Replaced group_ and ring_simps by algebra_simps;
removed compare_rls - use algebra_simps now
(* Title: HOL/Divides.thy
ID: $Id$
Author: Lawrence C Paulson, Cambridge University Computer Laboratory
Copyright 1999 University of Cambridge
*)
header {* The division operators div and mod *}
theory Divides
imports Nat Power Product_Type
uses "~~/src/Provers/Arith/cancel_div_mod.ML"
begin
subsection {* Syntactic division operations *}
class div = dvd +
fixes div :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "div" 70)
and mod :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "mod" 70)
subsection {* Abstract division in commutative semirings. *}
class semiring_div = comm_semiring_1_cancel + div +
assumes mod_div_equality: "a div b * b + a mod b = a"
and div_by_0 [simp]: "a div 0 = 0"
and div_0 [simp]: "0 div a = 0"
and div_mult_self1 [simp]: "b \<noteq> 0 \<Longrightarrow> (a + c * b) div b = c + a div b"
begin
text {* @{const div} and @{const mod} *}
lemma mod_div_equality2: "b * (a div b) + a mod b = a"
unfolding mult_commute [of b]
by (rule mod_div_equality)
lemma mod_div_equality': "a mod b + a div b * b = a"
using mod_div_equality [of a b]
by (simp only: add_ac)
lemma div_mod_equality: "((a div b) * b + a mod b) + c = a + c"
by (simp add: mod_div_equality)
lemma div_mod_equality2: "(b * (a div b) + a mod b) + c = a + c"
by (simp add: mod_div_equality2)
lemma mod_by_0 [simp]: "a mod 0 = a"
using mod_div_equality [of a zero] by simp
lemma mod_0 [simp]: "0 mod a = 0"
using mod_div_equality [of zero a] div_0 by simp
lemma div_mult_self2 [simp]:
assumes "b \<noteq> 0"
shows "(a + b * c) div b = c + a div b"
using assms div_mult_self1 [of b a c] by (simp add: mult_commute)
lemma mod_mult_self1 [simp]: "(a + c * b) mod b = a mod b"
proof (cases "b = 0")
case True then show ?thesis by simp
next
case False
have "a + c * b = (a + c * b) div b * b + (a + c * b) mod b"
by (simp add: mod_div_equality)
also from False div_mult_self1 [of b a c] have
"\<dots> = (c + a div b) * b + (a + c * b) mod b"
by (simp add: algebra_simps)
finally have "a = a div b * b + (a + c * b) mod b"
by (simp add: add_commute [of a] add_assoc left_distrib)
then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"
by (simp add: mod_div_equality)
then show ?thesis by simp
qed
lemma mod_mult_self2 [simp]: "(a + b * c) mod b = a mod b"
by (simp add: mult_commute [of b])
lemma div_mult_self1_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> b * a div b = a"
using div_mult_self2 [of b 0 a] by simp
lemma div_mult_self2_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> a * b div b = a"
using div_mult_self1 [of b 0 a] by simp
lemma mod_mult_self1_is_0 [simp]: "b * a mod b = 0"
using mod_mult_self2 [of 0 b a] by simp
lemma mod_mult_self2_is_0 [simp]: "a * b mod b = 0"
using mod_mult_self1 [of 0 a b] by simp
lemma div_by_1 [simp]: "a div 1 = a"
using div_mult_self2_is_id [of 1 a] zero_neq_one by simp
lemma mod_by_1 [simp]: "a mod 1 = 0"
proof -
from mod_div_equality [of a one] div_by_1 have "a + a mod 1 = a" by simp
then have "a + a mod 1 = a + 0" by simp
then show ?thesis by (rule add_left_imp_eq)
qed
lemma mod_self [simp]: "a mod a = 0"
using mod_mult_self2_is_0 [of 1] by simp
lemma div_self [simp]: "a \<noteq> 0 \<Longrightarrow> a div a = 1"
using div_mult_self2_is_id [of _ 1] by simp
lemma div_add_self1 [simp]:
assumes "b \<noteq> 0"
shows "(b + a) div b = a div b + 1"
using assms div_mult_self1 [of b a 1] by (simp add: add_commute)
lemma div_add_self2 [simp]:
assumes "b \<noteq> 0"
shows "(a + b) div b = a div b + 1"
using assms div_add_self1 [of b a] by (simp add: add_commute)
lemma mod_add_self1 [simp]:
"(b + a) mod b = a mod b"
using mod_mult_self1 [of a 1 b] by (simp add: add_commute)
lemma mod_add_self2 [simp]:
"(a + b) mod b = a mod b"
using mod_mult_self1 [of a 1 b] by simp
lemma mod_div_decomp:
fixes a b
obtains q r where "q = a div b" and "r = a mod b"
and "a = q * b + r"
proof -
from mod_div_equality have "a = a div b * b + a mod b" by simp
moreover have "a div b = a div b" ..
moreover have "a mod b = a mod b" ..
note that ultimately show thesis by blast
qed
lemma dvd_eq_mod_eq_0 [code unfold]: "a dvd b \<longleftrightarrow> b mod a = 0"
proof
assume "b mod a = 0"
with mod_div_equality [of b a] have "b div a * a = b" by simp
then have "b = a * (b div a)" unfolding mult_commute ..
then have "\<exists>c. b = a * c" ..
then show "a dvd b" unfolding dvd_def .
next
assume "a dvd b"
then have "\<exists>c. b = a * c" unfolding dvd_def .
then obtain c where "b = a * c" ..
then have "b mod a = a * c mod a" by simp
then have "b mod a = c * a mod a" by (simp add: mult_commute)
then show "b mod a = 0" by simp
qed
lemma mod_div_trivial [simp]: "a mod b div b = 0"
proof (cases "b = 0")
assume "b = 0"
thus ?thesis by simp
next
assume "b \<noteq> 0"
hence "a div b + a mod b div b = (a mod b + a div b * b) div b"
by (rule div_mult_self1 [symmetric])
also have "\<dots> = a div b"
by (simp only: mod_div_equality')
also have "\<dots> = a div b + 0"
by simp
finally show ?thesis
by (rule add_left_imp_eq)
qed
lemma mod_mod_trivial [simp]: "a mod b mod b = a mod b"
proof -
have "a mod b mod b = (a mod b + a div b * b) mod b"
by (simp only: mod_mult_self1)
also have "\<dots> = a mod b"
by (simp only: mod_div_equality')
finally show ?thesis .
qed
text {* Addition respects modular equivalence. *}
lemma mod_add_left_eq: "(a + b) mod c = (a mod c + b) mod c"
proof -
have "(a + b) mod c = (a div c * c + a mod c + b) mod c"
by (simp only: mod_div_equality)
also have "\<dots> = (a mod c + b + a div c * c) mod c"
by (simp only: add_ac)
also have "\<dots> = (a mod c + b) mod c"
by (rule mod_mult_self1)
finally show ?thesis .
qed
lemma mod_add_right_eq: "(a + b) mod c = (a + b mod c) mod c"
proof -
have "(a + b) mod c = (a + (b div c * c + b mod c)) mod c"
by (simp only: mod_div_equality)
also have "\<dots> = (a + b mod c + b div c * c) mod c"
by (simp only: add_ac)
also have "\<dots> = (a + b mod c) mod c"
by (rule mod_mult_self1)
finally show ?thesis .
qed
lemma mod_add_eq: "(a + b) mod c = (a mod c + b mod c) mod c"
by (rule trans [OF mod_add_left_eq mod_add_right_eq])
lemma mod_add_cong:
assumes "a mod c = a' mod c"
assumes "b mod c = b' mod c"
shows "(a + b) mod c = (a' + b') mod c"
proof -
have "(a mod c + b mod c) mod c = (a' mod c + b' mod c) mod c"
unfolding assms ..
thus ?thesis
by (simp only: mod_add_eq [symmetric])
qed
text {* Multiplication respects modular equivalence. *}
lemma mod_mult_left_eq: "(a * b) mod c = ((a mod c) * b) mod c"
proof -
have "(a * b) mod c = ((a div c * c + a mod c) * b) mod c"
by (simp only: mod_div_equality)
also have "\<dots> = (a mod c * b + a div c * b * c) mod c"
by (simp only: algebra_simps)
also have "\<dots> = (a mod c * b) mod c"
by (rule mod_mult_self1)
finally show ?thesis .
qed
lemma mod_mult_right_eq: "(a * b) mod c = (a * (b mod c)) mod c"
proof -
have "(a * b) mod c = (a * (b div c * c + b mod c)) mod c"
by (simp only: mod_div_equality)
also have "\<dots> = (a * (b mod c) + a * (b div c) * c) mod c"
by (simp only: algebra_simps)
also have "\<dots> = (a * (b mod c)) mod c"
by (rule mod_mult_self1)
finally show ?thesis .
qed
lemma mod_mult_eq: "(a * b) mod c = ((a mod c) * (b mod c)) mod c"
by (rule trans [OF mod_mult_left_eq mod_mult_right_eq])
lemma mod_mult_cong:
assumes "a mod c = a' mod c"
assumes "b mod c = b' mod c"
shows "(a * b) mod c = (a' * b') mod c"
proof -
have "(a mod c * (b mod c)) mod c = (a' mod c * (b' mod c)) mod c"
unfolding assms ..
thus ?thesis
by (simp only: mod_mult_eq [symmetric])
qed
lemma mod_mod_cancel:
assumes "c dvd b"
shows "a mod b mod c = a mod c"
proof -
from `c dvd b` obtain k where "b = c * k"
by (rule dvdE)
have "a mod b mod c = a mod (c * k) mod c"
by (simp only: `b = c * k`)
also have "\<dots> = (a mod (c * k) + a div (c * k) * k * c) mod c"
by (simp only: mod_mult_self1)
also have "\<dots> = (a div (c * k) * (c * k) + a mod (c * k)) mod c"
by (simp only: add_ac mult_ac)
also have "\<dots> = a mod c"
by (simp only: mod_div_equality)
finally show ?thesis .
qed
end
class ring_div = semiring_div + comm_ring_1
begin
text {* Negation respects modular equivalence. *}
lemma mod_minus_eq: "(- a) mod b = (- (a mod b)) mod b"
proof -
have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
by (simp only: mod_div_equality)
also have "\<dots> = (- (a mod b) + - (a div b) * b) mod b"
by (simp only: minus_add_distrib minus_mult_left add_ac)
also have "\<dots> = (- (a mod b)) mod b"
by (rule mod_mult_self1)
finally show ?thesis .
qed
lemma mod_minus_cong:
assumes "a mod b = a' mod b"
shows "(- a) mod b = (- a') mod b"
proof -
have "(- (a mod b)) mod b = (- (a' mod b)) mod b"
unfolding assms ..
thus ?thesis
by (simp only: mod_minus_eq [symmetric])
qed
text {* Subtraction respects modular equivalence. *}
lemma mod_diff_left_eq: "(a - b) mod c = (a mod c - b) mod c"
unfolding diff_minus
by (intro mod_add_cong mod_minus_cong) simp_all
lemma mod_diff_right_eq: "(a - b) mod c = (a - b mod c) mod c"
unfolding diff_minus
by (intro mod_add_cong mod_minus_cong) simp_all
lemma mod_diff_eq: "(a - b) mod c = (a mod c - b mod c) mod c"
unfolding diff_minus
by (intro mod_add_cong mod_minus_cong) simp_all
lemma mod_diff_cong:
assumes "a mod c = a' mod c"
assumes "b mod c = b' mod c"
shows "(a - b) mod c = (a' - b') mod c"
unfolding diff_minus using assms
by (intro mod_add_cong mod_minus_cong)
end
subsection {* Division on @{typ nat} *}
text {*
We define @{const div} and @{const mod} on @{typ nat} by means
of a characteristic relation with two input arguments
@{term "m\<Colon>nat"}, @{term "n\<Colon>nat"} and two output arguments
@{term "q\<Colon>nat"}(uotient) and @{term "r\<Colon>nat"}(emainder).
*}
definition divmod_rel :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool" where
"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)"
text {* @{const divmod_rel} is total: *}
lemma divmod_rel_ex:
obtains q r where "divmod_rel m n q r"
proof (cases "n = 0")
case True with that show thesis
by (auto simp add: divmod_rel_def)
next
case False
have "\<exists>q r. m = q * n + r \<and> r < n"
proof (induct m)
case 0 with `n \<noteq> 0`
have "(0\<Colon>nat) = 0 * n + 0 \<and> 0 < n" by simp
then show ?case by blast
next
case (Suc m) then obtain q' r'
where m: "m = q' * n + r'" and n: "r' < n" by auto
then show ?case proof (cases "Suc r' < n")
case True
from m n have "Suc m = q' * n + Suc r'" by simp
with True show ?thesis by blast
next
case False then have "n \<le> Suc r'" by auto
moreover from n have "Suc r' \<le> n" by auto
ultimately have "n = Suc r'" by auto
with m have "Suc m = Suc q' * n + 0" by simp
with `n \<noteq> 0` show ?thesis by blast
qed
qed
with that show thesis
using `n \<noteq> 0` by (auto simp add: divmod_rel_def)
qed
text {* @{const divmod_rel} is injective: *}
lemma divmod_rel_unique_div:
assumes "divmod_rel m n q r"
and "divmod_rel m n q' r'"
shows "q = q'"
proof (cases "n = 0")
case True with assms show ?thesis
by (simp add: divmod_rel_def)
next
case False
have aux: "\<And>q r q' r'. q' * n + r' = q * n + r \<Longrightarrow> r < n \<Longrightarrow> q' \<le> (q\<Colon>nat)"
apply (rule leI)
apply (subst less_iff_Suc_add)
apply (auto simp add: add_mult_distrib)
done
from `n \<noteq> 0` assms show ?thesis
by (auto simp add: divmod_rel_def
intro: order_antisym dest: aux sym)
qed
lemma divmod_rel_unique_mod:
assumes "divmod_rel m n q r"
and "divmod_rel m n q' r'"
shows "r = r'"
proof -
from assms have "q = q'" by (rule divmod_rel_unique_div)
with assms show ?thesis by (simp add: divmod_rel_def)
qed
text {*
We instantiate divisibility on the natural numbers by
means of @{const divmod_rel}:
*}
instantiation nat :: semiring_div
begin
definition divmod :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where
[code del]: "divmod m n = (THE (q, r). divmod_rel m n q r)"
definition div_nat where
"m div n = fst (divmod m n)"
definition mod_nat where
"m mod n = snd (divmod m n)"
lemma divmod_div_mod:
"divmod m n = (m div n, m mod n)"
unfolding div_nat_def mod_nat_def by simp
lemma divmod_eq:
assumes "divmod_rel m n q r"
shows "divmod m n = (q, r)"
using assms by (auto simp add: divmod_def
dest: divmod_rel_unique_div divmod_rel_unique_mod)
lemma div_eq:
assumes "divmod_rel m n q r"
shows "m div n = q"
using assms by (auto dest: divmod_eq simp add: div_nat_def)
lemma mod_eq:
assumes "divmod_rel m n q r"
shows "m mod n = r"
using assms by (auto dest: divmod_eq simp add: mod_nat_def)
lemma divmod_rel: "divmod_rel m n (m div n) (m mod n)"
proof -
from divmod_rel_ex
obtain q r where rel: "divmod_rel m n q r" .
moreover with div_eq mod_eq have "m div n = q" and "m mod n = r"
by simp_all
ultimately show ?thesis by simp
qed
lemma divmod_zero:
"divmod m 0 = (0, m)"
proof -
from divmod_rel [of m 0] show ?thesis
unfolding divmod_div_mod divmod_rel_def by simp
qed
lemma divmod_base:
assumes "m < n"
shows "divmod m n = (0, m)"
proof -
from divmod_rel [of m n] show ?thesis
unfolding divmod_div_mod divmod_rel_def
using assms by (cases "m div n = 0")
(auto simp add: gr0_conv_Suc [of "m div n"])
qed
lemma divmod_step:
assumes "0 < n" and "n \<le> m"
shows "divmod m n = (Suc ((m - n) div n), (m - n) mod n)"
proof -
from divmod_rel have divmod_m_n: "divmod_rel m n (m div n) (m mod n)" .
with assms have m_div_n: "m div n \<ge> 1"
by (cases "m div n") (auto simp add: divmod_rel_def)
from assms divmod_m_n have "divmod_rel (m - n) n (m div n - 1) (m mod n)"
by (cases "m div n") (auto simp add: divmod_rel_def)
with divmod_eq have "divmod (m - n) n = (m div n - 1, m mod n)" by simp
moreover from divmod_div_mod have "divmod (m - n) n = ((m - n) div n, (m - n) mod n)" .
ultimately have "m div n = Suc ((m - n) div n)"
and "m mod n = (m - n) mod n" using m_div_n by simp_all
then show ?thesis using divmod_div_mod by simp
qed
text {* The ''recursion'' equations for @{const div} and @{const mod} *}
lemma div_less [simp]:
fixes m n :: nat
assumes "m < n"
shows "m div n = 0"
using assms divmod_base divmod_div_mod by simp
lemma le_div_geq:
fixes m n :: nat
assumes "0 < n" and "n \<le> m"
shows "m div n = Suc ((m - n) div n)"
using assms divmod_step divmod_div_mod by simp
lemma mod_less [simp]:
fixes m n :: nat
assumes "m < n"
shows "m mod n = m"
using assms divmod_base divmod_div_mod by simp
lemma le_mod_geq:
fixes m n :: nat
assumes "n \<le> m"
shows "m mod n = (m - n) mod n"
using assms divmod_step divmod_div_mod by (cases "n = 0") simp_all
instance proof
fix m n :: nat show "m div n * n + m mod n = m"
using divmod_rel [of m n] by (simp add: divmod_rel_def)
next
fix n :: nat show "n div 0 = 0"
using divmod_zero divmod_div_mod [of n 0] by simp
next
fix n :: nat show "0 div n = 0"
using divmod_rel [of 0 n] by (cases n) (simp_all add: divmod_rel_def)
next
fix m n q :: nat assume "n \<noteq> 0" then show "(q + m * n) div n = m + q div n"
by (induct m) (simp_all add: le_div_geq)
qed
end
text {* Simproc for cancelling @{const div} and @{const mod} *}
(*lemmas mod_div_equality_nat = semiring_div_class.times_div_mod_plus_zero_one.mod_div_equality [of "m\<Colon>nat" n, standard]
lemmas mod_div_equality2_nat = mod_div_equality2 [of "n\<Colon>nat" m, standard*)
ML {*
structure CancelDivModData =
struct
val div_name = @{const_name div};
val mod_name = @{const_name mod};
val mk_binop = HOLogic.mk_binop;
val mk_sum = ArithData.mk_sum;
val dest_sum = ArithData.dest_sum;
(*logic*)
val div_mod_eqs = map mk_meta_eq [@{thm div_mod_equality}, @{thm div_mod_equality2}]
val trans = trans
val prove_eq_sums =
let val simps = @{thm add_0} :: @{thm add_0_right} :: @{thms add_ac}
in ArithData.prove_conv all_tac (ArithData.simp_all_tac simps) end;
end;
structure CancelDivMod = CancelDivModFun(CancelDivModData);
val cancel_div_mod_proc = Simplifier.simproc (the_context ())
"cancel_div_mod" ["(m::nat) + n"] (K CancelDivMod.proc);
Addsimprocs[cancel_div_mod_proc];
*}
text {* code generator setup *}
lemma divmod_if [code]: "divmod m n = (if n = 0 \<or> m < n then (0, m) else
let (q, r) = divmod (m - n) n in (Suc q, r))"
by (simp add: divmod_zero divmod_base divmod_step)
(simp add: divmod_div_mod)
code_modulename SML
Divides Nat
code_modulename OCaml
Divides Nat
code_modulename Haskell
Divides Nat
subsubsection {* Quotient *}
lemma div_geq: "0 < n \<Longrightarrow> \<not> m < n \<Longrightarrow> m div n = Suc ((m - n) div n)"
by (simp add: le_div_geq linorder_not_less)
lemma div_if: "0 < n \<Longrightarrow> m div n = (if m < n then 0 else Suc ((m - n) div n))"
by (simp add: div_geq)
lemma div_mult_self_is_m [simp]: "0<n ==> (m*n) div n = (m::nat)"
by simp
lemma div_mult_self1_is_m [simp]: "0<n ==> (n*m) div n = (m::nat)"
by simp
subsubsection {* Remainder *}
lemma mod_less_divisor [simp]:
fixes m n :: nat
assumes "n > 0"
shows "m mod n < (n::nat)"
using assms divmod_rel unfolding divmod_rel_def by auto
lemma mod_less_eq_dividend [simp]:
fixes m n :: nat
shows "m mod n \<le> m"
proof (rule add_leD2)
from mod_div_equality have "m div n * n + m mod n = m" .
then show "m div n * n + m mod n \<le> m" by auto
qed
lemma mod_geq: "\<not> m < (n\<Colon>nat) \<Longrightarrow> m mod n = (m - n) mod n"
by (simp add: le_mod_geq linorder_not_less)
lemma mod_if: "m mod (n\<Colon>nat) = (if m < n then m else (m - n) mod n)"
by (simp add: le_mod_geq)
lemma mod_1 [simp]: "m mod Suc 0 = 0"
by (induct m) (simp_all add: mod_geq)
lemma mod_mult_distrib: "(m mod n) * (k\<Colon>nat) = (m * k) mod (n * k)"
apply (cases "n = 0", simp)
apply (cases "k = 0", simp)
apply (induct m rule: nat_less_induct)
apply (subst mod_if, simp)
apply (simp add: mod_geq diff_mult_distrib)
done
lemma mod_mult_distrib2: "(k::nat) * (m mod n) = (k*m) mod (k*n)"
by (simp add: mult_commute [of k] mod_mult_distrib)
(* a simple rearrangement of mod_div_equality: *)
lemma mult_div_cancel: "(n::nat) * (m div n) = m - (m mod n)"
by (cut_tac a = m and b = n in mod_div_equality2, arith)
lemma mod_le_divisor[simp]: "0 < n \<Longrightarrow> m mod n \<le> (n::nat)"
apply (drule mod_less_divisor [where m = m])
apply simp
done
subsubsection {* Quotient and Remainder *}
lemma divmod_rel_mult1_eq:
"[| divmod_rel b c q r; c > 0 |]
==> divmod_rel (a*b) c (a*q + a*r div c) (a*r mod c)"
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
lemma div_mult1_eq: "(a*b) div c = a*(b div c) + a*(b mod c) div (c::nat)"
apply (cases "c = 0", simp)
apply (blast intro: divmod_rel [THEN divmod_rel_mult1_eq, THEN div_eq])
done
lemma mod_mult1_eq: "(a*b) mod c = a*(b mod c) mod (c::nat)"
by (rule mod_mult_right_eq)
lemma mod_mult1_eq': "(a*b) mod (c::nat) = ((a mod c) * b) mod c"
by (rule mod_mult_left_eq)
lemma mod_mult_distrib_mod:
"(a*b) mod (c::nat) = ((a mod c) * (b mod c)) mod c"
by (rule mod_mult_eq)
lemma divmod_rel_add1_eq:
"[| divmod_rel a c aq ar; divmod_rel b c bq br; c > 0 |]
==> divmod_rel (a + b) c (aq + bq + (ar+br) div c) ((ar + br) mod c)"
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
(*NOT suitable for rewriting: the RHS has an instance of the LHS*)
lemma div_add1_eq:
"(a+b) div (c::nat) = a div c + b div c + ((a mod c + b mod c) div c)"
apply (cases "c = 0", simp)
apply (blast intro: divmod_rel_add1_eq [THEN div_eq] divmod_rel)
done
lemma mod_add1_eq: "(a+b) mod (c::nat) = (a mod c + b mod c) mod c"
by (rule mod_add_eq)
lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c"
apply (cut_tac m = q and n = c in mod_less_divisor)
apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto)
apply (erule_tac P = "%x. ?lhs < ?rhs x" in ssubst)
apply (simp add: add_mult_distrib2)
done
lemma divmod_rel_mult2_eq: "[| divmod_rel a b q r; 0 < b; 0 < c |]
==> divmod_rel a (b*c) (q div c) (b*(q mod c) + r)"
by (auto simp add: mult_ac divmod_rel_def add_mult_distrib2 [symmetric] mod_lemma)
lemma div_mult2_eq: "a div (b*c) = (a div b) div (c::nat)"
apply (cases "b = 0", simp)
apply (cases "c = 0", simp)
apply (force simp add: divmod_rel [THEN divmod_rel_mult2_eq, THEN div_eq])
done
lemma mod_mult2_eq: "a mod (b*c) = b*(a div b mod c) + a mod (b::nat)"
apply (cases "b = 0", simp)
apply (cases "c = 0", simp)
apply (auto simp add: mult_commute divmod_rel [THEN divmod_rel_mult2_eq, THEN mod_eq])
done
subsubsection{*Cancellation of Common Factors in Division*}
lemma div_mult_mult_lemma:
"[| (0::nat) < b; 0 < c |] ==> (c*a) div (c*b) = a div b"
by (auto simp add: div_mult2_eq)
lemma div_mult_mult1 [simp]: "(0::nat) < c ==> (c*a) div (c*b) = a div b"
apply (cases "b = 0")
apply (auto simp add: linorder_neq_iff [of b] div_mult_mult_lemma)
done
lemma div_mult_mult2 [simp]: "(0::nat) < c ==> (a*c) div (b*c) = a div b"
apply (drule div_mult_mult1)
apply (auto simp add: mult_commute)
done
subsubsection{*Further Facts about Quotient and Remainder*}
lemma div_1 [simp]: "m div Suc 0 = m"
by (induct m) (simp_all add: div_geq)
(* Monotonicity of div in first argument *)
lemma div_le_mono [rule_format (no_asm)]:
"\<forall>m::nat. m \<le> n --> (m div k) \<le> (n div k)"
apply (case_tac "k=0", simp)
apply (induct "n" rule: nat_less_induct, clarify)
apply (case_tac "n<k")
(* 1 case n<k *)
apply simp
(* 2 case n >= k *)
apply (case_tac "m<k")
(* 2.1 case m<k *)
apply simp
(* 2.2 case m>=k *)
apply (simp add: div_geq diff_le_mono)
done
(* Antimonotonicity of div in second argument *)
lemma div_le_mono2: "!!m::nat. [| 0<m; m\<le>n |] ==> (k div n) \<le> (k div m)"
apply (subgoal_tac "0<n")
prefer 2 apply simp
apply (induct_tac k rule: nat_less_induct)
apply (rename_tac "k")
apply (case_tac "k<n", simp)
apply (subgoal_tac "~ (k<m) ")
prefer 2 apply simp
apply (simp add: div_geq)
apply (subgoal_tac "(k-n) div n \<le> (k-m) div n")
prefer 2
apply (blast intro: div_le_mono diff_le_mono2)
apply (rule le_trans, simp)
apply (simp)
done
lemma div_le_dividend [simp]: "m div n \<le> (m::nat)"
apply (case_tac "n=0", simp)
apply (subgoal_tac "m div n \<le> m div 1", simp)
apply (rule div_le_mono2)
apply (simp_all (no_asm_simp))
done
(* Similar for "less than" *)
lemma div_less_dividend [rule_format]:
"!!n::nat. 1<n ==> 0 < m --> m div n < m"
apply (induct_tac m rule: nat_less_induct)
apply (rename_tac "m")
apply (case_tac "m<n", simp)
apply (subgoal_tac "0<n")
prefer 2 apply simp
apply (simp add: div_geq)
apply (case_tac "n<m")
apply (subgoal_tac "(m-n) div n < (m-n) ")
apply (rule impI less_trans_Suc)+
apply assumption
apply (simp_all)
done
declare div_less_dividend [simp]
text{*A fact for the mutilated chess board*}
lemma mod_Suc: "Suc(m) mod n = (if Suc(m mod n) = n then 0 else Suc(m mod n))"
apply (case_tac "n=0", simp)
apply (induct "m" rule: nat_less_induct)
apply (case_tac "Suc (na) <n")
(* case Suc(na) < n *)
apply (frule lessI [THEN less_trans], simp add: less_not_refl3)
(* case n \<le> Suc(na) *)
apply (simp add: linorder_not_less le_Suc_eq mod_geq)
apply (auto simp add: Suc_diff_le le_mod_geq)
done
lemma nat_mod_div_trivial: "m mod n div n = (0 :: nat)"
by simp
lemma nat_mod_mod_trivial: "m mod n mod n = (m mod n :: nat)"
by simp
subsubsection {* The Divides Relation *}
lemma dvd_1_left [iff]: "Suc 0 dvd k"
unfolding dvd_def by simp
lemma dvd_1_iff_1 [simp]: "(m dvd Suc 0) = (m = Suc 0)"
by (simp add: dvd_def)
lemma dvd_anti_sym: "[| m dvd n; n dvd m |] ==> m = (n::nat)"
unfolding dvd_def
by (force dest: mult_eq_self_implies_10 simp add: mult_assoc mult_eq_1_iff)
text {* @{term "op dvd"} is a partial order *}
interpretation dvd!: order "op dvd" "\<lambda>n m \<Colon> nat. n dvd m \<and> \<not> m dvd n"
proof qed (auto intro: dvd_refl dvd_trans dvd_anti_sym)
lemma dvd_diff: "[| k dvd m; k dvd n |] ==> k dvd (m-n :: nat)"
unfolding dvd_def
by (blast intro: diff_mult_distrib2 [symmetric])
lemma dvd_diffD: "[| k dvd m-n; k dvd n; n\<le>m |] ==> k dvd (m::nat)"
apply (erule linorder_not_less [THEN iffD2, THEN add_diff_inverse, THEN subst])
apply (blast intro: dvd_add)
done
lemma dvd_diffD1: "[| k dvd m-n; k dvd m; n\<le>m |] ==> k dvd (n::nat)"
by (drule_tac m = m in dvd_diff, auto)
lemma dvd_reduce: "(k dvd n + k) = (k dvd (n::nat))"
apply (rule iffI)
apply (erule_tac [2] dvd_add)
apply (rule_tac [2] dvd_refl)
apply (subgoal_tac "n = (n+k) -k")
prefer 2 apply simp
apply (erule ssubst)
apply (erule dvd_diff)
apply (rule dvd_refl)
done
lemma dvd_mod: "!!n::nat. [| f dvd m; f dvd n |] ==> f dvd m mod n"
unfolding dvd_def
apply (case_tac "n = 0", auto)
apply (blast intro: mod_mult_distrib2 [symmetric])
done
lemma dvd_mod_imp_dvd: "[| (k::nat) dvd m mod n; k dvd n |] ==> k dvd m"
apply (subgoal_tac "k dvd (m div n) *n + m mod n")
apply (simp add: mod_div_equality)
apply (simp only: dvd_add dvd_mult)
done
lemma dvd_mod_iff: "k dvd n ==> ((k::nat) dvd m mod n) = (k dvd m)"
by (blast intro: dvd_mod_imp_dvd dvd_mod)
lemma dvd_mult_cancel: "!!k::nat. [| k*m dvd k*n; 0<k |] ==> m dvd n"
unfolding dvd_def
apply (erule exE)
apply (simp add: mult_ac)
done
lemma dvd_mult_cancel1: "0<m ==> (m*n dvd m) = (n = (1::nat))"
apply auto
apply (subgoal_tac "m*n dvd m*1")
apply (drule dvd_mult_cancel, auto)
done
lemma dvd_mult_cancel2: "0<m ==> (n*m dvd m) = (n = (1::nat))"
apply (subst mult_commute)
apply (erule dvd_mult_cancel1)
done
lemma dvd_imp_le: "[| k dvd n; 0 < n |] ==> k \<le> (n::nat)"
apply (unfold dvd_def, clarify)
apply (simp_all (no_asm_use) add: zero_less_mult_iff)
apply (erule conjE)
apply (rule le_trans)
apply (rule_tac [2] le_refl [THEN mult_le_mono])
apply (erule_tac [2] Suc_leI, simp)
done
lemma dvd_mult_div_cancel: "n dvd m ==> n * (m div n) = (m::nat)"
apply (subgoal_tac "m mod n = 0")
apply (simp add: mult_div_cancel)
apply (simp only: dvd_eq_mod_eq_0)
done
lemma le_imp_power_dvd: "!!i::nat. m \<le> n ==> i^m dvd i^n"
apply (unfold dvd_def)
apply (erule linorder_not_less [THEN iffD2, THEN add_diff_inverse, THEN subst])
apply (simp add: power_add)
done
lemma nat_zero_less_power_iff [simp]: "(x^n > 0) = (x > (0::nat) | n=0)"
by (induct n) auto
lemma power_le_dvd [rule_format]: "k^j dvd n --> i\<le>j --> k^i dvd (n::nat)"
apply (induct j)
apply (simp_all add: le_Suc_eq)
apply (blast dest!: dvd_mult_right)
done
lemma power_dvd_imp_le: "[|i^m dvd i^n; (1::nat) < i|] ==> m \<le> n"
apply (rule power_le_imp_le_exp, assumption)
apply (erule dvd_imp_le, simp)
done
lemma mod_eq_0_iff: "(m mod d = 0) = (\<exists>q::nat. m = d*q)"
by (auto simp add: dvd_eq_mod_eq_0 [symmetric] dvd_def)
lemmas mod_eq_0D [dest!] = mod_eq_0_iff [THEN iffD1]
(*Loses information, namely we also have r<d provided d is nonzero*)
lemma mod_eqD: "(m mod d = r) ==> \<exists>q::nat. m = r + q*d"
apply (cut_tac a = m in mod_div_equality)
apply (simp only: add_ac)
apply (blast intro: sym)
done
lemma split_div:
"P(n div k :: nat) =
((k = 0 \<longrightarrow> P 0) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P i)))"
(is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
proof
assume P: ?P
show ?Q
proof (cases)
assume "k = 0"
with P show ?Q by simp
next
assume not0: "k \<noteq> 0"
thus ?Q
proof (simp, intro allI impI)
fix i j
assume n: "n = k*i + j" and j: "j < k"
show "P i"
proof (cases)
assume "i = 0"
with n j P show "P i" by simp
next
assume "i \<noteq> 0"
with not0 n j P show "P i" by(simp add:add_ac)
qed
qed
qed
next
assume Q: ?Q
show ?P
proof (cases)
assume "k = 0"
with Q show ?P by simp
next
assume not0: "k \<noteq> 0"
with Q have R: ?R by simp
from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
show ?P by simp
qed
qed
lemma split_div_lemma:
assumes "0 < n"
shows "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> q = ((m\<Colon>nat) div n)" (is "?lhs \<longleftrightarrow> ?rhs")
proof
assume ?rhs
with mult_div_cancel have nq: "n * q = m - (m mod n)" by simp
then have A: "n * q \<le> m" by simp
have "n - (m mod n) > 0" using mod_less_divisor assms by auto
then have "m < m + (n - (m mod n))" by simp
then have "m < n + (m - (m mod n))" by simp
with nq have "m < n + n * q" by simp
then have B: "m < n * Suc q" by simp
from A B show ?lhs ..
next
assume P: ?lhs
then have "divmod_rel m n q (m - n * q)"
unfolding divmod_rel_def by (auto simp add: mult_ac)
then show ?rhs using divmod_rel by (rule divmod_rel_unique_div)
qed
theorem split_div':
"P ((m::nat) div n) = ((n = 0 \<and> P 0) \<or>
(\<exists>q. (n * q \<le> m \<and> m < n * (Suc q)) \<and> P q))"
apply (case_tac "0 < n")
apply (simp only: add: split_div_lemma)
apply simp_all
done
lemma split_mod:
"P(n mod k :: nat) =
((k = 0 \<longrightarrow> P n) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P j)))"
(is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
proof
assume P: ?P
show ?Q
proof (cases)
assume "k = 0"
with P show ?Q by simp
next
assume not0: "k \<noteq> 0"
thus ?Q
proof (simp, intro allI impI)
fix i j
assume "n = k*i + j" "j < k"
thus "P j" using not0 P by(simp add:add_ac mult_ac)
qed
qed
next
assume Q: ?Q
show ?P
proof (cases)
assume "k = 0"
with Q show ?P by simp
next
assume not0: "k \<noteq> 0"
with Q have R: ?R by simp
from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
show ?P by simp
qed
qed
theorem mod_div_equality': "(m::nat) mod n = m - (m div n) * n"
apply (rule_tac P="%x. m mod n = x - (m div n) * n" in
subst [OF mod_div_equality [of _ n]])
apply arith
done
lemma div_mod_equality':
fixes m n :: nat
shows "m div n * n = m - m mod n"
proof -
have "m mod n \<le> m mod n" ..
from div_mod_equality have
"m div n * n + m mod n - m mod n = m - m mod n" by simp
with diff_add_assoc [OF `m mod n \<le> m mod n`, of "m div n * n"] have
"m div n * n + (m mod n - m mod n) = m - m mod n"
by simp
then show ?thesis by simp
qed
subsubsection {*An ``induction'' law for modulus arithmetic.*}
lemma mod_induct_0:
assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
and base: "P i" and i: "i<p"
shows "P 0"
proof (rule ccontr)
assume contra: "\<not>(P 0)"
from i have p: "0<p" by simp
have "\<forall>k. 0<k \<longrightarrow> \<not> P (p-k)" (is "\<forall>k. ?A k")
proof
fix k
show "?A k"
proof (induct k)
show "?A 0" by simp -- "by contradiction"
next
fix n
assume ih: "?A n"
show "?A (Suc n)"
proof (clarsimp)
assume y: "P (p - Suc n)"
have n: "Suc n < p"
proof (rule ccontr)
assume "\<not>(Suc n < p)"
hence "p - Suc n = 0"
by simp
with y contra show "False"
by simp
qed
hence n2: "Suc (p - Suc n) = p-n" by arith
from p have "p - Suc n < p" by arith
with y step have z: "P ((Suc (p - Suc n)) mod p)"
by blast
show "False"
proof (cases "n=0")
case True
with z n2 contra show ?thesis by simp
next
case False
with p have "p-n < p" by arith
with z n2 False ih show ?thesis by simp
qed
qed
qed
qed
moreover
from i obtain k where "0<k \<and> i+k=p"
by (blast dest: less_imp_add_positive)
hence "0<k \<and> i=p-k" by auto
moreover
note base
ultimately
show "False" by blast
qed
lemma mod_induct:
assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
and base: "P i" and i: "i<p" and j: "j<p"
shows "P j"
proof -
have "\<forall>j<p. P j"
proof
fix j
show "j<p \<longrightarrow> P j" (is "?A j")
proof (induct j)
from step base i show "?A 0"
by (auto elim: mod_induct_0)
next
fix k
assume ih: "?A k"
show "?A (Suc k)"
proof
assume suc: "Suc k < p"
hence k: "k<p" by simp
with ih have "P k" ..
with step k have "P (Suc k mod p)"
by blast
moreover
from suc have "Suc k mod p = Suc k"
by simp
ultimately
show "P (Suc k)" by simp
qed
qed
qed
with j show ?thesis by blast
qed
end