src/HOL/Integ/IntArith.thy

HOL-Real -> HOL-Complex

header {* Integer arithmetic *}

theory IntArith = Bin
files ("int_arith1.ML") ("int_arith2.ML"):

use "int_arith1.ML"
setup int_arith_setup
use "int_arith2.ML"

declare zabs_split [arith_split]

lemma zabs_eq_iff:
    "(abs (z::int) = w) = (z = w \<and> 0 <= z \<or> z = -w \<and> z < 0)"
  by (auto simp add: zabs_def)

lemma int_nat_eq [simp]: "int (nat z) = (if 0 \<le> z then z else 0)"
  by simp

lemma split_nat[arith_split]:
  "P(nat(i::int)) = ((ALL 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

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 <= 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
  }
  from this 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)
 apply 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 <= 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
  }
  from this 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)
 apply simp+
done

end