: mod_div_equality)
   212   also have "\<dots> = (a mod c + b + a div c * c) mod c"
   213     by (simp only: add_ac)
   214   also have "\<dots> = (a mod c + b) mod c"
   215     by (rule mod_mult_self1)
   216   finally show ?thesis .
   217 qed
   218 
   219 lemma mod_add_right_eq: "(a + b) mod c = (a + b mod c) mod c"
   220 proof -
   221   have "(a + b) mod c = (a + (b div c * c + b mod c)) mod c"
   222     by (simp only: mod_div_equality)
   223   also have "\<dots> = (a + b mod c + b div c * c) mod c"
   224     by (simp only: add_ac)
   225   also have "\<dots> = (a + b mod c) mod c"
   226     by (rule mod_mult_self1)
   227   finally show ?thesis .
   228 qed
   229 
   230 lemma mod_add_eq: "(a + b) mod c = (a mod c + b mod c) mod c"
   231 by (rule trans [OF mod_add_left_eq mod_add_right_eq])
   232 
   233 lemma mod_add_cong:
   234   assumes "a mod c = a' mod c"
   235   assumes "b mod c = b' mod c"
   236   shows "(a + b) mod c = (a' + b') mod c"
   237 proof -
   238   have "(a mod c + b mod c) mod c = (a' mod c + b' mod c) mod c"
   239     unfolding assms ..
   240   thus ?thesis
   241     by (simp only: mod_add_eq [symmetric])
   242 qed
   243 
   244 lemma div_add [simp]: "z dvd x \<Longrightarrow> z dvd y
   245   \<Longrightarrow> (x + y) div z = x div z + y div z"
   246 by (cases "z = 0", simp, unfold dvd_def, auto simp add: algebra_simps)
   247 
   248 text {* Multiplication respects modular equivalence. *}
   249 
   250 lemma mod_mult_left_eq: "(a * b) mod c = ((a mod c) * b) mod c"
   251 proof -
   252   have "(a * b) mod c = ((a div c * c + a mod c) * b) mod c"
   253     by (simp only: mod_div_equality)
   254   also have "\<dots> = (a mod c * b + a div c * b * c) mod c"
   255     by (simp only: algebra_simps)
   256   also have "\<dots> = (a mod c * b) mod c"
   257     by (rule mod_mult_self1)
   258   finally show ?thesis .
   259 qed
   260 
   261 lemma mod_mult_right_eq: "(a * b) mod c = (a * (b mod c)) mod c"
   262 proof -
   263   have "(a * b) mod c = (a * (b div c * c + b mod c)) mod c"
   264     by (simp only: mod_div_equality)
   265   also have "\<dots> = (a * (b mod c) + a * (b div c) * c) mod c"
   266     by (simp only: algebra_simps)
   267   also have "\<dots> = (a * (b mod c)) mod c"
   268     by (rule mod_mult_self1)
   269   finally show ?thesis .
   270 qed
   271 
   272 lemma mod_mult_eq: "(a * b) mod c = ((a mod c) * (b mod c)) mod c"
   273 by (rule trans [OF mod_mult_left_eq mod_mult_right_eq])
   274 
   275 lemma mod_mult_cong:
   276   assumes "a mod c = a' mod c"
   277   assumes "b mod c = b' mod c"
   278   shows "(a * b) mod c = (a' * b') mod c"
   279 proof -
   280   have "(a mod c * (b mod c)) mod c = (a' mod c * (b' mod c)) mod c"
   281     unfolding assms ..
   282   thus ?thesis
   283     by (simp only: mod_mult_eq [symmetric])
   284 qed
   285 
   286 lemma mod_mod_cancel:
   287   assumes "c dvd b"
   288   shows "a mod b mod c = a mod c"
   289 proof -
   290   from `c dvd b` obtain k where "b = c * k"
   291     by (rule dvdE)
   292   have "a mod b mod c = a mod (c * k) mod c"
   293     by (simp only: `b = c * k`)
   294   also have "\<dots> = (a mod (c * k) + a div (c * k) * k * c) mod c"
   295     by (simp only: mod_mult_self1)
   296   also have "\<dots> = (a div (c * k) * (c * k) + a mod (c * k)) mod c"
   297     by (simp only: add_ac mult_ac)
   298   also have "\<dots> = a mod c"
   299     by (simp only: mod_div_equality)
   300   finally show ?thesis .
   301 qed
   302 
   303 lemma div_mult_div_if_dvd:
   304   "y dvd x \<Longrightarrow> z dvd w \<Longrightarrow> (x div y) * (w div z) = (x * w) div (y * z)"
   305   apply (cases "y = 0", simp)
   306   apply (cases "z = 0", simp)
   307   apply (auto elim!: dvdE simp add: algebra_simps)
   308   apply (subst mult_assoc [symmetric])
   309   apply (simp add: no_zero_divisors)
   310   done
   311 
   312 lemma div_mult_mult2 [simp]:
   313   "c \<noteq> 0 \<Longrightarrow> (a * c) div (b * c) = a div b"
   314   by (drule div_mult_mult1) (simp add: mult_commute)
   315 
   316 lemma div_mult_mult1_if [simp]:
   317   "(c * a) div (c * b) = (if c = 0 then 0 else a div b)"
   318   by simp_all
   319 
   320 lemma mod_mult_mult1:
   321   "(c * a) mod (c * b) = c * (a mod b)"
   322 proof (cases "c = 0")
   323   case True then show ?thesis by simp
   324 next
   325   case False
   326   from mod_div_equality
   327   have "((c * a) div (c * b)) * (c * b) + (c * a) mod (c * b) = c * a" .
   328   with False have "c * ((a div b) * b + a mod b) + (c * a) mod (c * b)
   329     = c * a + c * (a mod b)" by (simp add: algebra_simps)
   330   with mod_div_equality show ?thesis by simp 
   331 qed
   332   
   333 lemma mod_mult_mult2:
   334   "(a * c) mod (b * c) = (a mod b) * c"
   335   using mod_mult_mult1 [of c a b] by (simp add: mult_commute)
   336 
   337 lemma dvd_mod: "k dvd m \<Longrightarrow> k dvd n \<Longrightarrow> k dvd (m mod n)"
   338   unfolding dvd_def by (auto simp add: mod_mult_mult1)
   339 
   340 lemma dvd_mod_iff: "k dvd n \<Longrightarrow> k dvd (m mod n) \<longleftrightarrow> k dvd m"
   341 by (blast intro: dvd_mod_imp_dvd dvd_mod)
   342 
   343 lemma div_power:
   344   "y dvd x \<Longrightarrow> (x div y) ^ n = x ^ n div y ^ n"
   345 apply (induct n)
   346  apply simp
   347 apply(simp add: div_mult_div_if_dvd dvd_power_same)
   348 done
   349 
   350 end
   351 
   352 class ring_div = semiring_div + idom
   353 begin
   354 
   355 text {* Negation respects modular equivalence. *}
   356 
   357 lemma mod_minus_eq: "(- a) mod b = (- (a mod b)) mod b"
   358 proof -
   359   have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
   360     by (simp only: mod_div_equality)
   361   also have "\<dots> = (- (a mod b) + - (a div b) * b) mod b"
   362     by (simp only: minus_add_distrib minus_mult_left add_ac)
   363   also have "\<dots> = (- (a mod b)) mod b"
   364     by (rule mod_mult_self1)
   365   finally show ?thesis .
   366 qed
   367 
   368 lemma mod_minus_cong:
   369   assumes "a mod b = a' mod b"
   370   shows "(- a) mod b = (- a') mod b"
   371 proof -
   372   have "(- (a mod b)) mod b = (- (a' mod b)) mod b"
   373     unfolding assms ..
   374   thus ?thesis
   375     by (simp only: mod_minus_eq [symmetric])
   376 qed
   377 
   378 text {* Subtraction respects modular equivalence. *}
   379 
   380 lemma mod_diff_left_eq: "(a - b) mod c = (a mod c - b) mod c"
   381   unfolding diff_minus
   382   by (intro mod_add_cong mod_minus_cong) simp_all
   383 
   384 lemma mod_diff_right_eq: "(a - b) mod c = (a - b mod c) mod c"
   385   unfolding diff_minus
   386   by (intro mod_add_cong mod_minus_cong) simp_all
   387 
   388 lemma mod_diff_eq: "(a - b) mod c = (a mod c - b mod c) mod c"
   389   unfolding diff_minus
   390   by (intro mod_add_cong mod_minus_cong) simp_all
   391 
   392 lemma mod_diff_cong:
   393   assumes "a mod c = a' mod c"
   394   assumes "b mod c = b' mod c"
   395   shows "(a - b) mod c = (a' - b') mod c"
   396   unfolding diff_minus using assms
   397   by (intro mod_add_cong mod_minus_cong)
   398 
   399 lemma dvd_neg_div: "y dvd x \<Longrightarrow> -x div y = - (x div y)"
   400 apply (case_tac "y = 0") apply simp
   401 apply (auto simp add: dvd_def)
   402 apply (subgoal_tac "-(y * k) = y * - k")
   403  apply (erule ssubst)
   404  apply (erule div_mult_self1_is_id)
   405 apply simp
   406 done
   407 
   408 lemma dvd_div_neg: "y dvd x \<Longrightarrow> x div -y = - (x div y)"
   409 apply (case_tac "y = 0") apply simp
   410 apply (auto simp add: dvd_def)
   411 apply (subgoal_tac "y * k = -y * -k")
   412  apply (erule ssubst)
   413  apply (rule div_mult_self1_is_id)
   414  apply simp
   415 apply simp
   416 done
   417 
   418 end
   419 
   420 
   421 subsection {* Division on @{typ nat} *}
   422 
   423 text {*
   424   We define @{const div} and @{const mod} on @{typ nat} by means
   425   of a characteristic relation with two input arguments
   426   @{term "m\<Colon>nat"}, @{term "n\<Colon>nat"} and two output arguments
   427   @{term "q\<Colon>nat"}(uotient) and @{term "r\<Colon>nat"}(emainder).
   428 *}
   429 
   430 definition divmod_nat_rel :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat \<Rightarrow> bool" where
   431   "divmod_nat_rel m n qr \<longleftrightarrow>
   432     m = fst qr * n + snd qr \<and>
   433       (if n = 0 then fst qr = 0 else if n > 0 then 0 \<le> snd qr \<and> snd qr < n else n < snd qr \<and> snd qr \<le> 0)"
   434 
   435 text {* @{const divmod_nat_rel} is total: *}
   436 
   437 lemma divmod_nat_rel_ex:
   438   obtains q r where "divmod_nat_rel m n (q, r)"
   439 proof (cases "n = 0")
   440   case True  with that show thesis
   441     by (auto simp add: divmod_nat_rel_def)
   442 next
   443   case False
   444   have "\<exists>q r. m = q * n + r \<and> r < n"
   445   proof (induct m)
   446     case 0 with `n \<noteq> 0`
   447     have "(0\<Colon>nat) = 0 * n + 0 \<and> 0 < n" by simp
   448     then show ?case by blast
   449   next
   450     case (Suc m) then obtain q' r'
   451       where m: "m = q' * n + r'" and n: "r' < n" by auto
   452     then show ?case proof (cases "Suc r' < n")
   453       case True
   454       from m n have "Suc m = q' * n + Suc r'" by simp
   455       with True show ?thesis by blast
   456     next
   457       case False then have "n \<le> Suc r'" by auto
   458       moreover from n have "Suc r' \<le> n" by auto
   459       ultimately have "n = Suc r'" by auto
   460       with m have "Suc m = Suc q' * n + 0" by simp
   461       with `n \<noteq> 0` show ?thesis by blast
   462     qed
   463   qed
   464   with that show thesis
   465     using `n \<noteq> 0` by (auto simp add: divmod_nat_rel_def)
   466 qed
   467 
   468 text {* @{const divmod_nat_rel} is injective: *}
   469 
   470 lemma divmod_nat_rel_unique:
   471   assumes "divmod_nat_rel m n qr"
   472     and "divmod_nat_rel m n qr'"
   473   shows "qr = qr'"
   474 proof (cases "n = 0")
   475   case True with assms show ?thesis
   476     by (cases qr, cases qr')
   477       (simp add: divmod_nat_rel_def)
   478 next
   479   case False
   480   have aux: "\<And>q r q' r'. q' * n + r' = q * n + r \<Longrightarrow> r < n \<Longrightarrow> q' \<le> (q\<Colon>nat)"
   481   apply (rule leI)
   482   apply (subst less_iff_Suc_add)
   483   apply (auto simp add: add_mult_distrib)
   484   done
   485   from `n \<noteq> 0` assms have "fst qr = fst qr'"
   486     by (auto simp add: divmod_nat_rel_def intro: order_antisym dest: aux sym)
   487   moreover from this assms have "snd qr = snd qr'"
   488     by (simp add: divmod_nat_rel_def)
   489   ultimately show ?thesis by (cases qr, cases qr') simp
   490 qed
   491 
   492 text {*
   493   We instantiate divisibility on the natural numbers by
   494   means of @{const divmod_nat_rel}:
   495 *}
   496 
   497 instantiation nat :: semiring_div
   498 begin
   499 
   500 definition divmod_nat :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where
   501   [code del]: "divmod_nat m n = (THE qr. divmod_nat_rel m n qr)"
   502 
   503 lemma divmod_nat_rel_divmod_nat:
   504   "divmod_nat_rel m n (divmod_nat m n)"
   505 proof -
   506   from divmod_nat_rel_ex
   507     obtain qr where rel: "divmod_nat_rel m n qr" .
   508   then show ?thesis
   509   by (auto simp add: divmod_nat_def intro: theI elim: divmod_nat_rel_unique)
   510 qed
   511 
   512 lemma divmod_nat_eq:
   513   assumes "divmod_nat_rel m n qr" 
   514   shows "divmod_nat m n = qr"
   515   using assms by (auto intro: divmod_nat_rel_unique divmod_nat_rel_divmod_nat)
   516 
   517 definition div_nat where
   518   "m div n = fst (divmod_nat m n)"
   519 
   520 definition mod_nat where
   521   "m mod n = snd (divmod_nat m n)"
   522 
   523 lemma divmod_nat_div_mod:
   524   "divmod_nat m n = (m div n, m mod n)"
   525   unfolding div_nat_def mod_nat_def by simp
   526 
   527 lemma div_eq:
   528   assumes "divmod_nat_rel m n (q, r)" 
   529   shows "m div n = q"
   530   using assms by (auto dest: divmod_nat_eq simp add: divmod_nat_div_mod)
   531 
   532 lemma mod_eq:
   533   assumes "divmod_nat_rel m n (q, r)" 
   534   shows "m mod n = r"
   535   using assms by (auto dest: divmod_nat_eq simp add: divmod_nat_div_mod)
   536 
   537 lemma divmod_nat_rel: "divmod_nat_rel m n (m div n, m mod n)"
   538   by (simp add: div_nat_def mod_nat_def divmod_nat_rel_divmod_nat)
   539 
   540 lemma divmod_nat_zero:
   541   "divmod_nat m 0 = (0, m)"
   542 proof -
   543   from divmod_nat_rel [of m 0] show ?thesis
   544     unfolding divmod_nat_div_mod divmod_nat_rel_def by simp
   545 qed
   546 
   547 lemma divmod_nat_base:
   548   assumes "m < n"
   549   shows "divmod_nat m n = (0, m)"
   550 proof -
   551   from divmod_nat_rel [of m n] show ?thesis
   552     unfolding divmod_nat_div_mod divmod_nat_rel_def
   553     using assms by (cases "m div n = 0")
   554       (auto simp add: gr0_conv_Suc [of "m div n"])
   555 qed
   556 
   557 lemma divmod_nat_step:
   558   assumes "0 < n" and "n \<le> m"
   559   shows "divmod_nat m n = (Suc ((m - n) div n), (m - n) mod n)"
   560 proof -
   561   from divmod_nat_rel have divmod_nat_m_n: "divmod_nat_rel m n (m div n, m mod n)" .
   562   with assms have m_div_n: "m div n \<ge> 1"
   563     by (cases "m div n") (auto simp add: divmod_nat_rel_def)
   564   from assms divmod_nat_m_n have "divmod_nat_rel (m - n) n (m div n - Suc 0, m mod n)"
   565     by (cases "m div n") (auto simp add: divmod_nat_rel_def)
   566   with divmod_nat_eq have "divmod_nat (m - n) n = (m div n - Suc 0, m mod n)" by simp
   567   moreover from divmod_nat_div_mod have "divmod_nat (m - n) n = ((m - n) div n, (m - n) mod n)" .
   568   ultimately have "m div n = Suc ((m - n) div n)"
   569     and "m mod n = (m - n) mod n" using m_div_n by simp_all
   570   then show ?thesis using divmod_nat_div_mod by simp
   571 qed
   572 
   573 text {* The ''recursion'' equations for @{const div} and @{const mod} *}
   574 
   575 lemma div_less [simp]:
   576   fixes m n :: nat
   577   assumes "m < n"
   578   shows "m div n = 0"
   579   using assms divmod_nat_base divmod_nat_div_mod by simp
   580 
   581 lemma le_div_geq:
   582   fixes m n :: nat
   583   assumes "0 < n" and "n \<le> m"
   584   shows "m div n = Suc ((m - n) div n)"
   585   using assms divmod_nat_step divmod_nat_div_mod by simp
   586 
   587 lemma mod_less [simp]:
   588   fixes m n :: nat
   589   assumes "m < n"
   590   shows "m mod n = m"
   591   using assms divmod_nat_base divmod_nat_div_mod by simp
   592 
   593 lemma le_mod_geq:
   594   fixes m n :: nat
   595   assumes "n \<le> m"
   596   shows "m mod n = (m - n) mod n"
   597   using assms divmod_nat_step divmod_nat_div_mod by (cases "n = 0") simp_all
   598 
   599 instance proof -
   600   have [simp]: "\<And>n::nat. n div 0 = 0"
   601     by (simp add: div_nat_def divmod_nat_zero)
   602   have [simp]: "\<And>n::nat. 0 div n = 0"
   603   proof -
   604     fix n :: nat
   605     show "0 div n = 0"
   606       by (cases "n = 0") simp_all
   607   qed
   608   show "OFCLASS(nat, semiring_div_class)" proof
   609     fix m n :: nat
   610     show "m div n * n + m mod n = m"
   611       using divmod_nat_rel [of m n] by (simp add: divmod_nat_rel_def)
   612   next
   613     fix m n q :: nat
   614     assume "n \<noteq> 0"
   615     then show "(q + m * n) div n = m + q div n"
   616       by (induct m) (simp_all add: le_div_geq)
   617   next
   618     fix m n q :: nat
   619     assume "m \<noteq> 0"
   620     then show "(m * n) div (m * q) = n div q"
   621     proof (cases "n \<noteq> 0 \<and> q \<noteq> 0")
   622       case False then show ?thesis by auto
   623     next
   624       case True with `m \<noteq> 0`
   625         have "m > 0" and "n > 0" and "q > 0" by auto
   626       then have "\<And>a b. divmod_nat_rel n q (a, b) \<Longrightarrow> divmod_nat_rel (m * n) (m * q) (a, m * b)"
   627         by (auto simp add: divmod_nat_rel_def) (simp_all add: algebra_simps)
   628       moreover from divmod_nat_rel have "divmod_nat_rel n q (n div q, n mod q)" .
   629       ultimately have "divmod_nat_rel (m * n) (m * q) (n div q, m * (n mod q))" .
   630       then show ?thesis by (simp add: div_eq)
   631     qed
   632   qed simp_all
   633 qed
   634 
   635 end
   636 
   637 text {* Simproc for cancelling @{const div} and @{const mod} *}
   638 
   639 ML {*
   640 local
   641 
   642 structure CancelDivMod = CancelDivModFun(struct
   643 
   644   val div_name = @{const_name div};
   645   val mod_name = @{const_name mod};
   646   val mk_binop = HOLogic.mk_binop;
   647   val mk_sum = Nat_Arith.mk_sum;
   648   val dest_sum = Nat_Arith.dest_sum;
   649 
   650   val div_mod_eqs = map mk_meta_eq [@{thm div_mod_equality}, @{thm div_mod_equality2}];
   651 
   652   val trans = trans;
   653 
   654   val prove_eq_sums = Arith_Data.prove_conv2 all_tac (Arith_Data.simp_all_tac
   655     (@{thm monoid_add_class.add_0_left} :: @{thm monoid_add_class.add_0_right} :: @{thms add_ac}))
   656 
   657 end)
   658 
   659 in
   660 
   661 val cancel_div_mod_nat_proc = Simplifier.simproc @{theory}
   662   "cancel_div_mod" ["(m::nat) + n"] (K CancelDivMod.proc);
   663 
   664 val _ = Addsimprocs [cancel_div_mod_nat_proc];
   665 
   666 end
   667 *}
   668 
   669 text {* code generator setup *}
   670 
   671 lemma divmod_nat_if [code]: "divmod_nat m n = (if n = 0 \<or> m < n then (0, m) else
   672   let (q, r) = divmod_nat (m - n) n in (Suc q, r))"
   673 by (simp add: divmod_nat_zero divmod_nat_base divmod_nat_step)
   674     (simp add: divmod_nat_div_mod)
   675 
   676 code_modulename SML
   677   Divides Nat
   678 
   679 code_modulename OCaml
   680   Divides Nat
   681 
   682 code_modulename Haskell
   683   Divides Nat
   684 
   685 
   686 subsubsection {* Quotient *}
   687 
   688 lemma div_geq: "0 < n \<Longrightarrow>  \<not> m < n \<Longrightarrow> m div n = Suc ((m - n) div n)"
   689 by (simp add: le_div_geq linorder_not_less)
   690 
   691 lemma div_if: "0 < n \<Longrightarrow> m div n = (if m < n then 0 else Suc ((m - n) div n))"
   692 by (simp add: div_geq)
   693 
   694 lemma div_mult_self_is_m [simp]: "0<n ==> (m*n) div n = (m::nat)"
   695 by simp
   696 
   697 lemma div_mult_self1_is_m [simp]: "0<n ==> (n*m) div n = (m::nat)"
   698 by simp
   699 
   700 
   701 subsubsection {* Remainder *}
   702 
   703 lemma mod_less_divisor [simp]:
   704   fixes m n :: nat
   705   assumes "n > 0"
   706   shows "m mod n < (n::nat)"
   707   using assms divmod_nat_rel [of m n] unfolding divmod_nat_rel_def by auto
   708 
   709 lemma mod_less_eq_dividend [simp]:
   710   fixes m n :: nat
   711   shows "m mod n \<le> m"
   712 proof (rule add_leD2)
   713   from mod_div_equality have "m div n * n + m mod n = m" .
   714   then show "m div n * n + m mod n \<le> m" by auto
   715 qed
   716 
   717 lemma mod_geq: "\<not> m < (n\<Colon>nat) \<Longrightarrow> m mod n = (m - n) mod n"
   718 by (simp add: le_mod_geq linorder_not_less)
   719 
   720 lemma mod_if: "m mod (n\<Colon>nat) = (if m < n then m else (m - n) mod n)"
   721 by (simp add: le_mod_geq)
   722 
   723 lemma mod_1 [simp]: "m mod Suc 0 = 0"
   724 by (induct m) (simp_all add: mod_geq)
   725 
   726 lemma mod_mult_distrib: "(m mod n) * (k\<Colon>nat) = (m * k) mod (n * k)"
   727   apply (cases "n = 0", simp)
   728   apply (cases "k = 0", simp)
   729   apply (induct m rule: nat_less_induct)
   730   apply (subst mod_if, simp)
   731   apply (simp add: mod_geq diff_mult_distrib)
   732   done
   733 
   734 lemma mod_mult_distrib2: "(k::nat) * (m mod n) = (k*m) mod (k*n)"
   735 by (simp add: mult_commute [of k] mod_mult_distrib)
   736 
   737 (* a simple rearrangement of mod_div_equality: *)
   738 lemma mult_div_cancel: "(n::nat) * (m div n) = m - (m mod n)"
   739 by (cut_tac a = m and b = n in mod_div_equality2, arith)
   740 
   741 lemma mod_le_divisor[simp]: "0 < n \<Longrightarrow> m mod n \<le> (n::nat)"
   742   apply (drule mod_less_divisor [where m = m])
   743   apply simp
   744   done
   745 
   746 subsubsection {* Quotient and Remainder *}
   747 
   748 lemma divmod_nat_rel_mult1_eq:
   749   "divmod_nat_rel b c (q, r) \<Longrightarrow> c > 0
   750    \<Longrightarrow> divmod_nat_rel (a * b) c (a * q + a * r div c, a * r mod c)"
   751 by (auto simp add: split_ifs divmod_nat_rel_def algebra_simps)
   752 
   753 lemma div_mult1_eq:
   754   "(a * b) div c = a * (b div c) + a * (b mod c) div (c::nat)"
   755 apply (cases "c = 0", simp)
   756 apply (blast intro: divmod_nat_rel [THEN divmod_nat_rel_mult1_eq, THEN div_eq])
   757 done
   758 
   759 lemma divmod_nat_rel_add1_eq:
   760   "divmod_nat_rel a c (aq, ar) \<Longrightarrow> divmod_nat_rel b c (bq, br) \<Longrightarrow>  c > 0
   761    \<Longrightarrow> divmod_nat_rel (a + b) c (aq + bq + (ar + br) div c, (ar + br) mod c)"
   762 by (auto simp add: split_ifs divmod_nat_rel_def algebra_simps)
   763 
   764 (*NOT suitable for rewriting: the RHS has an instance of the LHS*)
   765 lemma div_add1_eq:
   766   "(a+b) div (c::nat) = a div c + b div c + ((a mod c + b mod c) div c)"
   767 apply (cases "c = 0", simp)
   768 apply (blast intro: divmod_nat_rel_add1_eq [THEN div_eq] divmod_nat_rel)
   769 done
   770 
   771 lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c"
   772   apply (cut_tac m = q and n = c in mod_less_divisor)
   773   apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto)
   774   apply (erule_tac P = "%x. ?lhs < ?rhs x" in ssubst)
   775   apply (simp add: add_mult_distrib2)
   776   done
   777 
   778 lemma divmod_nat_rel_mult2_eq:
   779   "divmod_nat_rel a b (q, r) \<Longrightarrow> 0 < b \<Longrightarrow> 0 < c
   780    \<Longrightarrow> divmod_nat_rel a (b * c) (q div c, b *(q mod c) + r)"
   781 by (auto simp add: mult_ac divmod_nat_rel_def add_mult_distrib2 [symmetric] mod_lemma)
   782 
   783 lemma div_mult2_eq: "a div (b*c) = (a div b) div (c::nat)"
   784   apply (cases "b = 0", simp)
   785   apply (cases "c = 0", simp)
   786   apply (force simp add: divmod_nat_rel [THEN divmod_nat_rel_mult2_eq, THEN div_eq])
   787   done
   788 
   789 lemma mod_mult2_eq: "a mod (b*c) = b*(a div b mod c) + a mod (b::nat)"
   790   apply (cases "b = 0", simp)
   791   apply (cases "c = 0", simp)
   792   apply (auto simp add: mult_commute divmod_nat_rel [THEN divmod_nat_rel_mult2_eq, THEN mod_eq])
   793   done
   794 
   795 
   796 subsubsection{*Further Facts about Quotient and Remainder*}
   797 
   798 lemma div_1 [simp]: "m div Suc 0 = m"
   799 by (induct m) (simp_all add: div_geq)
   800 
   801 
   802 (* Monotonicity of div in first argument *)
   803 lemma div_le_mono [rule_format (no_asm)]:
   804     "\<forall>m::nat. m \<le> n --> (m div k) \<le> (n div k)"
   805 apply (case_tac "k=0", simp)
   806 apply (induct "n" rule: nat_less_induct, clarify)
   807 apply (case_tac "n<k")
   808 (* 1  case n<k *)
   809 apply simp
   810 (* 2  case n >= k *)
   811 apply (case_tac "m<k")
   812 (* 2.1  case m<k *)
   813 apply simp
   814 (* 2.2  case m>=k *)
   815 apply (simp add: div_geq diff_le_mono)
   816 done
   817 
   818 (* Antimonotonicity of div in second argument *)
   819 lemma div_le_mono2: "!!m::nat. [| 0<m; m\<le>n |] ==> (k div n) \<le> (k div m)"
   820 apply (subgoal_tac "0<n")
   821  prefer 2 apply simp
   822 apply (induct_tac k rule: nat_less_induct)
   823 apply (rename_tac "k")
   824 apply (case_tac "k<n", simp)
   825 apply (subgoal_tac "~ (k<m) ")
   826  prefer 2 apply simp
   827 apply (simp add: div_geq)
   828 apply (subgoal_tac "(k-n) div n \<le> (k-m) div n")
   829  prefer 2
   830  apply (blast intro: div_le_mono diff_le_mono2)
   831 apply (rule le_trans, simp)
   832 apply (simp)
   833 done
   834 
   835 lemma div_le_dividend [simp]: "m div n \<le> (m::nat)"
   836 apply (case_tac "n=0", simp)
   837 apply (subgoal_tac "m div n \<le> m div 1", simp)
   838 apply (rule div_le_mono2)
   839 apply (simp_all (no_asm_simp))
   840 done
   841 
   842 (* Similar for "less than" *)
   843 lemma div_less_dividend [rule_format]:
   844      "!!n::nat. 1<n ==> 0 < m --> m div n < m"
   845 apply (induct_tac m rule: nat_less_induct)
   846 apply (rename_tac "m")
   847 apply (case_tac "m<n", simp)
   848 apply (subgoal_tac "0<n")
   849  prefer 2 apply simp
   850 apply (simp add: div_geq)
   851 apply (case_tac "n<m")
   852  apply (subgoal_tac "(m-n) div n < (m-n) ")
   853   apply (rule impI less_trans_Suc)+
   854 apply assumption
   855   apply (simp_all)
   856 done
   857 
   858 declare div_less_dividend [simp]
   859 
   860 text{*A fact for the mutilated chess board*}
   861 lemma mod_Suc: "Suc(m) mod n = (if Suc(m mod n) = n then 0 else Suc(m mod n))"
   862 apply (case_tac "n=0", simp)
   863 apply (induct "m" rule: nat_less_induct)
   864 apply (case_tac "Suc (na) <n")
   865 (* case Suc(na) < n *)
   866 apply (frule lessI [THEN less_trans], simp add: less_not_refl3)
   867 (* case n \<le> Suc(na) *)
   868 apply (simp add: linorder_not_less le_Suc_eq mod_geq)
   869 apply (auto simp add: Suc_diff_le le_mod_geq)
   870 done
   871 
   872 lemma mod_eq_0_iff: "(m mod d = 0) = (\<exists>q::nat. m = d*q)"
   873 by (auto simp add: dvd_eq_mod_eq_0 [symmetric] dvd_def)
   874 
   875 lemmas mod_eq_0D [dest!] = mod_eq_0_iff [THEN iffD1]
   876 
   877 (*Loses information, namely we also have r<d provided d is nonzero*)
   878 lemma mod_eqD: "(m mod d = r) ==> \<exists>q::nat. m = r + q*d"
   879   apply (cut_tac a = m in mod_div_equality)
   880   apply (simp only: add_ac)
   881   apply (blast intro: sym)
   882   done
   883 
   884 lemma split_div:
   885  "P(n div k :: nat) =
   886  ((k = 0 \<longrightarrow> P 0) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P i)))"
   887  (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
   888 proof
   889   assume P: ?P
   890   show ?Q
   891   proof (cases)
   892     assume "k = 0"
   893     with P show ?Q by simp
   894   next
   895     assume not0: "k \<noteq> 0"
   896     thus ?Q
   897     proof (simp, intro allI impI)
   898       fix i j
   899       assume n: "n = k*i + j" and j: "j < k"
   900       show "P i"
   901       proof (cases)
   902         assume "i = 0"
   903         with n j P show "P i" by simp
   904       next
   905         assume "i \<noteq> 0"
   906         with not0 n j P show "P i" by(simp add:add_ac)
   907       qed
   908     qed
   909   qed
   910 next
   911   assume Q: ?Q
   912   show ?P
   913   proof (cases)
   914     assume "k = 0"
   915     with Q show ?P by simp
   916   next
   917     assume not0: "k \<noteq> 0"
   918     with Q have R: ?R by simp
   919     from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
   920     show ?P by simp
   921   qed
   922 qed
   923 
   924 lemma split_div_lemma:
   925   assumes "0 < n"
   926   shows "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> q = ((m\<Colon>nat) div n)" (is "?lhs \<longleftrightarrow> ?rhs")
   927 proof
   928   assume ?rhs
   929   with mult_div_cancel have nq: "n * q = m - (m mod n)" by simp
   930   then have A: "n * q \<le> m" by simp
   931   have "n - (m mod n) > 0" using mod_less_divisor assms by auto
   932   then have "m < m + (n - (m mod n))" by simp
   933   then have "m < n + (m - (m mod n))" by simp
   934   with nq have "m < n + n * q" by simp
   935   then have B: "m < n * Suc q" by simp
   936   from A B show ?lhs ..
   937 next
   938   assume P: ?lhs
   939   then have "divmod_nat_rel m n (q, m - n * q)"
   940     unfolding divmod_nat_rel_def by (auto simp add: mult_ac)
   941   with divmod_nat_rel_unique divmod_nat_rel [of m n]
   942   have "(q, m - n * q) = (m div n, m mod n)" by auto
   943   then show ?rhs by simp
   944 qed
   945 
   946 theorem split_div':
   947   "P ((m::nat) div n) = ((n = 0 \<and> P 0) \<or>
   948    (\<exists>q. (n * q \<le> m \<and> m < n * (Suc q)) \<and> P q))"
   949   apply (case_tac "0 < n")
   950   apply (simp only: add: split_div_lemma)
   951   apply simp_all
   952   done
   953 
   954 lemma split_mod:
   955  "P(n mod k :: nat) =
   956  ((k = 0 \<longrightarrow> P n) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P j)))"
   957  (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
   958 proof
   959   assume P: ?P
   960   show ?Q
   961   proof (cases)
   962     assume "k = 0"
   963     with P show ?Q by simp
   964   next
   965     assume not0: "k \<noteq> 0"
   966     thus ?Q
   967     proof (simp, intro allI impI)
   968       fix i j
   969       assume "n = k*i + j" "j < k"
   970       thus "P j" using not0 P by(simp add:add_ac mult_ac)
   971     qed
   972   qed
   973 next
   974   assume Q: ?Q
   975   show ?P
   976   proof (cases)
   977     assume "k = 0"
   978     with Q show ?P by simp
   979   next
   980     assume not0: "k \<noteq> 0"
   981     with Q have R: ?R by simp
   982     from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
   983     show ?P by simp
   984   qed
   985 qed
   986 
   987 theorem mod_div_equality': "(m::nat) mod n = m - (m div n) * n"
   988   apply (rule_tac P="%x. m mod n = x - (m div n) * n" in
   989     subst [OF mod_div_equality [of _ n]])
   990   apply arith
   991   done
   992 
   993 lemma div_mod_equality':
   994   fixes m n :: nat
   995   shows "m div n * n = m - m mod n"
   996 proof -
   997   have "m mod n \<le> m mod n" ..
   998   from div_mod_equality have 
   999     "m div n * n + m mod n - m mod n = m - m mod n" by simp
  1000   with diff_add_assoc [OF `m mod n \<le> m mod n`, of "m div n * n"] have
  1001     "m div n * n + (m mod n - m mod n) = m - m mod n"
  1002     by simp
  1003   then show ?thesis by simp
  1004 qed
  1005 
  1006 
  1007 subsubsection {*An ``induction'' law for modulus arithmetic.*}
  1008 
  1009 lemma mod_induct_0:
  1010   assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
  1011   and base: "P i" and i: "i<p"
  1012   shows "P 0"
  1013 proof (rule ccontr)
  1014   assume contra: "\<not>(P 0)"
  1015   from i have p: "0<p" by simp
  1016   have "\<forall>k. 0<k \<longrightarrow> \<not> P (p-k)" (is "\<forall>k. ?A k")
  1017   proof
  1018     fix k
  1019     show "?A k"
  1020     proof (induct k)
  1021       show "?A 0" by simp  -- "by contradiction"
  1022     next
  1023       fix n
  1024       assume ih: "?A n"
  1025       show "?A (Suc n)"
  1026       proof (clarsimp)
  1027         assume y: "P (p - Suc n)"
  1028         have n: "Suc n < p"
  1029         proof (rule ccontr)
  1030           assume "\<not>(Suc n < p)"
  1031           hence "p - Suc n = 0"
  1032             by simp
  1033           with y contra show "False"
  1034             by simp
  1035         qed
  1036         hence n2: "Suc (p - Suc n) = p-n" by arith
  1037         from p have "p - Suc n < p" by arith
  1038         with y step have z: "P ((Suc (p - Suc n)) mod p)"
  1039           by blast
  1040         show "False"
  1041         proof (cases "n=0")
  1042           case True
  1043           with z n2 contra show ?thesis by simp
  1044         next
  1045           case False
  1046           with p have "p-n < p" by arith
  1047           with z n2 False ih show ?thesis by simp
  1048         qed
  1049       qed
  1050     qed
  1051   qed
  1052   moreover
  1053   from i obtain k where "0<k \<and> i+k=p"
  1054     by (blast dest: less_imp_add_positive)
  1055   hence "0<k \<and> i=p-k" by auto
  1056   moreover
  1057   note base
  1058   ultimately
  1059   show "False" by blast
  1060 qed
  1061 
  1062 lemma mod_induct:
  1063   assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
  1064   and base: "P i" and i: "i<p" and j: "j<p"
  1065   shows "P j"
  1066 proof -
  1067   have "\<forall>j<p. P j"
  1068   proof
  1069     fix j
  1070     show "j<p \<longrightarrow> P j" (is "?A j")
  1071     proof (induct j)
  1072       from step base i show "?A 0"
  1073         by (auto elim: mod_induct_0)
  1074     next
  1075       fix k
  1076       assume ih: "?A k"
  1077       show "?A (Suc k)"
  1078       proof
  1079         assume suc: "Suc k < p"
  1080         hence k: "k<p" by simp
  1081         with ih have "P k" ..
  1082         with step k have "P (Suc k mod p)"
  1083           by blast
  1084         moreover
  1085         from suc have "Suc k mod p = Suc k"
  1086           by simp
  1087         ultimately
  1088         show "P (Suc k)" by simp
  1089       qed
  1090     qed
  1091   qed
  1092   with j show ?thesis by blast
  1093 qed
  1094 
  1095 lemma div2_Suc_Suc [simp]: "Suc (Suc m) div 2 = Suc (m div 2)"
  1096 by (auto simp add: numeral_2_eq_2 le_div_geq)
  1097 
  1098 lemma add_self_div_2 [simp]: "(m + m) div 2 = (m::nat)"
  1099 by (simp add: nat_mult_2 [symmetric])
  1100 
  1101 lemma mod2_Suc_Suc [simp]: "Suc(Suc(m)) mod 2 = m mod 2"
  1102 apply (subgoal_tac "m mod 2 < 2")
  1103 apply (erule less_2_cases [THEN disjE])
  1104 apply (simp_all (no_asm_simp) add: Let_def mod_Suc nat_1)
  1105 done
  1106 
  1107 lemma mod2_gr_0 [simp]: "0 < (m\<Colon>nat) mod 2 \<longleftrightarrow> m mod 2 = 1"
  1108 proof -
  1109   { fix n :: nat have  "(n::nat) < 2 \<Longrightarrow> n = 0 \<or> n = 1" by (induct n) simp_all }
  1110   moreover have "m mod 2 < 2" by simp
  1111   ultimately have "m mod 2 = 0 \<or> m mod 2 = 1" .
  1112   then show ?thesis by auto
  1113 qed
  1114 
  1115 text{*These lemmas collapse some needless occurrences of Suc:
  1116     at least three Sucs, since two and fewer are rewritten back to Suc again!
  1117     We already have some rules to simplify operands smaller than 3.*}
  1118 
  1119 lemma div_Suc_eq_div_add3 [simp]: "m div (Suc (Suc (Suc n))) = m div (3+n)"
  1120 by (simp add: Suc3_eq_add_3)
  1121 
  1122 lemma mod_Suc_eq_mod_add3 [simp]: "m mod (Suc (Suc (Suc n))) = m mod (3+n)"
  1123 by (simp add: Suc3_eq_add_3)
  1124 
  1125 lemma Suc_div_eq_add3_div: "(Suc (Suc (Suc m))) div n = (3+m) div n"
  1126 by (simp add: Suc3_eq_add_3)
  1127 
  1128 lemma Suc_mod_eq_add3_mod: "(Suc (Suc (Suc m))) mod n = (3+m) mod n"
  1129 by (simp add: Suc3_eq_add_3)
  1130 
  1131 lemmas Suc_div_eq_add3_div_number_of =
  1132     Suc_div_eq_add3_div [of _ "number_of v", standard]
  1133 declare Suc_div_eq_add3_div_number_of [simp]
  1134 
  1135 lemmas Suc_mod_eq_add3_mod_number_of =
  1136     Suc_mod_eq_add3_mod [of _ "number_of v", standard]
  1137 declare Suc_mod_eq_add3_mod_number_of [simp]
  1138 
  1139 end