(* Title: HOL/Integ/IntArith.thy
ID: $Id$
Authors: Larry Paulson and Tobias Nipkow
*)
header {* Integer arithmetic *}
theory IntArith = Bin
files ("int_arith1.ML"):
text{*Duplicate: can't understand why it's necessary*}
declare numeral_0_eq_0 [simp]
subsection{*Instantiating Binary Arithmetic for the Integers*}
instance
int :: number ..
primrec (*the type constraint is essential!*)
number_of_Pls: "number_of bin.Pls = 0"
number_of_Min: "number_of bin.Min = - (1::int)"
number_of_BIT: "number_of(w BIT x) = (if x then 1 else 0) +
(number_of w) + (number_of w)"
declare number_of_Pls [simp del]
number_of_Min [simp del]
number_of_BIT [simp del]
instance int :: number_ring
proof
show "Numeral0 = (0::int)" by (rule number_of_Pls)
show "-1 = - (1::int)" by (rule number_of_Min)
fix w :: bin and x :: bool
show "(number_of (w BIT x) :: int) =
(if x then 1 else 0) + number_of w + number_of w"
by (rule number_of_BIT)
qed
subsection{*Inequality Reasoning for the Arithmetic Simproc*}
lemma add_numeral_0: "Numeral0 + a = (a::'a::number_ring)"
by simp
lemma add_numeral_0_right: "a + Numeral0 = (a::'a::number_ring)"
by simp
lemma mult_numeral_1: "Numeral1 * a = (a::'a::number_ring)"
by simp
lemma mult_numeral_1_right: "a * Numeral1 = (a::'a::number_ring)"
by simp
text{*Theorem lists for the cancellation simprocs. The use of binary numerals
for 0 and 1 reduces the number of special cases.*}
lemmas add_0s = add_numeral_0 add_numeral_0_right
lemmas mult_1s = mult_numeral_1 mult_numeral_1_right
mult_minus1 mult_minus1_right
subsection{*Special Arithmetic Rules for Abstract 0 and 1*}
text{*Arithmetic computations are defined for binary literals, which leaves 0
and 1 as special cases. Addition already has rules for 0, but not 1.
Multiplication and unary minus already have rules for both 0 and 1.*}
lemma binop_eq: "[|f x y = g x y; x = x'; y = y'|] ==> f x' y' = g x' y'"
by simp
lemmas add_number_of_eq = number_of_add [symmetric]
text{*Allow 1 on either or both sides*}
lemma one_add_one_is_two: "1 + 1 = (2::'a::number_ring)"
by (simp del: numeral_1_eq_1 add: numeral_1_eq_1 [symmetric] add_number_of_eq)
lemmas add_special =
one_add_one_is_two
binop_eq [of "op +", OF add_number_of_eq numeral_1_eq_1 refl, standard]
binop_eq [of "op +", OF add_number_of_eq refl numeral_1_eq_1, standard]
text{*Allow 1 on either or both sides (1-1 already simplifies to 0)*}
lemmas diff_special =
binop_eq [of "op -", OF diff_number_of_eq numeral_1_eq_1 refl, standard]
binop_eq [of "op -", OF diff_number_of_eq refl numeral_1_eq_1, standard]
text{*Allow 0 or 1 on either side with a binary numeral on the other*}
lemmas eq_special =
binop_eq [of "op =", OF eq_number_of_eq numeral_0_eq_0 refl, standard]
binop_eq [of "op =", OF eq_number_of_eq numeral_1_eq_1 refl, standard]
binop_eq [of "op =", OF eq_number_of_eq refl numeral_0_eq_0, standard]
binop_eq [of "op =", OF eq_number_of_eq refl numeral_1_eq_1, standard]
text{*Allow 0 or 1 on either side with a binary numeral on the other*}
lemmas less_special =
binop_eq [of "op <", OF less_number_of_eq_neg numeral_0_eq_0 refl, standard]
binop_eq [of "op <", OF less_number_of_eq_neg numeral_1_eq_1 refl, standard]
binop_eq [of "op <", OF less_number_of_eq_neg refl numeral_0_eq_0, standard]
binop_eq [of "op <", OF less_number_of_eq_neg refl numeral_1_eq_1, standard]
text{*Allow 0 or 1 on either side with a binary numeral on the other*}
lemmas le_special =
binop_eq [of "op \<le>", OF le_number_of_eq numeral_0_eq_0 refl, standard]
binop_eq [of "op \<le>", OF le_number_of_eq numeral_1_eq_1 refl, standard]
binop_eq [of "op \<le>", OF le_number_of_eq refl numeral_0_eq_0, standard]
binop_eq [of "op \<le>", OF le_number_of_eq refl numeral_1_eq_1, standard]
lemmas arith_special =
add_special diff_special eq_special less_special le_special
use "int_arith1.ML"
setup int_arith_setup
subsection{*Lemmas About Small Numerals*}
lemma of_int_m1 [simp]: "of_int -1 = (-1 :: 'a :: number_ring)"
proof -
have "(of_int -1 :: 'a) = of_int (- 1)" by simp
also have "... = - of_int 1" by (simp only: of_int_minus)
also have "... = -1" by simp
finally show ?thesis .
qed
lemma abs_minus_one [simp]: "abs (-1) = (1::'a::{ordered_idom,number_ring})"
by (simp add: abs_if)
lemma abs_power_minus_one [simp]:
"abs(-1 ^ n) = (1::'a::{ordered_idom,number_ring,ringpower})"
by (simp add: power_abs)
lemma of_int_number_of_eq:
"of_int (number_of v) = (number_of v :: 'a :: number_ring)"
apply (induct v)
apply (simp_all only: number_of of_int_add, simp_all)
done
text{*Lemmas for specialist use, NOT as default simprules*}
lemma mult_2: "2 * z = (z+z::'a::number_ring)"
proof -
have "2*z = (1 + 1)*z" by simp
also have "... = z+z" by (simp add: left_distrib)
finally show ?thesis .
qed
lemma mult_2_right: "z * 2 = (z+z::'a::number_ring)"
by (subst mult_commute, rule mult_2)
subsection{*More Inequality Reasoning*}
lemma zless_add1_eq: "(w < z + (1::int)) = (w<z | w=z)"
by arith
lemma add1_zle_eq: "(w + (1::int) \<le> z) = (w<z)"
by arith
lemma zle_diff1_eq [simp]: "(w \<le> z - (1::int)) = (w<z)"
by arith
lemma zle_add1_eq_le [simp]: "(w < z + (1::int)) = (w\<le>z)"
by arith
lemma int_one_le_iff_zero_less: "((1::int) \<le> z) = (0 < z)"
by arith
subsection{*The Functions @{term nat} and @{term int}*}
text{*Simplify the terms @{term "int 0"}, @{term "int(Suc 0)"} and
@{term "w + - z"}*}
declare Zero_int_def [symmetric, simp]
declare One_int_def [symmetric, simp]
text{*cooper.ML refers to this theorem*}
lemmas diff_int_def_symmetric = diff_int_def [symmetric, simp]
lemma nat_0: "nat 0 = 0"
by (simp add: nat_eq_iff)
lemma nat_1: "nat 1 = Suc 0"
by (subst nat_eq_iff, simp)
lemma nat_2: "nat 2 = Suc (Suc 0)"
by (subst nat_eq_iff, simp)
text{*This simplifies expressions of the form @{term "int n = z"} where
z is an integer literal.*}
declare int_eq_iff [of _ "number_of v", standard, simp]
lemma split_nat [arith_split]:
"P(nat(i::int)) = ((\<forall>n. i = int n \<longrightarrow> P n) & (i < 0 \<longrightarrow> P 0))"
(is "?P = (?L & ?R)")
proof (cases "i < 0")
case True thus ?thesis by simp
next
case False
have "?P = ?L"
proof
assume ?P thus ?L using False by clarsimp
next
assume ?L thus ?P using False by simp
qed
with False show ?thesis by simp
qed
(*Analogous to zadd_int*)
lemma zdiff_int: "n \<le> m ==> int m - int n = int (m-n)"
by (induct m n rule: diff_induct, simp_all)
lemma nat_mult_distrib: "(0::int) \<le> z ==> nat (z*z') = nat z * nat z'"
apply (case_tac "0 \<le> z'")
apply (rule inj_int [THEN injD])
apply (simp add: zmult_int [symmetric] zero_le_mult_iff)
apply (simp add: mult_le_0_iff)
done
lemma nat_mult_distrib_neg: "z \<le> (0::int) ==> nat(z*z') = nat(-z) * nat(-z')"
apply (rule trans)
apply (rule_tac [2] nat_mult_distrib, auto)
done
lemma nat_abs_mult_distrib: "nat (abs (w * z)) = nat (abs w) * nat (abs z)"
apply (case_tac "z=0 | w=0")
apply (auto simp add: zabs_def nat_mult_distrib [symmetric]
nat_mult_distrib_neg [symmetric] mult_less_0_iff)
done
subsubsection "Induction principles for int"
(* `set:int': dummy construction *)
theorem int_ge_induct[case_names base step,induct set:int]:
assumes ge: "k \<le> (i::int)" and
base: "P(k)" and
step: "\<And>i. \<lbrakk>k \<le> i; P i\<rbrakk> \<Longrightarrow> P(i+1)"
shows "P i"
proof -
{ fix n have "\<And>i::int. n = nat(i-k) \<Longrightarrow> k \<le> i \<Longrightarrow> P i"
proof (induct n)
case 0
hence "i = k" by arith
thus "P i" using base by simp
next
case (Suc n)
hence "n = nat((i - 1) - k)" by arith
moreover
have ki1: "k \<le> i - 1" using Suc.prems by arith
ultimately
have "P(i - 1)" by(rule Suc.hyps)
from step[OF ki1 this] show ?case by simp
qed
}
with ge show ?thesis by fast
qed
(* `set:int': dummy construction *)
theorem int_gr_induct[case_names base step,induct set:int]:
assumes gr: "k < (i::int)" and
base: "P(k+1)" and
step: "\<And>i. \<lbrakk>k < i; P i\<rbrakk> \<Longrightarrow> P(i+1)"
shows "P i"
apply(rule int_ge_induct[of "k + 1"])
using gr apply arith
apply(rule base)
apply (rule step, simp+)
done
theorem int_le_induct[consumes 1,case_names base step]:
assumes le: "i \<le> (k::int)" and
base: "P(k)" and
step: "\<And>i. \<lbrakk>i \<le> k; P i\<rbrakk> \<Longrightarrow> P(i - 1)"
shows "P i"
proof -
{ fix n have "\<And>i::int. n = nat(k-i) \<Longrightarrow> i \<le> k \<Longrightarrow> P i"
proof (induct n)
case 0
hence "i = k" by arith
thus "P i" using base by simp
next
case (Suc n)
hence "n = nat(k - (i+1))" by arith
moreover
have ki1: "i + 1 \<le> k" using Suc.prems by arith
ultimately
have "P(i+1)" by(rule Suc.hyps)
from step[OF ki1 this] show ?case by simp
qed
}
with le show ?thesis by fast
qed
theorem int_less_induct [consumes 1,case_names base step]:
assumes less: "(i::int) < k" and
base: "P(k - 1)" and
step: "\<And>i. \<lbrakk>i < k; P i\<rbrakk> \<Longrightarrow> P(i - 1)"
shows "P i"
apply(rule int_le_induct[of _ "k - 1"])
using less apply arith
apply(rule base)
apply (rule step, simp+)
done
subsection{*Intermediate value theorems*}
lemma int_val_lemma:
"(\<forall>i<n::nat. abs(f(i+1) - f i) \<le> 1) -->
f 0 \<le> k --> k \<le> f n --> (\<exists>i \<le> n. f i = (k::int))"
apply (induct_tac "n", simp)
apply (intro strip)
apply (erule impE, simp)
apply (erule_tac x = n in allE, simp)
apply (case_tac "k = f (n+1) ")
apply force
apply (erule impE)
apply (simp add: zabs_def split add: split_if_asm)
apply (blast intro: le_SucI)
done
lemmas nat0_intermed_int_val = int_val_lemma [rule_format (no_asm)]
lemma nat_intermed_int_val:
"[| \<forall>i. m \<le> i & i < n --> abs(f(i + 1::nat) - f i) \<le> 1; m < n;
f m \<le> k; k \<le> f n |] ==> ? i. m \<le> i & i \<le> n & f i = (k::int)"
apply (cut_tac n = "n-m" and f = "%i. f (i+m) " and k = k
in int_val_lemma)
apply simp
apply (erule impE)
apply (intro strip)
apply (erule_tac x = "i+m" in allE, arith)
apply (erule exE)
apply (rule_tac x = "i+m" in exI, arith)
done
subsection{*Products and 1, by T. M. Rasmussen*}
lemma zmult_eq_self_iff: "(m = m*(n::int)) = (n = 1 | m = 0)"
apply auto
apply (subgoal_tac "m*1 = m*n")
apply (drule mult_cancel_left [THEN iffD1], auto)
done
text{*FIXME: tidy*}
lemma pos_zmult_eq_1_iff: "0 < (m::int) ==> (m * n = 1) = (m = 1 & n = 1)"
apply auto
apply (case_tac "m=1")
apply (case_tac [2] "n=1")
apply (case_tac [4] "m=1")
apply (case_tac [5] "n=1", auto)
apply (tactic"distinct_subgoals_tac")
apply (subgoal_tac "1<m*n", simp)
apply (rule less_1_mult, arith)
apply (subgoal_tac "0<n", arith)
apply (subgoal_tac "0<m*n")
apply (drule zero_less_mult_iff [THEN iffD1], auto)
done
lemma zmult_eq_1_iff: "(m*n = (1::int)) = ((m = 1 & n = 1) | (m = -1 & n = -1))"
apply (case_tac "0<m")
apply (simp add: pos_zmult_eq_1_iff)
apply (case_tac "m=0")
apply (simp del: number_of_reorient)
apply (subgoal_tac "0 < -m")
apply (drule_tac n = "-n" in pos_zmult_eq_1_iff, auto)
done
ML
{*
val zle_diff1_eq = thm "zle_diff1_eq";
val zle_add1_eq_le = thm "zle_add1_eq_le";
val nonneg_eq_int = thm "nonneg_eq_int";
val abs_minus_one = thm "abs_minus_one";
val of_int_number_of_eq = thm"of_int_number_of_eq";
val nat_eq_iff = thm "nat_eq_iff";
val nat_eq_iff2 = thm "nat_eq_iff2";
val nat_less_iff = thm "nat_less_iff";
val int_eq_iff = thm "int_eq_iff";
val nat_0 = thm "nat_0";
val nat_1 = thm "nat_1";
val nat_2 = thm "nat_2";
val nat_less_eq_zless = thm "nat_less_eq_zless";
val nat_le_eq_zle = thm "nat_le_eq_zle";
val nat_intermed_int_val = thm "nat_intermed_int_val";
val zmult_eq_self_iff = thm "zmult_eq_self_iff";
val pos_zmult_eq_1_iff = thm "pos_zmult_eq_1_iff";
val zmult_eq_1_iff = thm "zmult_eq_1_iff";
val nat_add_distrib = thm "nat_add_distrib";
val nat_diff_distrib = thm "nat_diff_distrib";
val nat_mult_distrib = thm "nat_mult_distrib";
val nat_mult_distrib_neg = thm "nat_mult_distrib_neg";
val nat_abs_mult_distrib = thm "nat_abs_mult_distrib";
*}
end