Theory Sqrt_Babylonian.NthRoot_Impl

(*  Title:       Computing Square Roots using the Babylonian Method
    Author:      René Thiemann       <rene.thiemann@uibk.ac.at>
    Maintainer:  René Thiemann
    License:     LGPL
*)

(*
Copyright 2009-2014 René Thiemann

This file is part of IsaFoR/CeTA.

IsaFoR/CeTA is free software: you can redistribute it and/or modify it under the
terms of the GNU Lesser General Public License as published by the Free Software
Foundation, either version 3 of the License, or (at your option) any later
version.

IsaFoR/CeTA is distributed in the hope that it will be useful, but WITHOUT ANY
WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A
PARTICULAR PURPOSE.  See the GNU Lesser General Public License for more details.

You should have received a copy of the GNU Lesser General Public License along
with IsaFoR/CeTA. If not, see <http://www.gnu.org/licenses/>.
*)
section ‹Executable algorithms for $p$-th roots›

theory NthRoot_Impl
imports
  Log_Impl
  Cauchy.CauchysMeanTheorem
begin

text ‹
We implemented algorithms to decide $\sqrt[p]{n} \in \rats$ and to compute $\lfloor \sqrt[p]{n} \rfloor$.
To this end, we use a variant of Newton iteration which works with integer division instead of floating
point or rational division. To get suitable starting values for the Newton iteration, we also implemented
a function to approximate logarithms.
›

subsection ‹Logarithm›

text ‹For computing the $p$-th root of a number $n$, we must choose a starting value
  in the iteration. Here, we use @{term "2 ^ (nat of_int log 2 n / p)"}.
›

text ‹We use a partial efficient algorithm, which does not terminate on
  corner-cases, like $b = 0$ or $p = 1$, and invoke it properly afterwards.
  Then there is a second algorithm which terminates on these corner-cases by additional
  guards and on which we can perform induction.
›

subsection ‹Computing the $p$-th root of an integer number›

text ‹Using the logarithm, we can define an executable version of the
  intended  starting value. Its main property is the inequality
  @{term "(start_value x p) ^ p  x"}, i.e., the start value is larger
  than the p-th root. This property is essential, since our algorithm will abort
  as soon as we fall below the p-th root.›

definition start_value :: "int  nat  int" where
  "start_value n p = 2 ^ (nat of_nat (log_ceiling 2 n) / rat_of_nat p)"

lemma start_value_main: assumes x: "x  0" and p: "p > 0"
  shows "x  (start_value x p)^p  start_value x p  0"
proof (cases "x = 0")
  case True
  with p show ?thesis unfolding start_value_def True by simp
next
  case False
  with x have x: "x > 0" by auto
  define l2x where "l2x = log 2 x"
  define pow where "pow = nat rat_of_int l2x / of_nat p"
  have "root p x = x powr (1 / p)" by (rule root_powr_inverse, insert x p, auto)
  also have " = (2 powr (log 2 x)) powr (1 / p)" using powr_log_cancel[of 2 x] x by auto
  also have " = 2 powr (log 2 x * (1 / p))" by (rule powr_powr)
  also have "log 2 x * (1 / p) = log 2 x / p" using p by auto
  finally have r: "root p x = 2 powr (log 2 x / p)" .
  have lp: "log 2 x  0" using x by auto
  hence l2pos: "l2x  0" by (auto simp: l2x_def)
  have "log 2 x / p  l2x / p" using x p unfolding l2x_def
    by (metis divide_right_mono le_of_int_ceiling of_nat_0_le_iff)
  also have "  l2x / (p :: real)" by (simp add: ceiling_correct)
  also have "l2x / real p = l2x / real_of_rat (of_nat p)"
    by (metis of_rat_of_nat_eq)
  also have "of_int l2x = real_of_rat (of_int l2x)"
    by (metis of_rat_of_int_eq)
  also have "real_of_rat (of_int l2x) / real_of_rat (of_nat p) = real_of_rat (rat_of_int l2x / of_nat p)"
    by (metis of_rat_divide)
  also have "real_of_rat (rat_of_int l2x / rat_of_nat p) = rat_of_int l2x / of_nat p"
    by simp
  also have "rat_of_int l2x / of_nat p  real pow" unfolding pow_def by auto
  finally have le: "log 2 x / p  pow" .
  from powr_mono[OF le, of 2, folded r]
  have "root p x  2 powr pow" by auto
  also have " = 2 ^ pow" by (rule powr_realpow, auto)
  also have " = of_int ((2 :: int) ^ pow)" by simp
  also have "pow = (nat of_int (log_ceiling 2 x) / rat_of_nat p)"
    unfolding pow_def l2x_def using x by simp
  also have "real_of_int ((2 :: int) ^  ) = start_value x p" unfolding start_value_def  by simp
  finally have less: "root p x  start_value x p" .
  have "0  root p x" using p x by auto
  also have "  start_value x p" by (rule less)
  finally have start: "0  start_value x p" by simp
  from power_mono[OF less, of p] have "root p (of_int x) ^ p  of_int (start_value x p) ^ p" using p x by auto
  also have " = start_value x p ^ p" by simp
  also have "root p (of_int x) ^ p = x" using p x by force
  finally have "x  (start_value x p) ^ p" by presburger
  with start show ?thesis by auto
qed

lemma start_value: assumes x: "x  0" and p: "p > 0" shows "x  (start_value x p) ^ p" "start_value x p  0"
  using start_value_main[OF x p] by auto

text ‹We now define the Newton iteration to compute the $p$-th root. We are working on the integers,
  where every @{term "(/)"} is replaced by @{term "(div)"}. We are proving several things within
  a locale which ensures that $p > 0$, and where $pm = p - 1$.
›

locale fixed_root =
  fixes p pm :: nat
  assumes p: "p = Suc pm"
begin

function root_newton_int_main :: "int  int  int × bool" where
  "root_newton_int_main x n = (if (x < 0  n < 0) then (0,False) else (if x ^ p  n then (x, x ^ p = n)
    else root_newton_int_main ((n div (x ^ pm) + x * int pm) div (int p)) n))"
    by pat_completeness auto
end

text ‹For the executable algorithm we omit the guard and use a let-construction›

partial_function (tailrec) root_int_main' :: "nat  int  int  int  int  int × bool" where
  [code]: "root_int_main' pm ipm ip x n = (let xpm = x^pm; xp = xpm * x in if xp  n then (x, xp = n)
    else root_int_main' pm ipm ip ((n div xpm + x * ipm) div ip) n)"

text ‹In the following algorithm, we
  start the iteration.
  It will compute @{term "root p n"} and a boolean to indicate whether the root is exact.›

definition root_int_main :: "nat  int  int × bool" where
  "root_int_main p n  if p = 0 then (1,n = 1) else
     let pm = p - 1
       in root_int_main' pm (int pm) (int p) (start_value n p) n"

text ‹Once we have proven soundness of @{const fixed_root.root_newton_int_main} and equivalence
  to @{const root_int_main}, it
  is easy to assemble the following algorithm which computes all roots for arbitrary integers.›

definition root_int :: "nat  int  int list" where
  "root_int p x  if p = 0 then [] else
    if x = 0 then [0] else
      let e = even p; s = sgn x; x' = abs x
      in if x < 0  e then [] else case root_int_main p x' of (y,True)  if e then [y,-y] else [s * y] | _  []"

text ‹We start with proving termination of @{const fixed_root.root_newton_int_main}.›

context fixed_root
begin
lemma iteration_mono_eq: assumes xn: "x ^ p = (n :: int)"
  shows "(n div x ^ pm + x * int pm) div int p = x"
proof -
  have [simp]: " n. (x + x * n) = x * (1 + n)" by (auto simp: field_simps)
  show ?thesis unfolding xn[symmetric] p by simp
qed

lemma p0: "p  0" unfolding p by auto

text ‹The following property is the essential property for
  proving termination of @{const "root_newton_int_main"}.
›
lemma iteration_mono_less: assumes x: "x  0"
  and n: "n  0"
  and xn: "x ^ p > (n :: int)"
  shows "(n div x ^ pm + x * int pm) div int p < x"
proof -
  let ?sx = "(n div x ^ pm + x * int pm) div int p"
  from xn have xn_le: "x ^ p  n" by auto
  from xn x n have x0: "x > 0"
    using not_le p by fastforce
  from p have xp: "x ^ p = x * x ^ pm" by auto
  from x n have "n div x ^ pm * x ^ pm  n"
    by (auto simp add: minus_mod_eq_div_mult [symmetric] mod_int_pos_iff not_less power_le_zero_eq)
  also have "  x ^ p" using xn by auto
  finally have le: "n div x ^ pm  x" using x x0 unfolding xp by simp
  have "?sx  (x^p div x ^ pm + x * int pm) div int p"
    by (rule zdiv_mono1, insert le p0, unfold xp, auto)
  also have "x^p div x ^ pm = x" unfolding xp by auto
  also have "x + x * int pm = x * int p" unfolding p by (auto simp: field_simps)
  also have "x * int p div int p = x" using p by force
  finally have le: "?sx  x" .
  {
    assume "?sx = x"
    from arg_cong[OF this, of "λ x. x * int p"]
    have "x * int p  (n div x ^ pm + x * int pm) div (int p) * int p" using p0 by simp
    also have "  n div x ^ pm + x * int pm"
      unfolding mod_div_equality_int using p by auto
    finally have "n div x^pm  x" by (auto simp: p field_simps)
    from mult_right_mono[OF this, of "x ^ pm"]
    have ge: "n div x^pm * x^pm  x^p" unfolding xp using x by auto
    from div_mult_mod_eq[of n "x^pm"] have "n div x^pm * x^pm = n - n mod x^pm" by arith
    from ge[unfolded this]
    have le: "x^p  n - n mod x^pm" .
    from x n have ge: "n mod x ^ pm  0"
      by (auto simp add: mod_int_pos_iff not_less power_le_zero_eq)
    from le ge
    have "n  x^p" by auto
    with xn have False by auto
  }
  with le show ?thesis unfolding p by fastforce
qed

lemma iteration_mono_lesseq: assumes x: "x  0" and n: "n  0" and xn: "x ^ p  (n :: int)"
  shows "(n div x ^ pm + x * int pm) div int p  x"
proof (cases "x ^ p = n")
  case True
  from iteration_mono_eq[OF this] show ?thesis by simp
next
  case False
  with assms have "x ^ p > n" by auto
  from iteration_mono_less[OF x n this]
  show ?thesis by simp
qed

termination (* of root_newton_int_main *)
proof -
  let ?mm = "λ x  n :: int. nat x"
  let ?m1 = "λ (x,n). ?mm x n"
  let ?m = "measures [?m1]"
  show ?thesis
  proof (relation ?m)
    fix x n :: int
    assume "¬ x ^ p  n"
    hence x: "x ^ p > n" by auto
    assume "¬ (x < 0  n < 0)"
    hence x_n: "x  0" "n  0" by auto
    from x x_n have x0: "x > 0" using p by (cases "x = 0", auto)
    from iteration_mono_less[OF x_n x] x0
    show "(((n div x ^ pm + x * int pm) div int p, n), x, n)  ?m" by auto
  qed auto
qed

text ‹We next prove that @{const root_int_main'} is a correct implementation of @{const root_newton_int_main}.
We additionally prove that the result is always positive, a lower bound, and that the returned boolean indicates
whether the result has a root or not. We prove all these results in one go, so that we can share the
inductive proof.
›

abbreviation root_main' where "root_main'  root_int_main' pm (int pm) (int p)"

lemmas root_main'_simps = root_int_main'.simps[of pm "int pm" "int p"]

lemma root_main'_newton_pos: "x  0  n  0 
  root_main' x n = root_newton_int_main x n  (root_main' x n = (y,b)  y  0  y^p  n  b = (y^p = n))"
proof (induct x n rule: root_newton_int_main.induct)
  case (1 x n)
  have pm_x[simp]: "x ^ pm * x = x ^ p" unfolding p by simp
  from 1 have id: "(x < 0  n < 0) = False" by auto
  note d = root_main'_simps[of x n] root_newton_int_main.simps[of x n] id if_False Let_def
  show ?case
  proof (cases "x ^ p  n")
    case True
    thus ?thesis unfolding d using 1(2) by auto
  next
    case False
    hence id: "(x ^ p  n) = False" by simp
    from 1(3) 1(2) have not: "¬ (x < 0  n < 0)" by auto
    then have x: "x > 0  x = 0"
      by auto
    with 0  n have "0  (n div x ^ pm + x * int pm) div int p"
      by (auto simp add: p algebra_simps pos_imp_zdiv_nonneg_iff power_0_left)
    then show ?thesis unfolding d id pm_x
      by (rule 1(1)[OF not False _ 1(3)])
  qed
qed

lemma root_main': "x  0  n  0  root_main' x n = root_newton_int_main x n"
  using root_main'_newton_pos by blast

lemma root_main'_pos: "x  0  n  0  root_main' x n = (y,b)  y  0"
  using root_main'_newton_pos by blast

lemma root_main'_sound: "x  0  n  0  root_main' x n = (y,b)  b = (y ^ p = n)"
  using root_main'_newton_pos by blast

text ‹In order to prove completeness of the algorithms, we provide sharp upper and lower bounds
  for @{const root_main'}. For the upper bounds, we use Cauchy's mean theorem where we added
  the non-strict variant to Porter's formalization of this theorem.›

lemma root_main'_lower: "x  0  n  0  root_main' x n = (y,b)  y ^ p  n"
  using root_main'_newton_pos by blast

lemma root_newton_int_main_upper:
  shows "y ^ p  n  y  0  n  0  root_newton_int_main y n = (x,b)  n < (x + 1) ^ p"
proof (induct y n rule: root_newton_int_main.induct)
  case (1 y n)
  from 1(3) have y0: "y  0" .
  then have "y > 0  y = 0"
    by auto
  from 1(4) have n0: "n  0" .
  define y' where "y' = (n div (y ^ pm) + y * int pm) div (int p)"
  from y > 0  y = 0 n  0 have y'0: "y'  0"
    by (auto simp add: y'_def p algebra_simps pos_imp_zdiv_nonneg_iff power_0_left)
  let ?rt = "root_newton_int_main"
  from 1(5) have rt: "?rt y n = (x,b)" by auto
  from y0 n0 have not: "¬ (y < 0  n < 0)" "(y < 0  n < 0) = False" by auto
  note rt = rt[unfolded root_newton_int_main.simps[of y n] not(2) if_False, folded y'_def]
  note IH = 1(1)[folded y'_def, OF not(1) _ _ y'0 n0]
  show ?case
  proof (cases "y ^ p  n")
    case False note yyn = this
    with rt have rt: "?rt y' n = (x,b)" by simp
    show ?thesis
    proof (cases "n  y' ^ p")
      case True
      show ?thesis
        by (rule IH[OF False True rt])
    next
      case False
      with rt have x: "x = y'" unfolding root_newton_int_main.simps[of y' n]
        using n0 y'0 by simp
      from yyn have yyn: "y^p > n" by simp
      from False have yyn': "n > y' ^ p" by auto
      {
        assume pm: "pm = 0"
        have y': "y' = n" unfolding y'_def p pm by simp
        with yyn' have False unfolding p pm by auto
      }
      hence pm0: "pm > 0" by auto
      show ?thesis
      proof (cases "n = 0")
        case True
        thus ?thesis unfolding p
          by (metis False y'0 zero_le_power)
      next
        case False note n00 = this
        let ?y = "of_int y :: real"
        let ?n = "of_int n :: real"
        from yyn n0 have y00: "y  0" unfolding p by auto
        from y00 y0 have y0: "?y > 0" by auto
        from n0 False have n0: "?n > 0" by auto
        define Y where "Y = ?y * of_int pm"
        define NY where "NY = ?n / ?y ^ pm"
        note pos_intro = divide_nonneg_pos add_nonneg_nonneg mult_nonneg_nonneg
        have NY0: "NY > 0" unfolding NY_def using y0 n0
          by (metis NY_def zero_less_divide_iff zero_less_power)
        let ?ls = "NY # replicate pm ?y"
        have prod: "∏:replicate pm ?y = ?y ^ pm "
          by (induct pm, auto)
        have sum: "∑:replicate pm ?y = Y" unfolding Y_def
          by (induct pm, auto simp: field_simps)
        have pos: "pos ?ls" unfolding pos_def using NY0 y0 by auto
        have "root p ?n = gmean ?ls" unfolding gmean_def using y0
          by (auto simp: p NY_def prod)
        also have " < mean ?ls"
        proof (rule CauchysMeanTheorem_Less[OF pos het_gt_0I])
          show "NY  set ?ls" by simp
          from pm0 show "?y  set ?ls" by simp
          have "NY < ?y"
          proof -
            from yyn have less: "?n < ?y ^ Suc pm" unfolding p
              by (metis of_int_less_iff of_int_power)
            have "NY < ?y ^ Suc pm / ?y ^ pm" unfolding NY_def
              by (rule divide_strict_right_mono[OF less], insert y0, auto)
            thus ?thesis using y0 by auto
          qed
          thus "NY  ?y" by blast
        qed
        also have " = (NY + Y) / real p"
          by (simp add: mean_def sum p)
        finally have *: "root p ?n < (NY + Y) / real p" .
        have "?n = (root p ?n)^p" using n0
          by (metis neq0_conv p0 real_root_pow_pos)
        also have " < ((NY + Y) / real p)^p"
          by (rule power_strict_mono[OF *], insert n0 p, auto)
        finally have ineq1: "?n < ((NY + Y) / real p)^p" by auto
        {
          define s where "s = n div y ^ pm + y * int pm"
          define S where "S = NY + Y"
          have Y0: "Y  0" using y0 unfolding Y_def
            by (metis "1.prems"(2) mult_nonneg_nonneg of_int_0_le_iff of_nat_0_le_iff)
          have S0: "S > 0" using NY0 Y0 unfolding S_def by auto
          from p have p0: "p > 0" by auto
          have "?n / ?y ^ pm  < of_int (floor (?n / ?y^pm)) + 1"
            by (rule divide_less_floor1)
          also have "floor (?n / ?y ^ pm) = n div y^pm"
            unfolding div_is_floor_divide_real by (metis of_int_power)
          finally have "NY < of_int (n div y ^ pm) + 1" unfolding NY_def by simp
          hence less: "S < of_int s + 1" unfolding Y_def s_def S_def by simp
          { (* by sledgehammer *)
            have f1: "x0. rat_of_int rat_of_nat x0 = rat_of_nat x0"
              using of_int_of_nat_eq by simp
            have f2: "x0. real_of_int rat_of_nat x0 = real x0"
              using of_int_of_nat_eq by auto
            have f3: "x0 x1. rat_of_int x0 / rat_of_int x1 = real_of_int x0 / real_of_int x1"
              using div_is_floor_divide_rat div_is_floor_divide_real by simp
            have f4: "0 < rat_of_nat p"
              using p by simp
            have "S  s" using less floor_le_iff by auto
            hence "rat_of_int S / rat_of_nat p  rat_of_int s / rat_of_nat p"
              using f1 f3 f4 by (metis div_is_floor_divide_real zdiv_mono1)
            hence "S / real p  rat_of_int s / rat_of_nat p"
              using f1 f2 f3 f4 by (metis div_is_floor_divide_real floor_div_pos_int)
            hence "S / real p  real_of_int (s div int p) + 1"
              using f1 f3 by (metis div_is_floor_divide_real floor_le_iff floor_of_nat less_eq_real_def)
          }
          hence "S / real p  of_int(s div p) + 1" .
          note this[unfolded S_def s_def]
        }
        hence ge: "of_int y' + 1  (NY + Y) / p" unfolding y'_def
          by simp
        have pos1: "(NY + Y) / p  0" unfolding Y_def NY_def
          by (intro divide_nonneg_pos add_nonneg_nonneg mult_nonneg_nonneg,
          insert y0 n0 p0, auto)
        have pos2: "of_int y' + (1 :: rat)  0" using y'0 by auto
        have ineq2: "(of_int y' + 1) ^ p  ((NY + Y) / p) ^ p"
          by (rule power_mono[OF ge pos1])
        from order.strict_trans2[OF ineq1 ineq2]
        have "?n < of_int ((x + 1) ^ p)" unfolding x
          by (metis of_int_1 of_int_add of_int_power)
        thus "n < (x + 1) ^ p" using of_int_less_iff by blast
      qed
    qed
  next
    case True
    with rt have x: "x = y" by simp
    with 1(2) True have n: "n = y ^ p" by auto
    show ?thesis unfolding n x using y0 unfolding p
      by (metis add_le_less_mono add_less_cancel_left lessI less_add_one add.right_neutral le_iff_add power_strict_mono)
  qed
qed

lemma root_main'_upper:
  "x ^ p  n  x  0  n  0  root_main' x n = (y,b)  n < (y + 1) ^ p"
  using root_newton_int_main_upper[of n x y b] root_main'[of x n] by auto
end

text ‹Now we can prove all the nice properties of @{const root_int_main}.›

lemma root_int_main_all: assumes n: "n  0"
  and rm: "root_int_main p n = (y,b)"
  shows "y  0  b = (y ^ p = n)  (p > 0  y ^ p  n  n < (y + 1)^p)
     (p > 0  x  0  x ^ p = n  y = x  b)"
proof (cases "p = 0")
  case True
  with rm[unfolded root_int_main_def]
  have y: "y = 1" and b: "b = (n = 1)" by auto
  show ?thesis unfolding True y b using n by auto
next
  case False
  from False have p_0: "p > 0" by auto
  from False have "(p = 0) = False" by simp
  from rm[unfolded root_int_main_def this Let_def]
  have rm: "root_int_main' (p - 1) (int (p - 1)) (int p) (start_value n p) n = (y,b)" by simp
  from start_value[OF n p_0] have start: "n  (start_value n p)^p" "0  start_value n p" by auto
  interpret fixed_root p "p - 1"
    by (unfold_locales, insert False, auto)
  from root_main'_pos[OF start(2) n rm] have y: "y  0" .
  from root_main'_sound[OF start(2) n rm] have b: "b = (y ^ p = n)" .
  from root_main'_lower[OF start(2) n rm] have low: "y ^ p  n" .
  from root_main'_upper[OF start n rm] have up: "n < (y + 1) ^ p" .
  {
    assume n: "x ^ p = n" and x: "x  0"
    with low up have low: "y ^ p  x ^ p" and up: "x ^ p < (y+1) ^ p" by auto
    from power_strict_mono[of x y, OF _ x p_0] low have x: "x  y" by arith
    from power_mono[of "(y + 1)" x p] y up have y: "y  x" by arith
    from x y have "x = y" by auto
    with b n
    have "y = x  b" by auto
  }
  thus ?thesis using b low up y by auto
qed

lemma root_int_main: assumes n: "n  0"
  and rm: "root_int_main p n = (y,b)"
  shows "y  0" "b = (y ^ p = n)" "p > 0  y ^ p  n" "p > 0  n < (y + 1)^p"
    "p > 0  x  0  x ^ p = n  y = x  b"
  using root_int_main_all[OF n rm, of x] by blast+

lemma root_int[simp]: assumes p: "p  0  x  1"
  shows "set (root_int p x) = {y . y ^ p = x}"
proof (cases "p = 0")
  case True
  with p have "x  1" by auto
  thus ?thesis unfolding root_int_def True by auto
next
  case False
  hence p: "(p = 0) = False" and p0: "p > 0" by auto
  note d = root_int_def p if_False Let_def
  show ?thesis
  proof (cases "x = 0")
    case True
    thus ?thesis unfolding d using p0 by auto
  next
    case False
    hence x: "(x = 0) = False" by auto
    show ?thesis
    proof (cases "x < 0  even p")
      case True
      hence left: "set (root_int p x) = {}" unfolding d by auto
      {
        fix y
        assume x: "y ^ p = x"
        with True have "y ^ p < 0  even p" by auto
        hence False by presburger
      }
      with left show ?thesis by auto
    next
      case False
      with x p have cond: "(x = 0) = False" "(x < 0  even p) = False" by auto
      obtain y b where rt: "root_int_main p ¦x¦ = (y,b)" by force
      have "abs x  0" by auto
      note rm = root_int_main[OF this rt]
      have "?thesis =
        (set (case root_int_main p ¦x¦ of (y, True)  if even p then [y, - y] else [sgn x * y] | (y, False)  []) =
        {y. y ^ p = x})" unfolding d cond by blast
      also have "(case root_int_main p ¦x¦ of (y, True)  if even p then [y, - y] else [sgn x * y] | (y, False)  [])
        = (if b then if even p then [y, - y] else [sgn x * y] else [])" (is "_ = ?lhs")
        unfolding rt by auto
      also have "set ?lhs = {y. y ^ p = x}" (is "_ = ?rhs")
      proof -
        {
          fix z
          assume idx: "z ^ p = x"
          hence eq: "(abs z) ^ p = abs x" by (metis power_abs)
          from idx x p0 have z: "z  0" unfolding p by auto
          have "(y, b) = (¦z¦, True)"
            using rm(5)[OF p0 _ eq] by auto
          hence id: "y = abs z" "b = True" by auto
          have "z  set ?lhs" unfolding id using z by (auto simp: idx[symmetric], cases "z < 0", auto)
        }
        moreover
        {
          fix z
          assume z: "z  set ?lhs"
          hence b: "b = True" by (cases b, auto)
          note z = z[unfolded b if_True]
          from rm(2) b have yx: "y ^ p = ¦x¦" by auto
          from rm(1) have y: "y  0" .
          from False have "odd p  even p  x  0" by auto
          hence "z  ?rhs"
          proof
            assume odd: "odd p"
            with z have "z = sgn x * y" by auto
            hence "z ^ p = (sgn x * y) ^ p" by auto
            also have " = sgn x ^ p * y ^ p" unfolding power_mult_distrib by auto
            also have " = sgn x ^ p * abs x" unfolding yx by simp
            also have "sgn x ^ p = sgn x" using x odd by auto
            also have "sgn x * abs x = x" by (rule mult_sgn_abs)
            finally show "z  ?rhs" by auto
          next
            assume even: "even p  x  0"
            from z even have "z = y  z = -y" by auto
            hence id: "abs z = y" using y by auto
            with yx x even have z: "z  0" using p0 by (cases "y = 0", auto)
            have "z ^ p = (sgn z * abs z) ^ p" by (simp add: mult_sgn_abs)
            also have " = (sgn z * y) ^ p" using id by auto
            also have " = (sgn z)^p * y ^ p" unfolding power_mult_distrib  by simp
            also have " = sgn z ^ p * x" unfolding yx using even by auto
            also have "sgn z ^ p = 1" using even z by (auto)
            finally show "z  ?rhs" by auto
          qed
        }
        ultimately show ?thesis by blast
      qed
      finally show ?thesis by auto
    qed
  qed
qed

lemma root_int_pos: assumes x: "x  0" and ri: "root_int p x = y # ys"
  shows "y  0"
proof -
  from x have abs: "abs x = x" by auto
  note ri = ri[unfolded root_int_def Let_def abs]
  from ri have p: "(p = 0) = False" by (cases p, auto)
  note ri = ri[unfolded p if_False]
  show ?thesis
  proof (cases "x = 0")
    case True
    with ri show ?thesis by auto
  next
    case False
    hence "(x = 0) = False" "(x < 0  even p) = False" using x by auto
    note ri = ri[unfolded this if_False]
    obtain y' b' where r: "root_int_main p x = (y',b')" by force
    note ri = ri[unfolded this]
    hence y: "y = (if even p then y' else sgn x * y')" by (cases b', auto)
    from root_int_main(1)[OF x r] have y': "0  y'" .
    thus ?thesis unfolding y using x False by auto
  qed
qed

subsection ‹Floor and ceiling of roots›

text ‹Using the bounds for @{const root_int_main} we can easily design
  algorithms which compute @{term "floor (root p x)"} and @{term "ceiling (root p x)"}.
  To this end, we first develop algorithms for non-negative @{term x}, and later on
  these are used for the general case.›

definition "root_int_floor_pos p x = (if p = 0 then 0 else fst (root_int_main p x))"
definition "root_int_ceiling_pos p x = (if p = 0 then 0 else (case root_int_main p x of (y,b)  if b then y else y + 1))"

lemma root_int_floor_pos_lower: assumes p0: "p  0" and x: "x  0"
  shows "root_int_floor_pos p x ^ p  x"
  using root_int_main(3)[OF x, of p] p0 unfolding root_int_floor_pos_def
  by (cases "root_int_main p x", auto)

lemma root_int_floor_pos_pos: assumes x: "x  0"
  shows "root_int_floor_pos p x  0"
  using root_int_main(1)[OF x, of p]
  unfolding root_int_floor_pos_def
  by (cases "root_int_main p x", auto)

lemma root_int_floor_pos_upper: assumes p0: "p  0" and x: "x  0"
  shows "(root_int_floor_pos p x + 1) ^ p > x"
  using root_int_main(4)[OF x, of p] p0 unfolding root_int_floor_pos_def
  by (cases "root_int_main p x", auto)

lemma root_int_floor_pos: assumes x: "x  0"
  shows "root_int_floor_pos p x = floor (root p (of_int x))"
proof (cases "p = 0")
  case True
  thus ?thesis by (simp add: root_int_floor_pos_def)
next
  case False
  hence p: "p > 0" by auto
  let ?s1 = "real_of_int (root_int_floor_pos p x)"
  let ?s2 = "root p (of_int x)"
  from x have s1: "?s1  0"
    by (metis of_int_0_le_iff root_int_floor_pos_pos)
  from x have s2: "?s2  0"
    by (metis of_int_0_le_iff real_root_pos_pos_le)
  from s1 have s11: "?s1 + 1  0" by auto
  have id: "?s2 ^ p = of_int x" using x
    by (metis p of_int_0_le_iff real_root_pow_pos2)
  show ?thesis
  proof (rule floor_unique[symmetric])
    show "?s1  ?s2"
      unfolding compare_pow_le_iff[OF p s1 s2, symmetric]
      unfolding id
      using root_int_floor_pos_lower[OF False x]
      by (metis of_int_le_iff of_int_power)
    show "?s2 < ?s1 + 1"
      unfolding compare_pow_less_iff[OF p s2 s11, symmetric]
      unfolding id
      using root_int_floor_pos_upper[OF False x]
      by (metis of_int_add of_int_less_iff of_int_power of_int_1)
  qed
qed

lemma root_int_ceiling_pos: assumes x: "x  0"
  shows "root_int_ceiling_pos p x = ceiling (root p (of_int x))"
proof (cases "p = 0")
  case True
  thus ?thesis by (simp add: root_int_ceiling_pos_def)
next
  case False
  hence p: "p > 0" by auto
  obtain y b where s: "root_int_main p x = (y,b)" by force
  note rm = root_int_main[OF x s]
  note rm = rm(1-2) rm(3-5)[OF p]
  from rm(1) have y: "y  0" by simp
  let ?s = "root_int_ceiling_pos p x"
  let ?sx = "root p (of_int x)"
  note d = root_int_ceiling_pos_def
  show ?thesis
  proof (cases b)
    case True
    hence id: "?s = y" unfolding s d using p by auto
    from rm(2) True have xy: "x = y ^ p" by auto
    show ?thesis unfolding id unfolding xy using y
      by (simp add: p real_root_power_cancel)
  next
    case False
    hence id: "?s = root_int_floor_pos p x + 1" unfolding d root_int_floor_pos_def
      using s p by simp
    from False have x0: "x  0" using rm(5)[of 0] using s unfolding root_int_main_def Let_def using p
      by (cases "x = 0", auto)
    show ?thesis unfolding id root_int_floor_pos[OF x]
    proof (rule ceiling_unique[symmetric])
      show "?sx  real_of_int (root p (of_int x) + 1)"
        by (metis of_int_add real_of_int_floor_add_one_ge of_int_1)
      let ?l = "real_of_int (root p (of_int x) + 1) - 1"
      let ?m = "real_of_int root p (of_int x)"
      have "?l = ?m" by simp
      also have " < ?sx"
      proof -
        have le: "?m  ?sx" by (rule of_int_floor_le)
        have neq: "?m  ?sx"
        proof
          assume "?m = ?sx"
          hence "?m ^ p = ?sx ^ p" by auto
          also have " = of_int x" using x False
            by (metis p real_root_ge_0_iff real_root_pow_pos2 root_int_floor_pos root_int_floor_pos_pos zero_le_floor zero_less_Suc)
          finally have xs: "x = root p (of_int x) ^ p"
            by (metis floor_power floor_of_int)
          hence "root p (of_int x)  set (root_int p x)" using p by simp
          hence "root_int p x  []" by force
          with s False p  0 x x0 show False unfolding root_int_def
            by (cases p, auto)
        qed
        from le neq show ?thesis by arith
      qed
      finally show "?l < ?sx" .
    qed
  qed
qed


definition "root_int_floor p x = (if x  0 then root_int_floor_pos p x else - root_int_ceiling_pos p (- x))"
definition "root_int_ceiling p x = (if x  0 then root_int_ceiling_pos p x else - root_int_floor_pos p (- x))"

lemma root_int_floor[simp]: "root_int_floor p x = floor (root p (of_int x))"
proof -
  note d = root_int_floor_def
  show ?thesis
  proof (cases "x  0")
    case True
    with root_int_floor_pos[OF True, of p] show ?thesis unfolding d by simp
  next
    case False
    hence "- x  0" by auto
    from False root_int_ceiling_pos[OF this] show ?thesis unfolding d
      by (simp add: real_root_minus ceiling_minus)
  qed
qed

lemma root_int_ceiling[simp]: "root_int_ceiling p x = ceiling (root p (of_int x))"
proof -
  note d = root_int_ceiling_def
  show ?thesis
  proof (cases "x  0")
    case True
    with root_int_ceiling_pos[OF True] show ?thesis unfolding d by simp
  next
    case False
    hence "- x  0" by auto
    from False root_int_floor_pos[OF this, of p] show ?thesis unfolding d
      by (simp add: real_root_minus floor_minus)
  qed
qed

subsection ‹Downgrading algorithms to the naturals›

definition root_nat_floor :: "nat  nat  int" where
  "root_nat_floor p x = root_int_floor_pos p (int x)"

definition root_nat_ceiling :: "nat  nat  int" where
  "root_nat_ceiling p x = root_int_ceiling_pos p (int x)"

definition root_nat :: "nat  nat  nat list" where
  "root_nat p x = map nat (take 1 (root_int p x))"

lemma root_nat_floor [simp]: "root_nat_floor p x = floor (root p (real x))"
  unfolding root_nat_floor_def using root_int_floor_pos[of "int x" p]
  by auto

lemma root_nat_floor_lower: assumes p0: "p  0"
  shows "root_nat_floor p x ^ p  x"
  using root_int_floor_pos_lower[OF p0, of x] unfolding root_nat_floor_def by auto

lemma root_nat_floor_upper: assumes p0: "p  0"
  shows "(root_nat_floor p x + 1) ^ p > x"
  using root_int_floor_pos_upper[OF p0, of x] unfolding root_nat_floor_def by auto

lemma root_nat_ceiling [simp]: "root_nat_ceiling p x = ceiling (root p x)"
  unfolding root_nat_ceiling_def using root_int_ceiling_pos[of x p]
  by auto

lemma root_nat: assumes p0: "p  0  x  1"
  shows "set (root_nat p x) = { y. y ^ p = x}"
proof -
  {
    fix y
    assume "y  set (root_nat p x)"
    note y = this[unfolded root_nat_def]
    then obtain yi ys where ri: "root_int p x = yi # ys" by (cases "root_int p x", auto)
    with y have y: "y = nat yi" by auto
    from root_int_pos[OF _ ri] have yi: "0  yi" by auto
    from root_int[of p "int x"] p0 ri have "yi ^ p = x" by auto
    from arg_cong[OF this, of nat] yi have "nat yi ^ p = x"
      by (metis nat_int nat_power_eq)
    hence "y  {y. y ^ p = x}" using y by auto
  }
  moreover
  {
    fix y
    assume yx: "y ^ p = x"
    hence y: "int y ^ p = int x"
      by (metis of_nat_power)
    hence "set (root_int p (int x))  {}" using root_int[of p "int x"] p0
      by (metis (mono_tags) One_nat_def y ^ p = x empty_Collect_eq nat_power_eq_Suc_0_iff)
    then obtain yi ys where ri: "root_int p (int x) = yi # ys"
      by (cases "root_int p (int x)", auto)
    from root_int_pos[OF _ this] have yip: "yi  0" by auto
    from root_int[of p "int x", unfolded ri] p0 have yi: "yi ^ p = int x" by auto
    with y have "int y ^ p = yi ^ p" by auto
    from arg_cong[OF this, of nat] have id: "y ^ p = nat yi ^ p"
      by (metis y ^ p = x nat_int nat_power_eq yi yip)
    {
      assume p: "p  0"
      hence p0: "p > 0" by auto
      obtain yy b where rm: "root_int_main p (int x) = (yy,b)" by force
      from root_int_main(5)[OF _ rm p0 _ y] have "yy = int y" and "b = True" by auto
      note rm = rm[unfolded this]
      hence "y  set (root_nat p x)"
        unfolding root_nat_def p root_int_def using p0 p yx
        by auto
    }
    moreover
    {
      assume p: "p = 0"
      with p0 have "x  1" by auto
      with y p have False by auto
    }
    ultimately have "y  set (root_nat p x)" by auto
  }
  ultimately show ?thesis by blast
qed

subsection ‹Upgrading algorithms to the rationals›

text ‹The main observation to lift everything from the integers to the rationals is the fact, that one
  can reformulate $\frac{a}{b}^{1/p}$ as $\frac{(ab^{p-1})^{1/p}}b$.›

definition root_rat_floor :: "nat  rat  int" where
  "root_rat_floor p x  case quotient_of x of (a,b)  root_int_floor p (a * b^(p - 1)) div b"

definition root_rat_ceiling :: "nat  rat  int" where
  "root_rat_ceiling p x  - (root_rat_floor p (-x))"

definition root_rat :: "nat  rat  rat list" where
  "root_rat p x  case quotient_of x of (a,b)  concat
  (map (λ rb. map (λ ra. of_int ra / rat_of_int rb) (root_int p a)) (take 1 (root_int p b)))"


lemma root_rat_reform: assumes q: "quotient_of x = (a,b)"
  shows "root p (real_of_rat x) = root p (of_int (a * b ^ (p - 1))) / of_int b"
proof (cases "p = 0")
  case False
  from quotient_of_denom_pos[OF q] have b: "0 < b" by auto
  hence b: "0 < real_of_int b" by auto
  from quotient_of_div[OF q] have x: "root p (real_of_rat x) = root p (a / b)"
    by (metis of_rat_divide of_rat_of_int_eq)
  also have "a / b = a * real_of_int b ^ (p - 1) / of_int b ^ p" using b False
    by (cases p, auto simp: field_simps)
  also have "root p  = root p (a * real_of_int b ^ (p - 1)) / root p (of_int b ^ p)" by (rule real_root_divide)
  also have "root p (of_int b ^ p) = of_int b" using False b
    by (metis neq0_conv real_root_pow_pos real_root_power)
  also have "a * real_of_int b ^ (p - 1) = of_int (a * b ^ (p - 1))"
    by (metis of_int_mult of_int_power)
  finally show ?thesis .
qed auto

lemma root_rat_floor [simp]: "root_rat_floor p x = floor (root p (of_rat x))"
proof -
  obtain a b where q: "quotient_of x = (a,b)" by force
  from quotient_of_denom_pos[OF q] have b: "b > 0" .
  show ?thesis
    unfolding root_rat_floor_def q split root_int_floor
    unfolding root_rat_reform[OF q] floor_div_pos_int[OF b] ..
qed

lemma root_rat_ceiling [simp]: "root_rat_ceiling p x = ceiling (root p (of_rat x))"
  unfolding
    root_rat_ceiling_def
    ceiling_def
    real_root_minus
    root_rat_floor
    of_rat_minus
    ..

lemma root_rat[simp]: assumes p: "p  0  x  1"
  shows "set (root_rat p x) = { y. y ^ p = x}"
proof (cases "p = 0")
  case False
  note p = this
  obtain a b where q: "quotient_of x = (a,b)" by force
  note x = quotient_of_div[OF q]
  have b: "b > 0" by (rule quotient_of_denom_pos[OF q])
  note d = root_rat_def q split set_concat set_map
  {
    fix q
    assume "q  set (root_rat p x)"
    note mem = this[unfolded d]
    from mem obtain rb xs where rb: "root_int p b = Cons rb xs" by (cases "root_int p b", auto)
    note mem = mem[unfolded this]
    from mem obtain ra where ra: "ra  set (root_int p a)" and q: "q = of_int ra / of_int rb"
      by (cases "root_int p a", auto)
    from rb have "rb  set (root_int p b)" by auto
    with ra p have rb: "b = rb ^ p" and ra: "a = ra ^ p" by auto
    have "q  {y. y ^ p = x}" unfolding q x ra rb
      by (auto simp: power_divide)
  }
  moreover
  {
    fix q
    assume "q  {y. y ^ p = x}"
    hence "q ^ p = of_int a / of_int b" unfolding x by auto
    hence eq: "of_int b * q ^ p = of_int a" using b by auto
    obtain z n where quo: "quotient_of q = (z,n)" by force
    note qzn = quotient_of_div[OF quo]
    have n: "n > 0" using quotient_of_denom_pos[OF quo] .
    from eq[unfolded qzn] have "rat_of_int b * of_int z^p / of_int n^p = of_int a"
      unfolding power_divide by simp
    from arg_cong[OF this, of "λ x. x * of_int n^p"] n have "rat_of_int b * of_int z^p = of_int a * of_int n ^ p"
      by auto
    also have "rat_of_int b * of_int z^p = rat_of_int (b * z^p)" unfolding of_int_mult of_int_power ..
    also have "of_int a * rat_of_int n ^ p = of_int (a * n ^ p)" unfolding of_int_mult of_int_power ..
    finally have id: "a * n ^ p = b * z ^ p" by linarith
    from quotient_of_coprime[OF quo] have cop: "coprime (z ^ p) (n ^ p)"
      by simp
    from coprime_crossproduct_int[OF quotient_of_coprime[OF q] this] arg_cong[OF id, of abs]
    have "¦n ^ p¦ = ¦b¦"
      by (simp add: field_simps abs_mult)
    with n b have bnp: "b = n ^ p" by auto
    hence rn: "n  set (root_int p b)" using p by auto
    then obtain rb rs where rb: "root_int p b = Cons rb rs" by (cases "root_int p b", auto)
    from id[folded bnp] b have "a = z ^ p" by auto
    hence a: "z  set (root_int p a)" using p by auto
    from root_int_pos[OF _ rb] b have rb0: "rb  0" by auto
    from root_int[OF disjI1[OF p], of b] rb have "rb ^ p = b" by auto
    with bnp have id: "rb ^ p = n ^ p" by auto
    have "rb = n" by (rule power_eq_imp_eq_base[OF id], insert n rb0 p, auto)
    with rb have b: "n  set (take 1 (root_int p b))" by auto
    have "q  set (root_rat p x)" unfolding d qzn using b a by auto
  }
  ultimately show ?thesis by blast
next
  case True
  with p have x: "x  1" by auto
  obtain a b where q: "quotient_of x = (a,b)" by force
  show ?thesis unfolding True root_rat_def q split root_int_def using x
    by auto
qed

end