Session Sqrt_Babylonian

Theory Sqrt_Babylonian_Auxiliary

(*  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 ‹Auxiliary lemmas which might be moved into the Isabelle distribution.›

theory Sqrt_Babylonian_Auxiliary
imports 
  Complex_Main
begin

lemma mod_div_equality_int: "(n :: int) div x * x = n - n mod x"
  using div_mult_mod_eq[of n x] by arith

lemma div_is_floor_divide_rat: "n div y = rat_of_int n / rat_of_int y"
  unfolding Fract_of_int_quotient[symmetric] floor_Fract by simp

lemma div_is_floor_divide_real: "n div y = real_of_int n / of_int y"
  unfolding div_is_floor_divide_rat[of n y]
  by (metis Ratreal_def of_rat_divide of_rat_of_int_eq real_floor_code)

lemma floor_div_pos_int: 
  fixes r :: "'a :: floor_ceiling"
  assumes n: "n > 0" 
  shows "r / of_int n = r div n" (is "?l = ?r")
proof -
  let ?of_int = "of_int :: int  'a"
  define rhs where "rhs = r div n"
  let ?n = "?of_int n"
  define m where "m = r mod n"
  let ?m = "?of_int m"
  from div_mult_mod_eq[of "floor r" n] have dm: "rhs * n + m = r" unfolding rhs_def m_def by simp
  have mn: "m < n" and m0: "m  0" using n m_def by auto
  define e where "e = r - ?of_int r"
  have e0: "e  0" unfolding e_def 
    by (metis diff_self eq_iff floor_diff_of_int zero_le_floor)
  have e1: "e < 1" unfolding e_def 
    by (metis diff_self dual_order.refl floor_diff_of_int floor_le_zero)
  have "r = ?of_int r + e" unfolding e_def by simp
  also have "r = rhs * n + m" using dm by simp
  finally have "r = ?of_int (rhs * n + m) + e" .
  hence "r / ?n = ?of_int (rhs * n) / ?n + (e + ?m) / ?n" using n by (simp add: field_simps)
  also have "?of_int (rhs * n) / ?n = ?of_int rhs" using n by auto
  finally have *: "r / ?of_int n = (e + ?of_int m) / ?of_int n + ?of_int rhs" by simp
  have "?l = rhs + floor ((e + ?m) / ?n)" unfolding * by simp
  also have "floor ((e + ?m) / ?n) = 0"
  proof (rule floor_unique)
    show "?of_int 0  (e + ?m) / ?n" using e0 m0 n 
      by (metis add_increasing2 divide_nonneg_pos of_int_0 of_int_0_le_iff of_int_0_less_iff)
    show "(e + ?m) / ?n < ?of_int 0 + 1"
    proof (rule ccontr)
      from n have n': "?n > 0" "?n  0" by simp_all
      assume "¬ ?thesis"
      hence "(e + ?m) / ?n  1" by auto
      from mult_right_mono[OF this n'(2)]
      have "?n  e + ?m" using n'(1) by simp
      also have "?m  ?n - 1" using mn 
        by (metis of_int_1 of_int_diff of_int_le_iff zle_diff1_eq)
      finally have "?n  e + ?n - 1" by auto
      with e1 show False by arith
    qed
  qed
  finally show ?thesis unfolding rhs_def by simp
qed


lemma floor_div_neg_int: 
  fixes r :: "'a :: floor_ceiling"
  assumes n: "n < 0" 
  shows "r / of_int n = r div n"
proof -
  from n have n': "- n > 0" by auto
  have "r / of_int n =  - r / of_int (- n)" using n
    by (metis floor_of_int floor_zero less_int_code(1) minus_divide_left minus_minus nonzero_minus_divide_right of_int_minus)
  also have " =  - r  div (- n)" by (rule floor_div_pos_int[OF n'])
  also have " =  r  div n" using n 
  by (metis ceiling_def div_minus_right)
  finally show ?thesis .
qed

lemma divide_less_floor1: "n / y < of_int (floor (n / y)) + 1" 
  by (metis floor_less_iff less_add_one of_int_1 of_int_add)

context linordered_idom
begin

lemma sgn_int_pow_if [simp]:
  "sgn x ^ p = (if even p then 1 else sgn x)" if "x  0"
  using that by (induct p) simp_all

lemma compare_pow_le_iff: "p > 0  (x :: 'a)  0  y  0  (x ^ p  y ^ p) = (x  y)"
  by (rule power_mono_iff)

lemma compare_pow_less_iff: "p > 0  (x :: 'a)  0  y  0  (x ^ p < y ^ p) = (x < y)"
  using compare_pow_le_iff [of p x y]
  using local.dual_order.order_iff_strict local.power_strict_mono by blast
    
end

lemma quotient_of_int[simp]: "quotient_of (of_int i) = (i,1)" 
  by (metis Rat.of_int_def quotient_of_int)

lemma quotient_of_nat[simp]: "quotient_of (of_nat i) = (int i,1)" 
  by (metis Rat.of_int_def Rat.quotient_of_int of_int_of_nat_eq)

lemma square_lesseq_square: " x y. 0  (x :: 'a :: linordered_field)  0  y  (x * x  y * y) = (x  y)"
  by (metis mult_mono mult_strict_mono' not_less)

lemma square_less_square: " x y. 0  (x :: 'a :: linordered_field)  0  y  (x * x < y * y) = (x < y)"
  by (metis mult_mono mult_strict_mono' not_less)

lemma sqrt_sqrt[simp]: "x  0  sqrt x * sqrt x = x"
  by (metis real_sqrt_pow2 power2_eq_square)

lemma abs_lesseq_square: "abs (x :: real)  abs y  x * x  y * y"
  using square_lesseq_square[of "abs x" "abs y"] by auto

end

Theory Log_Impl

(*  Title:       Computing Square Roots using the Babylonian Method
    Author:      René Thiemann       <rene.thiemann@uibk.ac.at>
    Maintainer:  René Thiemann
    License:     LGPL
*)
section ‹A Fast Logarithm Algorithm›

theory Log_Impl
imports 
  Sqrt_Babylonian_Auxiliary
begin

text ‹We implement the discrete logarithm function in a manner similar to
  a repeated squaring exponentiation algorithm.›

text ‹In order to prove termination of the algorithm without intermediate checks 
  we need to ensure that we only use proper bases, 
  i.e., values of at least 2. This will be encoded into a separate type.›

typedef proper_base = "{x :: int. x  2}" by auto

setup_lifting type_definition_proper_base

lift_definition get_base :: "proper_base  int" is "λ x. x" .

lift_definition square_base :: "proper_base  proper_base" is "λ x. x * x" 
proof -
  fix i :: int
  assume i: "2  i"
  have "2 * 2  i * i" 
    by (rule mult_mono[OF i i], insert i, auto)
  thus "2  i * i" by auto
qed

lift_definition into_base :: "int  proper_base" is "λ x. if x  2 then x else 2" by auto

lemma square_base: "get_base (square_base b) = get_base b * get_base b" 
  by (transfer, auto)

lemma get_base_2: "get_base b  2"
  by (transfer, auto)

lemma b_less_square_base_b: "get_base b < get_base (square_base b)" 
  unfolding square_base using get_base_2[of b] by simp

lemma b_less_div_base_b: assumes xb: "¬ x < get_base b"
  shows "x div get_base b < x"
proof -
  from get_base_2[of b] have b: "get_base b  2" .
  with xb have x2: "x  2" by auto
  with b int_div_less_self[of x "(get_base b)"] 
  show ?thesis by auto
qed
    
text ‹We now state the main algorithm.›
    
function log_main :: "proper_base  int  nat × int" where
  "log_main b x = (if x < get_base b then (0,1) else
    case log_main (square_base b) x of 
      (z, bz)  
    let l = 2 * z; bz1 = bz * get_base b
      in if x < bz1 then (l,bz) else (Suc l,bz1))" 
  by pat_completeness auto

termination by (relation "measure (λ (b,x). nat (1 + x - get_base b))",
  insert b_less_square_base_b, auto)   

lemma log_main: "x > 0  log_main b x = (y,by)  by = (get_base b)^y  (get_base b)^y  x  x < (get_base b)^(Suc y)" 
proof (induct b x arbitrary: y "by" rule: log_main.induct)
  case (1 b x y "by")
  note x = 1(2)
  note y = 1(3)
  note IH = 1(1)
  let ?b = "get_base b" 
  show ?case
  proof (cases "x < ?b")
    case True
    with x y show ?thesis by auto
  next
    case False
    obtain z bz where zz: "log_main (square_base b) x = (z,bz)" 
      by (cases "log_main (square_base b) x", auto)
    have id: "get_base (square_base b) ^ k = ?b ^ (2 * k)" for k unfolding square_base
      by (simp add: power_mult semiring_normalization_rules(29))
    from IH[OF False x zz, unfolded id] 
    have z: "?b ^ (2 * z)  x" "x < ?b ^ (2 * Suc z)" and bz: "bz = get_base b ^ (2 * z)" by auto
    from y[unfolded log_main.simps[of b x] Let_def zz split] bz False
    have yy: "(if x < bz * ?b then (2 * z, bz) else (Suc (2 * z), bz * ?b)) =
      (y, by)" by auto
    show ?thesis
    proof (cases "x < bz * ?b")
      case True
      with yy have yz: "y = 2 * z" "by = bz" by auto
      from True z(1) bz show ?thesis unfolding yz by (auto simp: ac_simps)
    next
      case False
      with yy have yz: "y = Suc (2 * z)" "by = ?b * bz" by auto
      from False have "?b ^ Suc (2 * z)  x" by (auto simp: bz ac_simps)
      with z(2) bz show ?thesis unfolding yz by auto
    qed
  qed
qed
    
text ‹We then derive the floor- and ceiling-log functions.›

definition log_floor :: "int  int  nat" where
  "log_floor b x = fst (log_main (into_base b) x)" 

definition log_ceiling :: "int  int  nat" where
  "log_ceiling b x = (case log_main (into_base b) x of
     (y,by)  if x = by then y else Suc y)" 

lemma log_floor_sound: assumes "b > 1" "x > 0" "log_floor b x = y"  
  shows "b^y  x" "x < b^(Suc y)" 
proof -
  from assms(1,3) have id: "get_base (into_base b) = b" by transfer auto
  obtain yy bb where log: "log_main (into_base b) x = (yy,bb)" 
    by (cases "log_main (into_base b) x", auto)
  from log_main[OF assms(2) log] assms(3)[unfolded log_floor_def log] id
  show "b^y  x" "x < b^(Suc y)" by auto
qed

lemma log_ceiling_sound: assumes "b > 1" "x > 0" "log_ceiling b x = y"  
  shows "x  b^y" "y  0  b^(y - 1) < x" 
proof -
  from assms(1,3) have id: "get_base (into_base b) = b" by transfer auto
  obtain yy bb where log: "log_main (into_base b) x = (yy,bb)" 
    by (cases "log_main (into_base b) x", auto)
  from log_main[OF assms(2) log, unfolded id] assms(3)[unfolded log_ceiling_def log split]
  have bnd: "b ^ yy  x" "x < b ^ Suc yy" and
    y: "y = (if x = b ^ yy then yy else Suc yy)" by auto
  have "x  b^y  (y  0  b^(y - 1) < x)"
  proof (cases "x = b ^ yy")
    case True
    with y bnd assms(1) show ?thesis by (cases yy, auto)
  next
    case False
    with y bnd show ?thesis by auto
  qed
  thus "x  b^y" "y  0  b^(y - 1) < x" by auto
qed

text ‹Finally, we connect it to the @{const log} function working on real numbers.›

lemma log_floor[simp]: assumes b: "b > 1" and x: "x > 0"
  shows "log_floor b x = log b x"
proof -
  obtain y where y: "log_floor b x = y" by auto
  note main = log_floor_sound[OF assms y]
  from b x have *: "1 < real_of_int b" "0 < real_of_int (b ^ y)" "0 < real_of_int x" 
    and **: "1 < real_of_int b" "0 < real_of_int x" "0 < real_of_int (b ^ Suc y)" 
    by auto
  show ?thesis unfolding y
  proof (rule sym, rule floor_unique)
    show "real_of_int (int y)  log (real_of_int b) (real_of_int x)" 
      using main(1)[folded log_le_cancel_iff[OF *, unfolded of_int_le_iff]]
      using log_pow_cancel[of b y] b by auto
    show "log (real_of_int b) (real_of_int x) < real_of_int (int y) + 1" 
      using main(2)[folded log_less_cancel_iff[OF **, unfolded of_int_less_iff]]
      using log_pow_cancel[of b "Suc y"] b by auto
  qed
qed
    
lemma log_ceiling[simp]: assumes b: "b > 1" and x: "x > 0"
  shows "log_ceiling b x = log b x"
proof -
  obtain y where y: "log_ceiling b x = y" by auto
  note main = log_ceiling_sound[OF assms y]
  from b x have *: "1 < real_of_int b" "0 < real_of_int (b ^ (y - 1))" "0 < real_of_int x" 
    and **: "1 < real_of_int b" "0 < real_of_int x" "0 < real_of_int (b ^ y)" 
    by auto
  show ?thesis unfolding y
  proof (rule sym, rule ceiling_unique)
    show "log (real_of_int b) (real_of_int x)  real_of_int (int y)" 
      using main(1)[folded log_le_cancel_iff[OF **, unfolded of_int_le_iff]]
      using log_pow_cancel[of b y] b by auto
    from x have x: "x  1" by auto
    show "real_of_int (int y) - 1 < log (real_of_int b) (real_of_int x)" 
    proof (cases "y = 0")
      case False
      thus ?thesis 
        using main(2)[folded log_less_cancel_iff[OF *, unfolded of_int_less_iff]]
        using log_pow_cancel[of b "y - 1"] b x by auto
    next
      case True
      have "real_of_int (int y) - 1 = log b (1/b)" using True b 
        by (subst log_divide, auto)
      also have " < log b 1"
        by (subst log_less_cancel_iff, insert b, auto) 
      also have "  log b x" 
        by (subst log_le_cancel_iff, insert b x, auto) 
      finally show "real_of_int (int y) - 1 < log (real_of_int b) (real_of_int x)" .
    qed
  qed
qed

end

Theory 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

Theory Sqrt_Babylonian

(*  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/>.
*)

theory Sqrt_Babylonian
imports 
  Sqrt_Babylonian_Auxiliary
  NthRoot_Impl
begin

section ‹Executable algorithms for square roots›

text ‹
  This theory provides executable algorithms for computing square-roots of numbers which
  are all based on the Babylonian method (which is also known as Heron's method or Newton's method).
  
  For integers / naturals / rationals precise algorithms are given, i.e., here $sqrt\ x$ delivers
  a list of all integers / naturals / rationals $y$ where $y^2 = x$. 
  To this end, the Babylonian method has been adapted by using integer-divisions.

  In addition to the precise algorithms, we also provide approximation algorithms. One works for 
  arbitrary linear ordered fields, where some number $y$ is computed such that
  @{term "abs(y^2 - x) < ε"}. Moreover, for the naturals, integers, and rationals we provide algorithms to compute
  @{term "floor (sqrt x)"} and @{term "ceiling (sqrt x)"} which are all based
  on the underlying algorithm that is used to compute the precise square-roots on integers, if these 
  exist.

  The major motivation for developing the precise algorithms was given by \ceta{} \cite{CeTA},
  a tool for certifiying termination proofs. Here, non-linear equations of the form
  $(a_1x_1 + \dots a_nx_n)^2 = p$ had to be solved over the integers, where $p$ is a concrete polynomial.
  For example, for the equation $(ax + by)^2 = 4x^2 - 12xy + 9y^2$ one easily figures out that
  $a^2 = 4, b^2 = 9$, and $ab = -6$, which results in a possible solution $a = \sqrt 4 = 2, b = - \sqrt 9 = -3$.
›

subsection ‹The Babylonian method›

text ‹
The Babylonian method for computing $\sqrt n$ iteratively computes 
\[
x_{i+1} = \frac{\frac n{x_i} + x_i}2
\]
until $x_i^2 \approx n$. Note that if $x_0^2 \geq n$, then for all $i$ we have both
$x_i^2 \geq n$ and $x_i \geq x_{i+1}$. 
›

subsection ‹The Babylonian method using integer division›
text ‹
  First, the algorithm is developed for the non-negative integers.
  Here, the division operation $\frac xy$ is replaced by @{term "x div y = of_int x / of_int y"}.
  Note that replacing @{term "of_int x / of_int y"} by @{term "of_int x / of_int y"} would lead to non-termination
  in the following algorithm.

  We explicititly develop the algorithm on the integers and not on the naturals, as the calculations
  on the integers have been much easier. For example, $y - x + x = y$ on the integers, which would require
  the side-condition $y \geq x$ for the naturals. These conditions will make the reasoning much more tedious---as
  we have experienced in an earlier state of this development where everything was based on naturals.

  Since the elements
  $x_0, x_1, x_2,\dots$ are monotone decreasing, in the main algorithm we abort as soon as $x_i^2 \leq n$.›


text ‹\textbf{Since in the meantime, all of these algorithms have been generalized to arbitrary
  $p$-th roots in @{theory Sqrt_Babylonian.NthRoot_Impl}, we just instantiate the general algorithms by $p = 2$ and then provide 
  specialized code equations which are more efficient than the general purpose algorithms.}›

definition sqrt_int_main' :: "int  int  int × bool" where
  [simp]: "sqrt_int_main' x n = root_int_main' 1 1 2 x n"

lemma sqrt_int_main'_code[code]: "sqrt_int_main' x n = (let x2 = x * x in if x2  n then (x, x2 = n)
    else sqrt_int_main' ((n div x + x) div 2) n)"
  using root_int_main'.simps[of 1 1 2 x n]
  unfolding Let_def by auto

definition sqrt_int_main :: "int  int × bool" where
  [simp]: "sqrt_int_main x = root_int_main 2 x"

lemma sqrt_int_main_code[code]: "sqrt_int_main x = sqrt_int_main' (start_value x 2) x"
  by (simp add: root_int_main_def Let_def)

definition sqrt_int :: "int  int list" where
  "sqrt_int x = root_int 2 x"

lemma sqrt_int_code[code]: "sqrt_int x = (if x < 0 then [] else case sqrt_int_main x of (y,True)  if y = 0 then [0] else [y,-y] | _  [])"
proof -
  interpret fixed_root 2 1 by (unfold_locales, auto)
  obtain b y where res: "root_int_main 2 x = (b,y)" by force
  show ?thesis
    unfolding sqrt_int_def root_int_def Let_def
    using root_int_main[OF _ res]
    using res
    by simp
qed

lemma sqrt_int[simp]: "set (sqrt_int x) = {y. y * y = x}"
  unfolding sqrt_int_def by (simp add: power2_eq_square)


lemma sqrt_int_pos: assumes res: "sqrt_int x = Cons s ms"
  shows "s  0"
proof -
  note res = res[unfolded sqrt_int_code Let_def, simplified]
  from res have x0: "x  0" by (cases ?thesis, auto)
  obtain ss b where call: "sqrt_int_main x = (ss,b)" by force
  from res[unfolded call] x0 have "ss = s" 
    by (cases b, cases "ss = 0", auto)
  from root_int_main(1)[OF x0 call[unfolded this sqrt_int_main_def]]
  show ?thesis .
qed

definition [simp]: "sqrt_int_floor_pos x = root_int_floor_pos 2 x"

lemma sqrt_int_floor_pos_code[code]: "sqrt_int_floor_pos x = fst (sqrt_int_main x)"
  by (simp add: root_int_floor_pos_def)

lemma sqrt_int_floor_pos: assumes x: "x  0" 
  shows "sqrt_int_floor_pos x =  sqrt (of_int x) "
  using root_int_floor_pos[OF x, of 2] by (simp add: sqrt_def)

definition [simp]: "sqrt_int_ceiling_pos x = root_int_ceiling_pos 2 x"

lemma sqrt_int_ceiling_pos_code[code]: "sqrt_int_ceiling_pos x = (case sqrt_int_main x of (y,b)  if b then y else y + 1)"
  by (simp add: root_int_ceiling_pos_def)

lemma sqrt_int_ceiling_pos: assumes x: "x  0" 
  shows "sqrt_int_ceiling_pos x =  sqrt (of_int x) "
  using root_int_ceiling_pos[OF x, of 2] by (simp add: sqrt_def)

definition "sqrt_int_floor x = root_int_floor 2 x"

lemma sqrt_int_floor_code[code]: "sqrt_int_floor x = (if x  0 then sqrt_int_floor_pos x else - sqrt_int_ceiling_pos (- x))"
  unfolding sqrt_int_floor_def root_int_floor_def by simp

lemma sqrt_int_floor[simp]: "sqrt_int_floor x =  sqrt (of_int x) "
  by (simp add: sqrt_int_floor_def sqrt_def)

definition "sqrt_int_ceiling x = root_int_ceiling 2 x"

lemma sqrt_int_ceiling_code[code]: "sqrt_int_ceiling x = (if x  0 then sqrt_int_ceiling_pos x else - sqrt_int_floor_pos (- x))"
  unfolding sqrt_int_ceiling_def root_int_ceiling_def by simp

lemma sqrt_int_ceiling[simp]: "sqrt_int_ceiling x =  sqrt (of_int x) "
  by (simp add: sqrt_int_ceiling_def sqrt_def)

lemma sqrt_int_ceiling_bound: "0  x  x  (sqrt_int_ceiling x)^2"
  unfolding sqrt_int_ceiling using le_of_int_ceiling sqrt_le_D
  by (metis of_int_power_le_of_int_cancel_iff)


subsection ‹Square roots for the naturals›


definition sqrt_nat :: "nat  nat list"
  where "sqrt_nat x = root_nat 2 x"
 
lemma sqrt_nat_code[code]: "sqrt_nat x  map nat (take 1 (sqrt_int (int x)))"
  unfolding sqrt_nat_def root_nat_def sqrt_int_def by simp

lemma sqrt_nat[simp]: "set (sqrt_nat x) = { y. y * y = x}" 
  unfolding sqrt_nat_def using root_nat[of 2 x] by (simp add: power2_eq_square)

definition sqrt_nat_floor :: "nat  int" where
  "sqrt_nat_floor x = root_nat_floor 2 x"

lemma sqrt_nat_floor_code[code]: "sqrt_nat_floor x = sqrt_int_floor_pos (int x)"
  unfolding sqrt_nat_floor_def root_nat_floor_def by simp

lemma sqrt_nat_floor[simp]: "sqrt_nat_floor x =  sqrt (real x) "
  unfolding sqrt_nat_floor_def by (simp add: sqrt_def)

definition sqrt_nat_ceiling :: "nat  int" where
  "sqrt_nat_ceiling x = root_nat_ceiling 2 x"

lemma sqrt_nat_ceiling_code[code]: "sqrt_nat_ceiling x = sqrt_int_ceiling_pos (int x)"
  unfolding sqrt_nat_ceiling_def root_nat_ceiling_def by simp

lemma sqrt_nat_ceiling[simp]: "sqrt_nat_ceiling x =  sqrt (real x) "
  unfolding sqrt_nat_ceiling_def by (simp add: sqrt_def)

subsection ‹Square roots for the rationals›

definition sqrt_rat :: "rat  rat list" where
  "sqrt_rat x = root_rat 2 x"

lemma sqrt_rat_code[code]: "sqrt_rat x = (case quotient_of x of (z,n)  (case sqrt_int n of 
    []  [] 
  | sn # xs  map (λ sz. of_int sz / of_int sn) (sqrt_int z)))"
proof -
  obtain z n where q: "quotient_of x = (z,n)" by force
  show ?thesis
  unfolding sqrt_rat_def root_rat_def q split sqrt_int_def
  by (cases "root_int 2 n", auto)
qed

lemma sqrt_rat[simp]: "set (sqrt_rat x) = { y. y * y = x}"
  unfolding sqrt_rat_def using root_rat[of 2 x]
  by (simp add: power2_eq_square)

lemma sqrt_rat_pos: assumes sqrt: "sqrt_rat x = Cons s ms" 
  shows "s  0"
proof -
  obtain z n where q: "quotient_of x = (z,n)" by force
  note sqrt = sqrt[unfolded sqrt_rat_code q, simplified]
  let ?sz = "sqrt_int z"
  let ?sn = "sqrt_int n"
  from q have n: "n > 0" by (rule quotient_of_denom_pos)
  from sqrt obtain sz mz where sz: "?sz = sz # mz" by (cases ?sn, auto)
  from sqrt obtain sn mn where sn: "?sn = sn # mn" by (cases ?sn, auto)
  from sqrt_int_pos[OF sz] sqrt_int_pos[OF sn] have pos: "0  sz" "0  sn" by auto
  from sqrt sz sn have s: "s = of_int sz / of_int sn" by auto
  show ?thesis unfolding s using pos
    by (metis of_int_0_le_iff zero_le_divide_iff)
qed

definition sqrt_rat_floor :: "rat  int" where
  "sqrt_rat_floor x = root_rat_floor 2 x"

lemma sqrt_rat_floor_code[code]: "sqrt_rat_floor x = (case quotient_of x of (a,b)  sqrt_int_floor (a * b) div b)"
  unfolding sqrt_rat_floor_def root_rat_floor_def by (simp add: sqrt_def)

lemma sqrt_rat_floor[simp]: "sqrt_rat_floor x =  sqrt (of_rat x) "
  unfolding sqrt_rat_floor_def by (simp add: sqrt_def)

definition sqrt_rat_ceiling :: "rat  int" where
  "sqrt_rat_ceiling x = root_rat_ceiling 2 x"

lemma sqrt_rat_ceiling_code[code]: "sqrt_rat_ceiling x = - (sqrt_rat_floor (-x))"
  unfolding sqrt_rat_ceiling_def sqrt_rat_floor_def root_rat_ceiling_def by simp

lemma sqrt_rat_ceiling: "sqrt_rat_ceiling x =  sqrt (of_rat x) "
  unfolding sqrt_rat_ceiling_def by (simp add: sqrt_def)

lemma sqr_rat_of_int: assumes x: "x * x = rat_of_int i"
  shows " j :: int. j * j = i"
proof -
  from x have mem: "x  set (sqrt_rat (rat_of_int i))" by simp
  from x have "rat_of_int i  0" by (metis zero_le_square)
  hence *: "quotient_of (rat_of_int i) = (i,1)" by (metis quotient_of_int)
  have 1: "sqrt_int 1 = [1,-1]" by code_simp
  from mem sqrt_rat_code * split 1 
  have x: "x  rat_of_int ` {y. y * y = i}" by auto
  thus ?thesis by auto
qed

subsection ‹Approximating square roots›

text ‹
  The difference to the previous algorithms is that now we abort, once the distance is below
  $\epsilon$.  
  Moreover, here we use standard division and not integer division.
  This part is not yet generalized by @{theory Sqrt_Babylonian.NthRoot_Impl}.

  We first provide the executable version without guard @{term "x > 0"} as partial function,
  and afterwards prove termination and soundness for a similar algorithm that is defined within the upcoming
locale.
›

partial_function (tailrec) sqrt_approx_main_impl :: "'a :: linordered_field  'a  'a  'a" where 
  [code]: "sqrt_approx_main_impl ε n x = (if x * x - n < ε then x else sqrt_approx_main_impl ε n 
    ((n / x + x) / 2))"

text ‹We setup a locale where we ensure that we have standard assumptions: positive $\epsilon$ and
  positive $n$. We require sort @{term floor_ceiling}, since @{term " x "} is used for the termination
  argument.›
locale sqrt_approximation = 
  fixes ε :: "'a :: {linordered_field,floor_ceiling}"
  and n :: 'a
  assumes ε : "ε > 0"
  and n: "n > 0"
begin

function sqrt_approx_main :: "'a  'a" where 
  "sqrt_approx_main x = (if x > 0 then (if x * x - n < ε then x else sqrt_approx_main 
    ((n / x + x) / 2)) else 0)"
    by pat_completeness auto

text ‹Termination essentially is a proof of convergence. Here, one complication is the fact
  that the limit is not always defined. E.g., if @{typ "'a"} is @{typ rat} then there is no
  square root of 2. Therefore, the error-rate $\frac x{\sqrt n} - 1$ is not expressible. 
  Instead we use the expression $\frac{x^2}n - 1$ as error-rate which
  does not require any square-root operation.›
termination
proof -
  define er where "er x = (x * x / n - 1)" for x
  define c where "c = 2 * n / ε"
  define m where "m x = nat  c * er x " for x
  have c: "c > 0" unfolding c_def using n ε by auto
  show ?thesis
  proof
    show "wf (measures [m])" by simp
  next
    fix x 
    assume x: "0 < x" and xe: "¬ x * x - n < ε"
    define y where "y = (n / x + x) / 2"    
    show "((n / x + x) / 2,x)  measures [m]" unfolding y_def[symmetric]
    proof (rule measures_less)
      from n have inv_n: "1 / n > 0" by auto
      from xe have "x * x - n  ε" by simp
      from this[unfolded mult_le_cancel_left_pos[OF inv_n, of ε, symmetric]]
      have erxen: "er x  ε / n" unfolding er_def using n by (simp add: field_simps)
      have en: "ε / n > 0" and ne: "n / ε > 0" using ε n by auto
      from en erxen have erx: "er x > 0" by linarith
      have pos: "er x * 4 + er x * (er x * 4) > 0" using erx
        by (auto intro: add_pos_nonneg)
      have "er y = 1 / 4 * (n / (x * x) - 2  + x * x / n)" unfolding er_def y_def using x n
        by (simp add: field_simps)
      also have " = 1 / 4 * er x * er x / (1 + er x)" unfolding er_def using x n
        by (simp add: field_simps)
      finally have "er y = 1 / 4 * er x * er x / (1 + er x)" .
      also have " < 1 / 4 * (1 + er x) * er x / (1 + er x)" using erx erx pos
        by (auto simp: field_simps)
      also have " = er x / 4" using erx by (simp add: field_simps)
      finally have er_y_x: "er y  er x / 4" by linarith
      from erxen have "c * er x  2" unfolding c_def mult_le_cancel_left_pos[OF ne, of _ "er x", symmetric]
        using n ε by (auto simp: field_simps)
      hence pos: "c * er x > 0" "c * er x  2" by auto
      show "m y < m x" unfolding m_def nat_mono_iff[OF pos(1)]
      proof -      
        have "c * er y  c * (er x / 4)"
          by (rule floor_mono, unfold mult_le_cancel_left_pos[OF c], rule er_y_x)
        also have " < c * er x / 4 + 1" by auto
        also have "  c * er x"
          by (rule floor_mono, insert pos(2), simp add: field_simps)
        finally show "c * er y < c * er x" .
      qed
    qed
  qed
qed

text ‹Once termination is proven, it is easy to show equivalence of 
  @{const sqrt_approx_main_impl} and @{const sqrt_approx_main}.›
lemma sqrt_approx_main_impl: "x > 0  sqrt_approx_main_impl ε n x = sqrt_approx_main x"
proof (induct x rule: sqrt_approx_main.induct)
  case (1 x)
  hence x: "x > 0" by auto
  hence nx: "0 < (n / x + x) / 2" using n by (auto intro: pos_add_strict)
  note simps = sqrt_approx_main_impl.simps[of _ _ x] sqrt_approx_main.simps[of x]
  show ?case 
  proof (cases "x * x - n < ε")
    case True
    thus ?thesis unfolding simps using x by auto
  next
    case False
    show ?thesis using 1(1)[OF x False nx] unfolding simps using x False by auto
  qed
qed

text ‹Also soundness is not complicated.›

lemma sqrt_approx_main_sound: assumes x: "x > 0" and xx: "x * x > n"
  shows "sqrt_approx_main x * sqrt_approx_main x > n  sqrt_approx_main x * sqrt_approx_main x - n < ε"
  using assms
proof (induct x rule: sqrt_approx_main.induct)
  case (1 x)
  from 1 have x:  "x > 0" "(x > 0) = True" by auto
  note simp = sqrt_approx_main.simps[of x, unfolded x if_True]
  show ?case
  proof (cases "x * x - n < ε")
    case True
    with 1 show ?thesis unfolding simp by simp
  next
    case False
    let ?y = "(n / x + x) / 2"
    from False simp have simp: "sqrt_approx_main x = sqrt_approx_main ?y" by simp
    from n x have y: "?y > 0" by (auto intro: pos_add_strict)
    note IH = 1(1)[OF x(1) False y]
    from x have x4: "4 * x * x > 0" by (auto intro: mult_sign_intros)
    show ?thesis unfolding simp
    proof (rule IH)
      show "n < ?y * ?y"
        unfolding mult_less_cancel_left_pos[OF x4, of n, symmetric]
      proof -
        have id: "4 * x * x * (?y * ?y) = 4 * x * x * n + (n - x * x) * (n - x * x)" using x(1)
          by (simp add: field_simps)
        from 1(3) have "x * x - n > 0" by auto
        from mult_pos_pos[OF this this]
        show "4 * x * x * n < 4 * x * x * (?y * ?y)" unfolding id 
          by (simp add: field_simps)
      qed
    qed
  qed
qed   

end

text ‹It remains to assemble everything into one algorithm.›

definition sqrt_approx :: "'a :: {linordered_field,floor_ceiling}  'a  'a" where
  "sqrt_approx ε x  if ε > 0 then (if x = 0 then 0 else let xpos = abs x in sqrt_approx_main_impl ε xpos (xpos + 1)) else 0"


lemma sqrt_approx: assumes ε: "ε > 0"
  shows "¦sqrt_approx ε x * sqrt_approx ε x - ¦x¦¦ < ε"
proof (cases "x = 0")
  case True
  with ε show ?thesis unfolding sqrt_approx_def by auto
next
  case False
  let ?x = "¦x¦" 
  let ?sqrti = "sqrt_approx_main_impl ε ?x (?x + 1)"
  let ?sqrt = "sqrt_approximation.sqrt_approx_main ε ?x (?x + 1)"
  define sqrt where "sqrt = ?sqrt"
  from False have x: "?x > 0" "?x + 1 > 0" by auto
  interpret sqrt_approximation ε ?x
    by (unfold_locales, insert x ε, auto)
  from False ε have "sqrt_approx ε x = ?sqrti" unfolding sqrt_approx_def by (simp add: Let_def)
  also have "?sqrti = ?sqrt"
    by (rule sqrt_approx_main_impl, auto)
  finally have id: "sqrt_approx ε x = sqrt" unfolding sqrt_def .
  have sqrt: "sqrt * sqrt > ?x  sqrt * sqrt - ?x < ε" unfolding sqrt_def
    by (rule sqrt_approx_main_sound[OF x(2)], insert x mult_pos_pos[OF x(1) x(1)], auto simp: field_simps)
  show ?thesis unfolding id using sqrt by auto
qed

subsection ‹Some tests›

text ‹Testing executabity and show that sqrt 2 is irrational›
lemma "¬ ( i :: rat. i * i = 2)"
proof -
  have "set (sqrt_rat 2) = {}" by eval
  thus ?thesis by simp
qed

text ‹Testing speed›
lemma "¬ ( i :: int. i * i = 1234567890123456789012345678901234567890)"
proof -
  have "set (sqrt_int 1234567890123456789012345678901234567890) = {}" by eval
  thus ?thesis by simp
qed

text ‹The following test›

value "let ε = 1 / 100000000 :: rat; s = sqrt_approx ε 2 in (s, s * s - 2, ¦s * s - 2¦ < ε)"

text ‹results in (1.4142135623731116, 4.738200762148612e-14, True).›
 
end