src/HOL/ex/Simproc_Tests.thy
author hoelzl
Thu Jan 31 11:31:27 2013 +0100 (2013-01-31)
changeset 50999 3de230ed0547
parent 48559 686cc7c47589
child 51717 9e7d1c139569
permissions -rw-r--r--
introduce order topology
     1 (*  Title:      HOL/ex/Simproc_Tests.thy
     2     Author:     Brian Huffman
     3 *)
     4 
     5 header {* Testing of arithmetic simprocs *}
     6 
     7 theory Simproc_Tests
     8 imports Main
     9 begin
    10 
    11 text {*
    12   This theory tests the various simprocs defined in @{file
    13   "~~/src/HOL/Nat.thy"} and @{file "~~/src/HOL/Numeral_Simprocs.thy"}.
    14   Many of the tests are derived from commented-out code originally
    15   found in @{file "~~/src/HOL/Tools/numeral_simprocs.ML"}.
    16 *}
    17 
    18 subsection {* ML bindings *}
    19 
    20 ML {*
    21   fun test ps = CHANGED (asm_simp_tac (HOL_basic_ss addsimprocs ps) 1)
    22 *}
    23 
    24 subsection {* Cancellation simprocs from @{text Nat.thy} *}
    25 
    26 notepad begin
    27   fix a b c d :: nat
    28   {
    29     assume "b = Suc c" have "a + b = Suc (c + a)"
    30       by (tactic {* test [@{simproc nateq_cancel_sums}] *}) fact
    31   next
    32     assume "b < Suc c" have "a + b < Suc (c + a)"
    33       by (tactic {* test [@{simproc natless_cancel_sums}] *}) fact
    34   next
    35     assume "b \<le> Suc c" have "a + b \<le> Suc (c + a)"
    36       by (tactic {* test [@{simproc natle_cancel_sums}] *}) fact
    37   next
    38     assume "b - Suc c = d" have "a + b - Suc (c + a) = d"
    39       by (tactic {* test [@{simproc natdiff_cancel_sums}] *}) fact
    40   }
    41 end
    42 
    43 schematic_lemma "\<And>(y::?'b::size). size (?x::?'a::size) \<le> size y + size ?x"
    44   by (tactic {* test [@{simproc natle_cancel_sums}] *}) (rule le0)
    45 (* TODO: test more simprocs with schematic variables *)
    46 
    47 subsection {* Abelian group cancellation simprocs *}
    48 
    49 notepad begin
    50   fix a b c u :: "'a::ab_group_add"
    51   {
    52     assume "(a + 0) - (b + 0) = u" have "(a + c) - (b + c) = u"
    53       by (tactic {* test [@{simproc group_cancel_diff}] *}) fact
    54   next
    55     assume "a + 0 = b + 0" have "a + c = b + c"
    56       by (tactic {* test [@{simproc group_cancel_eq}] *}) fact
    57   }
    58 end
    59 (* TODO: more tests for Groups.group_cancel_{add,diff,eq,less,le} *)
    60 
    61 subsection {* @{text int_combine_numerals} *}
    62 
    63 (* FIXME: int_combine_numerals often unnecessarily regroups addition
    64 and rewrites subtraction to negation. Ideally it should behave more
    65 like Groups.abel_cancel_sum, preserving the shape of terms as much as
    66 possible. *)
    67 
    68 notepad begin
    69   fix a b c d oo uu i j k l u v w x y z :: "'a::comm_ring_1"
    70   {
    71     assume "a + - b = u" have "(a + c) - (b + c) = u"
    72       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
    73   next
    74     assume "10 + (2 * l + oo) = uu"
    75     have "l + 2 + 2 + 2 + (l + 2) + (oo + 2) = uu"
    76       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
    77   next
    78     assume "-3 + (i + (j + k)) = y"
    79     have "(i + j + 12 + k) - 15 = y"
    80       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
    81   next
    82     assume "7 + (i + (j + k)) = y"
    83     have "(i + j + 12 + k) - 5 = y"
    84       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
    85   next
    86     assume "-4 * (u * v) + (2 * x + y) = w"
    87     have "(2*x - (u*v) + y) - v*3*u = w"
    88       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
    89   next
    90     assume "2 * x * u * v + y = w"
    91     have "(2*x*u*v + (u*v)*4 + y) - v*u*4 = w"
    92       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
    93   next
    94     assume "3 * (u * v) + (2 * x * u * v + y) = w"
    95     have "(2*x*u*v + (u*v)*4 + y) - v*u = w"
    96       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
    97   next
    98     assume "-3 * (u * v) + (- (x * u * v) + - y) = w"
    99     have "u*v - (x*u*v + (u*v)*4 + y) = w"
   100       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
   101   next
   102     assume "a + - c = d"
   103     have "a + -(b+c) + b = d"
   104       apply (simp only: minus_add_distrib)
   105       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
   106   next
   107     assume "-2 * b + (a + - c) = d"
   108     have "a + -(b+c) - b = d"
   109       apply (simp only: minus_add_distrib)
   110       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
   111   next
   112     assume "-7 + (i + (j + (k + (- u + - y)))) = z"
   113     have "(i + j + -2 + k) - (u + 5 + y) = z"
   114       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
   115   next
   116     assume "-27 + (i + (j + k)) = y"
   117     have "(i + j + -12 + k) - 15 = y"
   118       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
   119   next
   120     assume "27 + (i + (j + k)) = y"
   121     have "(i + j + 12 + k) - -15 = y"
   122       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
   123   next
   124     assume "3 + (i + (j + k)) = y"
   125     have "(i + j + -12 + k) - -15 = y"
   126       by (tactic {* test [@{simproc int_combine_numerals}] *}) fact
   127   }
   128 end
   129 
   130 subsection {* @{text inteq_cancel_numerals} *}
   131 
   132 notepad begin
   133   fix i j k u vv w y z w' y' z' :: "'a::comm_ring_1"
   134   {
   135     assume "u = 0" have "2*u = u"
   136       by (tactic {* test [@{simproc inteq_cancel_numerals}] *}) fact
   137 (* conclusion matches Rings.ring_1_no_zero_divisors_class.mult_cancel_right2 *)
   138   next
   139     assume "i + (j + k) = 3 + (u + y)"
   140     have "(i + j + 12 + k) = u + 15 + y"
   141       by (tactic {* test [@{simproc inteq_cancel_numerals}] *}) fact
   142   next
   143     assume "7 + (j + (i + k)) = y"
   144     have "(i + j*2 + 12 + k) = j + 5 + y"
   145       by (tactic {* test [@{simproc inteq_cancel_numerals}] *}) fact
   146   next
   147     assume "u + (6*z + (4*y + 6*w)) = 6*z' + (4*y' + (6*w' + vv))"
   148     have "2*y + 3*z + 6*w + 2*y + 3*z + 2*u = 2*y' + 3*z' + 6*w' + 2*y' + 3*z' + u + vv"
   149       by (tactic {* test [@{simproc int_combine_numerals}, @{simproc inteq_cancel_numerals}] *}) fact
   150   }
   151 end
   152 
   153 subsection {* @{text intless_cancel_numerals} *}
   154 
   155 notepad begin
   156   fix b c i j k u y :: "'a::linordered_idom"
   157   {
   158     assume "y < 2 * b" have "y - b < b"
   159       by (tactic {* test [@{simproc intless_cancel_numerals}] *}) fact
   160   next
   161     assume "c + y < 4 * b" have "y - (3*b + c) < b - 2*c"
   162       by (tactic {* test [@{simproc intless_cancel_numerals}] *}) fact
   163   next
   164     assume "i + (j + k) < 8 + (u + y)"
   165     have "(i + j + -3 + k) < u + 5 + y"
   166       by (tactic {* test [@{simproc intless_cancel_numerals}] *}) fact
   167   next
   168     assume "9 + (i + (j + k)) < u + y"
   169     have "(i + j + 3 + k) < u + -6 + y"
   170       by (tactic {* test [@{simproc intless_cancel_numerals}] *}) fact
   171   }
   172 end
   173 
   174 subsection {* @{text ring_eq_cancel_numeral_factor} *}
   175 
   176 notepad begin
   177   fix x y :: "'a::{idom,ring_char_0}"
   178   {
   179     assume "3*x = 4*y" have "9*x = 12 * y"
   180       by (tactic {* test [@{simproc ring_eq_cancel_numeral_factor}] *}) fact
   181   next
   182     assume "-3*x = 4*y" have "-99*x = 132 * y"
   183       by (tactic {* test [@{simproc ring_eq_cancel_numeral_factor}] *}) fact
   184   next
   185     assume "111*x = -44*y" have "999*x = -396 * y"
   186       by (tactic {* test [@{simproc ring_eq_cancel_numeral_factor}] *}) fact
   187   next
   188     assume "11*x = 9*y" have "-99*x = -81 * y"
   189       by (tactic {* test [@{simproc ring_eq_cancel_numeral_factor}] *}) fact
   190   next
   191     assume "2*x = y" have "-2 * x = -1 * y"
   192       by (tactic {* test [@{simproc ring_eq_cancel_numeral_factor}] *}) fact
   193   next
   194     assume "2*x = y" have "-2 * x = -y"
   195       by (tactic {* test [@{simproc ring_eq_cancel_numeral_factor}] *}) fact
   196   }
   197 end
   198 
   199 subsection {* @{text int_div_cancel_numeral_factors} *}
   200 
   201 notepad begin
   202   fix x y z :: "'a::{semiring_div,comm_ring_1,ring_char_0}"
   203   {
   204     assume "(3*x) div (4*y) = z" have "(9*x) div (12*y) = z"
   205       by (tactic {* test [@{simproc int_div_cancel_numeral_factors}] *}) fact
   206   next
   207     assume "(-3*x) div (4*y) = z" have "(-99*x) div (132*y) = z"
   208       by (tactic {* test [@{simproc int_div_cancel_numeral_factors}] *}) fact
   209   next
   210     assume "(111*x) div (-44*y) = z" have "(999*x) div (-396*y) = z"
   211       by (tactic {* test [@{simproc int_div_cancel_numeral_factors}] *}) fact
   212   next
   213     assume "(11*x) div (9*y) = z" have "(-99*x) div (-81*y) = z"
   214       by (tactic {* test [@{simproc int_div_cancel_numeral_factors}] *}) fact
   215   next
   216     assume "(2*x) div y = z"
   217     have "(-2 * x) div (-1 * y) = z"
   218       by (tactic {* test [@{simproc int_div_cancel_numeral_factors}] *}) fact
   219   }
   220 end
   221 
   222 subsection {* @{text ring_less_cancel_numeral_factor} *}
   223 
   224 notepad begin
   225   fix x y :: "'a::linordered_idom"
   226   {
   227     assume "3*x < 4*y" have "9*x < 12 * y"
   228       by (tactic {* test [@{simproc ring_less_cancel_numeral_factor}] *}) fact
   229   next
   230     assume "-3*x < 4*y" have "-99*x < 132 * y"
   231       by (tactic {* test [@{simproc ring_less_cancel_numeral_factor}] *}) fact
   232   next
   233     assume "111*x < -44*y" have "999*x < -396 * y"
   234       by (tactic {* test [@{simproc ring_less_cancel_numeral_factor}] *}) fact
   235   next
   236     assume "9*y < 11*x" have "-99*x < -81 * y"
   237       by (tactic {* test [@{simproc ring_less_cancel_numeral_factor}] *}) fact
   238   next
   239     assume "y < 2*x" have "-2 * x < -y"
   240       by (tactic {* test [@{simproc ring_less_cancel_numeral_factor}] *}) fact
   241   next
   242     assume "23*y < x" have "-x < -23 * y"
   243       by (tactic {* test [@{simproc ring_less_cancel_numeral_factor}] *}) fact
   244   }
   245 end
   246 
   247 subsection {* @{text ring_le_cancel_numeral_factor} *}
   248 
   249 notepad begin
   250   fix x y :: "'a::linordered_idom"
   251   {
   252     assume "3*x \<le> 4*y" have "9*x \<le> 12 * y"
   253       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   254   next
   255     assume "-3*x \<le> 4*y" have "-99*x \<le> 132 * y"
   256       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   257   next
   258     assume "111*x \<le> -44*y" have "999*x \<le> -396 * y"
   259       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   260   next
   261     assume "9*y \<le> 11*x" have "-99*x \<le> -81 * y"
   262       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   263   next
   264     assume "y \<le> 2*x" have "-2 * x \<le> -1 * y"
   265       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   266   next
   267     assume "23*y \<le> x" have "-x \<le> -23 * y"
   268       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   269   next
   270     assume "y \<le> 0" have "0 \<le> y * -2"
   271       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   272   next
   273     assume "- x \<le> y" have "- (2 * x) \<le> 2*y"
   274       by (tactic {* test [@{simproc ring_le_cancel_numeral_factor}] *}) fact
   275   }
   276 end
   277 
   278 subsection {* @{text divide_cancel_numeral_factor} *}
   279 
   280 notepad begin
   281   fix x y z :: "'a::{field_inverse_zero,ring_char_0}"
   282   {
   283     assume "(3*x) / (4*y) = z" have "(9*x) / (12 * y) = z"
   284       by (tactic {* test [@{simproc divide_cancel_numeral_factor}] *}) fact
   285   next
   286     assume "(-3*x) / (4*y) = z" have "(-99*x) / (132 * y) = z"
   287       by (tactic {* test [@{simproc divide_cancel_numeral_factor}] *}) fact
   288   next
   289     assume "(111*x) / (-44*y) = z" have "(999*x) / (-396 * y) = z"
   290       by (tactic {* test [@{simproc divide_cancel_numeral_factor}] *}) fact
   291   next
   292     assume "(11*x) / (9*y) = z" have "(-99*x) / (-81 * y) = z"
   293       by (tactic {* test [@{simproc divide_cancel_numeral_factor}] *}) fact
   294   next
   295     assume "(2*x) / y = z" have "(-2 * x) / (-1 * y) = z"
   296       by (tactic {* test [@{simproc divide_cancel_numeral_factor}] *}) fact
   297   }
   298 end
   299 
   300 subsection {* @{text ring_eq_cancel_factor} *}
   301 
   302 notepad begin
   303   fix a b c d k x y :: "'a::idom"
   304   {
   305     assume "k = 0 \<or> x = y" have "x*k = k*y"
   306       by (tactic {* test [@{simproc ring_eq_cancel_factor}] *}) fact
   307   next
   308     assume "k = 0 \<or> 1 = y" have "k = k*y"
   309       by (tactic {* test [@{simproc ring_eq_cancel_factor}] *}) fact
   310   next
   311     assume "b = 0 \<or> a*c = 1" have "a*(b*c) = b"
   312       by (tactic {* test [@{simproc ring_eq_cancel_factor}] *}) fact
   313   next
   314     assume "a = 0 \<or> b = 0 \<or> c = d*x" have "a*(b*c) = d*b*(x*a)"
   315       by (tactic {* test [@{simproc ring_eq_cancel_factor}] *}) fact
   316   next
   317     assume "k = 0 \<or> x = y" have "x*k = k*y"
   318       by (tactic {* test [@{simproc ring_eq_cancel_factor}] *}) fact
   319   next
   320     assume "k = 0 \<or> 1 = y" have "k = k*y"
   321       by (tactic {* test [@{simproc ring_eq_cancel_factor}] *}) fact
   322   }
   323 end
   324 
   325 subsection {* @{text int_div_cancel_factor} *}
   326 
   327 notepad begin
   328   fix a b c d k uu x y :: "'a::semiring_div"
   329   {
   330     assume "(if k = 0 then 0 else x div y) = uu"
   331     have "(x*k) div (k*y) = uu"
   332       by (tactic {* test [@{simproc int_div_cancel_factor}] *}) fact
   333   next
   334     assume "(if k = 0 then 0 else 1 div y) = uu"
   335     have "(k) div (k*y) = uu"
   336       by (tactic {* test [@{simproc int_div_cancel_factor}] *}) fact
   337   next
   338     assume "(if b = 0 then 0 else a * c) = uu"
   339     have "(a*(b*c)) div b = uu"
   340       by (tactic {* test [@{simproc int_div_cancel_factor}] *}) fact
   341   next
   342     assume "(if a = 0 then 0 else if b = 0 then 0 else c div (d * x)) = uu"
   343     have "(a*(b*c)) div (d*b*(x*a)) = uu"
   344       by (tactic {* test [@{simproc int_div_cancel_factor}] *}) fact
   345   }
   346 end
   347 
   348 lemma shows "a*(b*c)/(y*z) = d*(b::'a::linordered_field_inverse_zero)*(x*a)/z"
   349 oops -- "FIXME: need simproc to cover this case"
   350 
   351 subsection {* @{text divide_cancel_factor} *}
   352 
   353 notepad begin
   354   fix a b c d k uu x y :: "'a::field_inverse_zero"
   355   {
   356     assume "(if k = 0 then 0 else x / y) = uu"
   357     have "(x*k) / (k*y) = uu"
   358       by (tactic {* test [@{simproc divide_cancel_factor}] *}) fact
   359   next
   360     assume "(if k = 0 then 0 else 1 / y) = uu"
   361     have "(k) / (k*y) = uu"
   362       by (tactic {* test [@{simproc divide_cancel_factor}] *}) fact
   363   next
   364     assume "(if b = 0 then 0 else a * c / 1) = uu"
   365     have "(a*(b*c)) / b = uu"
   366       by (tactic {* test [@{simproc divide_cancel_factor}] *}) fact
   367   next
   368     assume "(if a = 0 then 0 else if b = 0 then 0 else c / (d * x)) = uu"
   369     have "(a*(b*c)) / (d*b*(x*a)) = uu"
   370       by (tactic {* test [@{simproc divide_cancel_factor}] *}) fact
   371   }
   372 end
   373 
   374 lemma
   375   fixes a b c d x y z :: "'a::linordered_field_inverse_zero"
   376   shows "a*(b*c)/(y*z) = d*(b)*(x*a)/z"
   377 oops -- "FIXME: need simproc to cover this case"
   378 
   379 subsection {* @{text linordered_ring_less_cancel_factor} *}
   380 
   381 notepad begin
   382   fix x y z :: "'a::linordered_idom"
   383   {
   384     assume "0 < z \<Longrightarrow> x < y" have "0 < z \<Longrightarrow> x*z < y*z"
   385       by (tactic {* test [@{simproc linordered_ring_less_cancel_factor}] *}) fact
   386   next
   387     assume "0 < z \<Longrightarrow> x < y" have "0 < z \<Longrightarrow> x*z < z*y"
   388       by (tactic {* test [@{simproc linordered_ring_less_cancel_factor}] *}) fact
   389   next
   390     assume "0 < z \<Longrightarrow> x < y" have "0 < z \<Longrightarrow> z*x < y*z"
   391       by (tactic {* test [@{simproc linordered_ring_less_cancel_factor}] *}) fact
   392   next
   393     assume "0 < z \<Longrightarrow> x < y" have "0 < z \<Longrightarrow> z*x < z*y"
   394       by (tactic {* test [@{simproc linordered_ring_less_cancel_factor}] *}) fact
   395   next
   396     txt "This simproc now uses the simplifier to prove that terms to
   397       be canceled are positive/negative."
   398     assume z_pos: "0 < z"
   399     assume "x < y" have "z*x < z*y"
   400       by (tactic {* CHANGED (asm_simp_tac (HOL_basic_ss
   401         addsimprocs [@{simproc linordered_ring_less_cancel_factor}]
   402         addsimps [@{thm z_pos}]) 1) *}) fact
   403   }
   404 end
   405 
   406 subsection {* @{text linordered_ring_le_cancel_factor} *}
   407 
   408 notepad begin
   409   fix x y z :: "'a::linordered_idom"
   410   {
   411     assume "0 < z \<Longrightarrow> x \<le> y" have "0 < z \<Longrightarrow> x*z \<le> y*z"
   412       by (tactic {* test [@{simproc linordered_ring_le_cancel_factor}] *}) fact
   413   next
   414     assume "0 < z \<Longrightarrow> x \<le> y" have "0 < z \<Longrightarrow> z*x \<le> z*y"
   415       by (tactic {* test [@{simproc linordered_ring_le_cancel_factor}] *}) fact
   416   }
   417 end
   418 
   419 subsection {* @{text field_combine_numerals} *}
   420 
   421 notepad begin
   422   fix x y z uu :: "'a::{field_inverse_zero,ring_char_0}"
   423   {
   424     assume "5 / 6 * x = uu" have "x / 2 + x / 3 = uu"
   425       by (tactic {* test [@{simproc field_combine_numerals}] *}) fact
   426   next
   427     assume "6 / 9 * x + y = uu" have "x / 3 + y + x / 3 = uu"
   428       by (tactic {* test [@{simproc field_combine_numerals}] *}) fact
   429   next
   430     assume "9 / 9 * x = uu" have "2 * x / 3 + x / 3 = uu"
   431       by (tactic {* test [@{simproc field_combine_numerals}] *}) fact
   432   next
   433     assume "y + z = uu"
   434     have "x / 2 + y - 3 * x / 6 + z = uu"
   435       by (tactic {* test [@{simproc field_combine_numerals}] *}) fact
   436   next
   437     assume "1 / 15 * x + y = uu"
   438     have "7 * x / 5 + y - 4 * x / 3 = uu"
   439       by (tactic {* test [@{simproc field_combine_numerals}] *}) fact
   440   }
   441 end
   442 
   443 lemma
   444   fixes x :: "'a::{linordered_field_inverse_zero}"
   445   shows "2/3 * x + x / 3 = uu"
   446 apply (tactic {* test [@{simproc field_combine_numerals}] *})?
   447 oops -- "FIXME: test fails"
   448 
   449 subsection {* @{text nat_combine_numerals} *}
   450 
   451 notepad begin
   452   fix i j k m n u :: nat
   453   {
   454     assume "4*k = u" have "k + 3*k = u"
   455       by (tactic {* test [@{simproc nat_combine_numerals}] *}) fact
   456   next
   457     (* FIXME "Suc (i + 3) \<equiv> i + 4" *)
   458     assume "4 * Suc 0 + i = u" have "Suc (i + 3) = u"
   459       by (tactic {* test [@{simproc nat_combine_numerals}] *}) fact
   460   next
   461     (* FIXME "Suc (i + j + 3 + k) \<equiv> i + j + 4 + k" *)
   462     assume "4 * Suc 0 + (i + (j + k)) = u" have "Suc (i + j + 3 + k) = u"
   463       by (tactic {* test [@{simproc nat_combine_numerals}] *}) fact
   464   next
   465     assume "2 * j + 4 * k = u" have "k + j + 3*k + j = u"
   466       by (tactic {* test [@{simproc nat_combine_numerals}] *}) fact
   467   next
   468     assume "6 * Suc 0 + (5 * (i * j) + (4 * k + i)) = u"
   469     have "Suc (j*i + i + k + 5 + 3*k + i*j*4) = u"
   470       by (tactic {* test [@{simproc nat_combine_numerals}] *}) fact
   471   next
   472     assume "5 * (m * n) = u" have "(2*n*m) + (3*(m*n)) = u"
   473       by (tactic {* test [@{simproc nat_combine_numerals}] *}) fact
   474   }
   475 end
   476 
   477 subsection {* @{text nateq_cancel_numerals} *}
   478 
   479 notepad begin
   480   fix i j k l oo u uu vv w y z w' y' z' :: "nat"
   481   {
   482     assume "Suc 0 * u = 0" have "2*u = (u::nat)"
   483       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   484   next
   485     assume "Suc 0 * u = Suc 0" have "2*u = Suc (u)"
   486       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   487   next
   488     assume "i + (j + k) = 3 * Suc 0 + (u + y)"
   489     have "(i + j + 12 + k) = u + 15 + y"
   490       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   491   next
   492     assume "7 * Suc 0 + (i + (j + k)) = u + y"
   493     have "(i + j + 12 + k) = u + 5 + y"
   494       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   495   next
   496     assume "11 * Suc 0 + (i + (j + k)) = u + y"
   497     have "(i + j + 12 + k) = Suc (u + y)"
   498       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   499   next
   500     assume "i + (j + k) = 2 * Suc 0 + (u + y)"
   501     have "(i + j + 5 + k) = Suc (Suc (Suc (Suc (Suc (Suc (Suc (u + y)))))))"
   502       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   503   next
   504     assume "Suc 0 * u + (2 * y + 3 * z) = Suc 0"
   505     have "2*y + 3*z + 2*u = Suc (u)"
   506       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   507   next
   508     assume "Suc 0 * u + (2 * y + (3 * z + (6 * w + (2 * y + 3 * z)))) = Suc 0"
   509     have "2*y + 3*z + 6*w + 2*y + 3*z + 2*u = Suc (u)"
   510       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   511   next
   512     assume "Suc 0 * u + (2 * y + (3 * z + (6 * w + (2 * y + 3 * z)))) =
   513       2 * y' + (3 * z' + (6 * w' + (2 * y' + (3 * z' + vv))))"
   514     have "2*y + 3*z + 6*w + 2*y + 3*z + 2*u =
   515       2*y' + 3*z' + 6*w' + 2*y' + 3*z' + u + vv"
   516       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   517   next
   518     assume "2 * u + (2 * z + (5 * Suc 0 + 2 * y)) = vv"
   519     have "6 + 2*y + 3*z + 4*u = Suc (vv + 2*u + z)"
   520       by (tactic {* test [@{simproc nateq_cancel_numerals}] *}) fact
   521   }
   522 end
   523 
   524 subsection {* @{text natless_cancel_numerals} *}
   525 
   526 notepad begin
   527   fix length :: "'a \<Rightarrow> nat" and l1 l2 xs :: "'a" and f :: "nat \<Rightarrow> 'a"
   528   fix c i j k l m oo u uu vv w y z w' y' z' :: "nat"
   529   {
   530     assume "0 < j" have "(2*length xs < 2*length xs + j)"
   531       by (tactic {* test [@{simproc natless_cancel_numerals}] *}) fact
   532   next
   533     assume "0 < j" have "(2*length xs < length xs * 2 + j)"
   534       by (tactic {* test [@{simproc natless_cancel_numerals}] *}) fact
   535   next
   536     assume "i + (j + k) < u + y"
   537     have "(i + j + 5 + k) < Suc (Suc (Suc (Suc (Suc (u + y)))))"
   538       by (tactic {* test [@{simproc natless_cancel_numerals}] *}) fact
   539   next
   540     assume "0 < Suc 0 * (m * n) + u" have "(2*n*m) < (3*(m*n)) + u"
   541       by (tactic {* test [@{simproc natless_cancel_numerals}] *}) fact
   542   }
   543 end
   544 
   545 subsection {* @{text natle_cancel_numerals} *}
   546 
   547 notepad begin
   548   fix length :: "'a \<Rightarrow> nat" and l2 l3 :: "'a" and f :: "nat \<Rightarrow> 'a"
   549   fix c e i j k l oo u uu vv w y z w' y' z' :: "nat"
   550   {
   551     assume "u + y \<le> 36 * Suc 0 + (i + (j + k))"
   552     have "Suc (Suc (Suc (Suc (Suc (u + y))))) \<le> ((i + j) + 41 + k)"
   553       by (tactic {* test [@{simproc natle_cancel_numerals}] *}) fact
   554   next
   555     assume "5 * Suc 0 + (case length (f c) of 0 \<Rightarrow> 0 | Suc k \<Rightarrow> k) = 0"
   556     have "(Suc (Suc (Suc (Suc (Suc (Suc (case length (f c) of 0 => 0 | Suc k => k)))))) \<le> Suc 0)"
   557       by (tactic {* test [@{simproc natle_cancel_numerals}] *}) fact
   558   next
   559     assume "6 + length l2 = 0" have "Suc (Suc (Suc (Suc (Suc (Suc (length l1 + length l2)))))) \<le> length l1"
   560       by (tactic {* test [@{simproc natle_cancel_numerals}] *}) fact
   561   next
   562     assume "5 + length l3 = 0"
   563     have "( (Suc (Suc (Suc (Suc (Suc (length (compT P E A ST mxr e) + length l3)))))) \<le> length (compT P E A ST mxr e))"
   564       by (tactic {* test [@{simproc natle_cancel_numerals}] *}) fact
   565   next
   566     assume "5 + length (compT P E (A \<union> A' e) ST mxr c) = 0"
   567     have "( (Suc (Suc (Suc (Suc (Suc (length (compT P E A ST mxr e) + length (compT P E (A Un A' e) ST mxr c))))))) \<le> length (compT P E A ST mxr e))"
   568       by (tactic {* test [@{simproc natle_cancel_numerals}] *}) fact
   569   }
   570 end
   571 
   572 subsection {* @{text natdiff_cancel_numerals} *}
   573 
   574 notepad begin
   575   fix length :: "'a \<Rightarrow> nat" and l2 l3 :: "'a" and f :: "nat \<Rightarrow> 'a"
   576   fix c e i j k l oo u uu vv v w x y z zz w' y' z' :: "nat"
   577   {
   578     assume "i + (j + k) - 3 * Suc 0 = y" have "(i + j + 12 + k) - 15 = y"
   579       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   580   next
   581     assume "7 * Suc 0 + (i + (j + k)) - 0 = y" have "(i + j + 12 + k) - 5 = y"
   582       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   583   next
   584     assume "u - Suc 0 * Suc 0 = y" have "Suc u - 2 = y"
   585       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   586   next
   587     assume "Suc 0 * Suc 0 + u - 0 = y" have "Suc (Suc (Suc u)) - 2 = y"
   588       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   589   next
   590     assume "Suc 0 * Suc 0 + (i + (j + k)) - 0 = y"
   591     have "(i + j + 2 + k) - 1 = y"
   592       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   593   next
   594     assume "i + (j + k) - Suc 0 * Suc 0 = y"
   595     have "(i + j + 1 + k) - 2 = y"
   596       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   597   next
   598     assume "2 * x + y - 2 * (u * v) = w"
   599     have "(2*x + (u*v) + y) - v*3*u = w"
   600       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   601   next
   602     assume "2 * x * u * v + (5 + y) - 0 = w"
   603     have "(2*x*u*v + 5 + (u*v)*4 + y) - v*u*4 = w"
   604       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   605   next
   606     assume "3 * (u * v) + (2 * x * u * v + y) - 0 = w"
   607     have "(2*x*u*v + (u*v)*4 + y) - v*u = w"
   608       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   609   next
   610     assume "3 * u + (2 + (2 * x * u * v + y)) - 0 = w"
   611     have "Suc (Suc (2*x*u*v + u*4 + y)) - u = w"
   612       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   613   next
   614     assume "Suc (Suc 0 * (u * v)) - 0 = w"
   615     have "Suc ((u*v)*4) - v*3*u = w"
   616       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   617   next
   618     assume "2 - 0 = w" have "Suc (Suc ((u*v)*3)) - v*3*u = w"
   619       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   620   next
   621     assume "17 * Suc 0 + (i + (j + k)) - (u + y) = zz"
   622     have "(i + j + 32 + k) - (u + 15 + y) = zz"
   623       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   624   next
   625     assume "u + y - 0 = v" have "Suc (Suc (Suc (Suc (Suc (u + y))))) - 5 = v"
   626       by (tactic {* test [@{simproc natdiff_cancel_numerals}] *}) fact
   627   }
   628 end
   629 
   630 subsection {* Factor-cancellation simprocs for type @{typ nat} *}
   631 
   632 text {* @{text nat_eq_cancel_factor}, @{text nat_less_cancel_factor},
   633 @{text nat_le_cancel_factor}, @{text nat_divide_cancel_factor}, and
   634 @{text nat_dvd_cancel_factor}. *}
   635 
   636 notepad begin
   637   fix a b c d k x y uu :: nat
   638   {
   639     assume "k = 0 \<or> x = y" have "x*k = k*y"
   640       by (tactic {* test [@{simproc nat_eq_cancel_factor}] *}) fact
   641   next
   642     assume "k = 0 \<or> Suc 0 = y" have "k = k*y"
   643       by (tactic {* test [@{simproc nat_eq_cancel_factor}] *}) fact
   644   next
   645     assume "b = 0 \<or> a * c = Suc 0" have "a*(b*c) = b"
   646       by (tactic {* test [@{simproc nat_eq_cancel_factor}] *}) fact
   647   next
   648     assume "a = 0 \<or> b = 0 \<or> c = d * x" have "a*(b*c) = d*b*(x*a)"
   649       by (tactic {* test [@{simproc nat_eq_cancel_factor}] *}) fact
   650   next
   651     assume "0 < k \<and> x < y" have "x*k < k*y"
   652       by (tactic {* test [@{simproc nat_less_cancel_factor}] *}) fact
   653   next
   654     assume "0 < k \<and> Suc 0 < y" have "k < k*y"
   655       by (tactic {* test [@{simproc nat_less_cancel_factor}] *}) fact
   656   next
   657     assume "0 < b \<and> a * c < Suc 0" have "a*(b*c) < b"
   658       by (tactic {* test [@{simproc nat_less_cancel_factor}] *}) fact
   659   next
   660     assume "0 < a \<and> 0 < b \<and> c < d * x" have "a*(b*c) < d*b*(x*a)"
   661       by (tactic {* test [@{simproc nat_less_cancel_factor}] *}) fact
   662   next
   663     assume "0 < k \<longrightarrow> x \<le> y" have "x*k \<le> k*y"
   664       by (tactic {* test [@{simproc nat_le_cancel_factor}] *}) fact
   665   next
   666     assume "0 < k \<longrightarrow> Suc 0 \<le> y" have "k \<le> k*y"
   667       by (tactic {* test [@{simproc nat_le_cancel_factor}] *}) fact
   668   next
   669     assume "0 < b \<longrightarrow> a * c \<le> Suc 0" have "a*(b*c) \<le> b"
   670       by (tactic {* test [@{simproc nat_le_cancel_factor}] *}) fact
   671   next
   672     assume "0 < a \<longrightarrow> 0 < b \<longrightarrow> c \<le> d * x" have "a*(b*c) \<le> d*b*(x*a)"
   673       by (tactic {* test [@{simproc nat_le_cancel_factor}] *}) fact
   674   next
   675     assume "(if k = 0 then 0 else x div y) = uu" have "(x*k) div (k*y) = uu"
   676       by (tactic {* test [@{simproc nat_div_cancel_factor}] *}) fact
   677   next
   678     assume "(if k = 0 then 0 else Suc 0 div y) = uu" have "k div (k*y) = uu"
   679       by (tactic {* test [@{simproc nat_div_cancel_factor}] *}) fact
   680   next
   681     assume "(if b = 0 then 0 else a * c) = uu" have "(a*(b*c)) div (b) = uu"
   682       by (tactic {* test [@{simproc nat_div_cancel_factor}] *}) fact
   683   next
   684     assume "(if a = 0 then 0 else if b = 0 then 0 else c div (d * x)) = uu"
   685     have "(a*(b*c)) div (d*b*(x*a)) = uu"
   686       by (tactic {* test [@{simproc nat_div_cancel_factor}] *}) fact
   687   next
   688     assume "k = 0 \<or> x dvd y" have "(x*k) dvd (k*y)"
   689       by (tactic {* test [@{simproc nat_dvd_cancel_factor}] *}) fact
   690   next
   691     assume "k = 0 \<or> Suc 0 dvd y" have "k dvd (k*y)"
   692       by (tactic {* test [@{simproc nat_dvd_cancel_factor}] *}) fact
   693   next
   694     assume "b = 0 \<or> a * c dvd Suc 0" have "(a*(b*c)) dvd (b)"
   695       by (tactic {* test [@{simproc nat_dvd_cancel_factor}] *}) fact
   696   next
   697     assume "b = 0 \<or> Suc 0 dvd a * c" have "b dvd (a*(b*c))"
   698       by (tactic {* test [@{simproc nat_dvd_cancel_factor}] *}) fact
   699   next
   700     assume "a = 0 \<or> b = 0 \<or> c dvd d * x" have "(a*(b*c)) dvd (d*b*(x*a))"
   701       by (tactic {* test [@{simproc nat_dvd_cancel_factor}] *}) fact
   702   }
   703 end
   704 
   705 subsection {* Numeral-cancellation simprocs for type @{typ nat} *}
   706 
   707 notepad begin
   708   fix x y z :: nat
   709   {
   710     assume "3 * x = 4 * y" have "9*x = 12 * y"
   711       by (tactic {* test [@{simproc nat_eq_cancel_numeral_factor}] *}) fact
   712   next
   713     assume "3 * x < 4 * y" have "9*x < 12 * y"
   714       by (tactic {* test [@{simproc nat_less_cancel_numeral_factor}] *}) fact
   715   next
   716     assume "3 * x \<le> 4 * y" have "9*x \<le> 12 * y"
   717       by (tactic {* test [@{simproc nat_le_cancel_numeral_factor}] *}) fact
   718   next
   719     assume "(3 * x) div (4 * y) = z" have "(9*x) div (12 * y) = z"
   720       by (tactic {* test [@{simproc nat_div_cancel_numeral_factor}] *}) fact
   721   next
   722     assume "(3 * x) dvd (4 * y)" have "(9*x) dvd (12 * y)"
   723       by (tactic {* test [@{simproc nat_dvd_cancel_numeral_factor}] *}) fact
   724   }
   725 end
   726 
   727 subsection {* Integer numeral div/mod simprocs *}
   728 
   729 notepad begin
   730   have "(10::int) div 3 = 3"
   731     by (tactic {* test [@{simproc binary_int_div}] *})
   732   have "(10::int) mod 3 = 1"
   733     by (tactic {* test [@{simproc binary_int_mod}] *})
   734   have "(10::int) div -3 = -4"
   735     by (tactic {* test [@{simproc binary_int_div}] *})
   736   have "(10::int) mod -3 = -2"
   737     by (tactic {* test [@{simproc binary_int_mod}] *})
   738   have "(-10::int) div 3 = -4"
   739     by (tactic {* test [@{simproc binary_int_div}] *})
   740   have "(-10::int) mod 3 = 2"
   741     by (tactic {* test [@{simproc binary_int_mod}] *})
   742   have "(-10::int) div -3 = 3"
   743     by (tactic {* test [@{simproc binary_int_div}] *})
   744   have "(-10::int) mod -3 = -1"
   745     by (tactic {* test [@{simproc binary_int_mod}] *})
   746   have "(8452::int) mod 3 = 1"
   747     by (tactic {* test [@{simproc binary_int_mod}] *})
   748   have "(59485::int) div 434 = 137"
   749     by (tactic {* test [@{simproc binary_int_div}] *})
   750   have "(1000006::int) mod 10 = 6"
   751     by (tactic {* test [@{simproc binary_int_mod}] *})
   752   have "10000000 div 2 = (5000000::int)"
   753     by (tactic {* test [@{simproc binary_int_div}] *})
   754   have "10000001 mod 2 = (1::int)"
   755     by (tactic {* test [@{simproc binary_int_mod}] *})
   756   have "10000055 div 32 = (312501::int)"
   757     by (tactic {* test [@{simproc binary_int_div}] *})
   758   have "10000055 mod 32 = (23::int)"
   759     by (tactic {* test [@{simproc binary_int_mod}] *})
   760   have "100094 div 144 = (695::int)"
   761     by (tactic {* test [@{simproc binary_int_div}] *})
   762   have "100094 mod 144 = (14::int)"
   763     by (tactic {* test [@{simproc binary_int_mod}] *})
   764 end
   765 
   766 end