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products/sources/formale Sprachen/Isabelle/HOL/Library/ (Beweissystem Isabelle Version 2025-1^©) Datei vom 16.11.2025 mit Größe 16 kB

Quelle Log_Nat.thy Sprache: Isabelle

(* Title: HOL/Library/Log_Nat.thy
 Author: Johannes Hölzl, Fabian Immler, Manuel Eberl
 Copyright 2012 TU München
*)

section \<open>Logarithm of Natural Numbers\<close>

theory Log_Nat
imports Complex_Main
begin

subsection \<open>Preliminaries\<close>

lemma divide_nat_diff_div_nat_less_one:
 "real x / real b - real (x div b) < 1" for x b :: nat
proof (cases "b = 0")
 case True
 then show ?thesis
 by simp
next
 case False
 then have "real (x div b) + real (x mod b) / real b - real (x div b) < 1"
 by (simp add: field_simps)
 then show ?thesis
 by (metis of_nat_of_nat_div_aux)
qed

subsection \<open>Floorlog\<close>

definition floorlog :: "nat \ nat \ nat"
 where "floorlog b a = (if a > 0 \ b > 1 then nat \log b a\ + 1 else 0)"

lemma floorlog_mono: "x \ y \ floorlog b x \ floorlog b y"
 by (auto simp: floorlog_def floor_mono nat_mono)

lemma floorlog_bounds:
 "b ^ (floorlog b x - 1) \ x \ x 0" "b > 1"
proof
 show "b ^ (floorlog b x - 1) \ x"
 proof -
 have "b ^ nat \log b x\ = b powr \log b x\"
 using powr_realpow[symmetric, of b "nat \log b x\"] \x > 0\ \b > 1\
 by simp
 also have "\ \ b powr log b x" using \b > 1\ by simp
 also have "\ = real_of_int x" using \0 < x\ \b > 1\ by simp
 finally have "b ^ nat \log b x\ \ real_of_int x" by simp
 then show ?thesis
 using \<open>0 < x\<close> \<open>b > 1\<close> of_nat_le_iff
 by (fastforce simp add: floorlog_def)
 qed
 show "x 0\ \b > 1\ by simp
 also have "\ < b powr (\log b x\ + 1)"
 using that by (intro powr_less_mono) auto
 also have "\ = b ^ nat (\log b (real_of_int x)\ + 1)"
 using that by (simp flip: powr_realpow)
 finally
 have "x x > 0\<close> \<open>b > 1\<close> by (simp add: floorlog_def nat_add_distrib)
 qed
qed

lemma floorlog_power [simp]:
 "floorlog b (a * b ^ c) = floorlog b a + c" if "a > 0" "b > 1"
proof -
 have "\log b a + real c\ = \log b a\ + c" by arith
 then show ?thesis using that
 by (auto simp: floorlog_def log_mult powr_realpow[symmetric] nat_add_distrib)
qed

lemma floor_log_add_eqI:
 "\log b (a + r)\ = \log b a\" if "b > 1" "a \ 1" "0 \ r" "r < 1"
 for a b :: nat and r :: real
proof (rule floor_eq2)
 have "log b a \ log b (a + r)" using that by force
 then show "\log b a\ \ log b (a + r)" by arith
next
 define l::int where "l = int b ^ (nat \log b a\ + 1)"
 have l_def_real: "l = b powr (\log b a\ + 1)"
 using that by (simp add: l_def powr_add powr_real_of_int)
 have "a < l"
 proof -
 have "a = b powr (log b a)" using that by simp
 also have "\ < b powr floor ((log b a) + 1)"
 using that(1) by auto
 also have "\ = l"
 using that by (simp add: l_def powr_real_of_int powr_add)
 finally show ?thesis by simp
 qed
 then have "a + r < l" using that by simp
 then have "log b (a + r) < log b l" using that by simp
 also have "\ = real_of_int \log b a\ + 1"
 using that by (simp add: l_def_real)
 finally show "log b (a + r) < real_of_int \log b a\ + 1" .
qed

lemma floor_log_div:
 "\log b x\ = \log b (x div b)\ + 1" if "b > 1" "x > 0" "x div b > 0"
 for b x :: nat
proof-
 have "\log b x\ = \log b (x / b * b)\" using that by simp
 also have "\ = \log b (x / b) + log b b\"
 using that by (subst log_mult) auto
 also have "\ = \log b (x / b)\ + 1" using that by simp
 also have "\log b (x / b)\ = \log b (x div b + (x / b - x div b))\" by simp
 also have "\ = \log b (x div b)\"
 using that of_nat_div_le_of_nat divide_nat_diff_div_nat_less_one
 by (intro floor_log_add_eqI) auto
 finally show ?thesis .
qed

lemma compute_floorlog [code]:
 "floorlog b x = (if x > 0 \ b > 1 then floorlog b (x div b) + 1 else 0)"
 by (auto simp: floorlog_def floor_log_div[of b x] div_eq_0_iff nat_add_distrib
 intro!: floor_eq2)

lemma floor_log_eq_if:
 "\log b x\ = \log b y\" if "x div b = y div b" "b > 1" "x > 0" "x div b \ 1"
 for b x y :: nat
proof -
 have "y > 0" using that by (auto intro: ccontr)
 thus ?thesis using that by (simp add: floor_log_div)
qed

lemma floorlog_eq_if:
 "floorlog b x = floorlog b y" if "x div b = y div b" "b > 1" "x > 0" "x div b \ 1"
 for b x y :: nat
proof -
 have "y > 0" using that by (auto intro: ccontr)
 then show ?thesis using that
 by (auto simp add: floorlog_def eq_nat_nat_iff intro: floor_log_eq_if)
qed

lemma floorlog_leD:
 "floorlog b x \ w \ b > 1 \ x 1 \ floorlog b x \ w"
 by (drule less_imp_of_nat_less[where 'a=real])
 (auto simp: floorlog_def Suc_le_eq nat_less_iff floor_less_iff log_of_power_less)

lemma floorlog_eq_zero_iff:
 "floorlog b x = 0 \ b \ 1 \ x \ 0"
 by (auto simp: floorlog_def)

lemma floorlog_le_iff:
 "floorlog b x \ w \ b \ 1 \ b > 1 \ 0 \ w \ x 1"
 using that le_log_of_power[of b w x] power_not_zero
 by (force simp: floorlog_def Suc_le_eq powr_realpow not_less Suc_nat_eq_nat_zadd1
 zless_nat_eq_int_zless int_add_floor less_floor_iff
 simp del: floor_add2)

lemma floorlog_geI:
 "w \ floorlog b x" if "b ^ (w - 1) \ x" "b > 1"
 using floorlog_ge_SucI[of b "w - 1" x] that
 by auto

lemma floorlog_geD:
 "b ^ (w - 1) \ x" if "w \ floorlog b x" "w > 0"
proof -
 have "b > 1" "0 < x"
 using that by (auto simp: floorlog_def split: if_splits)
 have "b ^ (w - 1) \ x" if "b ^ w \ x"
 proof -
 have "b ^ (w - 1) \ b ^ w"
 using \<open>b > 1\<close>
 by (auto intro!: power_increasing)
 also note that
 finally show ?thesis .
 qed
 moreover have "b ^ nat \log (real b) (real x)\ \ x" (is "?l \ _")
 proof -
 have "0 \ log (real b) (real x)"
 using \<open>b > 1\<close> \<open>0 < x\<close>
 by auto
 then have "?l \ b powr log (real b) (real x)"
 using \<open>b > 1\<close>
 by (auto simp flip: powr_realpow intro!: powr_mono of_nat_floor)
 also have "\ = x" using \b > 1\ \0 < x\
 by auto
 finally show ?thesis
 unfolding of_nat_le_iff .
 qed
 ultimately show ?thesis
 using that
 by (auto simp: floorlog_def le_nat_iff le_floor_iff le_log_iff powr_realpow
 split: if_splits elim!: le_SucE)
qed

subsection \<open>\<close>

definition ceillog2 :: "nat \ nat" where
 "ceillog2 n = (if n = 0 then 0 else nat \log 2 (real n)\)"

lemma ceillog2_0 [simp]: "ceillog2 0 = 0"
 and ceillog2_Suc_0 [simp]: "ceillog2 (Suc 0) = 0"
 and ceillog2_2 [simp]: "ceillog2 2 = 1"
 by (auto simp: ceillog2_def)

lemma ceillog2_le1_eq_0 [simp]: "n \ 1 \ ceillog2 n = 0"
 by (cases n) auto

lemma ceillog2_2_power [simp]: "ceillog2 (2 ^ n) = n"
 by (auto simp: ceillog2_def)

lemma ceillog2_ge_log:
 assumes "n > 0"
 shows "real (ceillog2 n) \ log 2 (real n)"
proof -
 have "real_of_int \log 2 (real n)\ \ log 2 (real n)"
 by linarith
 thus ?thesis
 using assms unfolding ceillog2_def by auto
qed

lemma ceillog2_less_log:
 assumes "n > 0"
 shows "real (ceillog2 n) < log 2 (real n) + 1"
proof -
 have "real_of_int \log 2 (real n)\ < log 2 (real n) + 1"
 by linarith
 thus ?thesis
 using assms unfolding ceillog2_def by auto
qed

lemma ceillog2_le_iff:
 assumes "n > 0"
 shows "ceillog2 n \ l \ n \ 2 ^ l"
proof -
 have "ceillog2 n \ l \ real n \ 2 ^ l"
 unfolding ceillog2_def using assms by (auto simp: log_le_iff powr_realpow)
 also have "2 ^ l = real (2 ^ l)"
 by simp
 also have "real n \ real (2 ^ l) \ n \ 2 ^ l"
 by linarith
 finally show ?thesis .
qed

lemma ceillog2_ge_iff:
 assumes "n > 0"
 shows "ceillog2 n \ l \ 2 ^ l < 2 * n"
proof -
 have "-1 < (0 :: real)"
 by auto
 also have "\ \ log 2 (real n)"
 using assms by auto
 finally have "ceillog2 n \ l \ real l - 1 < log 2 (real n)"
 unfolding ceillog2_def using assms by (auto simp: le_nat_iff le_ceiling_iff)
 also have "\ \ real l < log 2 (real (2 * n))"
 using assms by (auto simp: log_mult)
 also have "\ \ 2 ^ l < real (2 * n)"
 using assms by (subst less_log_iff) (auto simp: powr_realpow)
 also have "2 ^ l = real (2 ^ l)"
 by simp
 also have "real (2 ^ l) < real (2 * n) \ 2 ^ l < 2 * n"
 by linarith
 finally show ?thesis .
qed

lemma le_two_power_ceillog2: "n \ 2 ^ ceillog2 n"
 using neq0_conv ceillog2_le_iff by blast

lemma two_power_ceillog2_gt:
 assumes "n > 0"
 shows "2 * n > 2 ^ ceillog2 n"
 using ceillog2_ge_iff[of n "ceillog2 n"] assms by simp

lemma ceillog2_eqI:
 assumes "n \ 2 ^ l" "2 ^ l < 2 * n"
 shows "ceillog2 n = l"
 by (metis Suc_leI assms bot_nat_0.not_eq_extremum ceillog2_ge_iff ceillog2_le_iff le_antisym mult_is_0
 not_less_eq_eq)

lemma ceillog2_rec_even:
 assumes "k > 0"
 shows "ceillog2 (2 * k) = Suc (ceillog2 k)"
 by (rule ceillog2_eqI) (auto simp: le_two_power_ceillog2 two_power_ceillog2_gt assms)

lemma ceillog2_mono:
 assumes "m \ n"
 shows "ceillog2 m \ ceillog2 n"
proof (cases "m = 0")
 case False
 have "\log 2 (real m)\ \ \log 2 (real n)\"
 by (intro ceiling_mono) (use False assms in auto)
 hence "nat \log 2 (real m)\ \ nat \log 2 (real n)\"
 by linarith
 thus ?thesis using False assms
 unfolding ceillog2_def by simp
qed auto

lemma ceillog2_rec_odd:
 assumes "k > 0"
 shows "ceillog2 (Suc (2 * k)) = Suc (ceillog2 (Suc k))"
proof -
 have "2 ^ ceillog2 (Suc (2 * k)) > Suc (2 * k)"
 by (metis assms diff_Suc_1 dvd_triv_left le_two_power_ceillog2 mult_pos_pos nat_power_eq_Suc_0_iff
 order_less_le pos2 semiring_parity_class.even_mask_iff)
 then have "ceillog2 (2 * k + 2) \ ceillog2 (2 * k + 1)"
 by (simp add: ceillog2_le_iff)
 moreover have "ceillog2 (2 * k + 2) \ ceillog2 (2 * k + 1)"
 by (rule ceillog2_mono) auto
 ultimately have "ceillog2 (2 * k + 2) = ceillog2 (2 * k + 1)"
 by (rule antisym)
 also have "2 * k + 2 = 2 * Suc k"
 by simp
 also have "ceillog2 (2 * Suc k) = Suc (ceillog2 (Suc k))"
 by (rule ceillog2_rec_even) auto
 finally show ?thesis
 by simp
qed

(* TODO: better code is possible using bitlen and "count trailing 0 bits" *)
lemma ceillog2_rec:
 "ceillog2 n = (if n \ 1 then 0 else 1 + ceillog2 ((n + 1) div 2))"
proof (cases "n \ 1")
 case True
 thus ?thesis
 by (cases n) auto
next
 case False
 thus ?thesis
 by (cases "even n") (auto elim!: evenE oddE simp: ceillog2_rec_even ceillog2_rec_odd)
qed

lemma funpow_div2_ceillog2_le_1:
 "((\n. (n + 1) div 2) ^^ ceillog2 n) n \ 1"
proof (induction n rule: less_induct)
 case (less n)
 show ?case
 proof (cases "n \ 1")
 case True
 thus ?thesis by (cases n) auto
 next
 case False
 have "((\n. (n + 1) div 2) ^^ Suc (ceillog2 ((n + 1) div 2))) n \ 1"
 using less.IH[of "(n+1) div 2"] False by (subst funpow_Suc_right) auto
 also have "Suc (ceillog2 ((n + 1) div 2)) = ceillog2 n"
 using False by (subst ceillog2_rec[of n]) auto
 finally show ?thesis .
 qed
qed

fun ceillog2_aux :: "nat \ nat \ nat" where
 "ceillog2_aux acc n = (if n \ 1 then acc else ceillog2_aux (acc + 1) ((n + 1) div 2))"

lemmas [simp del] = ceillog2_aux.simps

lemma ceillog2_aux_correct: "ceillog2_aux acc n = ceillog2 n + acc"
proof (induction acc n rule: ceillog2_aux.induct)
 case (1 acc n)
 show ?case
 proof (cases "n \ 1")
 case False
 thus ?thesis using ceillog2_rec[of n] "1.IH"
 by (auto simp: ceillog2_aux.simps[of acc n])
 qed (auto simp: ceillog2_aux.simps[of acc n])
qed

(* TODO: better code equation using bit operations *)
lemma ceillog2_code [code]: "ceillog2 n = ceillog2_aux 0 n"
 by (simp add: ceillog2_aux_correct)

subsection \<open>Bitlen\<close>

definition bitlen :: "int \ int"
 where "bitlen a = floorlog 2 (nat a)"

lemma bitlen_alt_def:
 "bitlen a = (if a > 0 then \log 2 a\ + 1 else 0)"
 by (simp add: bitlen_def floorlog_def)

lemma bitlen_zero [simp]:
 "bitlen 0 = 0"
 by (auto simp: bitlen_def floorlog_def)

lemma bitlen_nonneg:
 "0 \ bitlen x"
 by (simp add: bitlen_def)

lemma bitlen_bounds:
 "2 ^ nat (bitlen x - 1) \ x \ x < 2 ^ nat (bitlen x)" if "x > 0"
proof -
 from that have "bitlen x \ 1" by (auto simp: bitlen_alt_def)
 with that floorlog_bounds[of "nat x" 2] show ?thesis
 by (auto simp add: bitlen_def le_nat_iff nat_less_iff nat_diff_distrib)
qed

lemma bitlen_pow2 [simp]:
 "bitlen (b * 2 ^ c) = bitlen b + c" if "b > 0"
 using that by (simp add: bitlen_def nat_mult_distrib nat_power_eq)

lemma compute_bitlen [code]:
 "bitlen x = (if x > 0 then bitlen (x div 2) + 1 else 0)"
 by (simp add: bitlen_def nat_div_distrib compute_floorlog)

lemma bitlen_eq_zero_iff:
 "bitlen x = 0 \ x \ 0"
 by (auto simp add: bitlen_alt_def)
 (metis compute_bitlen add.commute bitlen_alt_def bitlen_nonneg less_add_same_cancel2
 not_less zero_less_one)

lemma bitlen_div:
 "1 \ real_of_int m / 2^nat (bitlen m - 1)"
 and "real_of_int m / 2^nat (bitlen m - 1) < 2" if "0 < m"
proof -
 let ?B = "2^nat (bitlen m - 1)"

 have "?B \ m" using bitlen_bounds[OF \0 ] ..
 then have "1 * ?B \ real_of_int m"
 unfolding of_int_le_iff[symmetric] by auto
 then show "1 \ real_of_int m / ?B" by auto

 from that have "0 \ bitlen m - 1" by (auto simp: bitlen_alt_def)

 have "m < 2^nat(bitlen m)" using bitlen_bounds[OF that] ..
 also from that have "\ = 2^nat(bitlen m - 1 + 1)"
 by (auto simp: bitlen_def)
 also have "\ = ?B * 2"
 unfolding nat_add_distrib[OF \<open>0 \<le> bitlen m - 1\<close> zero_le_one] by auto
 finally have "real_of_int m < 2 * ?B"
 by (metis (full_types) mult.commute power.simps(2) of_int_less_numeral_power_cancel_iff)
 then have "real_of_int m / ?B < 2 * ?B / ?B"
 by (rule divide_strict_right_mono) auto
 then show "real_of_int m / ?B < 2" by auto
qed

lemma bitlen_le_iff_floorlog:
 "bitlen x \ w \ w \ 0 \ floorlog 2 (nat x) \ nat w"
 by (auto simp: bitlen_def)

lemma bitlen_le_iff_power:
 "bitlen x \ w \ w \ 0 \ x < 2 ^ nat w"
 by (auto simp: bitlen_le_iff_floorlog floorlog_le_iff)

lemma less_power_nat_iff_bitlen:
 "x < 2 ^ w \ bitlen (int x) \ w"
 using bitlen_le_iff_power[of x w]
 by auto

lemma bitlen_ge_iff_power:
 "w \ bitlen x \ w \ 0 \ 2 ^ (nat w - 1) \ x"
 unfolding bitlen_def
 by (auto simp flip: nat_le_iff intro: floorlog_geI dest: floorlog_geD)

lemma bitlen_twopow_add_eq:
 "bitlen (2 ^ w + b) = w + 1" if "0 \ b" "b < 2 ^ w"
 by (auto simp: that nat_add_distrib bitlen_le_iff_power bitlen_ge_iff_power intro!: antisym)

end

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