src/HOL/Divides.thy
author wenzelm
Mon Mar 16 18:24:30 2009 +0100 (2009-03-16)
changeset 30549 d2d7874648bd
parent 30499 1a1a9ca977d6
child 30653 fbd548c4bb6a
permissions -rw-r--r--
simplified method setup;
paulson@3366
     1
(*  Title:      HOL/Divides.thy
paulson@3366
     2
    ID:         $Id$
paulson@3366
     3
    Author:     Lawrence C Paulson, Cambridge University Computer Laboratory
paulson@6865
     4
    Copyright   1999  University of Cambridge
huffman@18154
     5
*)
paulson@3366
     6
haftmann@27651
     7
header {* The division operators div and mod *}
paulson@3366
     8
nipkow@15131
     9
theory Divides
krauss@26748
    10
imports Nat Power Product_Type
haftmann@22993
    11
uses "~~/src/Provers/Arith/cancel_div_mod.ML"
nipkow@15131
    12
begin
paulson@3366
    13
haftmann@25942
    14
subsection {* Syntactic division operations *}
haftmann@25942
    15
haftmann@27651
    16
class div = dvd +
haftmann@27540
    17
  fixes div :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "div" 70)
haftmann@27651
    18
    and mod :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "mod" 70)
haftmann@27540
    19
haftmann@27540
    20
haftmann@27651
    21
subsection {* Abstract division in commutative semirings. *}
haftmann@25942
    22
haftmann@29509
    23
class semiring_div = comm_semiring_1_cancel + div +
haftmann@25942
    24
  assumes mod_div_equality: "a div b * b + a mod b = a"
haftmann@27651
    25
    and div_by_0 [simp]: "a div 0 = 0"
haftmann@27651
    26
    and div_0 [simp]: "0 div a = 0"
haftmann@27651
    27
    and div_mult_self1 [simp]: "b \<noteq> 0 \<Longrightarrow> (a + c * b) div b = c + a div b"
haftmann@25942
    28
begin
haftmann@25942
    29
haftmann@26100
    30
text {* @{const div} and @{const mod} *}
haftmann@26100
    31
haftmann@26062
    32
lemma mod_div_equality2: "b * (a div b) + a mod b = a"
haftmann@26062
    33
  unfolding mult_commute [of b]
haftmann@26062
    34
  by (rule mod_div_equality)
haftmann@26062
    35
huffman@29403
    36
lemma mod_div_equality': "a mod b + a div b * b = a"
huffman@29403
    37
  using mod_div_equality [of a b]
huffman@29403
    38
  by (simp only: add_ac)
huffman@29403
    39
haftmann@26062
    40
lemma div_mod_equality: "((a div b) * b + a mod b) + c = a + c"
nipkow@29667
    41
by (simp add: mod_div_equality)
haftmann@26062
    42
haftmann@26062
    43
lemma div_mod_equality2: "(b * (a div b) + a mod b) + c = a + c"
nipkow@29667
    44
by (simp add: mod_div_equality2)
haftmann@26062
    45
haftmann@27651
    46
lemma mod_by_0 [simp]: "a mod 0 = a"
nipkow@30180
    47
using mod_div_equality [of a zero] by simp
haftmann@27651
    48
haftmann@27651
    49
lemma mod_0 [simp]: "0 mod a = 0"
nipkow@30180
    50
using mod_div_equality [of zero a] div_0 by simp
haftmann@27651
    51
haftmann@27651
    52
lemma div_mult_self2 [simp]:
haftmann@27651
    53
  assumes "b \<noteq> 0"
haftmann@27651
    54
  shows "(a + b * c) div b = c + a div b"
haftmann@27651
    55
  using assms div_mult_self1 [of b a c] by (simp add: mult_commute)
haftmann@26100
    56
haftmann@27651
    57
lemma mod_mult_self1 [simp]: "(a + c * b) mod b = a mod b"
haftmann@27651
    58
proof (cases "b = 0")
haftmann@27651
    59
  case True then show ?thesis by simp
haftmann@27651
    60
next
haftmann@27651
    61
  case False
haftmann@27651
    62
  have "a + c * b = (a + c * b) div b * b + (a + c * b) mod b"
haftmann@27651
    63
    by (simp add: mod_div_equality)
haftmann@27651
    64
  also from False div_mult_self1 [of b a c] have
haftmann@27651
    65
    "\<dots> = (c + a div b) * b + (a + c * b) mod b"
nipkow@29667
    66
      by (simp add: algebra_simps)
haftmann@27651
    67
  finally have "a = a div b * b + (a + c * b) mod b"
haftmann@27651
    68
    by (simp add: add_commute [of a] add_assoc left_distrib)
haftmann@27651
    69
  then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"
haftmann@27651
    70
    by (simp add: mod_div_equality)
haftmann@27651
    71
  then show ?thesis by simp
haftmann@27651
    72
qed
haftmann@27651
    73
haftmann@27651
    74
lemma mod_mult_self2 [simp]: "(a + b * c) mod b = a mod b"
nipkow@29667
    75
by (simp add: mult_commute [of b])
haftmann@27651
    76
haftmann@27651
    77
lemma div_mult_self1_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> b * a div b = a"
haftmann@27651
    78
  using div_mult_self2 [of b 0 a] by simp
haftmann@27651
    79
haftmann@27651
    80
lemma div_mult_self2_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> a * b div b = a"
haftmann@27651
    81
  using div_mult_self1 [of b 0 a] by simp
haftmann@27651
    82
haftmann@27651
    83
lemma mod_mult_self1_is_0 [simp]: "b * a mod b = 0"
haftmann@27651
    84
  using mod_mult_self2 [of 0 b a] by simp
haftmann@27651
    85
haftmann@27651
    86
lemma mod_mult_self2_is_0 [simp]: "a * b mod b = 0"
haftmann@27651
    87
  using mod_mult_self1 [of 0 a b] by simp
haftmann@26062
    88
haftmann@27651
    89
lemma div_by_1 [simp]: "a div 1 = a"
haftmann@27651
    90
  using div_mult_self2_is_id [of 1 a] zero_neq_one by simp
haftmann@27651
    91
haftmann@27651
    92
lemma mod_by_1 [simp]: "a mod 1 = 0"
haftmann@27651
    93
proof -
haftmann@27651
    94
  from mod_div_equality [of a one] div_by_1 have "a + a mod 1 = a" by simp
haftmann@27651
    95
  then have "a + a mod 1 = a + 0" by simp
haftmann@27651
    96
  then show ?thesis by (rule add_left_imp_eq)
haftmann@27651
    97
qed
haftmann@27651
    98
haftmann@27651
    99
lemma mod_self [simp]: "a mod a = 0"
haftmann@27651
   100
  using mod_mult_self2_is_0 [of 1] by simp
haftmann@27651
   101
haftmann@27651
   102
lemma div_self [simp]: "a \<noteq> 0 \<Longrightarrow> a div a = 1"
haftmann@27651
   103
  using div_mult_self2_is_id [of _ 1] by simp
haftmann@27651
   104
haftmann@27676
   105
lemma div_add_self1 [simp]:
haftmann@27651
   106
  assumes "b \<noteq> 0"
haftmann@27651
   107
  shows "(b + a) div b = a div b + 1"
haftmann@27651
   108
  using assms div_mult_self1 [of b a 1] by (simp add: add_commute)
haftmann@26062
   109
haftmann@27676
   110
lemma div_add_self2 [simp]:
haftmann@27651
   111
  assumes "b \<noteq> 0"
haftmann@27651
   112
  shows "(a + b) div b = a div b + 1"
haftmann@27651
   113
  using assms div_add_self1 [of b a] by (simp add: add_commute)
haftmann@27651
   114
haftmann@27676
   115
lemma mod_add_self1 [simp]:
haftmann@27651
   116
  "(b + a) mod b = a mod b"
haftmann@27651
   117
  using mod_mult_self1 [of a 1 b] by (simp add: add_commute)
haftmann@27651
   118
haftmann@27676
   119
lemma mod_add_self2 [simp]:
haftmann@27651
   120
  "(a + b) mod b = a mod b"
haftmann@27651
   121
  using mod_mult_self1 [of a 1 b] by simp
haftmann@27651
   122
haftmann@27651
   123
lemma mod_div_decomp:
haftmann@27651
   124
  fixes a b
haftmann@27651
   125
  obtains q r where "q = a div b" and "r = a mod b"
haftmann@27651
   126
    and "a = q * b + r"
haftmann@27651
   127
proof -
haftmann@27651
   128
  from mod_div_equality have "a = a div b * b + a mod b" by simp
haftmann@27651
   129
  moreover have "a div b = a div b" ..
haftmann@27651
   130
  moreover have "a mod b = a mod b" ..
haftmann@27651
   131
  note that ultimately show thesis by blast
haftmann@27651
   132
qed
haftmann@27651
   133
nipkow@29108
   134
lemma dvd_eq_mod_eq_0 [code unfold]: "a dvd b \<longleftrightarrow> b mod a = 0"
haftmann@25942
   135
proof
haftmann@25942
   136
  assume "b mod a = 0"
haftmann@25942
   137
  with mod_div_equality [of b a] have "b div a * a = b" by simp
haftmann@25942
   138
  then have "b = a * (b div a)" unfolding mult_commute ..
haftmann@25942
   139
  then have "\<exists>c. b = a * c" ..
haftmann@25942
   140
  then show "a dvd b" unfolding dvd_def .
haftmann@25942
   141
next
haftmann@25942
   142
  assume "a dvd b"
haftmann@25942
   143
  then have "\<exists>c. b = a * c" unfolding dvd_def .
haftmann@25942
   144
  then obtain c where "b = a * c" ..
haftmann@25942
   145
  then have "b mod a = a * c mod a" by simp
haftmann@25942
   146
  then have "b mod a = c * a mod a" by (simp add: mult_commute)
haftmann@27651
   147
  then show "b mod a = 0" by simp
haftmann@25942
   148
qed
haftmann@25942
   149
huffman@29403
   150
lemma mod_div_trivial [simp]: "a mod b div b = 0"
huffman@29403
   151
proof (cases "b = 0")
huffman@29403
   152
  assume "b = 0"
huffman@29403
   153
  thus ?thesis by simp
huffman@29403
   154
next
huffman@29403
   155
  assume "b \<noteq> 0"
huffman@29403
   156
  hence "a div b + a mod b div b = (a mod b + a div b * b) div b"
huffman@29403
   157
    by (rule div_mult_self1 [symmetric])
huffman@29403
   158
  also have "\<dots> = a div b"
huffman@29403
   159
    by (simp only: mod_div_equality')
huffman@29403
   160
  also have "\<dots> = a div b + 0"
huffman@29403
   161
    by simp
huffman@29403
   162
  finally show ?thesis
huffman@29403
   163
    by (rule add_left_imp_eq)
huffman@29403
   164
qed
huffman@29403
   165
huffman@29403
   166
lemma mod_mod_trivial [simp]: "a mod b mod b = a mod b"
huffman@29403
   167
proof -
huffman@29403
   168
  have "a mod b mod b = (a mod b + a div b * b) mod b"
huffman@29403
   169
    by (simp only: mod_mult_self1)
huffman@29403
   170
  also have "\<dots> = a mod b"
huffman@29403
   171
    by (simp only: mod_div_equality')
huffman@29403
   172
  finally show ?thesis .
huffman@29403
   173
qed
huffman@29403
   174
nipkow@29925
   175
lemma dvd_imp_mod_0: "a dvd b \<Longrightarrow> b mod a = 0"
nipkow@29948
   176
by (rule dvd_eq_mod_eq_0[THEN iffD1])
nipkow@29925
   177
nipkow@29925
   178
lemma dvd_div_mult_self: "a dvd b \<Longrightarrow> (b div a) * a = b"
nipkow@29925
   179
by (subst (2) mod_div_equality [of b a, symmetric]) (simp add:dvd_imp_mod_0)
nipkow@29925
   180
nipkow@30052
   181
lemma dvd_div_mult: "a dvd b \<Longrightarrow> (b div a) * c = b * c div a"
nipkow@30052
   182
apply (cases "a = 0")
nipkow@30052
   183
 apply simp
nipkow@30052
   184
apply (auto simp: dvd_def mult_assoc)
nipkow@30052
   185
done
nipkow@30052
   186
nipkow@29925
   187
lemma div_dvd_div[simp]:
nipkow@29925
   188
  "a dvd b \<Longrightarrow> a dvd c \<Longrightarrow> (b div a dvd c div a) = (b dvd c)"
nipkow@29925
   189
apply (cases "a = 0")
nipkow@29925
   190
 apply simp
nipkow@29925
   191
apply (unfold dvd_def)
nipkow@29925
   192
apply auto
nipkow@29925
   193
 apply(blast intro:mult_assoc[symmetric])
nipkow@29925
   194
apply(fastsimp simp add: mult_assoc)
nipkow@29925
   195
done
nipkow@29925
   196
huffman@30078
   197
lemma dvd_mod_imp_dvd: "[| k dvd m mod n;  k dvd n |] ==> k dvd m"
huffman@30078
   198
  apply (subgoal_tac "k dvd (m div n) *n + m mod n")
huffman@30078
   199
   apply (simp add: mod_div_equality)
huffman@30078
   200
  apply (simp only: dvd_add dvd_mult)
huffman@30078
   201
  done
huffman@30078
   202
huffman@29403
   203
text {* Addition respects modular equivalence. *}
huffman@29403
   204
huffman@29403
   205
lemma mod_add_left_eq: "(a + b) mod c = (a mod c + b) mod c"
huffman@29403
   206
proof -
huffman@29403
   207
  have "(a + b) mod c = (a div c * c + a mod c + b) mod c"
huffman@29403
   208
    by (simp only: mod_div_equality)
huffman@29403
   209
  also have "\<dots> = (a mod c + b + a div c * c) mod c"
huffman@29403
   210
    by (simp only: add_ac)
huffman@29403
   211
  also have "\<dots> = (a mod c + b) mod c"
huffman@29403
   212
    by (rule mod_mult_self1)
huffman@29403
   213
  finally show ?thesis .
huffman@29403
   214
qed
huffman@29403
   215
huffman@29403
   216
lemma mod_add_right_eq: "(a + b) mod c = (a + b mod c) mod c"
huffman@29403
   217
proof -
huffman@29403
   218
  have "(a + b) mod c = (a + (b div c * c + b mod c)) mod c"
huffman@29403
   219
    by (simp only: mod_div_equality)
huffman@29403
   220
  also have "\<dots> = (a + b mod c + b div c * c) mod c"
huffman@29403
   221
    by (simp only: add_ac)
huffman@29403
   222
  also have "\<dots> = (a + b mod c) mod c"
huffman@29403
   223
    by (rule mod_mult_self1)
huffman@29403
   224
  finally show ?thesis .
huffman@29403
   225
qed
huffman@29403
   226
huffman@29403
   227
lemma mod_add_eq: "(a + b) mod c = (a mod c + b mod c) mod c"
huffman@29403
   228
by (rule trans [OF mod_add_left_eq mod_add_right_eq])
huffman@29403
   229
huffman@29403
   230
lemma mod_add_cong:
huffman@29403
   231
  assumes "a mod c = a' mod c"
huffman@29403
   232
  assumes "b mod c = b' mod c"
huffman@29403
   233
  shows "(a + b) mod c = (a' + b') mod c"
huffman@29403
   234
proof -
huffman@29403
   235
  have "(a mod c + b mod c) mod c = (a' mod c + b' mod c) mod c"
huffman@29403
   236
    unfolding assms ..
huffman@29403
   237
  thus ?thesis
huffman@29403
   238
    by (simp only: mod_add_eq [symmetric])
huffman@29403
   239
qed
huffman@29403
   240
huffman@29403
   241
text {* Multiplication respects modular equivalence. *}
huffman@29403
   242
huffman@29403
   243
lemma mod_mult_left_eq: "(a * b) mod c = ((a mod c) * b) mod c"
huffman@29403
   244
proof -
huffman@29403
   245
  have "(a * b) mod c = ((a div c * c + a mod c) * b) mod c"
huffman@29403
   246
    by (simp only: mod_div_equality)
huffman@29403
   247
  also have "\<dots> = (a mod c * b + a div c * b * c) mod c"
nipkow@29667
   248
    by (simp only: algebra_simps)
huffman@29403
   249
  also have "\<dots> = (a mod c * b) mod c"
huffman@29403
   250
    by (rule mod_mult_self1)
huffman@29403
   251
  finally show ?thesis .
huffman@29403
   252
qed
huffman@29403
   253
huffman@29403
   254
lemma mod_mult_right_eq: "(a * b) mod c = (a * (b mod c)) mod c"
huffman@29403
   255
proof -
huffman@29403
   256
  have "(a * b) mod c = (a * (b div c * c + b mod c)) mod c"
huffman@29403
   257
    by (simp only: mod_div_equality)
huffman@29403
   258
  also have "\<dots> = (a * (b mod c) + a * (b div c) * c) mod c"
nipkow@29667
   259
    by (simp only: algebra_simps)
huffman@29403
   260
  also have "\<dots> = (a * (b mod c)) mod c"
huffman@29403
   261
    by (rule mod_mult_self1)
huffman@29403
   262
  finally show ?thesis .
huffman@29403
   263
qed
huffman@29403
   264
huffman@29403
   265
lemma mod_mult_eq: "(a * b) mod c = ((a mod c) * (b mod c)) mod c"
huffman@29403
   266
by (rule trans [OF mod_mult_left_eq mod_mult_right_eq])
huffman@29403
   267
huffman@29403
   268
lemma mod_mult_cong:
huffman@29403
   269
  assumes "a mod c = a' mod c"
huffman@29403
   270
  assumes "b mod c = b' mod c"
huffman@29403
   271
  shows "(a * b) mod c = (a' * b') mod c"
huffman@29403
   272
proof -
huffman@29403
   273
  have "(a mod c * (b mod c)) mod c = (a' mod c * (b' mod c)) mod c"
huffman@29403
   274
    unfolding assms ..
huffman@29403
   275
  thus ?thesis
huffman@29403
   276
    by (simp only: mod_mult_eq [symmetric])
huffman@29403
   277
qed
huffman@29403
   278
huffman@29404
   279
lemma mod_mod_cancel:
huffman@29404
   280
  assumes "c dvd b"
huffman@29404
   281
  shows "a mod b mod c = a mod c"
huffman@29404
   282
proof -
huffman@29404
   283
  from `c dvd b` obtain k where "b = c * k"
huffman@29404
   284
    by (rule dvdE)
huffman@29404
   285
  have "a mod b mod c = a mod (c * k) mod c"
huffman@29404
   286
    by (simp only: `b = c * k`)
huffman@29404
   287
  also have "\<dots> = (a mod (c * k) + a div (c * k) * k * c) mod c"
huffman@29404
   288
    by (simp only: mod_mult_self1)
huffman@29404
   289
  also have "\<dots> = (a div (c * k) * (c * k) + a mod (c * k)) mod c"
huffman@29404
   290
    by (simp only: add_ac mult_ac)
huffman@29404
   291
  also have "\<dots> = a mod c"
huffman@29404
   292
    by (simp only: mod_div_equality)
huffman@29404
   293
  finally show ?thesis .
huffman@29404
   294
qed
huffman@29404
   295
haftmann@25942
   296
end
haftmann@25942
   297
nipkow@30476
   298
lemma div_mult_div_if_dvd: "(y::'a::{semiring_div,no_zero_divisors}) dvd x \<Longrightarrow> 
nipkow@30476
   299
  z dvd w \<Longrightarrow> (x div y) * (w div z) = (x * w) div (y * z)"
nipkow@30476
   300
unfolding dvd_def
nipkow@30476
   301
  apply clarify
nipkow@30476
   302
  apply (case_tac "y = 0")
nipkow@30476
   303
  apply simp
nipkow@30476
   304
  apply (case_tac "z = 0")
nipkow@30476
   305
  apply simp
nipkow@30476
   306
  apply (simp add: algebra_simps)
nipkow@30476
   307
  apply (subst mult_assoc [symmetric])
nipkow@30476
   308
  apply (simp add: no_zero_divisors)
nipkow@30476
   309
done
nipkow@30476
   310
nipkow@30476
   311
nipkow@30476
   312
lemma div_power: "(y::'a::{semiring_div,no_zero_divisors,recpower}) dvd x \<Longrightarrow>
nipkow@30476
   313
    (x div y)^n = x^n div y^n"
nipkow@30476
   314
apply (induct n)
nipkow@30476
   315
 apply simp
nipkow@30476
   316
apply(simp add: div_mult_div_if_dvd dvd_power_same)
nipkow@30476
   317
done
nipkow@30476
   318
huffman@29405
   319
class ring_div = semiring_div + comm_ring_1
huffman@29405
   320
begin
huffman@29405
   321
huffman@29405
   322
text {* Negation respects modular equivalence. *}
huffman@29405
   323
huffman@29405
   324
lemma mod_minus_eq: "(- a) mod b = (- (a mod b)) mod b"
huffman@29405
   325
proof -
huffman@29405
   326
  have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
huffman@29405
   327
    by (simp only: mod_div_equality)
huffman@29405
   328
  also have "\<dots> = (- (a mod b) + - (a div b) * b) mod b"
huffman@29405
   329
    by (simp only: minus_add_distrib minus_mult_left add_ac)
huffman@29405
   330
  also have "\<dots> = (- (a mod b)) mod b"
huffman@29405
   331
    by (rule mod_mult_self1)
huffman@29405
   332
  finally show ?thesis .
huffman@29405
   333
qed
huffman@29405
   334
huffman@29405
   335
lemma mod_minus_cong:
huffman@29405
   336
  assumes "a mod b = a' mod b"
huffman@29405
   337
  shows "(- a) mod b = (- a') mod b"
huffman@29405
   338
proof -
huffman@29405
   339
  have "(- (a mod b)) mod b = (- (a' mod b)) mod b"
huffman@29405
   340
    unfolding assms ..
huffman@29405
   341
  thus ?thesis
huffman@29405
   342
    by (simp only: mod_minus_eq [symmetric])
huffman@29405
   343
qed
huffman@29405
   344
huffman@29405
   345
text {* Subtraction respects modular equivalence. *}
huffman@29405
   346
huffman@29405
   347
lemma mod_diff_left_eq: "(a - b) mod c = (a mod c - b) mod c"
huffman@29405
   348
  unfolding diff_minus
huffman@29405
   349
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   350
huffman@29405
   351
lemma mod_diff_right_eq: "(a - b) mod c = (a - b mod c) mod c"
huffman@29405
   352
  unfolding diff_minus
huffman@29405
   353
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   354
huffman@29405
   355
lemma mod_diff_eq: "(a - b) mod c = (a mod c - b mod c) mod c"
huffman@29405
   356
  unfolding diff_minus
huffman@29405
   357
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   358
huffman@29405
   359
lemma mod_diff_cong:
huffman@29405
   360
  assumes "a mod c = a' mod c"
huffman@29405
   361
  assumes "b mod c = b' mod c"
huffman@29405
   362
  shows "(a - b) mod c = (a' - b') mod c"
huffman@29405
   363
  unfolding diff_minus using assms
huffman@29405
   364
  by (intro mod_add_cong mod_minus_cong)
huffman@29405
   365
nipkow@30180
   366
lemma dvd_neg_div: "y dvd x \<Longrightarrow> -x div y = - (x div y)"
nipkow@30180
   367
apply (case_tac "y = 0") apply simp
nipkow@30180
   368
apply (auto simp add: dvd_def)
nipkow@30180
   369
apply (subgoal_tac "-(y * k) = y * - k")
nipkow@30180
   370
 apply (erule ssubst)
nipkow@30180
   371
 apply (erule div_mult_self1_is_id)
nipkow@30180
   372
apply simp
nipkow@30180
   373
done
nipkow@30180
   374
nipkow@30180
   375
lemma dvd_div_neg: "y dvd x \<Longrightarrow> x div -y = - (x div y)"
nipkow@30180
   376
apply (case_tac "y = 0") apply simp
nipkow@30180
   377
apply (auto simp add: dvd_def)
nipkow@30180
   378
apply (subgoal_tac "y * k = -y * -k")
nipkow@30180
   379
 apply (erule ssubst)
nipkow@30180
   380
 apply (rule div_mult_self1_is_id)
nipkow@30180
   381
 apply simp
nipkow@30180
   382
apply simp
nipkow@30180
   383
done
nipkow@30180
   384
huffman@29405
   385
end
huffman@29405
   386
haftmann@25942
   387
haftmann@26100
   388
subsection {* Division on @{typ nat} *}
haftmann@26100
   389
haftmann@26100
   390
text {*
haftmann@26100
   391
  We define @{const div} and @{const mod} on @{typ nat} by means
haftmann@26100
   392
  of a characteristic relation with two input arguments
haftmann@26100
   393
  @{term "m\<Colon>nat"}, @{term "n\<Colon>nat"} and two output arguments
haftmann@26100
   394
  @{term "q\<Colon>nat"}(uotient) and @{term "r\<Colon>nat"}(emainder).
haftmann@26100
   395
*}
haftmann@26100
   396
haftmann@26100
   397
definition divmod_rel :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool" where
haftmann@26100
   398
  "divmod_rel m n q r \<longleftrightarrow> m = q * n + r \<and> (if n > 0 then 0 \<le> r \<and> r < n else q = 0)"
haftmann@26100
   399
haftmann@26100
   400
text {* @{const divmod_rel} is total: *}
haftmann@26100
   401
haftmann@26100
   402
lemma divmod_rel_ex:
haftmann@26100
   403
  obtains q r where "divmod_rel m n q r"
haftmann@26100
   404
proof (cases "n = 0")
haftmann@26100
   405
  case True with that show thesis
haftmann@26100
   406
    by (auto simp add: divmod_rel_def)
haftmann@26100
   407
next
haftmann@26100
   408
  case False
haftmann@26100
   409
  have "\<exists>q r. m = q * n + r \<and> r < n"
haftmann@26100
   410
  proof (induct m)
haftmann@26100
   411
    case 0 with `n \<noteq> 0`
haftmann@26100
   412
    have "(0\<Colon>nat) = 0 * n + 0 \<and> 0 < n" by simp
haftmann@26100
   413
    then show ?case by blast
haftmann@26100
   414
  next
haftmann@26100
   415
    case (Suc m) then obtain q' r'
haftmann@26100
   416
      where m: "m = q' * n + r'" and n: "r' < n" by auto
haftmann@26100
   417
    then show ?case proof (cases "Suc r' < n")
haftmann@26100
   418
      case True
haftmann@26100
   419
      from m n have "Suc m = q' * n + Suc r'" by simp
haftmann@26100
   420
      with True show ?thesis by blast
haftmann@26100
   421
    next
haftmann@26100
   422
      case False then have "n \<le> Suc r'" by auto
haftmann@26100
   423
      moreover from n have "Suc r' \<le> n" by auto
haftmann@26100
   424
      ultimately have "n = Suc r'" by auto
haftmann@26100
   425
      with m have "Suc m = Suc q' * n + 0" by simp
haftmann@26100
   426
      with `n \<noteq> 0` show ?thesis by blast
haftmann@26100
   427
    qed
haftmann@26100
   428
  qed
haftmann@26100
   429
  with that show thesis
haftmann@26100
   430
    using `n \<noteq> 0` by (auto simp add: divmod_rel_def)
haftmann@26100
   431
qed
haftmann@26100
   432
haftmann@26100
   433
text {* @{const divmod_rel} is injective: *}
haftmann@26100
   434
haftmann@26100
   435
lemma divmod_rel_unique_div:
haftmann@26100
   436
  assumes "divmod_rel m n q r"
haftmann@26100
   437
    and "divmod_rel m n q' r'"
haftmann@26100
   438
  shows "q = q'"
haftmann@26100
   439
proof (cases "n = 0")
haftmann@26100
   440
  case True with assms show ?thesis
haftmann@26100
   441
    by (simp add: divmod_rel_def)
haftmann@26100
   442
next
haftmann@26100
   443
  case False
haftmann@26100
   444
  have aux: "\<And>q r q' r'. q' * n + r' = q * n + r \<Longrightarrow> r < n \<Longrightarrow> q' \<le> (q\<Colon>nat)"
haftmann@26100
   445
  apply (rule leI)
haftmann@26100
   446
  apply (subst less_iff_Suc_add)
haftmann@26100
   447
  apply (auto simp add: add_mult_distrib)
haftmann@26100
   448
  done
haftmann@26100
   449
  from `n \<noteq> 0` assms show ?thesis
haftmann@26100
   450
    by (auto simp add: divmod_rel_def
haftmann@26100
   451
      intro: order_antisym dest: aux sym)
haftmann@26100
   452
qed
haftmann@26100
   453
haftmann@26100
   454
lemma divmod_rel_unique_mod:
haftmann@26100
   455
  assumes "divmod_rel m n q r"
haftmann@26100
   456
    and "divmod_rel m n q' r'"
haftmann@26100
   457
  shows "r = r'"
haftmann@26100
   458
proof -
haftmann@26100
   459
  from assms have "q = q'" by (rule divmod_rel_unique_div)
haftmann@26100
   460
  with assms show ?thesis by (simp add: divmod_rel_def)
haftmann@26100
   461
qed
haftmann@26100
   462
haftmann@26100
   463
text {*
haftmann@26100
   464
  We instantiate divisibility on the natural numbers by
haftmann@26100
   465
  means of @{const divmod_rel}:
haftmann@26100
   466
*}
haftmann@25942
   467
haftmann@25942
   468
instantiation nat :: semiring_div
haftmann@25571
   469
begin
haftmann@25571
   470
haftmann@26100
   471
definition divmod :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where
haftmann@28562
   472
  [code del]: "divmod m n = (THE (q, r). divmod_rel m n q r)"
haftmann@26100
   473
haftmann@26100
   474
definition div_nat where
haftmann@26100
   475
  "m div n = fst (divmod m n)"
haftmann@26100
   476
haftmann@26100
   477
definition mod_nat where
haftmann@26100
   478
  "m mod n = snd (divmod m n)"
haftmann@25571
   479
haftmann@26100
   480
lemma divmod_div_mod:
haftmann@26100
   481
  "divmod m n = (m div n, m mod n)"
haftmann@26100
   482
  unfolding div_nat_def mod_nat_def by simp
haftmann@26100
   483
haftmann@26100
   484
lemma divmod_eq:
haftmann@26100
   485
  assumes "divmod_rel m n q r" 
haftmann@26100
   486
  shows "divmod m n = (q, r)"
haftmann@26100
   487
  using assms by (auto simp add: divmod_def
haftmann@26100
   488
    dest: divmod_rel_unique_div divmod_rel_unique_mod)
haftmann@25942
   489
haftmann@26100
   490
lemma div_eq:
haftmann@26100
   491
  assumes "divmod_rel m n q r" 
haftmann@26100
   492
  shows "m div n = q"
haftmann@26100
   493
  using assms by (auto dest: divmod_eq simp add: div_nat_def)
haftmann@26100
   494
haftmann@26100
   495
lemma mod_eq:
haftmann@26100
   496
  assumes "divmod_rel m n q r" 
haftmann@26100
   497
  shows "m mod n = r"
haftmann@26100
   498
  using assms by (auto dest: divmod_eq simp add: mod_nat_def)
haftmann@25571
   499
haftmann@26100
   500
lemma divmod_rel: "divmod_rel m n (m div n) (m mod n)"
haftmann@26100
   501
proof -
haftmann@26100
   502
  from divmod_rel_ex
haftmann@26100
   503
    obtain q r where rel: "divmod_rel m n q r" .
haftmann@26100
   504
  moreover with div_eq mod_eq have "m div n = q" and "m mod n = r"
haftmann@26100
   505
    by simp_all
haftmann@26100
   506
  ultimately show ?thesis by simp
haftmann@26100
   507
qed
paulson@14267
   508
haftmann@26100
   509
lemma divmod_zero:
haftmann@26100
   510
  "divmod m 0 = (0, m)"
haftmann@26100
   511
proof -
haftmann@26100
   512
  from divmod_rel [of m 0] show ?thesis
haftmann@26100
   513
    unfolding divmod_div_mod divmod_rel_def by simp
haftmann@26100
   514
qed
haftmann@25942
   515
haftmann@26100
   516
lemma divmod_base:
haftmann@26100
   517
  assumes "m < n"
haftmann@26100
   518
  shows "divmod m n = (0, m)"
haftmann@26100
   519
proof -
haftmann@26100
   520
  from divmod_rel [of m n] show ?thesis
haftmann@26100
   521
    unfolding divmod_div_mod divmod_rel_def
haftmann@26100
   522
    using assms by (cases "m div n = 0")
haftmann@26100
   523
      (auto simp add: gr0_conv_Suc [of "m div n"])
haftmann@26100
   524
qed
haftmann@25942
   525
haftmann@26100
   526
lemma divmod_step:
haftmann@26100
   527
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   528
  shows "divmod m n = (Suc ((m - n) div n), (m - n) mod n)"
haftmann@26100
   529
proof -
haftmann@26100
   530
  from divmod_rel have divmod_m_n: "divmod_rel m n (m div n) (m mod n)" .
haftmann@26100
   531
  with assms have m_div_n: "m div n \<ge> 1"
haftmann@26100
   532
    by (cases "m div n") (auto simp add: divmod_rel_def)
huffman@30079
   533
  from assms divmod_m_n have "divmod_rel (m - n) n (m div n - Suc 0) (m mod n)"
haftmann@26100
   534
    by (cases "m div n") (auto simp add: divmod_rel_def)
huffman@30079
   535
  with divmod_eq have "divmod (m - n) n = (m div n - Suc 0, m mod n)" by simp
haftmann@26100
   536
  moreover from divmod_div_mod have "divmod (m - n) n = ((m - n) div n, (m - n) mod n)" .
haftmann@26100
   537
  ultimately have "m div n = Suc ((m - n) div n)"
haftmann@26100
   538
    and "m mod n = (m - n) mod n" using m_div_n by simp_all
haftmann@26100
   539
  then show ?thesis using divmod_div_mod by simp
haftmann@26100
   540
qed
haftmann@25942
   541
wenzelm@26300
   542
text {* The ''recursion'' equations for @{const div} and @{const mod} *}
haftmann@26100
   543
haftmann@26100
   544
lemma div_less [simp]:
haftmann@26100
   545
  fixes m n :: nat
haftmann@26100
   546
  assumes "m < n"
haftmann@26100
   547
  shows "m div n = 0"
haftmann@26100
   548
  using assms divmod_base divmod_div_mod by simp
haftmann@25942
   549
haftmann@26100
   550
lemma le_div_geq:
haftmann@26100
   551
  fixes m n :: nat
haftmann@26100
   552
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   553
  shows "m div n = Suc ((m - n) div n)"
haftmann@26100
   554
  using assms divmod_step divmod_div_mod by simp
paulson@14267
   555
haftmann@26100
   556
lemma mod_less [simp]:
haftmann@26100
   557
  fixes m n :: nat
haftmann@26100
   558
  assumes "m < n"
haftmann@26100
   559
  shows "m mod n = m"
haftmann@26100
   560
  using assms divmod_base divmod_div_mod by simp
haftmann@26100
   561
haftmann@26100
   562
lemma le_mod_geq:
haftmann@26100
   563
  fixes m n :: nat
haftmann@26100
   564
  assumes "n \<le> m"
haftmann@26100
   565
  shows "m mod n = (m - n) mod n"
haftmann@26100
   566
  using assms divmod_step divmod_div_mod by (cases "n = 0") simp_all
paulson@14267
   567
haftmann@25942
   568
instance proof
haftmann@26100
   569
  fix m n :: nat show "m div n * n + m mod n = m"
haftmann@26100
   570
    using divmod_rel [of m n] by (simp add: divmod_rel_def)
haftmann@25942
   571
next
haftmann@26100
   572
  fix n :: nat show "n div 0 = 0"
haftmann@26100
   573
    using divmod_zero divmod_div_mod [of n 0] by simp
haftmann@25942
   574
next
haftmann@27651
   575
  fix n :: nat show "0 div n = 0"
haftmann@27651
   576
    using divmod_rel [of 0 n] by (cases n) (simp_all add: divmod_rel_def)
haftmann@27651
   577
next
haftmann@27651
   578
  fix m n q :: nat assume "n \<noteq> 0" then show "(q + m * n) div n = m + q div n"
haftmann@25942
   579
    by (induct m) (simp_all add: le_div_geq)
haftmann@25942
   580
qed
haftmann@26100
   581
haftmann@25942
   582
end
paulson@14267
   583
haftmann@26100
   584
text {* Simproc for cancelling @{const div} and @{const mod} *}
haftmann@25942
   585
haftmann@27651
   586
(*lemmas mod_div_equality_nat = semiring_div_class.times_div_mod_plus_zero_one.mod_div_equality [of "m\<Colon>nat" n, standard]
haftmann@27651
   587
lemmas mod_div_equality2_nat = mod_div_equality2 [of "n\<Colon>nat" m, standard*)
haftmann@25942
   588
haftmann@25942
   589
ML {*
haftmann@25942
   590
structure CancelDivModData =
haftmann@25942
   591
struct
haftmann@25942
   592
haftmann@26100
   593
val div_name = @{const_name div};
haftmann@26100
   594
val mod_name = @{const_name mod};
haftmann@25942
   595
val mk_binop = HOLogic.mk_binop;
haftmann@30496
   596
val mk_sum = Nat_Arith.mk_sum;
haftmann@30496
   597
val dest_sum = Nat_Arith.dest_sum;
haftmann@25942
   598
haftmann@25942
   599
(*logic*)
paulson@14267
   600
haftmann@25942
   601
val div_mod_eqs = map mk_meta_eq [@{thm div_mod_equality}, @{thm div_mod_equality2}]
haftmann@25942
   602
haftmann@25942
   603
val trans = trans
haftmann@25942
   604
haftmann@25942
   605
val prove_eq_sums =
haftmann@25942
   606
  let val simps = @{thm add_0} :: @{thm add_0_right} :: @{thms add_ac}
haftmann@30496
   607
  in Arith_Data.prove_conv2 all_tac (Arith_Data.simp_all_tac simps) end;
haftmann@25942
   608
haftmann@25942
   609
end;
haftmann@25942
   610
haftmann@25942
   611
structure CancelDivMod = CancelDivModFun(CancelDivModData);
haftmann@25942
   612
wenzelm@28262
   613
val cancel_div_mod_proc = Simplifier.simproc (the_context ())
haftmann@26100
   614
  "cancel_div_mod" ["(m::nat) + n"] (K CancelDivMod.proc);
haftmann@25942
   615
haftmann@25942
   616
Addsimprocs[cancel_div_mod_proc];
haftmann@25942
   617
*}
haftmann@25942
   618
haftmann@26100
   619
text {* code generator setup *}
haftmann@26100
   620
haftmann@26100
   621
lemma divmod_if [code]: "divmod m n = (if n = 0 \<or> m < n then (0, m) else
haftmann@26100
   622
  let (q, r) = divmod (m - n) n in (Suc q, r))"
nipkow@29667
   623
by (simp add: divmod_zero divmod_base divmod_step)
haftmann@26100
   624
    (simp add: divmod_div_mod)
haftmann@26100
   625
haftmann@26100
   626
code_modulename SML
haftmann@26100
   627
  Divides Nat
haftmann@26100
   628
haftmann@26100
   629
code_modulename OCaml
haftmann@26100
   630
  Divides Nat
haftmann@26100
   631
haftmann@26100
   632
code_modulename Haskell
haftmann@26100
   633
  Divides Nat
haftmann@26100
   634
haftmann@26100
   635
haftmann@26100
   636
subsubsection {* Quotient *}
haftmann@26100
   637
haftmann@26100
   638
lemma div_geq: "0 < n \<Longrightarrow>  \<not> m < n \<Longrightarrow> m div n = Suc ((m - n) div n)"
nipkow@29667
   639
by (simp add: le_div_geq linorder_not_less)
haftmann@26100
   640
haftmann@26100
   641
lemma div_if: "0 < n \<Longrightarrow> m div n = (if m < n then 0 else Suc ((m - n) div n))"
nipkow@29667
   642
by (simp add: div_geq)
haftmann@26100
   643
haftmann@26100
   644
lemma div_mult_self_is_m [simp]: "0<n ==> (m*n) div n = (m::nat)"
nipkow@29667
   645
by simp
haftmann@26100
   646
haftmann@26100
   647
lemma div_mult_self1_is_m [simp]: "0<n ==> (n*m) div n = (m::nat)"
nipkow@29667
   648
by simp
haftmann@26100
   649
haftmann@25942
   650
haftmann@25942
   651
subsubsection {* Remainder *}
haftmann@25942
   652
haftmann@26100
   653
lemma mod_less_divisor [simp]:
haftmann@26100
   654
  fixes m n :: nat
haftmann@26100
   655
  assumes "n > 0"
haftmann@26100
   656
  shows "m mod n < (n::nat)"
haftmann@26100
   657
  using assms divmod_rel unfolding divmod_rel_def by auto
paulson@14267
   658
haftmann@26100
   659
lemma mod_less_eq_dividend [simp]:
haftmann@26100
   660
  fixes m n :: nat
haftmann@26100
   661
  shows "m mod n \<le> m"
haftmann@26100
   662
proof (rule add_leD2)
haftmann@26100
   663
  from mod_div_equality have "m div n * n + m mod n = m" .
haftmann@26100
   664
  then show "m div n * n + m mod n \<le> m" by auto
haftmann@26100
   665
qed
haftmann@26100
   666
haftmann@26100
   667
lemma mod_geq: "\<not> m < (n\<Colon>nat) \<Longrightarrow> m mod n = (m - n) mod n"
nipkow@29667
   668
by (simp add: le_mod_geq linorder_not_less)
paulson@14267
   669
haftmann@26100
   670
lemma mod_if: "m mod (n\<Colon>nat) = (if m < n then m else (m - n) mod n)"
nipkow@29667
   671
by (simp add: le_mod_geq)
haftmann@26100
   672
paulson@14267
   673
lemma mod_1 [simp]: "m mod Suc 0 = 0"
nipkow@29667
   674
by (induct m) (simp_all add: mod_geq)
paulson@14267
   675
haftmann@26100
   676
lemma mod_mult_distrib: "(m mod n) * (k\<Colon>nat) = (m * k) mod (n * k)"
wenzelm@22718
   677
  apply (cases "n = 0", simp)
wenzelm@22718
   678
  apply (cases "k = 0", simp)
wenzelm@22718
   679
  apply (induct m rule: nat_less_induct)
wenzelm@22718
   680
  apply (subst mod_if, simp)
wenzelm@22718
   681
  apply (simp add: mod_geq diff_mult_distrib)
wenzelm@22718
   682
  done
paulson@14267
   683
paulson@14267
   684
lemma mod_mult_distrib2: "(k::nat) * (m mod n) = (k*m) mod (k*n)"
nipkow@29667
   685
by (simp add: mult_commute [of k] mod_mult_distrib)
paulson@14267
   686
paulson@14267
   687
(* a simple rearrangement of mod_div_equality: *)
paulson@14267
   688
lemma mult_div_cancel: "(n::nat) * (m div n) = m - (m mod n)"
nipkow@29667
   689
by (cut_tac a = m and b = n in mod_div_equality2, arith)
paulson@14267
   690
nipkow@15439
   691
lemma mod_le_divisor[simp]: "0 < n \<Longrightarrow> m mod n \<le> (n::nat)"
wenzelm@22718
   692
  apply (drule mod_less_divisor [where m = m])
wenzelm@22718
   693
  apply simp
wenzelm@22718
   694
  done
paulson@14267
   695
haftmann@26100
   696
subsubsection {* Quotient and Remainder *}
paulson@14267
   697
haftmann@26100
   698
lemma divmod_rel_mult1_eq:
haftmann@26100
   699
  "[| divmod_rel b c q r; c > 0 |]
haftmann@26100
   700
   ==> divmod_rel (a*b) c (a*q + a*r div c) (a*r mod c)"
nipkow@29667
   701
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   702
paulson@14267
   703
lemma div_mult1_eq: "(a*b) div c = a*(b div c) + a*(b mod c) div (c::nat)"
nipkow@25134
   704
apply (cases "c = 0", simp)
haftmann@26100
   705
apply (blast intro: divmod_rel [THEN divmod_rel_mult1_eq, THEN div_eq])
nipkow@25134
   706
done
paulson@14267
   707
haftmann@26100
   708
lemma divmod_rel_add1_eq:
haftmann@26100
   709
  "[| divmod_rel a c aq ar; divmod_rel b c bq br;  c > 0 |]
haftmann@26100
   710
   ==> divmod_rel (a + b) c (aq + bq + (ar+br) div c) ((ar + br) mod c)"
nipkow@29667
   711
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   712
paulson@14267
   713
(*NOT suitable for rewriting: the RHS has an instance of the LHS*)
paulson@14267
   714
lemma div_add1_eq:
nipkow@25134
   715
  "(a+b) div (c::nat) = a div c + b div c + ((a mod c + b mod c) div c)"
nipkow@25134
   716
apply (cases "c = 0", simp)
haftmann@26100
   717
apply (blast intro: divmod_rel_add1_eq [THEN div_eq] divmod_rel)
nipkow@25134
   718
done
paulson@14267
   719
paulson@14267
   720
lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c"
wenzelm@22718
   721
  apply (cut_tac m = q and n = c in mod_less_divisor)
wenzelm@22718
   722
  apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto)
wenzelm@22718
   723
  apply (erule_tac P = "%x. ?lhs < ?rhs x" in ssubst)
wenzelm@22718
   724
  apply (simp add: add_mult_distrib2)
wenzelm@22718
   725
  done
paulson@10559
   726
haftmann@26100
   727
lemma divmod_rel_mult2_eq: "[| divmod_rel a b q r;  0 < b;  0 < c |]
haftmann@26100
   728
      ==> divmod_rel a (b*c) (q div c) (b*(q mod c) + r)"
nipkow@29667
   729
by (auto simp add: mult_ac divmod_rel_def add_mult_distrib2 [symmetric] mod_lemma)
paulson@14267
   730
paulson@14267
   731
lemma div_mult2_eq: "a div (b*c) = (a div b) div (c::nat)"
wenzelm@22718
   732
  apply (cases "b = 0", simp)
wenzelm@22718
   733
  apply (cases "c = 0", simp)
haftmann@26100
   734
  apply (force simp add: divmod_rel [THEN divmod_rel_mult2_eq, THEN div_eq])
wenzelm@22718
   735
  done
paulson@14267
   736
paulson@14267
   737
lemma mod_mult2_eq: "a mod (b*c) = b*(a div b mod c) + a mod (b::nat)"
wenzelm@22718
   738
  apply (cases "b = 0", simp)
wenzelm@22718
   739
  apply (cases "c = 0", simp)
haftmann@26100
   740
  apply (auto simp add: mult_commute divmod_rel [THEN divmod_rel_mult2_eq, THEN mod_eq])
wenzelm@22718
   741
  done
paulson@14267
   742
paulson@14267
   743
haftmann@25942
   744
subsubsection{*Cancellation of Common Factors in Division*}
paulson@14267
   745
paulson@14267
   746
lemma div_mult_mult_lemma:
wenzelm@22718
   747
    "[| (0::nat) < b;  0 < c |] ==> (c*a) div (c*b) = a div b"
nipkow@29667
   748
by (auto simp add: div_mult2_eq)
paulson@14267
   749
paulson@14267
   750
lemma div_mult_mult1 [simp]: "(0::nat) < c ==> (c*a) div (c*b) = a div b"
wenzelm@22718
   751
  apply (cases "b = 0")
wenzelm@22718
   752
  apply (auto simp add: linorder_neq_iff [of b] div_mult_mult_lemma)
wenzelm@22718
   753
  done
paulson@14267
   754
paulson@14267
   755
lemma div_mult_mult2 [simp]: "(0::nat) < c ==> (a*c) div (b*c) = a div b"
wenzelm@22718
   756
  apply (drule div_mult_mult1)
wenzelm@22718
   757
  apply (auto simp add: mult_commute)
wenzelm@22718
   758
  done
paulson@14267
   759
paulson@14267
   760
haftmann@25942
   761
subsubsection{*Further Facts about Quotient and Remainder*}
paulson@14267
   762
paulson@14267
   763
lemma div_1 [simp]: "m div Suc 0 = m"
nipkow@29667
   764
by (induct m) (simp_all add: div_geq)
paulson@14267
   765
paulson@14267
   766
paulson@14267
   767
(* Monotonicity of div in first argument *)
paulson@14267
   768
lemma div_le_mono [rule_format (no_asm)]:
wenzelm@22718
   769
    "\<forall>m::nat. m \<le> n --> (m div k) \<le> (n div k)"
paulson@14267
   770
apply (case_tac "k=0", simp)
paulson@15251
   771
apply (induct "n" rule: nat_less_induct, clarify)
paulson@14267
   772
apply (case_tac "n<k")
paulson@14267
   773
(* 1  case n<k *)
paulson@14267
   774
apply simp
paulson@14267
   775
(* 2  case n >= k *)
paulson@14267
   776
apply (case_tac "m<k")
paulson@14267
   777
(* 2.1  case m<k *)
paulson@14267
   778
apply simp
paulson@14267
   779
(* 2.2  case m>=k *)
nipkow@15439
   780
apply (simp add: div_geq diff_le_mono)
paulson@14267
   781
done
paulson@14267
   782
paulson@14267
   783
(* Antimonotonicity of div in second argument *)
paulson@14267
   784
lemma div_le_mono2: "!!m::nat. [| 0<m; m\<le>n |] ==> (k div n) \<le> (k div m)"
paulson@14267
   785
apply (subgoal_tac "0<n")
wenzelm@22718
   786
 prefer 2 apply simp
paulson@15251
   787
apply (induct_tac k rule: nat_less_induct)
paulson@14267
   788
apply (rename_tac "k")
paulson@14267
   789
apply (case_tac "k<n", simp)
paulson@14267
   790
apply (subgoal_tac "~ (k<m) ")
wenzelm@22718
   791
 prefer 2 apply simp
paulson@14267
   792
apply (simp add: div_geq)
paulson@15251
   793
apply (subgoal_tac "(k-n) div n \<le> (k-m) div n")
paulson@14267
   794
 prefer 2
paulson@14267
   795
 apply (blast intro: div_le_mono diff_le_mono2)
paulson@14267
   796
apply (rule le_trans, simp)
nipkow@15439
   797
apply (simp)
paulson@14267
   798
done
paulson@14267
   799
paulson@14267
   800
lemma div_le_dividend [simp]: "m div n \<le> (m::nat)"
paulson@14267
   801
apply (case_tac "n=0", simp)
paulson@14267
   802
apply (subgoal_tac "m div n \<le> m div 1", simp)
paulson@14267
   803
apply (rule div_le_mono2)
paulson@14267
   804
apply (simp_all (no_asm_simp))
paulson@14267
   805
done
paulson@14267
   806
wenzelm@22718
   807
(* Similar for "less than" *)
paulson@17085
   808
lemma div_less_dividend [rule_format]:
paulson@14267
   809
     "!!n::nat. 1<n ==> 0 < m --> m div n < m"
paulson@15251
   810
apply (induct_tac m rule: nat_less_induct)
paulson@14267
   811
apply (rename_tac "m")
paulson@14267
   812
apply (case_tac "m<n", simp)
paulson@14267
   813
apply (subgoal_tac "0<n")
wenzelm@22718
   814
 prefer 2 apply simp
paulson@14267
   815
apply (simp add: div_geq)
paulson@14267
   816
apply (case_tac "n<m")
paulson@15251
   817
 apply (subgoal_tac "(m-n) div n < (m-n) ")
paulson@14267
   818
  apply (rule impI less_trans_Suc)+
paulson@14267
   819
apply assumption
nipkow@15439
   820
  apply (simp_all)
paulson@14267
   821
done
paulson@14267
   822
paulson@17085
   823
declare div_less_dividend [simp]
paulson@17085
   824
paulson@14267
   825
text{*A fact for the mutilated chess board*}
paulson@14267
   826
lemma mod_Suc: "Suc(m) mod n = (if Suc(m mod n) = n then 0 else Suc(m mod n))"
paulson@14267
   827
apply (case_tac "n=0", simp)
paulson@15251
   828
apply (induct "m" rule: nat_less_induct)
paulson@14267
   829
apply (case_tac "Suc (na) <n")
paulson@14267
   830
(* case Suc(na) < n *)
paulson@14267
   831
apply (frule lessI [THEN less_trans], simp add: less_not_refl3)
paulson@14267
   832
(* case n \<le> Suc(na) *)
paulson@16796
   833
apply (simp add: linorder_not_less le_Suc_eq mod_geq)
nipkow@15439
   834
apply (auto simp add: Suc_diff_le le_mod_geq)
paulson@14267
   835
done
paulson@14267
   836
paulson@14267
   837
haftmann@27651
   838
subsubsection {* The Divides Relation *}
paulson@24286
   839
paulson@14267
   840
lemma dvd_1_left [iff]: "Suc 0 dvd k"
wenzelm@22718
   841
  unfolding dvd_def by simp
paulson@14267
   842
paulson@14267
   843
lemma dvd_1_iff_1 [simp]: "(m dvd Suc 0) = (m = Suc 0)"
nipkow@29667
   844
by (simp add: dvd_def)
paulson@14267
   845
huffman@30079
   846
lemma nat_dvd_1_iff_1 [simp]: "m dvd (1::nat) \<longleftrightarrow> m = 1"
huffman@30079
   847
by (simp add: dvd_def)
huffman@30079
   848
paulson@14267
   849
lemma dvd_anti_sym: "[| m dvd n; n dvd m |] ==> m = (n::nat)"
wenzelm@22718
   850
  unfolding dvd_def
wenzelm@22718
   851
  by (force dest: mult_eq_self_implies_10 simp add: mult_assoc mult_eq_1_iff)
paulson@14267
   852
haftmann@23684
   853
text {* @{term "op dvd"} is a partial order *}
haftmann@23684
   854
haftmann@29509
   855
interpretation dvd!: order "op dvd" "\<lambda>n m \<Colon> nat. n dvd m \<and> \<not> m dvd n"
haftmann@28823
   856
  proof qed (auto intro: dvd_refl dvd_trans dvd_anti_sym)
paulson@14267
   857
nipkow@30042
   858
lemma nat_dvd_diff[simp]: "[| k dvd m; k dvd n |] ==> k dvd (m-n :: nat)"
nipkow@30042
   859
unfolding dvd_def
nipkow@30042
   860
by (blast intro: diff_mult_distrib2 [symmetric])
paulson@14267
   861
paulson@14267
   862
lemma dvd_diffD: "[| k dvd m-n; k dvd n; n\<le>m |] ==> k dvd (m::nat)"
wenzelm@22718
   863
  apply (erule linorder_not_less [THEN iffD2, THEN add_diff_inverse, THEN subst])
wenzelm@22718
   864
  apply (blast intro: dvd_add)
wenzelm@22718
   865
  done
paulson@14267
   866
paulson@14267
   867
lemma dvd_diffD1: "[| k dvd m-n; k dvd m; n\<le>m |] ==> k dvd (n::nat)"
nipkow@30042
   868
by (drule_tac m = m in nat_dvd_diff, auto)
paulson@14267
   869
paulson@14267
   870
lemma dvd_reduce: "(k dvd n + k) = (k dvd (n::nat))"
wenzelm@22718
   871
  apply (rule iffI)
wenzelm@22718
   872
   apply (erule_tac [2] dvd_add)
wenzelm@22718
   873
   apply (rule_tac [2] dvd_refl)
wenzelm@22718
   874
  apply (subgoal_tac "n = (n+k) -k")
wenzelm@22718
   875
   prefer 2 apply simp
wenzelm@22718
   876
  apply (erule ssubst)
nipkow@30042
   877
  apply (erule nat_dvd_diff)
wenzelm@22718
   878
  apply (rule dvd_refl)
wenzelm@22718
   879
  done
paulson@14267
   880
paulson@14267
   881
lemma dvd_mod: "!!n::nat. [| f dvd m; f dvd n |] ==> f dvd m mod n"
wenzelm@22718
   882
  unfolding dvd_def
wenzelm@22718
   883
  apply (case_tac "n = 0", auto)
wenzelm@22718
   884
  apply (blast intro: mod_mult_distrib2 [symmetric])
wenzelm@22718
   885
  done
paulson@14267
   886
paulson@14267
   887
lemma dvd_mod_iff: "k dvd n ==> ((k::nat) dvd m mod n) = (k dvd m)"
nipkow@29667
   888
by (blast intro: dvd_mod_imp_dvd dvd_mod)
paulson@14267
   889
paulson@14267
   890
lemma dvd_mult_cancel: "!!k::nat. [| k*m dvd k*n; 0<k |] ==> m dvd n"
wenzelm@22718
   891
  unfolding dvd_def
wenzelm@22718
   892
  apply (erule exE)
wenzelm@22718
   893
  apply (simp add: mult_ac)
wenzelm@22718
   894
  done
paulson@14267
   895
paulson@14267
   896
lemma dvd_mult_cancel1: "0<m ==> (m*n dvd m) = (n = (1::nat))"
wenzelm@22718
   897
  apply auto
wenzelm@22718
   898
   apply (subgoal_tac "m*n dvd m*1")
wenzelm@22718
   899
   apply (drule dvd_mult_cancel, auto)
wenzelm@22718
   900
  done
paulson@14267
   901
paulson@14267
   902
lemma dvd_mult_cancel2: "0<m ==> (n*m dvd m) = (n = (1::nat))"
wenzelm@22718
   903
  apply (subst mult_commute)
wenzelm@22718
   904
  apply (erule dvd_mult_cancel1)
wenzelm@22718
   905
  done
paulson@14267
   906
paulson@14267
   907
lemma dvd_imp_le: "[| k dvd n; 0 < n |] ==> k \<le> (n::nat)"
wenzelm@22718
   908
  apply (unfold dvd_def, clarify)
wenzelm@22718
   909
  apply (simp_all (no_asm_use) add: zero_less_mult_iff)
wenzelm@22718
   910
  apply (erule conjE)
wenzelm@22718
   911
  apply (rule le_trans)
wenzelm@22718
   912
   apply (rule_tac [2] le_refl [THEN mult_le_mono])
wenzelm@22718
   913
   apply (erule_tac [2] Suc_leI, simp)
wenzelm@22718
   914
  done
paulson@14267
   915
paulson@14267
   916
lemma dvd_mult_div_cancel: "n dvd m ==> n * (m div n) = (m::nat)"
wenzelm@22718
   917
  apply (subgoal_tac "m mod n = 0")
wenzelm@22718
   918
   apply (simp add: mult_div_cancel)
wenzelm@22718
   919
  apply (simp only: dvd_eq_mod_eq_0)
wenzelm@22718
   920
  done
paulson@14267
   921
nipkow@25162
   922
lemma nat_zero_less_power_iff [simp]: "(x^n > 0) = (x > (0::nat) | n=0)"
wenzelm@22718
   923
  by (induct n) auto
haftmann@21408
   924
haftmann@21408
   925
lemma power_dvd_imp_le: "[|i^m dvd i^n;  (1::nat) < i|] ==> m \<le> n"
wenzelm@22718
   926
  apply (rule power_le_imp_le_exp, assumption)
wenzelm@22718
   927
  apply (erule dvd_imp_le, simp)
wenzelm@22718
   928
  done
haftmann@21408
   929
paulson@14267
   930
lemma mod_eq_0_iff: "(m mod d = 0) = (\<exists>q::nat. m = d*q)"
nipkow@29667
   931
by (auto simp add: dvd_eq_mod_eq_0 [symmetric] dvd_def)
paulson@17084
   932
wenzelm@22718
   933
lemmas mod_eq_0D [dest!] = mod_eq_0_iff [THEN iffD1]
paulson@14267
   934
paulson@14267
   935
(*Loses information, namely we also have r<d provided d is nonzero*)
paulson@14267
   936
lemma mod_eqD: "(m mod d = r) ==> \<exists>q::nat. m = r + q*d"
haftmann@27651
   937
  apply (cut_tac a = m in mod_div_equality)
wenzelm@22718
   938
  apply (simp only: add_ac)
wenzelm@22718
   939
  apply (blast intro: sym)
wenzelm@22718
   940
  done
paulson@14267
   941
nipkow@13152
   942
lemma split_div:
nipkow@13189
   943
 "P(n div k :: nat) =
nipkow@13189
   944
 ((k = 0 \<longrightarrow> P 0) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P i)))"
nipkow@13189
   945
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
   946
proof
nipkow@13189
   947
  assume P: ?P
nipkow@13189
   948
  show ?Q
nipkow@13189
   949
  proof (cases)
nipkow@13189
   950
    assume "k = 0"
haftmann@27651
   951
    with P show ?Q by simp
nipkow@13189
   952
  next
nipkow@13189
   953
    assume not0: "k \<noteq> 0"
nipkow@13189
   954
    thus ?Q
nipkow@13189
   955
    proof (simp, intro allI impI)
nipkow@13189
   956
      fix i j
nipkow@13189
   957
      assume n: "n = k*i + j" and j: "j < k"
nipkow@13189
   958
      show "P i"
nipkow@13189
   959
      proof (cases)
wenzelm@22718
   960
        assume "i = 0"
wenzelm@22718
   961
        with n j P show "P i" by simp
nipkow@13189
   962
      next
wenzelm@22718
   963
        assume "i \<noteq> 0"
wenzelm@22718
   964
        with not0 n j P show "P i" by(simp add:add_ac)
nipkow@13189
   965
      qed
nipkow@13189
   966
    qed
nipkow@13189
   967
  qed
nipkow@13189
   968
next
nipkow@13189
   969
  assume Q: ?Q
nipkow@13189
   970
  show ?P
nipkow@13189
   971
  proof (cases)
nipkow@13189
   972
    assume "k = 0"
haftmann@27651
   973
    with Q show ?P by simp
nipkow@13189
   974
  next
nipkow@13189
   975
    assume not0: "k \<noteq> 0"
nipkow@13189
   976
    with Q have R: ?R by simp
nipkow@13189
   977
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
   978
    show ?P by simp
nipkow@13189
   979
  qed
nipkow@13189
   980
qed
nipkow@13189
   981
berghofe@13882
   982
lemma split_div_lemma:
haftmann@26100
   983
  assumes "0 < n"
haftmann@26100
   984
  shows "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> q = ((m\<Colon>nat) div n)" (is "?lhs \<longleftrightarrow> ?rhs")
haftmann@26100
   985
proof
haftmann@26100
   986
  assume ?rhs
haftmann@26100
   987
  with mult_div_cancel have nq: "n * q = m - (m mod n)" by simp
haftmann@26100
   988
  then have A: "n * q \<le> m" by simp
haftmann@26100
   989
  have "n - (m mod n) > 0" using mod_less_divisor assms by auto
haftmann@26100
   990
  then have "m < m + (n - (m mod n))" by simp
haftmann@26100
   991
  then have "m < n + (m - (m mod n))" by simp
haftmann@26100
   992
  with nq have "m < n + n * q" by simp
haftmann@26100
   993
  then have B: "m < n * Suc q" by simp
haftmann@26100
   994
  from A B show ?lhs ..
haftmann@26100
   995
next
haftmann@26100
   996
  assume P: ?lhs
haftmann@26100
   997
  then have "divmod_rel m n q (m - n * q)"
haftmann@26100
   998
    unfolding divmod_rel_def by (auto simp add: mult_ac)
haftmann@26100
   999
  then show ?rhs using divmod_rel by (rule divmod_rel_unique_div)
haftmann@26100
  1000
qed
berghofe@13882
  1001
berghofe@13882
  1002
theorem split_div':
berghofe@13882
  1003
  "P ((m::nat) div n) = ((n = 0 \<and> P 0) \<or>
paulson@14267
  1004
   (\<exists>q. (n * q \<le> m \<and> m < n * (Suc q)) \<and> P q))"
berghofe@13882
  1005
  apply (case_tac "0 < n")
berghofe@13882
  1006
  apply (simp only: add: split_div_lemma)
haftmann@27651
  1007
  apply simp_all
berghofe@13882
  1008
  done
berghofe@13882
  1009
nipkow@13189
  1010
lemma split_mod:
nipkow@13189
  1011
 "P(n mod k :: nat) =
nipkow@13189
  1012
 ((k = 0 \<longrightarrow> P n) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P j)))"
nipkow@13189
  1013
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
  1014
proof
nipkow@13189
  1015
  assume P: ?P
nipkow@13189
  1016
  show ?Q
nipkow@13189
  1017
  proof (cases)
nipkow@13189
  1018
    assume "k = 0"
haftmann@27651
  1019
    with P show ?Q by simp
nipkow@13189
  1020
  next
nipkow@13189
  1021
    assume not0: "k \<noteq> 0"
nipkow@13189
  1022
    thus ?Q
nipkow@13189
  1023
    proof (simp, intro allI impI)
nipkow@13189
  1024
      fix i j
nipkow@13189
  1025
      assume "n = k*i + j" "j < k"
nipkow@13189
  1026
      thus "P j" using not0 P by(simp add:add_ac mult_ac)
nipkow@13189
  1027
    qed
nipkow@13189
  1028
  qed
nipkow@13189
  1029
next
nipkow@13189
  1030
  assume Q: ?Q
nipkow@13189
  1031
  show ?P
nipkow@13189
  1032
  proof (cases)
nipkow@13189
  1033
    assume "k = 0"
haftmann@27651
  1034
    with Q show ?P by simp
nipkow@13189
  1035
  next
nipkow@13189
  1036
    assume not0: "k \<noteq> 0"
nipkow@13189
  1037
    with Q have R: ?R by simp
nipkow@13189
  1038
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
  1039
    show ?P by simp
nipkow@13189
  1040
  qed
nipkow@13189
  1041
qed
nipkow@13189
  1042
berghofe@13882
  1043
theorem mod_div_equality': "(m::nat) mod n = m - (m div n) * n"
berghofe@13882
  1044
  apply (rule_tac P="%x. m mod n = x - (m div n) * n" in
berghofe@13882
  1045
    subst [OF mod_div_equality [of _ n]])
berghofe@13882
  1046
  apply arith
berghofe@13882
  1047
  done
berghofe@13882
  1048
haftmann@22800
  1049
lemma div_mod_equality':
haftmann@22800
  1050
  fixes m n :: nat
haftmann@22800
  1051
  shows "m div n * n = m - m mod n"
haftmann@22800
  1052
proof -
haftmann@22800
  1053
  have "m mod n \<le> m mod n" ..
haftmann@22800
  1054
  from div_mod_equality have 
haftmann@22800
  1055
    "m div n * n + m mod n - m mod n = m - m mod n" by simp
haftmann@22800
  1056
  with diff_add_assoc [OF `m mod n \<le> m mod n`, of "m div n * n"] have
haftmann@22800
  1057
    "m div n * n + (m mod n - m mod n) = m - m mod n"
haftmann@22800
  1058
    by simp
haftmann@22800
  1059
  then show ?thesis by simp
haftmann@22800
  1060
qed
haftmann@22800
  1061
haftmann@22800
  1062
haftmann@25942
  1063
subsubsection {*An ``induction'' law for modulus arithmetic.*}
paulson@14640
  1064
paulson@14640
  1065
lemma mod_induct_0:
paulson@14640
  1066
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1067
  and base: "P i" and i: "i<p"
paulson@14640
  1068
  shows "P 0"
paulson@14640
  1069
proof (rule ccontr)
paulson@14640
  1070
  assume contra: "\<not>(P 0)"
paulson@14640
  1071
  from i have p: "0<p" by simp
paulson@14640
  1072
  have "\<forall>k. 0<k \<longrightarrow> \<not> P (p-k)" (is "\<forall>k. ?A k")
paulson@14640
  1073
  proof
paulson@14640
  1074
    fix k
paulson@14640
  1075
    show "?A k"
paulson@14640
  1076
    proof (induct k)
paulson@14640
  1077
      show "?A 0" by simp  -- "by contradiction"
paulson@14640
  1078
    next
paulson@14640
  1079
      fix n
paulson@14640
  1080
      assume ih: "?A n"
paulson@14640
  1081
      show "?A (Suc n)"
paulson@14640
  1082
      proof (clarsimp)
wenzelm@22718
  1083
        assume y: "P (p - Suc n)"
wenzelm@22718
  1084
        have n: "Suc n < p"
wenzelm@22718
  1085
        proof (rule ccontr)
wenzelm@22718
  1086
          assume "\<not>(Suc n < p)"
wenzelm@22718
  1087
          hence "p - Suc n = 0"
wenzelm@22718
  1088
            by simp
wenzelm@22718
  1089
          with y contra show "False"
wenzelm@22718
  1090
            by simp
wenzelm@22718
  1091
        qed
wenzelm@22718
  1092
        hence n2: "Suc (p - Suc n) = p-n" by arith
wenzelm@22718
  1093
        from p have "p - Suc n < p" by arith
wenzelm@22718
  1094
        with y step have z: "P ((Suc (p - Suc n)) mod p)"
wenzelm@22718
  1095
          by blast
wenzelm@22718
  1096
        show "False"
wenzelm@22718
  1097
        proof (cases "n=0")
wenzelm@22718
  1098
          case True
wenzelm@22718
  1099
          with z n2 contra show ?thesis by simp
wenzelm@22718
  1100
        next
wenzelm@22718
  1101
          case False
wenzelm@22718
  1102
          with p have "p-n < p" by arith
wenzelm@22718
  1103
          with z n2 False ih show ?thesis by simp
wenzelm@22718
  1104
        qed
paulson@14640
  1105
      qed
paulson@14640
  1106
    qed
paulson@14640
  1107
  qed
paulson@14640
  1108
  moreover
paulson@14640
  1109
  from i obtain k where "0<k \<and> i+k=p"
paulson@14640
  1110
    by (blast dest: less_imp_add_positive)
paulson@14640
  1111
  hence "0<k \<and> i=p-k" by auto
paulson@14640
  1112
  moreover
paulson@14640
  1113
  note base
paulson@14640
  1114
  ultimately
paulson@14640
  1115
  show "False" by blast
paulson@14640
  1116
qed
paulson@14640
  1117
paulson@14640
  1118
lemma mod_induct:
paulson@14640
  1119
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1120
  and base: "P i" and i: "i<p" and j: "j<p"
paulson@14640
  1121
  shows "P j"
paulson@14640
  1122
proof -
paulson@14640
  1123
  have "\<forall>j<p. P j"
paulson@14640
  1124
  proof
paulson@14640
  1125
    fix j
paulson@14640
  1126
    show "j<p \<longrightarrow> P j" (is "?A j")
paulson@14640
  1127
    proof (induct j)
paulson@14640
  1128
      from step base i show "?A 0"
wenzelm@22718
  1129
        by (auto elim: mod_induct_0)
paulson@14640
  1130
    next
paulson@14640
  1131
      fix k
paulson@14640
  1132
      assume ih: "?A k"
paulson@14640
  1133
      show "?A (Suc k)"
paulson@14640
  1134
      proof
wenzelm@22718
  1135
        assume suc: "Suc k < p"
wenzelm@22718
  1136
        hence k: "k<p" by simp
wenzelm@22718
  1137
        with ih have "P k" ..
wenzelm@22718
  1138
        with step k have "P (Suc k mod p)"
wenzelm@22718
  1139
          by blast
wenzelm@22718
  1140
        moreover
wenzelm@22718
  1141
        from suc have "Suc k mod p = Suc k"
wenzelm@22718
  1142
          by simp
wenzelm@22718
  1143
        ultimately
wenzelm@22718
  1144
        show "P (Suc k)" by simp
paulson@14640
  1145
      qed
paulson@14640
  1146
    qed
paulson@14640
  1147
  qed
paulson@14640
  1148
  with j show ?thesis by blast
paulson@14640
  1149
qed
paulson@14640
  1150
paulson@3366
  1151
end