src/HOL/Binomial.thy
 author hoelzl Fri Feb 19 13:40:50 2016 +0100 (2016-02-19) changeset 62378 85ed00c1fe7c parent 62347 2230b7047376 child 62481 b5d8e57826df permissions -rw-r--r--
generalize more theorems to support enat and ennreal
     1 (*  Title       : Binomial.thy

     2     Author      : Jacques D. Fleuriot

     3     Copyright   : 1998  University of Cambridge

     4     Conversion to Isar and new proofs by Lawrence C Paulson, 2004

     5     The integer version of factorial and other additions by Jeremy Avigad.

     6     Additional binomial identities by Chaitanya Mangla and Manuel Eberl

     7 *)

     8

     9 section\<open>Factorial Function, Binomial Coefficients and Binomial Theorem\<close>

    10

    11 theory Binomial

    12 imports Main

    13 begin

    14

    15 subsection \<open>Factorial\<close>

    16

    17 fun (in semiring_char_0) fact :: "nat \<Rightarrow> 'a"

    18   where "fact 0 = 1" | "fact (Suc n) = of_nat (Suc n) * fact n"

    19

    20 lemmas fact_Suc = fact.simps(2)

    21

    22 lemma fact_1 [simp]: "fact 1 = 1"

    23   by simp

    24

    25 lemma fact_Suc_0 [simp]: "fact (Suc 0) = Suc 0"

    26   by simp

    27

    28 lemma of_nat_fact [simp]:

    29   "of_nat (fact n) = fact n"

    30   by (induct n) (auto simp add: algebra_simps of_nat_mult)

    31

    32 lemma of_int_fact [simp]:

    33   "of_int (fact n) = fact n"

    34 proof -

    35   have "of_int (of_nat (fact n)) = fact n"

    36     unfolding of_int_of_nat_eq by simp

    37   then show ?thesis

    38     by simp

    39 qed

    40

    41 lemma fact_reduce: "n > 0 \<Longrightarrow> fact n = of_nat n * fact (n - 1)"

    42   by (cases n) auto

    43

    44 lemma fact_nonzero [simp]: "fact n \<noteq> (0::'a::{semiring_char_0,semiring_no_zero_divisors})"

    45   apply (induct n)

    46   apply auto

    47   using of_nat_eq_0_iff by fastforce

    48

    49 lemma fact_mono_nat: "m \<le> n \<Longrightarrow> fact m \<le> (fact n :: nat)"

    50   by (induct n) (auto simp: le_Suc_eq)

    51

    52 lemma fact_in_Nats: "fact n \<in> \<nat>" by (induction n) auto

    53

    54 lemma fact_in_Ints: "fact n \<in> \<int>" by (induction n) auto

    55

    56 context

    57   assumes "SORT_CONSTRAINT('a::linordered_semidom)"

    58 begin

    59

    60   lemma fact_mono: "m \<le> n \<Longrightarrow> fact m \<le> (fact n :: 'a)"

    61     by (metis of_nat_fact of_nat_le_iff fact_mono_nat)

    62

    63   lemma fact_ge_1 [simp]: "fact n \<ge> (1 :: 'a)"

    64     by (metis le0 fact.simps(1) fact_mono)

    65

    66   lemma fact_gt_zero [simp]: "fact n > (0 :: 'a)"

    67     using fact_ge_1 less_le_trans zero_less_one by blast

    68

    69   lemma fact_ge_zero [simp]: "fact n \<ge> (0 :: 'a)"

    70     by (simp add: less_imp_le)

    71

    72   lemma fact_not_neg [simp]: "~ (fact n < (0 :: 'a))"

    73     by (simp add: not_less_iff_gr_or_eq)

    74

    75   lemma fact_le_power:

    76       "fact n \<le> (of_nat (n^n) ::'a)"

    77   proof (induct n)

    78     case (Suc n)

    79     then have *: "fact n \<le> (of_nat (Suc n ^ n) ::'a)"

    80       by (rule order_trans) (simp add: power_mono del: of_nat_power)

    81     have "fact (Suc n) = (of_nat (Suc n) * fact n ::'a)"

    82       by (simp add: algebra_simps)

    83     also have "... \<le> (of_nat (Suc n) * of_nat (Suc n ^ n) ::'a)"

    84       by (simp add: "*" ordered_comm_semiring_class.comm_mult_left_mono del: of_nat_power)

    85     also have "... \<le> (of_nat (Suc n ^ Suc n) ::'a)"

    86       by (metis of_nat_mult order_refl power_Suc)

    87     finally show ?case .

    88   qed simp

    89

    90 end

    91

    92 text\<open>Note that @{term "fact 0 = fact 1"}\<close>

    93 lemma fact_less_mono_nat: "\<lbrakk>0 < m; m < n\<rbrakk> \<Longrightarrow> fact m < (fact n :: nat)"

    94   by (induct n) (auto simp: less_Suc_eq)

    95

    96 lemma fact_less_mono:

    97   "\<lbrakk>0 < m; m < n\<rbrakk> \<Longrightarrow> fact m < (fact n :: 'a::linordered_semidom)"

    98   by (metis of_nat_fact of_nat_less_iff fact_less_mono_nat)

    99

   100 lemma fact_ge_Suc_0_nat [simp]: "fact n \<ge> Suc 0"

   101   by (metis One_nat_def fact_ge_1)

   102

   103 lemma dvd_fact:

   104   shows "1 \<le> m \<Longrightarrow> m \<le> n \<Longrightarrow> m dvd fact n"

   105   by (induct n) (auto simp: dvdI le_Suc_eq)

   106

   107 lemma fact_ge_self: "fact n \<ge> n"

   108   by (cases "n = 0") (simp_all add: dvd_imp_le dvd_fact)

   109

   110 lemma fact_altdef_nat: "fact n = \<Prod>{1..n}"

   111   by (induct n) (auto simp: atLeastAtMostSuc_conv)

   112

   113 lemma fact_altdef: "fact n = (\<Prod>i=1..n. of_nat i)"

   114   by (induct n) (auto simp: atLeastAtMostSuc_conv)

   115

   116 lemma fact_altdef': "fact n = of_nat (\<Prod>{1..n})"

   117   by (subst fact_altdef_nat [symmetric]) simp

   118

   119 lemma fact_dvd: "n \<le> m \<Longrightarrow> fact n dvd (fact m :: 'a :: {semiring_div,linordered_semidom})"

   120   by (induct m) (auto simp: le_Suc_eq)

   121

   122 lemma fact_mod: "m \<le> n \<Longrightarrow> fact n mod (fact m :: 'a :: {semiring_div,linordered_semidom}) = 0"

   123   by (auto simp add: fact_dvd)

   124

   125 lemma fact_div_fact:

   126   assumes "m \<ge> n"

   127   shows "(fact m) div (fact n) = \<Prod>{n + 1..m}"

   128 proof -

   129   obtain d where "d = m - n" by auto

   130   from assms this have "m = n + d" by auto

   131   have "fact (n + d) div (fact n) = \<Prod>{n + 1..n + d}"

   132   proof (induct d)

   133     case 0

   134     show ?case by simp

   135   next

   136     case (Suc d')

   137     have "fact (n + Suc d') div fact n = Suc (n + d') * fact (n + d') div fact n"

   138       by simp

   139     also from Suc.hyps have "... = Suc (n + d') * \<Prod>{n + 1..n + d'}"

   140       unfolding div_mult1_eq[of _ "fact (n + d')"] by (simp add: fact_mod)

   141     also have "... = \<Prod>{n + 1..n + Suc d'}"

   142       by (simp add: atLeastAtMostSuc_conv)

   143     finally show ?case .

   144   qed

   145   from this \<open>m = n + d\<close> show ?thesis by simp

   146 qed

   147

   148 lemma fact_num_eq_if:

   149     "fact m = (if m=0 then 1 else of_nat m * fact (m - 1))"

   150 by (cases m) auto

   151

   152 lemma fact_eq_rev_setprod_nat: "fact k = (\<Prod>i<k. k - i)"

   153   unfolding fact_altdef_nat

   154   by (rule setprod.reindex_bij_witness[where i="\<lambda>i. k - i" and j="\<lambda>i. k - i"]) auto

   155

   156 lemma fact_div_fact_le_pow:

   157   assumes "r \<le> n" shows "fact n div fact (n - r) \<le> n ^ r"

   158 proof -

   159   have "\<And>r. r \<le> n \<Longrightarrow> \<Prod>{n - r..n} = (n - r) * \<Prod>{Suc (n - r)..n}"

   160     by (subst setprod.insert[symmetric]) (auto simp: atLeastAtMost_insertL)

   161   with assms show ?thesis

   162     by (induct r rule: nat.induct) (auto simp add: fact_div_fact Suc_diff_Suc mult_le_mono)

   163 qed

   164

   165 lemma fact_numeral:  \<comment>\<open>Evaluation for specific numerals\<close>

   166   "fact (numeral k) = (numeral k) * (fact (pred_numeral k))"

   167   by (metis fact.simps(2) numeral_eq_Suc of_nat_numeral)

   168

   169

   170 text \<open>This development is based on the work of Andy Gordon and

   171   Florian Kammueller.\<close>

   172

   173 subsection \<open>Basic definitions and lemmas\<close>

   174

   175 primrec binomial :: "nat \<Rightarrow> nat \<Rightarrow> nat" (infixl "choose" 65)

   176 where

   177   "0 choose k = (if k = 0 then 1 else 0)"

   178 | "Suc n choose k = (if k = 0 then 1 else (n choose (k - 1)) + (n choose k))"

   179

   180 lemma binomial_n_0 [simp]: "(n choose 0) = 1"

   181   by (cases n) simp_all

   182

   183 lemma binomial_0_Suc [simp]: "(0 choose Suc k) = 0"

   184   by simp

   185

   186 lemma binomial_Suc_Suc [simp]: "(Suc n choose Suc k) = (n choose k) + (n choose Suc k)"

   187   by simp

   188

   189 lemma choose_reduce_nat:

   190   "0 < (n::nat) \<Longrightarrow> 0 < k \<Longrightarrow>

   191     (n choose k) = ((n - 1) choose (k - 1)) + ((n - 1) choose k)"

   192   by (metis Suc_diff_1 binomial.simps(2) neq0_conv)

   193

   194 lemma binomial_eq_0: "n < k \<Longrightarrow> n choose k = 0"

   195   by (induct n arbitrary: k) auto

   196

   197 declare binomial.simps [simp del]

   198

   199 lemma binomial_n_n [simp]: "n choose n = 1"

   200   by (induct n) (simp_all add: binomial_eq_0)

   201

   202 lemma binomial_Suc_n [simp]: "Suc n choose n = Suc n"

   203   by (induct n) simp_all

   204

   205 lemma binomial_1 [simp]: "n choose Suc 0 = n"

   206   by (induct n) simp_all

   207

   208 lemma zero_less_binomial: "k \<le> n \<Longrightarrow> n choose k > 0"

   209   by (induct n k rule: diff_induct) simp_all

   210

   211 lemma binomial_eq_0_iff [simp]: "n choose k = 0 \<longleftrightarrow> n < k"

   212   by (metis binomial_eq_0 less_numeral_extra(3) not_less zero_less_binomial)

   213

   214 lemma zero_less_binomial_iff [simp]: "n choose k > 0 \<longleftrightarrow> k \<le> n"

   215   by (metis binomial_eq_0_iff not_less0 not_less zero_less_binomial)

   216

   217 lemma Suc_times_binomial_eq:

   218   "Suc n * (n choose k) = (Suc n choose Suc k) * Suc k"

   219   apply (induct n arbitrary: k, simp add: binomial.simps)

   220   apply (case_tac k)

   221    apply (auto simp add: add_mult_distrib add_mult_distrib2 le_Suc_eq binomial_eq_0)

   222   done

   223

   224 lemma binomial_le_pow2: "n choose k \<le> 2^n"

   225   apply (induction n arbitrary: k)

   226   apply (simp add: binomial.simps)

   227   apply (case_tac k)

   228   apply (auto simp: power_Suc)

   229   by (simp add: add_le_mono mult_2)

   230

   231 text\<open>The absorption property\<close>

   232 lemma Suc_times_binomial:

   233   "Suc k * (Suc n choose Suc k) = Suc n * (n choose k)"

   234   using Suc_times_binomial_eq by auto

   235

   236 text\<open>This is the well-known version of absorption, but it's harder to use because of the

   237   need to reason about division.\<close>

   238 lemma binomial_Suc_Suc_eq_times:

   239     "(Suc n choose Suc k) = (Suc n * (n choose k)) div Suc k"

   240   by (simp add: Suc_times_binomial_eq del: mult_Suc mult_Suc_right)

   241

   242 text\<open>Another version of absorption, with -1 instead of Suc.\<close>

   243 lemma times_binomial_minus1_eq:

   244   "0 < k \<Longrightarrow> k * (n choose k) = n * ((n - 1) choose (k - 1))"

   245   using Suc_times_binomial_eq [where n = "n - 1" and k = "k - 1"]

   246   by (auto split add: nat_diff_split)

   247

   248

   249 subsection \<open>Combinatorial theorems involving \<open>choose\<close>\<close>

   250

   251 text \<open>By Florian Kamm\"uller, tidied by LCP.\<close>

   252

   253 lemma card_s_0_eq_empty: "finite A \<Longrightarrow> card {B. B \<subseteq> A & card B = 0} = 1"

   254   by (simp cong add: conj_cong add: finite_subset [THEN card_0_eq])

   255

   256 lemma choose_deconstruct: "finite M \<Longrightarrow> x \<notin> M \<Longrightarrow>

   257     {s. s \<subseteq> insert x M \<and> card s = Suc k} =

   258     {s. s \<subseteq> M \<and> card s = Suc k} \<union> {s. \<exists>t. t \<subseteq> M \<and> card t = k \<and> s = insert x t}"

   259   apply safe

   260      apply (auto intro: finite_subset [THEN card_insert_disjoint])

   261   by (metis (full_types) Diff_insert_absorb Set.set_insert Zero_neq_Suc card_Diff_singleton_if

   262      card_eq_0_iff diff_Suc_1 in_mono subset_insert_iff)

   263

   264 lemma finite_bex_subset [simp]:

   265   assumes "finite B"

   266     and "\<And>A. A \<subseteq> B \<Longrightarrow> finite {x. P x A}"

   267   shows "finite {x. \<exists>A \<subseteq> B. P x A}"

   268   by (metis (no_types) assms finite_Collect_bounded_ex finite_Collect_subsets)

   269

   270 text\<open>There are as many subsets of @{term A} having cardinality @{term k}

   271  as there are sets obtained from the former by inserting a fixed element

   272  @{term x} into each.\<close>

   273 lemma constr_bij:

   274    "finite A \<Longrightarrow> x \<notin> A \<Longrightarrow>

   275     card {B. \<exists>C. C \<subseteq> A \<and> card C = k \<and> B = insert x C} =

   276     card {B. B \<subseteq> A & card(B) = k}"

   277   apply (rule card_bij_eq [where f = "\<lambda>s. s - {x}" and g = "insert x"])

   278   apply (auto elim!: equalityE simp add: inj_on_def)

   279   apply (metis card_Diff_singleton_if finite_subset in_mono)

   280   done

   281

   282 text \<open>

   283   Main theorem: combinatorial statement about number of subsets of a set.

   284 \<close>

   285

   286 theorem n_subsets: "finite A \<Longrightarrow> card {B. B \<subseteq> A \<and> card B = k} = (card A choose k)"

   287 proof (induct k arbitrary: A)

   288   case 0 then show ?case by (simp add: card_s_0_eq_empty)

   289 next

   290   case (Suc k)

   291   show ?case using \<open>finite A\<close>

   292   proof (induct A)

   293     case empty show ?case by (simp add: card_s_0_eq_empty)

   294   next

   295     case (insert x A)

   296     then show ?case using Suc.hyps

   297       apply (simp add: card_s_0_eq_empty choose_deconstruct)

   298       apply (subst card_Un_disjoint)

   299          prefer 4 apply (force simp add: constr_bij)

   300         prefer 3 apply force

   301        prefer 2 apply (blast intro: finite_Pow_iff [THEN iffD2]

   302          finite_subset [of _ "Pow (insert x F)" for F])

   303       apply (blast intro: finite_Pow_iff [THEN iffD2, THEN [2] finite_subset])

   304       done

   305   qed

   306 qed

   307

   308

   309 subsection \<open>The binomial theorem (courtesy of Tobias Nipkow):\<close>

   310

   311 text\<open>Avigad's version, generalized to any commutative ring\<close>

   312 theorem binomial_ring: "(a+b::'a::{comm_ring_1,power})^n =

   313   (\<Sum>k=0..n. (of_nat (n choose k)) * a^k * b^(n-k))" (is "?P n")

   314 proof (induct n)

   315   case 0 then show "?P 0" by simp

   316 next

   317   case (Suc n)

   318   have decomp: "{0..n+1} = {0} Un {n+1} Un {1..n}"

   319     by auto

   320   have decomp2: "{0..n} = {0} Un {1..n}"

   321     by auto

   322   have "(a+b)^(n+1) =

   323       (a+b) * (\<Sum>k=0..n. of_nat (n choose k) * a^k * b^(n-k))"

   324     using Suc.hyps by simp

   325   also have "\<dots> = a*(\<Sum>k=0..n. of_nat (n choose k) * a^k * b^(n-k)) +

   326                    b*(\<Sum>k=0..n. of_nat (n choose k) * a^k * b^(n-k))"

   327     by (rule distrib_right)

   328   also have "\<dots> = (\<Sum>k=0..n. of_nat (n choose k) * a^(k+1) * b^(n-k)) +

   329                   (\<Sum>k=0..n. of_nat (n choose k) * a^k * b^(n-k+1))"

   330     by (auto simp add: setsum_right_distrib ac_simps)

   331   also have "\<dots> = (\<Sum>k=0..n. of_nat (n choose k) * a^k * b^(n+1-k)) +

   332                   (\<Sum>k=1..n+1. of_nat (n choose (k - 1)) * a^k * b^(n+1-k))"

   333     by (simp add:setsum_shift_bounds_cl_Suc_ivl Suc_diff_le field_simps

   334         del:setsum_cl_ivl_Suc)

   335   also have "\<dots> = a^(n+1) + b^(n+1) +

   336                   (\<Sum>k=1..n. of_nat (n choose (k - 1)) * a^k * b^(n+1-k)) +

   337                   (\<Sum>k=1..n. of_nat (n choose k) * a^k * b^(n+1-k))"

   338     by (simp add: decomp2)

   339   also have

   340       "\<dots> = a^(n+1) + b^(n+1) +

   341             (\<Sum>k=1..n. of_nat(n+1 choose k) * a^k * b^(n+1-k))"

   342     by (auto simp add: field_simps setsum.distrib [symmetric] choose_reduce_nat)

   343   also have "\<dots> = (\<Sum>k=0..n+1. of_nat (n+1 choose k) * a^k * b^(n+1-k))"

   344     using decomp by (simp add: field_simps)

   345   finally show "?P (Suc n)" by simp

   346 qed

   347

   348 text\<open>Original version for the naturals\<close>

   349 corollary binomial: "(a+b::nat)^n = (\<Sum>k=0..n. (of_nat (n choose k)) * a^k * b^(n-k))"

   350     using binomial_ring [of "int a" "int b" n]

   351   by (simp only: of_nat_add [symmetric] of_nat_mult [symmetric] of_nat_power [symmetric]

   352            of_nat_setsum [symmetric]

   353            of_nat_eq_iff of_nat_id)

   354

   355 lemma binomial_fact_lemma: "k \<le> n \<Longrightarrow> fact k * fact (n - k) * (n choose k) = fact n"

   356 proof (induct n arbitrary: k rule: nat_less_induct)

   357   fix n k assume H: "\<forall>m<n. \<forall>x\<le>m. fact x * fact (m - x) * (m choose x) =

   358                       fact m" and kn: "k \<le> n"

   359   let ?ths = "fact k * fact (n - k) * (n choose k) = fact n"

   360   { assume "n=0" then have ?ths using kn by simp }

   361   moreover

   362   { assume "k=0" then have ?ths using kn by simp }

   363   moreover

   364   { assume nk: "n=k" then have ?ths by simp }

   365   moreover

   366   { fix m h assume n: "n = Suc m" and h: "k = Suc h" and hm: "h < m"

   367     from n have mn: "m < n" by arith

   368     from hm have hm': "h \<le> m" by arith

   369     from hm h n kn have km: "k \<le> m" by arith

   370     have "m - h = Suc (m - Suc h)" using  h km hm by arith

   371     with km h have th0: "fact (m - h) = (m - h) * fact (m - k)"

   372       by simp

   373     from n h th0

   374     have "fact k * fact (n - k) * (n choose k) =

   375         k * (fact h * fact (m - h) * (m choose h)) +

   376         (m - h) * (fact k * fact (m - k) * (m choose k))"

   377       by (simp add: field_simps)

   378     also have "\<dots> = (k + (m - h)) * fact m"

   379       using H[rule_format, OF mn hm'] H[rule_format, OF mn km]

   380       by (simp add: field_simps)

   381     finally have ?ths using h n km by simp }

   382   moreover have "n=0 \<or> k = 0 \<or> k = n \<or> (\<exists>m h. n = Suc m \<and> k = Suc h \<and> h < m)"

   383     using kn by presburger

   384   ultimately show ?ths by blast

   385 qed

   386

   387 lemma binomial_fact:

   388   assumes kn: "k \<le> n"

   389   shows "(of_nat (n choose k) :: 'a::field_char_0) =

   390          (fact n) / (fact k * fact(n - k))"

   391   using binomial_fact_lemma[OF kn]

   392   apply (simp add: field_simps)

   393   by (metis mult.commute of_nat_fact of_nat_mult)

   394

   395 lemma choose_row_sum: "(\<Sum>k=0..n. n choose k) = 2^n"

   396   using binomial [of 1 "1" n]

   397   by (simp add: numeral_2_eq_2)

   398

   399 lemma sum_choose_lower: "(\<Sum>k=0..n. (r+k) choose k) = Suc (r+n) choose n"

   400   by (induct n) auto

   401

   402 lemma sum_choose_upper: "(\<Sum>k=0..n. k choose m) = Suc n choose Suc m"

   403   by (induct n) auto

   404

   405 lemma choose_alternating_sum:

   406   "n > 0 \<Longrightarrow> (\<Sum>i\<le>n. (-1)^i * of_nat (n choose i)) = (0 :: 'a :: comm_ring_1)"

   407   using binomial_ring[of "-1 :: 'a" 1 n] by (simp add: atLeast0AtMost mult_of_nat_commute zero_power)

   408

   409 lemma choose_even_sum:

   410   assumes "n > 0"

   411   shows   "2 * (\<Sum>i\<le>n. if even i then of_nat (n choose i) else 0) = (2 ^ n :: 'a :: comm_ring_1)"

   412 proof -

   413   have "2 ^ n = (\<Sum>i\<le>n. of_nat (n choose i)) + (\<Sum>i\<le>n. (-1) ^ i * of_nat (n choose i) :: 'a)"

   414     using choose_row_sum[of n]

   415     by (simp add: choose_alternating_sum assms atLeast0AtMost of_nat_setsum[symmetric] of_nat_power)

   416   also have "\<dots> = (\<Sum>i\<le>n. of_nat (n choose i) + (-1) ^ i * of_nat (n choose i))"

   417     by (simp add: setsum.distrib)

   418   also have "\<dots> = 2 * (\<Sum>i\<le>n. if even i then of_nat (n choose i) else 0)"

   419     by (subst setsum_right_distrib, intro setsum.cong) simp_all

   420   finally show ?thesis ..

   421 qed

   422

   423 lemma choose_odd_sum:

   424   assumes "n > 0"

   425   shows   "2 * (\<Sum>i\<le>n. if odd i then of_nat (n choose i) else 0) = (2 ^ n :: 'a :: comm_ring_1)"

   426 proof -

   427   have "2 ^ n = (\<Sum>i\<le>n. of_nat (n choose i)) - (\<Sum>i\<le>n. (-1) ^ i * of_nat (n choose i) :: 'a)"

   428     using choose_row_sum[of n]

   429     by (simp add: choose_alternating_sum assms atLeast0AtMost of_nat_setsum[symmetric] of_nat_power)

   430   also have "\<dots> = (\<Sum>i\<le>n. of_nat (n choose i) - (-1) ^ i * of_nat (n choose i))"

   431     by (simp add: setsum_subtractf)

   432   also have "\<dots> = 2 * (\<Sum>i\<le>n. if odd i then of_nat (n choose i) else 0)"

   433     by (subst setsum_right_distrib, intro setsum.cong) simp_all

   434   finally show ?thesis ..

   435 qed

   436

   437 lemma choose_row_sum': "(\<Sum>k\<le>n. (n choose k)) = 2 ^ n"

   438   using choose_row_sum[of n] by (simp add: atLeast0AtMost)

   439

   440 lemma natsum_reverse_index:

   441   fixes m::nat

   442   shows "(\<And>k. m \<le> k \<Longrightarrow> k \<le> n \<Longrightarrow> g k = f (m + n - k)) \<Longrightarrow> (\<Sum>k=m..n. f k) = (\<Sum>k=m..n. g k)"

   443   by (rule setsum.reindex_bij_witness[where i="\<lambda>k. m+n-k" and j="\<lambda>k. m+n-k"]) auto

   444

   445 text\<open>NW diagonal sum property\<close>

   446 lemma sum_choose_diagonal:

   447   assumes "m\<le>n" shows "(\<Sum>k=0..m. (n-k) choose (m-k)) = Suc n choose m"

   448 proof -

   449   have "(\<Sum>k=0..m. (n-k) choose (m-k)) = (\<Sum>k=0..m. (n-m+k) choose k)"

   450     by (rule natsum_reverse_index) (simp add: assms)

   451   also have "... = Suc (n-m+m) choose m"

   452     by (rule sum_choose_lower)

   453   also have "... = Suc n choose m" using assms

   454     by simp

   455   finally show ?thesis .

   456 qed

   457

   458 subsection\<open>Pochhammer's symbol : generalized rising factorial\<close>

   459

   460 text \<open>See @{url "http://en.wikipedia.org/wiki/Pochhammer_symbol"}\<close>

   461

   462 definition (in comm_semiring_1) "pochhammer (a :: 'a) n =

   463   (if n = 0 then 1 else setprod (\<lambda>n. a + of_nat n) {0 .. n - 1})"

   464

   465 lemma pochhammer_0 [simp]: "pochhammer a 0 = 1"

   466   by (simp add: pochhammer_def)

   467

   468 lemma pochhammer_1 [simp]: "pochhammer a 1 = a"

   469   by (simp add: pochhammer_def)

   470

   471 lemma pochhammer_Suc0 [simp]: "pochhammer a (Suc 0) = a"

   472   by (simp add: pochhammer_def)

   473

   474 lemma pochhammer_Suc_setprod: "pochhammer a (Suc n) = setprod (\<lambda>n. a + of_nat n) {0 .. n}"

   475   by (simp add: pochhammer_def)

   476

   477 lemma pochhammer_of_nat: "pochhammer (of_nat x) n = of_nat (pochhammer x n)"

   478   by (simp add: pochhammer_def of_nat_setprod)

   479

   480 lemma pochhammer_of_int: "pochhammer (of_int x) n = of_int (pochhammer x n)"

   481   by (simp add: pochhammer_def of_int_setprod)

   482

   483 lemma setprod_nat_ivl_Suc: "setprod f {0 .. Suc n} = setprod f {0..n} * f (Suc n)"

   484 proof -

   485   have "{0..Suc n} = {0..n} \<union> {Suc n}" by auto

   486   then show ?thesis by (simp add: field_simps)

   487 qed

   488

   489 lemma setprod_nat_ivl_1_Suc: "setprod f {0 .. Suc n} = f 0 * setprod f {1.. Suc n}"

   490 proof -

   491   have "{0..Suc n} = {0} \<union> {1 .. Suc n}" by auto

   492   then show ?thesis by simp

   493 qed

   494

   495

   496 lemma pochhammer_Suc: "pochhammer a (Suc n) = pochhammer a n * (a + of_nat n)"

   497 proof (cases n)

   498   case 0

   499   then show ?thesis by simp

   500 next

   501   case (Suc n)

   502   show ?thesis unfolding Suc pochhammer_Suc_setprod setprod_nat_ivl_Suc ..

   503 qed

   504

   505 lemma pochhammer_rec: "pochhammer a (Suc n) = a * pochhammer (a + 1) n"

   506 proof (cases "n = 0")

   507   case True

   508   then show ?thesis by (simp add: pochhammer_Suc_setprod)

   509 next

   510   case False

   511   have *: "finite {1 .. n}" "0 \<notin> {1 .. n}" by auto

   512   have eq: "insert 0 {1 .. n} = {0..n}" by auto

   513   have **: "(\<Prod>n\<in>{1::nat..n}. a + of_nat n) = (\<Prod>n\<in>{0::nat..n - 1}. a + 1 + of_nat n)"

   514     apply (rule setprod.reindex_cong [where l = Suc])

   515     using False

   516     apply (auto simp add: fun_eq_iff field_simps)

   517     done

   518   show ?thesis

   519     apply (simp add: pochhammer_def)

   520     unfolding setprod.insert [OF *, unfolded eq]

   521     using ** apply (simp add: field_simps)

   522     done

   523 qed

   524

   525 lemma pochhammer_rec': "pochhammer z (Suc n) = (z + of_nat n) * pochhammer z n"

   526 proof (induction n arbitrary: z)

   527   case (Suc n z)

   528   have "pochhammer z (Suc (Suc n)) = z * pochhammer (z + 1) (Suc n)"

   529     by (simp add: pochhammer_rec)

   530   also note Suc

   531   also have "z * ((z + 1 + of_nat n) * pochhammer (z + 1) n) =

   532                (z + of_nat (Suc n)) * pochhammer z (Suc n)"

   533     by (simp_all add: pochhammer_rec algebra_simps)

   534   finally show ?case .

   535 qed simp_all

   536

   537 lemma pochhammer_fact: "fact n = pochhammer 1 n"

   538   unfolding fact_altdef

   539   apply (cases n)

   540    apply (simp_all add: of_nat_setprod pochhammer_Suc_setprod)

   541   apply (rule setprod.reindex_cong [where l = Suc])

   542     apply (auto simp add: fun_eq_iff)

   543   done

   544

   545 lemma pochhammer_of_nat_eq_0_lemma:

   546   assumes "k > n"

   547   shows "pochhammer (- (of_nat n :: 'a:: idom)) k = 0"

   548 proof (cases "n = 0")

   549   case True

   550   then show ?thesis

   551     using assms by (cases k) (simp_all add: pochhammer_rec)

   552 next

   553   case False

   554   from assms obtain h where "k = Suc h" by (cases k) auto

   555   then show ?thesis

   556     by (simp add: pochhammer_Suc_setprod)

   557        (metis Suc_leI Suc_le_mono assms atLeastAtMost_iff less_eq_nat.simps(1))

   558 qed

   559

   560 lemma pochhammer_of_nat_eq_0_lemma':

   561   assumes kn: "k \<le> n"

   562   shows "pochhammer (- (of_nat n :: 'a:: {idom,ring_char_0})) k \<noteq> 0"

   563 proof (cases k)

   564   case 0

   565   then show ?thesis by simp

   566 next

   567   case (Suc h)

   568   then show ?thesis

   569     apply (simp add: pochhammer_Suc_setprod)

   570     using Suc kn apply (auto simp add: algebra_simps)

   571     done

   572 qed

   573

   574 lemma pochhammer_of_nat_eq_0_iff:

   575   shows "pochhammer (- (of_nat n :: 'a:: {idom,ring_char_0})) k = 0 \<longleftrightarrow> k > n"

   576   (is "?l = ?r")

   577   using pochhammer_of_nat_eq_0_lemma[of n k, where ?'a='a]

   578     pochhammer_of_nat_eq_0_lemma'[of k n, where ?'a = 'a]

   579   by (auto simp add: not_le[symmetric])

   580

   581 lemma pochhammer_eq_0_iff: "pochhammer a n = (0::'a::field_char_0) \<longleftrightarrow> (\<exists>k < n. a = - of_nat k)"

   582   apply (auto simp add: pochhammer_of_nat_eq_0_iff)

   583   apply (cases n)

   584    apply (auto simp add: pochhammer_def algebra_simps group_add_class.eq_neg_iff_add_eq_0)

   585   apply (metis leD not_less_eq)

   586   done

   587

   588 lemma pochhammer_eq_0_mono:

   589   "pochhammer a n = (0::'a::field_char_0) \<Longrightarrow> m \<ge> n \<Longrightarrow> pochhammer a m = 0"

   590   unfolding pochhammer_eq_0_iff by auto

   591

   592 lemma pochhammer_neq_0_mono:

   593   "pochhammer a m \<noteq> (0::'a::field_char_0) \<Longrightarrow> m \<ge> n \<Longrightarrow> pochhammer a n \<noteq> 0"

   594   unfolding pochhammer_eq_0_iff by auto

   595

   596 lemma pochhammer_minus:

   597     "pochhammer (- b) k = ((- 1) ^ k :: 'a::comm_ring_1) * pochhammer (b - of_nat k + 1) k"

   598 proof (cases k)

   599   case 0

   600   then show ?thesis by simp

   601 next

   602   case (Suc h)

   603   have eq: "((- 1) ^ Suc h :: 'a) = (\<Prod>i=0..h. - 1)"

   604     using setprod_constant[where A="{0 .. h}" and y="- 1 :: 'a"]

   605     by auto

   606   show ?thesis

   607     unfolding Suc pochhammer_Suc_setprod eq setprod.distrib[symmetric]

   608     by (rule setprod.reindex_bij_witness[where i="op - h" and j="op - h"])

   609        (auto simp: of_nat_diff)

   610 qed

   611

   612 lemma pochhammer_minus':

   613     "pochhammer (b - of_nat k + 1) k = ((- 1) ^ k :: 'a::comm_ring_1) * pochhammer (- b) k"

   614   unfolding pochhammer_minus[where b=b]

   615   unfolding mult.assoc[symmetric]

   616   unfolding power_add[symmetric]

   617   by simp

   618

   619 lemma pochhammer_same: "pochhammer (- of_nat n) n =

   620     ((- 1) ^ n :: 'a::{semiring_char_0,comm_ring_1,semiring_no_zero_divisors}) * (fact n)"

   621   unfolding pochhammer_minus

   622   by (simp add: of_nat_diff pochhammer_fact)

   623

   624 lemma pochhammer_product':

   625   "pochhammer z (n + m) = pochhammer z n * pochhammer (z + of_nat n) m"

   626 proof (induction n arbitrary: z)

   627   case (Suc n z)

   628   have "pochhammer z (Suc n) * pochhammer (z + of_nat (Suc n)) m =

   629             z * (pochhammer (z + 1) n * pochhammer (z + 1 + of_nat n) m)"

   630     by (simp add: pochhammer_rec ac_simps)

   631   also note Suc[symmetric]

   632   also have "z * pochhammer (z + 1) (n + m) = pochhammer z (Suc (n + m))"

   633     by (subst pochhammer_rec) simp

   634   finally show ?case by simp

   635 qed simp

   636

   637 lemma pochhammer_product:

   638   "m \<le> n \<Longrightarrow> pochhammer z n = pochhammer z m * pochhammer (z + of_nat m) (n - m)"

   639   using pochhammer_product'[of z m "n - m"] by simp

   640

   641 lemma pochhammer_times_pochhammer_half:

   642   fixes z :: "'a :: field_char_0"

   643   shows "pochhammer z (Suc n) * pochhammer (z + 1/2) (Suc n) = (\<Prod>k=0..2*n+1. z + of_nat k / 2)"

   644 proof (induction n)

   645   case (Suc n)

   646   def n' \<equiv> "Suc n"

   647   have "pochhammer z (Suc n') * pochhammer (z + 1 / 2) (Suc n') =

   648           (pochhammer z n' * pochhammer (z + 1 / 2) n') *

   649           ((z + of_nat n') * (z + 1/2 + of_nat n'))" (is "_ = _ * ?A")

   650      by (simp_all add: pochhammer_rec' mult_ac)

   651   also have "?A = (z + of_nat (Suc (2 * n + 1)) / 2) * (z + of_nat (Suc (Suc (2 * n + 1))) / 2)"

   652     (is "_ = ?A") by (simp add: field_simps n'_def of_nat_mult)

   653   also note Suc[folded n'_def]

   654   also have "(\<Prod>k = 0..2 * n + 1. z + of_nat k / 2) * ?A = (\<Prod>k = 0..2 * Suc n + 1. z + of_nat k / 2)"

   655     by (simp add: setprod_nat_ivl_Suc)

   656   finally show ?case by (simp add: n'_def)

   657 qed (simp add: setprod_nat_ivl_Suc)

   658

   659 lemma pochhammer_double:

   660   fixes z :: "'a :: field_char_0"

   661   shows "pochhammer (2 * z) (2 * n) = of_nat (2^(2*n)) * pochhammer z n * pochhammer (z+1/2) n"

   662 proof (induction n)

   663   case (Suc n)

   664   have "pochhammer (2 * z) (2 * (Suc n)) = pochhammer (2 * z) (2 * n) *

   665           (2 * (z + of_nat n)) * (2 * (z + of_nat n) + 1)"

   666     by (simp add: pochhammer_rec' ac_simps of_nat_mult)

   667   also note Suc

   668   also have "of_nat (2 ^ (2 * n)) * pochhammer z n * pochhammer (z + 1/2) n *

   669                  (2 * (z + of_nat n)) * (2 * (z + of_nat n) + 1) =

   670              of_nat (2 ^ (2 * (Suc n))) * pochhammer z (Suc n) * pochhammer (z + 1/2) (Suc n)"

   671     by (simp add: of_nat_mult field_simps pochhammer_rec')

   672   finally show ?case .

   673 qed simp

   674

   675 lemma pochhammer_absorb_comp:

   676   "((r :: 'a :: comm_ring_1) - of_nat k) * pochhammer (- r) k = r * pochhammer (-r + 1) k"

   677   (is "?lhs = ?rhs")

   678 proof -

   679   have "?lhs = -pochhammer (-r) (Suc k)" by (subst pochhammer_rec') (simp add: algebra_simps)

   680   also have "\<dots> = ?rhs" by (subst pochhammer_rec) simp

   681   finally show ?thesis .

   682 qed

   683

   684

   685 subsection\<open>Generalized binomial coefficients\<close>

   686

   687 definition (in field_char_0) gbinomial :: "'a \<Rightarrow> nat \<Rightarrow> 'a" (infixl "gchoose" 65)

   688   where "a gchoose n =

   689     (if n = 0 then 1 else (setprod (\<lambda>i. a - of_nat i) {0 .. n - 1}) / (fact n))"

   690

   691 lemma gbinomial_0 [simp]: "a gchoose 0 = 1" "0 gchoose (Suc n) = 0"

   692   by (simp_all add: gbinomial_def)

   693

   694 lemma gbinomial_pochhammer: "a gchoose n = (- 1) ^ n * pochhammer (- a) n / (fact n)"

   695 proof (cases "n = 0")

   696   case True

   697   then show ?thesis by simp

   698 next

   699   case False

   700   from this setprod_constant[of "{0 .. n - 1}" "- (1:: 'a)"]

   701   have eq: "(- (1::'a)) ^ n = setprod (\<lambda>i. - 1) {0 .. n - 1}"

   702     by auto

   703   from False show ?thesis

   704     by (simp add: pochhammer_def gbinomial_def field_simps

   705       eq setprod.distrib[symmetric])

   706 qed

   707

   708 lemma gbinomial_pochhammer':

   709   "(s :: 'a :: field_char_0) gchoose n = pochhammer (s - of_nat n + 1) n / fact n"

   710 proof -

   711   have "s gchoose n = ((-1)^n * (-1)^n) * pochhammer (s - of_nat n + 1) n / fact n"

   712     by (simp add: gbinomial_pochhammer pochhammer_minus mult_ac)

   713   also have "(-1 :: 'a)^n * (-1)^n = 1" by (subst power_add [symmetric]) simp

   714   finally show ?thesis by simp

   715 qed

   716

   717 lemma binomial_gbinomial:

   718     "of_nat (n choose k) = (of_nat n gchoose k :: 'a::field_char_0)"

   719 proof -

   720   { assume kn: "k > n"

   721     then have ?thesis

   722       by (subst binomial_eq_0[OF kn])

   723          (simp add: gbinomial_pochhammer field_simps  pochhammer_of_nat_eq_0_iff) }

   724   moreover

   725   { assume "k=0" then have ?thesis by simp }

   726   moreover

   727   { assume kn: "k \<le> n" and k0: "k\<noteq> 0"

   728     from k0 obtain h where h: "k = Suc h" by (cases k) auto

   729     from h

   730     have eq:"(- 1 :: 'a) ^ k = setprod (\<lambda>i. - 1) {0..h}"

   731       by (subst setprod_constant) auto

   732     have eq': "(\<Prod>i\<in>{0..h}. of_nat n + - (of_nat i :: 'a)) = (\<Prod>i\<in>{n - h..n}. of_nat i)"

   733         using h kn

   734       by (intro setprod.reindex_bij_witness[where i="op - n" and j="op - n"])

   735          (auto simp: of_nat_diff)

   736     have th0: "finite {1..n - Suc h}" "finite {n - h .. n}"

   737         "{1..n - Suc h} \<inter> {n - h .. n} = {}" and

   738         eq3: "{1..n - Suc h} \<union> {n - h .. n} = {1..n}"

   739       using h kn by auto

   740     from eq[symmetric]

   741     have ?thesis using kn

   742       apply (simp add: binomial_fact[OF kn, where ?'a = 'a]

   743         gbinomial_pochhammer field_simps pochhammer_Suc_setprod)

   744       apply (simp add: pochhammer_Suc_setprod fact_altdef h

   745         of_nat_setprod setprod.distrib[symmetric] eq' del: One_nat_def power_Suc)

   746       unfolding setprod.union_disjoint[OF th0, unfolded eq3, of "of_nat:: nat \<Rightarrow> 'a"] eq[unfolded h]

   747       unfolding mult.assoc

   748       unfolding setprod.distrib[symmetric]

   749       apply simp

   750       apply (intro setprod.reindex_bij_witness[where i="op - n" and j="op - n"])

   751       apply (auto simp: of_nat_diff)

   752       done

   753   }

   754   moreover

   755   have "k > n \<or> k = 0 \<or> (k \<le> n \<and> k \<noteq> 0)" by arith

   756   ultimately show ?thesis by blast

   757 qed

   758

   759 lemma gbinomial_1[simp]: "a gchoose 1 = a"

   760   by (simp add: gbinomial_def)

   761

   762 lemma gbinomial_Suc0[simp]: "a gchoose (Suc 0) = a"

   763   by (simp add: gbinomial_def)

   764

   765 lemma gbinomial_mult_1:

   766   fixes a :: "'a :: field_char_0"

   767   shows "a * (a gchoose n) =

   768     of_nat n * (a gchoose n) + of_nat (Suc n) * (a gchoose (Suc n))"  (is "?l = ?r")

   769 proof -

   770   have "?r = ((- 1) ^n * pochhammer (- a) n / (fact n)) * (of_nat n - (- a + of_nat n))"

   771     unfolding gbinomial_pochhammer

   772       pochhammer_Suc of_nat_mult right_diff_distrib power_Suc

   773     apply (simp del: of_nat_Suc fact.simps)

   774     apply (auto simp add: field_simps simp del: of_nat_Suc)

   775     done

   776   also have "\<dots> = ?l" unfolding gbinomial_pochhammer

   777     by (simp add: field_simps)

   778   finally show ?thesis ..

   779 qed

   780

   781 lemma gbinomial_mult_1':

   782   fixes a :: "'a :: field_char_0"

   783   shows "(a gchoose n) * a = of_nat n * (a gchoose n) + of_nat (Suc n) * (a gchoose (Suc n))"

   784   by (simp add: mult.commute gbinomial_mult_1)

   785

   786 lemma gbinomial_Suc:

   787     "a gchoose (Suc k) = (setprod (\<lambda>i. a - of_nat i) {0 .. k}) / (fact (Suc k))"

   788   by (simp add: gbinomial_def)

   789

   790 lemma gbinomial_mult_fact:

   791   fixes a :: "'a::field_char_0"

   792   shows

   793    "fact (Suc k) * (a gchoose (Suc k)) =

   794     (setprod (\<lambda>i. a - of_nat i) {0 .. k})"

   795   by (simp_all add: gbinomial_Suc field_simps del: fact.simps)

   796

   797 lemma gbinomial_mult_fact':

   798   fixes a :: "'a::field_char_0"

   799   shows "(a gchoose (Suc k)) * fact (Suc k) = (setprod (\<lambda>i. a - of_nat i) {0 .. k})"

   800   using gbinomial_mult_fact[of k a]

   801   by (subst mult.commute)

   802

   803 lemma gbinomial_Suc_Suc:

   804   fixes a :: "'a :: field_char_0"

   805   shows "(a + 1) gchoose (Suc k) = a gchoose k + (a gchoose (Suc k))"

   806 proof (cases k)

   807   case 0

   808   then show ?thesis by simp

   809 next

   810   case (Suc h)

   811   have eq0: "(\<Prod>i\<in>{1..k}. (a + 1) - of_nat i) = (\<Prod>i\<in>{0..h}. a - of_nat i)"

   812     apply (rule setprod.reindex_cong [where l = Suc])

   813       using Suc

   814       apply auto

   815     done

   816   have "fact (Suc k) * (a gchoose k + (a gchoose (Suc k))) =

   817         (a gchoose Suc h) * (fact (Suc (Suc h))) +

   818         (a gchoose Suc (Suc h)) * (fact (Suc (Suc h)))"

   819     by (simp add: Suc field_simps del: fact.simps)

   820   also have "... = (a gchoose Suc h) * of_nat (Suc (Suc h) * fact (Suc h)) +

   821                    (\<Prod>i = 0..Suc h. a - of_nat i)"

   822     by (metis fact.simps(2) gbinomial_mult_fact' of_nat_fact of_nat_id)

   823   also have "... = (fact (Suc h) * (a gchoose Suc h)) * of_nat (Suc (Suc h)) +

   824                    (\<Prod>i = 0..Suc h. a - of_nat i)"

   825     by (simp only: fact.simps(2) mult.commute mult.left_commute of_nat_fact of_nat_id of_nat_mult)

   826   also have "... =  of_nat (Suc (Suc h)) * (\<Prod>i = 0..h. a - of_nat i) +

   827                     (\<Prod>i = 0..Suc h. a - of_nat i)"

   828     by (metis gbinomial_mult_fact mult.commute)

   829   also have "... = (\<Prod>i = 0..Suc h. a - of_nat i) +

   830                    (of_nat h * (\<Prod>i = 0..h. a - of_nat i) + 2 * (\<Prod>i = 0..h. a - of_nat i))"

   831     by (simp add: field_simps)

   832   also have "... =

   833     ((a gchoose Suc h) * (fact (Suc h)) * of_nat (Suc k)) + (\<Prod>i\<in>{0::nat..Suc h}. a - of_nat i)"

   834     unfolding gbinomial_mult_fact'

   835     by (simp add: comm_semiring_class.distrib field_simps Suc)

   836   also have "\<dots> = (\<Prod>i\<in>{0..h}. a - of_nat i) * (a + 1)"

   837     unfolding gbinomial_mult_fact' setprod_nat_ivl_Suc

   838     by (simp add: field_simps Suc)

   839   also have "\<dots> = (\<Prod>i\<in>{0..k}. (a + 1) - of_nat i)"

   840     using eq0

   841     by (simp add: Suc setprod_nat_ivl_1_Suc)

   842   also have "\<dots> = (fact (Suc k)) * ((a + 1) gchoose (Suc k))"

   843     unfolding gbinomial_mult_fact ..

   844   finally show ?thesis

   845     by (metis fact_nonzero mult_cancel_left)

   846 qed

   847

   848 lemma gbinomial_reduce_nat:

   849   fixes a :: "'a :: field_char_0"

   850   shows "0 < k \<Longrightarrow> a gchoose k = (a - 1) gchoose (k - 1) + ((a - 1) gchoose k)"

   851   by (metis Suc_pred' diff_add_cancel gbinomial_Suc_Suc)

   852

   853 lemma gchoose_row_sum_weighted:

   854   fixes r :: "'a::field_char_0"

   855   shows "(\<Sum>k = 0..m. (r gchoose k) * (r/2 - of_nat k)) = of_nat(Suc m) / 2 * (r gchoose (Suc m))"

   856 proof (induct m)

   857   case 0 show ?case by simp

   858 next

   859   case (Suc m)

   860   from Suc show ?case

   861     by (simp add: field_simps distrib gbinomial_mult_1)

   862 qed

   863

   864 lemma binomial_symmetric:

   865   assumes kn: "k \<le> n"

   866   shows "n choose k = n choose (n - k)"

   867 proof-

   868   from kn have kn': "n - k \<le> n" by arith

   869   from binomial_fact_lemma[OF kn] binomial_fact_lemma[OF kn']

   870   have "fact k * fact (n - k) * (n choose k) =

   871     fact (n - k) * fact (n - (n - k)) * (n choose (n - k))" by simp

   872   then show ?thesis using kn by simp

   873 qed

   874

   875 lemma choose_rising_sum:

   876   "(\<Sum>j\<le>m. ((n + j) choose n)) = ((n + m + 1) choose (n + 1))"

   877   "(\<Sum>j\<le>m. ((n + j) choose n)) = ((n + m + 1) choose m)"

   878 proof -

   879   show "(\<Sum>j\<le>m. ((n + j) choose n)) = ((n + m + 1) choose (n + 1))" by (induction m) simp_all

   880   also have "... = ((n + m + 1) choose m)" by (subst binomial_symmetric) simp_all

   881   finally show "(\<Sum>j\<le>m. ((n + j) choose n)) = ((n + m + 1) choose m)" .

   882 qed

   883

   884 lemma choose_linear_sum:

   885   "(\<Sum>i\<le>n. i * (n choose i)) = n * 2 ^ (n - 1)"

   886 proof (cases n)

   887   case (Suc m)

   888   have "(\<Sum>i\<le>n. i * (n choose i)) = (\<Sum>i\<le>Suc m. i * (Suc m choose i))" by (simp add: Suc)

   889   also have "... = Suc m * 2 ^ m"

   890     by (simp only: setsum_atMost_Suc_shift Suc_times_binomial setsum_right_distrib[symmetric])

   891        (simp add: choose_row_sum')

   892   finally show ?thesis using Suc by simp

   893 qed simp_all

   894

   895 lemma choose_alternating_linear_sum:

   896   assumes "n \<noteq> 1"

   897   shows   "(\<Sum>i\<le>n. (-1)^i * of_nat i * of_nat (n choose i) :: 'a :: comm_ring_1) = 0"

   898 proof (cases n)

   899   case (Suc m)

   900   with assms have "m > 0" by simp

   901   have "(\<Sum>i\<le>n. (-1) ^ i * of_nat i * of_nat (n choose i) :: 'a) =

   902             (\<Sum>i\<le>Suc m. (-1) ^ i * of_nat i * of_nat (Suc m choose i))" by (simp add: Suc)

   903   also have "... = (\<Sum>i\<le>m. (-1) ^ (Suc i) * of_nat (Suc i * (Suc m choose Suc i)))"

   904     by (simp only: setsum_atMost_Suc_shift setsum_right_distrib[symmetric] of_nat_mult mult_ac) simp

   905   also have "... = -of_nat (Suc m) * (\<Sum>i\<le>m. (-1) ^ i * of_nat ((m choose i)))"

   906     by (subst setsum_right_distrib, rule setsum.cong[OF refl], subst Suc_times_binomial)

   907        (simp add: algebra_simps of_nat_mult)

   908   also have "(\<Sum>i\<le>m. (-1 :: 'a) ^ i * of_nat ((m choose i))) = 0"

   909     using choose_alternating_sum[OF \<open>m > 0\<close>] by simp

   910   finally show ?thesis by simp

   911 qed simp

   912

   913 lemma vandermonde:

   914   "(\<Sum>k\<le>r. (m choose k) * (n choose (r - k))) = (m + n) choose r"

   915 proof (induction n arbitrary: r)

   916   case 0

   917   have "(\<Sum>k\<le>r. (m choose k) * (0 choose (r - k))) = (\<Sum>k\<le>r. if k = r then (m choose k) else 0)"

   918     by (intro setsum.cong) simp_all

   919   also have "... = m choose r" by (simp add: setsum.delta)

   920   finally show ?case by simp

   921 next

   922   case (Suc n r)

   923   show ?case by (cases r) (simp_all add: Suc [symmetric] algebra_simps setsum.distrib Suc_diff_le)

   924 qed

   925

   926 lemma choose_square_sum:

   927   "(\<Sum>k\<le>n. (n choose k)^2) = ((2*n) choose n)"

   928   using vandermonde[of n n n] by (simp add: power2_eq_square mult_2 binomial_symmetric [symmetric])

   929

   930 lemma pochhammer_binomial_sum:

   931   fixes a b :: "'a :: comm_ring_1"

   932   shows "pochhammer (a + b) n = (\<Sum>k\<le>n. of_nat (n choose k) * pochhammer a k * pochhammer b (n - k))"

   933 proof (induction n arbitrary: a b)

   934   case (Suc n a b)

   935   have "(\<Sum>k\<le>Suc n. of_nat (Suc n choose k) * pochhammer a k * pochhammer b (Suc n - k)) =

   936             (\<Sum>i\<le>n. of_nat (n choose i) * pochhammer a (Suc i) * pochhammer b (n - i)) +

   937             ((\<Sum>i\<le>n. of_nat (n choose Suc i) * pochhammer a (Suc i) * pochhammer b (n - i)) +

   938             pochhammer b (Suc n))"

   939     by (subst setsum_atMost_Suc_shift) (simp add: ring_distribs setsum.distrib)

   940   also have "(\<Sum>i\<le>n. of_nat (n choose i) * pochhammer a (Suc i) * pochhammer b (n - i)) =

   941                a * pochhammer ((a + 1) + b) n"

   942     by (subst Suc) (simp add: setsum_right_distrib pochhammer_rec mult_ac)

   943   also have "(\<Sum>i\<le>n. of_nat (n choose Suc i) * pochhammer a (Suc i) * pochhammer b (n - i)) + pochhammer b (Suc n) =

   944                (\<Sum>i=0..Suc n. of_nat (n choose i) * pochhammer a i * pochhammer b (Suc n - i))"

   945     by (subst setsum_head_Suc, simp, subst setsum_shift_bounds_cl_Suc_ivl) (simp add: atLeast0AtMost)

   946   also have "... = (\<Sum>i\<le>n. of_nat (n choose i) * pochhammer a i * pochhammer b (Suc n - i))"

   947     using Suc by (intro setsum.mono_neutral_right) (auto simp: not_le binomial_eq_0)

   948   also have "... = (\<Sum>i\<le>n. of_nat (n choose i) * pochhammer a i * pochhammer b (Suc (n - i)))"

   949     by (intro setsum.cong) (simp_all add: Suc_diff_le)

   950   also have "... = b * pochhammer (a + (b + 1)) n"

   951     by (subst Suc) (simp add: setsum_right_distrib mult_ac pochhammer_rec)

   952   also have "a * pochhammer ((a + 1) + b) n + b * pochhammer (a + (b + 1)) n =

   953                pochhammer (a + b) (Suc n)" by (simp add: pochhammer_rec algebra_simps)

   954   finally show ?case ..

   955 qed simp_all

   956

   957

   958 text\<open>Contributed by Manuel Eberl, generalised by LCP.

   959   Alternative definition of the binomial coefficient as @{term "\<Prod>i<k. (n - i) / (k - i)"}\<close>

   960 lemma gbinomial_altdef_of_nat:

   961   fixes k :: nat

   962     and x :: "'a :: {field_char_0,field}"

   963   shows "x gchoose k = (\<Prod>i<k. (x - of_nat i) / of_nat (k - i) :: 'a)"

   964 proof -

   965   have "(x gchoose k) = (\<Prod>i<k. x - of_nat i) / of_nat (fact k)"

   966     unfolding gbinomial_def

   967     by (auto simp: gr0_conv_Suc lessThan_Suc_atMost atLeast0AtMost)

   968   also have "\<dots> = (\<Prod>i<k. (x - of_nat i) / of_nat (k - i) :: 'a)"

   969     unfolding fact_eq_rev_setprod_nat of_nat_setprod

   970     by (auto simp add: setprod_dividef intro!: setprod.cong of_nat_diff[symmetric])

   971   finally show ?thesis .

   972 qed

   973

   974 lemma gbinomial_ge_n_over_k_pow_k:

   975   fixes k :: nat

   976     and x :: "'a :: linordered_field"

   977   assumes "of_nat k \<le> x"

   978   shows "(x / of_nat k :: 'a) ^ k \<le> x gchoose k"

   979 proof -

   980   have x: "0 \<le> x"

   981     using assms of_nat_0_le_iff order_trans by blast

   982   have "(x / of_nat k :: 'a) ^ k = (\<Prod>i<k. x / of_nat k :: 'a)"

   983     by (simp add: setprod_constant)

   984   also have "\<dots> \<le> x gchoose k"

   985     unfolding gbinomial_altdef_of_nat

   986   proof (safe intro!: setprod_mono)

   987     fix i :: nat

   988     assume ik: "i < k"

   989     from assms have "x * of_nat i \<ge> of_nat (i * k)"

   990       by (metis mult.commute mult_le_cancel_right of_nat_less_0_iff of_nat_mult)

   991     then have "x * of_nat k - x * of_nat i \<le> x * of_nat k - of_nat (i * k)" by arith

   992     then have "x * of_nat (k - i) \<le> (x - of_nat i) * of_nat k"

   993       using ik

   994       by (simp add: algebra_simps zero_less_mult_iff of_nat_diff of_nat_mult)

   995     then have "x * of_nat (k - i) \<le> (x - of_nat i) * (of_nat k :: 'a)"

   996       unfolding of_nat_mult[symmetric] of_nat_le_iff .

   997     with assms show "x / of_nat k \<le> (x - of_nat i) / (of_nat (k - i) :: 'a)"

   998       using \<open>i < k\<close> by (simp add: field_simps)

   999   qed (simp add: x zero_le_divide_iff)

  1000   finally show ?thesis .

  1001 qed

  1002

  1003 lemma gbinomial_negated_upper: "(a gchoose b) = (-1) ^ b * ((of_nat b - a - 1) gchoose b)"

  1004   by (simp add: gbinomial_pochhammer pochhammer_minus algebra_simps)

  1005

  1006 lemma gbinomial_minus: "((-a) gchoose b) = (-1) ^ b * ((a + of_nat b - 1) gchoose b)"

  1007   by (subst gbinomial_negated_upper) (simp add: add_ac)

  1008

  1009 lemma Suc_times_gbinomial:

  1010   "of_nat (Suc b) * ((a + 1) gchoose (Suc b)) = (a + 1) * (a gchoose b)"

  1011 proof (cases b)

  1012   case (Suc b)

  1013   hence "((a + 1) gchoose (Suc (Suc b))) =

  1014              (\<Prod>i = 0..Suc b. a + (1 - of_nat i)) / fact (b + 2)"

  1015     by (simp add: field_simps gbinomial_def)

  1016   also have "(\<Prod>i = 0..Suc b. a + (1 - of_nat i)) = (a + 1) * (\<Prod>i = 0..b. a - of_nat i)"

  1017     by (simp add: setprod_nat_ivl_1_Suc setprod_shift_bounds_cl_Suc_ivl)

  1018   also have "... / fact (b + 2) = (a + 1) / of_nat (Suc (Suc b)) * (a gchoose Suc b)"

  1019     by (simp_all add: gbinomial_def setprod_nat_ivl_1_Suc setprod_shift_bounds_cl_Suc_ivl)

  1020   finally show ?thesis by (simp add: Suc field_simps del: of_nat_Suc)

  1021 qed simp

  1022

  1023 lemma gbinomial_factors:

  1024   "((a + 1) gchoose (Suc b)) = (a + 1) / of_nat (Suc b) * (a gchoose b)"

  1025 proof (cases b)

  1026   case (Suc b)

  1027   hence "((a + 1) gchoose (Suc (Suc b))) =

  1028              (\<Prod>i = 0..Suc b. a + (1 - of_nat i)) / fact (b + 2)"

  1029     by (simp add: field_simps gbinomial_def)

  1030   also have "(\<Prod>i = 0..Suc b. a + (1 - of_nat i)) = (a + 1) * (\<Prod>i = 0..b. a - of_nat i)"

  1031     by (simp add: setprod_nat_ivl_1_Suc setprod_shift_bounds_cl_Suc_ivl)

  1032   also have "... / fact (b + 2) = (a + 1) / of_nat (Suc (Suc b)) * (a gchoose Suc b)"

  1033     by (simp_all add: gbinomial_def setprod_nat_ivl_1_Suc setprod_shift_bounds_cl_Suc_ivl)

  1034   finally show ?thesis by (simp add: Suc)

  1035 qed simp

  1036

  1037 lemma gbinomial_rec: "((r + 1) gchoose (Suc k)) = (r gchoose k) * ((r + 1) / of_nat (Suc k))"

  1038   using gbinomial_mult_1[of r k]

  1039   by (subst gbinomial_Suc_Suc) (simp add: field_simps del: of_nat_Suc, simp add: algebra_simps)

  1040

  1041 lemma gbinomial_of_nat_symmetric: "k \<le> n \<Longrightarrow> (of_nat n) gchoose k = (of_nat n) gchoose (n - k)"

  1042   using binomial_symmetric[of k n] by (simp add: binomial_gbinomial [symmetric])

  1043

  1044

  1045 text \<open>The absorption identity (equation 5.5 \cite[p.~157]{GKP}):$  1046 {r \choose k} = \frac{r}{k}{r - 1 \choose k - 1},\quad \textnormal{integer } k \neq 0.   1047$\<close>

  1048 lemma gbinomial_absorption':

  1049   "k > 0 \<Longrightarrow> (r gchoose k) = (r / of_nat(k)) * (r - 1 gchoose (k - 1))"

  1050   using gbinomial_rec[of "r - 1" "k - 1"]

  1051   by (simp_all add: gbinomial_rec field_simps del: of_nat_Suc)

  1052

  1053 text \<open>The absorption identity is written in the following form to avoid

  1054 division by $k$ (the lower index) and therefore remove the $k \neq 0$

  1055 restriction\cite[p.~157]{GKP}:$  1056 k{r \choose k} = r{r - 1 \choose k - 1}, \quad \textnormal{integer } k.   1057$\<close>

  1058 lemma gbinomial_absorption:

  1059   "of_nat (Suc k) * (r gchoose Suc k) = r * ((r - 1) gchoose k)"

  1060   using gbinomial_absorption'[of "Suc k" r] by (simp add: field_simps del: of_nat_Suc)

  1061

  1062 text \<open>The absorption identity for natural number binomial coefficients:\<close>

  1063 lemma binomial_absorption:

  1064   "Suc k * (n choose Suc k) = n * ((n - 1) choose k)"

  1065   by (cases n) (simp_all add: binomial_eq_0 Suc_times_binomial del: binomial_Suc_Suc mult_Suc)

  1066

  1067 text \<open>The absorption companion identity for natural number coefficients,

  1068 following the proof by GKP \cite[p.~157]{GKP}:\<close>

  1069 lemma binomial_absorb_comp:

  1070   "(n - k) * (n choose k) = n * ((n - 1) choose k)" (is "?lhs = ?rhs")

  1071 proof (cases "n \<le> k")

  1072   case False

  1073   then have "?rhs = Suc ((n - 1) - k) * (n choose Suc ((n - 1) - k))"

  1074     using binomial_symmetric[of k "n - 1"] binomial_absorption[of "(n - 1) - k" n]

  1075     by simp

  1076   also from False have "Suc ((n - 1) - k) = n - k" by simp

  1077   also from False have "n choose \<dots> = n choose k" by (intro binomial_symmetric [symmetric]) simp_all

  1078   finally show ?thesis ..

  1079 qed auto

  1080

  1081 text \<open>The generalised absorption companion identity:\<close>

  1082 lemma gbinomial_absorb_comp: "(r - of_nat k) * (r gchoose k) = r * ((r - 1) gchoose k)"

  1083   using pochhammer_absorb_comp[of r k] by (simp add: gbinomial_pochhammer)

  1084

  1085 lemma gbinomial_addition_formula:

  1086   "r gchoose (Suc k) = ((r - 1) gchoose (Suc k)) + ((r - 1) gchoose k)"

  1087   using gbinomial_Suc_Suc[of "r - 1" k] by (simp add: algebra_simps)

  1088

  1089 lemma binomial_addition_formula:

  1090   "0 < n \<Longrightarrow> n choose (Suc k) = ((n - 1) choose (Suc k)) + ((n - 1) choose k)"

  1091   by (subst choose_reduce_nat) simp_all

  1092

  1093

  1094 text \<open>

  1095   Equation 5.9 of the reference material \cite[p.~159]{GKP} is a useful

  1096   summation formula, operating on both indices:$  1097 \sum\limits_{k \leq n}{r + k \choose k} = {r + n + 1 \choose n},   1098 \quad \textnormal{integer } n.   1099$

  1100 \<close>

  1101 lemma gbinomial_parallel_sum:

  1102 "(\<Sum>k\<le>n. (r + of_nat k) gchoose k) = (r + of_nat n + 1) gchoose n"

  1103 proof (induction n)

  1104   case (Suc m)

  1105   thus ?case using gbinomial_Suc_Suc[of "(r + of_nat m + 1)" m] by (simp add: add_ac)

  1106 qed auto

  1107

  1108 subsection \<open>Summation on the upper index\<close>

  1109 text \<open>

  1110   Another summation formula is equation 5.10 of the reference material \cite[p.~160]{GKP},

  1111   aptly named \emph{summation on the upper index}:$\sum_{0 \leq k \leq n} {k \choose m} =   1112 {n + 1 \choose m + 1}, \quad \textnormal{integers } m, n \geq 0.$

  1113 \<close>

  1114 lemma gbinomial_sum_up_index:

  1115   "(\<Sum>k = 0..n. (of_nat k gchoose m) :: 'a :: field_char_0) = (of_nat n + 1) gchoose (m + 1)"

  1116 proof (induction n)

  1117   case 0

  1118   show ?case using gbinomial_Suc_Suc[of 0 m] by (cases m) auto

  1119 next

  1120   case (Suc n)

  1121   thus ?case using gbinomial_Suc_Suc[of "of_nat (Suc n) :: 'a" m] by (simp add: add_ac)

  1122 qed

  1123

  1124 lemma gbinomial_index_swap:

  1125   "((-1) ^ m) * ((- (of_nat n) - 1) gchoose m) = ((-1) ^ n) * ((- (of_nat m) - 1) gchoose n)"

  1126   (is "?lhs = ?rhs")

  1127 proof -

  1128   have "?lhs = (of_nat (m + n) gchoose m)"

  1129     by (subst gbinomial_negated_upper) (simp add: power_mult_distrib [symmetric])

  1130   also have "\<dots> = (of_nat (m + n) gchoose n)" by (subst gbinomial_of_nat_symmetric) simp_all

  1131   also have "\<dots> = ?rhs" by (subst gbinomial_negated_upper) simp

  1132   finally show ?thesis .

  1133 qed

  1134

  1135 lemma gbinomial_sum_lower_neg:

  1136   "(\<Sum>k\<le>m. (r gchoose k) * (- 1) ^ k) = (- 1) ^ m * (r - 1 gchoose m)" (is "?lhs = ?rhs")

  1137 proof -

  1138   have "?lhs = (\<Sum>k\<le>m. -(r + 1) + of_nat k gchoose k)"

  1139     by (intro setsum.cong[OF refl]) (subst gbinomial_negated_upper, simp add: power_mult_distrib)

  1140   also have "\<dots>  = -r + of_nat m gchoose m" by (subst gbinomial_parallel_sum) simp

  1141   also have "\<dots> = ?rhs" by (subst gbinomial_negated_upper) (simp add: power_mult_distrib)

  1142   finally show ?thesis .

  1143 qed

  1144

  1145 lemma gbinomial_partial_row_sum:

  1146 "(\<Sum>k\<le>m. (r gchoose k) * ((r / 2) - of_nat k)) = ((of_nat m + 1)/2) * (r gchoose (m + 1))"

  1147 proof (induction m)

  1148   case (Suc mm)

  1149   hence "(\<Sum>k\<le>Suc mm. (r gchoose k) * (r / 2 - of_nat k)) =

  1150              (r - of_nat (Suc mm)) * (r gchoose Suc mm) / 2" by (simp add: field_simps)

  1151   also have "\<dots> = r * (r - 1 gchoose Suc mm) / 2" by (subst gbinomial_absorb_comp) (rule refl)

  1152   also have "\<dots> = (of_nat (Suc mm) + 1) / 2 * (r gchoose (Suc mm + 1))"

  1153     by (subst gbinomial_absorption [symmetric]) simp

  1154   finally show ?case .

  1155 qed simp_all

  1156

  1157 lemma setsum_bounds_lt_plus1: "(\<Sum>k<mm. f (Suc k)) = (\<Sum>k=1..mm. f k)"

  1158   by (induction mm) simp_all

  1159

  1160 lemma gbinomial_partial_sum_poly:

  1161   "(\<Sum>k\<le>m. (of_nat m + r gchoose k) * x^k * y^(m-k)) =

  1162        (\<Sum>k\<le>m. (-r gchoose k) * (-x)^k * (x + y)^(m-k))" (is "?lhs m = ?rhs m")

  1163 proof (induction m)

  1164   case (Suc mm)

  1165   def G \<equiv> "\<lambda>i k. (of_nat i + r gchoose k) * x^k * y^(i-k)" and S \<equiv> ?lhs

  1166   have SG_def: "S = (\<lambda>i. (\<Sum>k\<le>i. (G i k)))" unfolding S_def G_def ..

  1167

  1168   have "S (Suc mm) = G (Suc mm) 0 + (\<Sum>k=Suc 0..Suc mm. G (Suc mm) k)"

  1169     using SG_def by (simp add: setsum_head_Suc atLeast0AtMost [symmetric])

  1170   also have "(\<Sum>k=Suc 0..Suc mm. G (Suc mm) k) = (\<Sum>k=0..mm. G (Suc mm) (Suc k))"

  1171     by (subst setsum_shift_bounds_cl_Suc_ivl) simp

  1172   also have "\<dots> = (\<Sum>k=0..mm. ((of_nat mm + r gchoose (Suc k))

  1173                     + (of_nat mm + r gchoose k)) * x^(Suc k) * y^(mm - k))"

  1174     unfolding G_def by (subst gbinomial_addition_formula) simp

  1175   also have "\<dots> = (\<Sum>k=0..mm. (of_nat mm + r gchoose (Suc k)) * x^(Suc k) * y^(mm - k))

  1176                   + (\<Sum>k=0..mm. (of_nat mm + r gchoose k) * x^(Suc k) * y^(mm - k))"

  1177     by (subst setsum.distrib [symmetric]) (simp add: algebra_simps)

  1178   also have "(\<Sum>k=0..mm. (of_nat mm + r gchoose (Suc k)) * x^(Suc k) * y^(mm - k)) =

  1179                (\<Sum>k<Suc mm. (of_nat mm + r gchoose (Suc k)) * x^(Suc k) * y^(mm - k))"

  1180     by (simp only: atLeast0AtMost lessThan_Suc_atMost)

  1181   also have "\<dots> = (\<Sum>k<mm. (of_nat mm + r gchoose Suc k) * x^(Suc k) * y^(mm-k))

  1182                       + (of_nat mm + r gchoose (Suc mm)) * x^(Suc mm)" (is "_ = ?A + ?B")

  1183     by (subst setsum_lessThan_Suc) simp

  1184   also have "?A = (\<Sum>k=1..mm. (of_nat mm + r gchoose k) * x^k * y^(mm - k + 1))"

  1185   proof (subst setsum_bounds_lt_plus1 [symmetric], intro setsum.cong[OF refl], clarify)

  1186     fix k assume "k < mm"

  1187     hence "mm - k = mm - Suc k + 1" by linarith

  1188     thus "(of_nat mm + r gchoose Suc k) * x ^ Suc k * y ^ (mm - k) =

  1189             (of_nat mm + r gchoose Suc k) * x ^ Suc k * y ^ (mm - Suc k + 1)" by (simp only:)

  1190   qed

  1191   also have "\<dots> + ?B = y * (\<Sum>k=1..mm. (G mm k)) + (of_nat mm + r gchoose (Suc mm)) * x^(Suc mm)"

  1192     unfolding G_def by (subst setsum_right_distrib) (simp add: algebra_simps)

  1193   also have "(\<Sum>k=0..mm. (of_nat mm + r gchoose k) * x^(Suc k) * y^(mm - k)) = x * (S mm)"

  1194       unfolding S_def by (subst setsum_right_distrib) (simp add: atLeast0AtMost algebra_simps)

  1195   also have "(G (Suc mm) 0) = y * (G mm 0)" by (simp add: G_def)

  1196   finally have "S (Suc mm) = y * ((G mm 0) + (\<Sum>k=1..mm. (G mm k)))

  1197                 + (of_nat mm + r gchoose (Suc mm)) * x^(Suc mm) + x * (S mm)"

  1198     by (simp add: ring_distribs)

  1199   also have "(G mm 0) + (\<Sum>k=1..mm. (G mm k)) = S mm"

  1200     by (simp add: setsum_head_Suc[symmetric] SG_def atLeast0AtMost)

  1201   finally have "S (Suc mm) = (x + y) * (S mm) + (of_nat mm + r gchoose (Suc mm)) * x^(Suc mm)"

  1202     by (simp add: algebra_simps)

  1203   also have "(of_nat mm + r gchoose (Suc mm)) = (-1) ^ (Suc mm) * (-r gchoose (Suc mm))"

  1204     by (subst gbinomial_negated_upper) simp

  1205   also have "(-1) ^ Suc mm * (- r gchoose Suc mm) * x ^ Suc mm =

  1206                  (-r gchoose (Suc mm)) * (-x) ^ Suc mm" by (simp add: power_minus[of x])

  1207   also have "(x + y) * S mm + \<dots> = (x + y) * ?rhs mm + (-r gchoose (Suc mm)) * (-x)^Suc mm"

  1208     unfolding S_def by (subst Suc.IH) simp

  1209   also have "(x + y) * ?rhs mm = (\<Sum>n\<le>mm. ((- r gchoose n) * (- x) ^ n * (x + y) ^ (Suc mm - n)))"

  1210     by (subst setsum_right_distrib, rule setsum.cong) (simp_all add: Suc_diff_le)

  1211   also have "\<dots> + (-r gchoose (Suc mm)) * (-x)^Suc mm =

  1212                  (\<Sum>n\<le>Suc mm. (- r gchoose n) * (- x) ^ n * (x + y) ^ (Suc mm - n))" by simp

  1213   finally show ?case unfolding S_def .

  1214 qed simp_all

  1215

  1216 lemma gbinomial_partial_sum_poly_xpos:

  1217   "(\<Sum>k\<le>m. (of_nat m + r gchoose k) * x^k * y^(m-k)) =

  1218      (\<Sum>k\<le>m. (of_nat k + r - 1 gchoose k) * x^k * (x + y)^(m-k))"

  1219   apply (subst gbinomial_partial_sum_poly)

  1220   apply (subst gbinomial_negated_upper)

  1221   apply (intro setsum.cong, rule refl)

  1222   apply (simp add: power_mult_distrib [symmetric])

  1223   done

  1224

  1225 lemma setsum_nat_symmetry:

  1226   "(\<Sum>k = 0..(m::nat). f k) = (\<Sum>k = 0..m. f (m - k))"

  1227   by (rule setsum.reindex_bij_witness[where i="\<lambda>i. m - i" and j="\<lambda>i. m - i"]) auto

  1228

  1229 lemma binomial_r_part_sum: "(\<Sum>k\<le>m. (2 * m + 1 choose k)) = 2 ^ (2 * m)"

  1230 proof -

  1231   have "2 * 2^(2*m) = (\<Sum>k = 0..(2 * m + 1). (2 * m + 1 choose k))"

  1232     using choose_row_sum[where n="2 * m + 1"] by simp

  1233   also have "(\<Sum>k = 0..(2 * m + 1). (2 * m + 1 choose k)) = (\<Sum>k = 0..m. (2 * m + 1 choose k))

  1234                 + (\<Sum>k = m+1..2*m+1. (2 * m + 1 choose k))"

  1235     using setsum_ub_add_nat[of 0 m "\<lambda>k. 2 * m + 1 choose k" "m+1"] by (simp add: mult_2)

  1236   also have "(\<Sum>k = m+1..2*m+1. (2 * m + 1 choose k)) =

  1237                  (\<Sum>k = 0..m. (2 * m + 1 choose (k + (m + 1))))"

  1238     by (subst setsum_shift_bounds_cl_nat_ivl [symmetric]) (simp add: mult_2)

  1239   also have "\<dots> = (\<Sum>k = 0..m. (2 * m + 1 choose (m - k)))"

  1240     by (intro setsum.cong[OF refl], subst binomial_symmetric) simp_all

  1241   also have "\<dots> = (\<Sum>k = 0..m. (2 * m + 1 choose k))"

  1242     by (subst (2) setsum_nat_symmetry) (rule refl)

  1243   also have "\<dots> + \<dots> = 2 * \<dots>" by simp

  1244   finally show ?thesis by (subst (asm) mult_cancel1) (simp add: atLeast0AtMost)

  1245 qed

  1246

  1247 lemma gbinomial_r_part_sum:

  1248   "(\<Sum>k\<le>m. (2 * (of_nat m) + 1 gchoose k)) = 2 ^ (2 * m)" (is "?lhs = ?rhs")

  1249 proof -

  1250   have "?lhs = of_nat (\<Sum>k\<le>m. (2 * m + 1) choose k)"

  1251     by (simp add: binomial_gbinomial of_nat_mult add_ac of_nat_setsum)

  1252   also have "\<dots> = of_nat (2 ^ (2 * m))" by (subst binomial_r_part_sum) (rule refl)

  1253   finally show ?thesis by (simp add: of_nat_power)

  1254 qed

  1255

  1256 lemma gbinomial_sum_nat_pow2:

  1257    "(\<Sum>k\<le>m. (of_nat (m + k) gchoose k :: 'a :: field_char_0) / 2 ^ k) = 2 ^ m" (is "?lhs = ?rhs")

  1258 proof -

  1259   have "2 ^ m * 2 ^ m = (2 ^ (2*m) :: 'a)" by (induction m) simp_all

  1260   also have "\<dots> = (\<Sum>k\<le>m. (2 * (of_nat m) + 1 gchoose k))" using gbinomial_r_part_sum ..

  1261   also have "\<dots> = (\<Sum>k\<le>m. (of_nat (m + k) gchoose k) * 2 ^ (m - k))"

  1262     using gbinomial_partial_sum_poly_xpos[where x="1" and y="1" and r="of_nat m + 1" and m="m"]

  1263     by (simp add: add_ac)

  1264   also have "\<dots> = 2 ^ m * (\<Sum>k\<le>m. (of_nat (m + k) gchoose k) / 2 ^ k)"

  1265     by (subst setsum_right_distrib) (simp add: algebra_simps power_diff)

  1266   finally show ?thesis by (subst (asm) mult_left_cancel) simp_all

  1267 qed

  1268

  1269 lemma gbinomial_trinomial_revision:

  1270   assumes "k \<le> m"

  1271   shows   "(r gchoose m) * (of_nat m gchoose k) = (r gchoose k) * (r - of_nat k gchoose (m - k))"

  1272 proof -

  1273   have "(r gchoose m) * (of_nat m gchoose k) =

  1274             (r gchoose m) * fact m / (fact k * fact (m - k))"

  1275     using assms by (simp add: binomial_gbinomial [symmetric] binomial_fact)

  1276   also have "\<dots> = (r gchoose k) * (r - of_nat k gchoose (m - k))" using assms

  1277     by (simp add: gbinomial_pochhammer power_diff pochhammer_product)

  1278   finally show ?thesis .

  1279 qed

  1280

  1281

  1282 text\<open>Versions of the theorems above for the natural-number version of "choose"\<close>

  1283 lemma binomial_altdef_of_nat:

  1284   fixes n k :: nat

  1285     and x :: "'a :: {field_char_0,field}"  \<comment>\<open>the point is to constrain @{typ 'a}\<close>

  1286   assumes "k \<le> n"

  1287   shows "of_nat (n choose k) = (\<Prod>i<k. of_nat (n - i) / of_nat (k - i) :: 'a)"

  1288 using assms

  1289 by (simp add: gbinomial_altdef_of_nat binomial_gbinomial of_nat_diff)

  1290

  1291 lemma binomial_ge_n_over_k_pow_k:

  1292   fixes k n :: nat

  1293     and x :: "'a :: linordered_field"

  1294   assumes "k \<le> n"

  1295   shows "(of_nat n / of_nat k :: 'a) ^ k \<le> of_nat (n choose k)"

  1296 by (simp add: assms gbinomial_ge_n_over_k_pow_k binomial_gbinomial of_nat_diff)

  1297

  1298 lemma binomial_le_pow:

  1299   assumes "r \<le> n"

  1300   shows "n choose r \<le> n ^ r"

  1301 proof -

  1302   have "n choose r \<le> fact n div fact (n - r)"

  1303     using \<open>r \<le> n\<close> by (subst binomial_fact_lemma[symmetric]) auto

  1304   with fact_div_fact_le_pow [OF assms] show ?thesis by auto

  1305 qed

  1306

  1307 lemma binomial_altdef_nat: "(k::nat) \<le> n \<Longrightarrow>

  1308     n choose k = fact n div (fact k * fact (n - k))"

  1309  by (subst binomial_fact_lemma [symmetric]) auto

  1310

  1311 lemma choose_dvd: "k \<le> n \<Longrightarrow> fact k * fact (n - k) dvd (fact n :: 'a :: {semiring_div,linordered_semidom})"

  1312   unfolding dvd_def

  1313   apply (rule exI [where x="of_nat (n choose k)"])

  1314   using binomial_fact_lemma [of k n, THEN arg_cong [where f = of_nat and 'b='a]]

  1315   apply (auto simp: of_nat_mult)

  1316   done

  1317

  1318 lemma fact_fact_dvd_fact:

  1319     "fact k * fact n dvd (fact (k+n) :: 'a :: {semiring_div,linordered_semidom})"

  1320 by (metis add.commute add_diff_cancel_left' choose_dvd le_add2)

  1321

  1322 lemma choose_mult_lemma:

  1323      "((m+r+k) choose (m+k)) * ((m+k) choose k) = ((m+r+k) choose k) * ((m+r) choose m)"

  1324 proof -

  1325   have "((m+r+k) choose (m+k)) * ((m+k) choose k) =

  1326         fact (m+r + k) div (fact (m + k) * fact (m+r - m)) * (fact (m + k) div (fact k * fact m))"

  1327     by (simp add: assms binomial_altdef_nat)

  1328   also have "... = fact (m+r+k) div (fact r * (fact k * fact m))"

  1329     apply (subst div_mult_div_if_dvd)

  1330     apply (auto simp: algebra_simps fact_fact_dvd_fact)

  1331     apply (metis add.assoc add.commute fact_fact_dvd_fact)

  1332     done

  1333   also have "... = (fact (m+r+k) * fact (m+r)) div (fact r * (fact k * fact m) * fact (m+r))"

  1334     apply (subst div_mult_div_if_dvd [symmetric])

  1335     apply (auto simp add: algebra_simps)

  1336     apply (metis fact_fact_dvd_fact dvd_trans nat_mult_dvd_cancel_disj)

  1337     done

  1338   also have "... = (fact (m+r+k) div (fact k * fact (m+r)) * (fact (m+r) div (fact r * fact m)))"

  1339     apply (subst div_mult_div_if_dvd)

  1340     apply (auto simp: fact_fact_dvd_fact algebra_simps)

  1341     done

  1342   finally show ?thesis

  1343     by (simp add: binomial_altdef_nat mult.commute)

  1344 qed

  1345

  1346 text\<open>The "Subset of a Subset" identity\<close>

  1347 lemma choose_mult:

  1348   assumes "k\<le>m" "m\<le>n"

  1349     shows "(n choose m) * (m choose k) = (n choose k) * ((n-k) choose (m-k))"

  1350 using assms choose_mult_lemma [of "m-k" "n-m" k]

  1351 by simp

  1352

  1353

  1354 subsection \<open>Binomial coefficients\<close>

  1355

  1356 lemma choose_one: "(n::nat) choose 1 = n"

  1357   by simp

  1358

  1359 (*FIXME: messy and apparently unused*)

  1360 lemma binomial_induct [rule_format]: "(ALL (n::nat). P n n) \<longrightarrow>

  1361     (ALL n. P (Suc n) 0) \<longrightarrow> (ALL n. (ALL k < n. P n k \<longrightarrow> P n (Suc k) \<longrightarrow>

  1362     P (Suc n) (Suc k))) \<longrightarrow> (ALL k <= n. P n k)"

  1363   apply (induct n)

  1364   apply auto

  1365   apply (case_tac "k = 0")

  1366   apply auto

  1367   apply (case_tac "k = Suc n")

  1368   apply auto

  1369   apply (metis Suc_le_eq fact.cases le_Suc_eq le_eq_less_or_eq)

  1370   done

  1371

  1372 lemma card_UNION:

  1373   assumes "finite A" and "\<forall>k \<in> A. finite k"

  1374   shows "card (\<Union>A) = nat (\<Sum>I | I \<subseteq> A \<and> I \<noteq> {}. (- 1) ^ (card I + 1) * int (card (\<Inter>I)))"

  1375   (is "?lhs = ?rhs")

  1376 proof -

  1377   have "?rhs = nat (\<Sum>I | I \<subseteq> A \<and> I \<noteq> {}. (- 1) ^ (card I + 1) * (\<Sum>_\<in>\<Inter>I. 1))" by simp

  1378   also have "\<dots> = nat (\<Sum>I | I \<subseteq> A \<and> I \<noteq> {}. (\<Sum>_\<in>\<Inter>I. (- 1) ^ (card I + 1)))" (is "_ = nat ?rhs")

  1379     by(subst setsum_right_distrib) simp

  1380   also have "?rhs = (\<Sum>(I, _)\<in>Sigma {I. I \<subseteq> A \<and> I \<noteq> {}} Inter. (- 1) ^ (card I + 1))"

  1381     using assms by(subst setsum.Sigma)(auto)

  1382   also have "\<dots> = (\<Sum>(x, I)\<in>(SIGMA x:UNIV. {I. I \<subseteq> A \<and> I \<noteq> {} \<and> x \<in> \<Inter>I}). (- 1) ^ (card I + 1))"

  1383     by (rule setsum.reindex_cong [where l = "\<lambda>(x, y). (y, x)"]) (auto intro: inj_onI simp add: split_beta)

  1384   also have "\<dots> = (\<Sum>(x, I)\<in>(SIGMA x:\<Union>A. {I. I \<subseteq> A \<and> I \<noteq> {} \<and> x \<in> \<Inter>I}). (- 1) ^ (card I + 1))"

  1385     using assms by(auto intro!: setsum.mono_neutral_cong_right finite_SigmaI2 intro: finite_subset[where B="\<Union>A"])

  1386   also have "\<dots> = (\<Sum>x\<in>\<Union>A. (\<Sum>I|I \<subseteq> A \<and> I \<noteq> {} \<and> x \<in> \<Inter>I. (- 1) ^ (card I + 1)))"

  1387     using assms by(subst setsum.Sigma) auto

  1388   also have "\<dots> = (\<Sum>_\<in>\<Union>A. 1)" (is "setsum ?lhs _ = _")

  1389   proof(rule setsum.cong[OF refl])

  1390     fix x

  1391     assume x: "x \<in> \<Union>A"

  1392     def K \<equiv> "{X \<in> A. x \<in> X}"

  1393     with \<open>finite A\<close> have K: "finite K" by auto

  1394     let ?I = "\<lambda>i. {I. I \<subseteq> A \<and> card I = i \<and> x \<in> \<Inter>I}"

  1395     have "inj_on snd (SIGMA i:{1..card A}. ?I i)"

  1396       using assms by(auto intro!: inj_onI)

  1397     moreover have [symmetric]: "snd  (SIGMA i:{1..card A}. ?I i) = {I. I \<subseteq> A \<and> I \<noteq> {} \<and> x \<in> \<Inter>I}"

  1398       using assms by(auto intro!: rev_image_eqI[where x="(card a, a)" for a]

  1399         simp add: card_gt_0_iff[folded Suc_le_eq]

  1400         dest: finite_subset intro: card_mono)

  1401     ultimately have "?lhs x = (\<Sum>(i, I)\<in>(SIGMA i:{1..card A}. ?I i). (- 1) ^ (i + 1))"

  1402       by (rule setsum.reindex_cong [where l = snd]) fastforce

  1403     also have "\<dots> = (\<Sum>i=1..card A. (\<Sum>I|I \<subseteq> A \<and> card I = i \<and> x \<in> \<Inter>I. (- 1) ^ (i + 1)))"

  1404       using assms by(subst setsum.Sigma) auto

  1405     also have "\<dots> = (\<Sum>i=1..card A. (- 1) ^ (i + 1) * (\<Sum>I|I \<subseteq> A \<and> card I = i \<and> x \<in> \<Inter>I. 1))"

  1406       by(subst setsum_right_distrib) simp

  1407     also have "\<dots> = (\<Sum>i=1..card K. (- 1) ^ (i + 1) * (\<Sum>I|I \<subseteq> K \<and> card I = i. 1))" (is "_ = ?rhs")

  1408     proof(rule setsum.mono_neutral_cong_right[rule_format])

  1409       show "{1..card K} \<subseteq> {1..card A}" using \<open>finite A\<close>

  1410         by(auto simp add: K_def intro: card_mono)

  1411     next

  1412       fix i

  1413       assume "i \<in> {1..card A} - {1..card K}"

  1414       hence i: "i \<le> card A" "card K < i" by auto

  1415       have "{I. I \<subseteq> A \<and> card I = i \<and> x \<in> \<Inter>I} = {I. I \<subseteq> K \<and> card I = i}"

  1416         by(auto simp add: K_def)

  1417       also have "\<dots> = {}" using \<open>finite A\<close> i

  1418         by(auto simp add: K_def dest: card_mono[rotated 1])

  1419       finally show "(- 1) ^ (i + 1) * (\<Sum>I | I \<subseteq> A \<and> card I = i \<and> x \<in> \<Inter>I. 1 :: int) = 0"

  1420         by(simp only:) simp

  1421     next

  1422       fix i

  1423       have "(\<Sum>I | I \<subseteq> A \<and> card I = i \<and> x \<in> \<Inter>I. 1) = (\<Sum>I | I \<subseteq> K \<and> card I = i. 1 :: int)"

  1424         (is "?lhs = ?rhs")

  1425         by(rule setsum.cong)(auto simp add: K_def)

  1426       thus "(- 1) ^ (i + 1) * ?lhs = (- 1) ^ (i + 1) * ?rhs" by simp

  1427     qed simp

  1428     also have "{I. I \<subseteq> K \<and> card I = 0} = {{}}" using assms

  1429       by(auto simp add: card_eq_0_iff K_def dest: finite_subset)

  1430     hence "?rhs = (\<Sum>i = 0..card K. (- 1) ^ (i + 1) * (\<Sum>I | I \<subseteq> K \<and> card I = i. 1 :: int)) + 1"

  1431       by(subst (2) setsum_head_Suc)(simp_all )

  1432     also have "\<dots> = (\<Sum>i = 0..card K. (- 1) * ((- 1) ^ i * int (card K choose i))) + 1"

  1433       using K by(subst n_subsets[symmetric]) simp_all

  1434     also have "\<dots> = - (\<Sum>i = 0..card K. (- 1) ^ i * int (card K choose i)) + 1"

  1435       by(subst setsum_right_distrib[symmetric]) simp

  1436     also have "\<dots> =  - ((-1 + 1) ^ card K) + 1"

  1437       by(subst binomial_ring)(simp add: ac_simps)

  1438     also have "\<dots> = 1" using x K by(auto simp add: K_def card_gt_0_iff)

  1439     finally show "?lhs x = 1" .

  1440   qed

  1441   also have "nat \<dots> = card (\<Union>A)" by simp

  1442   finally show ?thesis ..

  1443 qed

  1444

  1445 text\<open>The number of nat lists of length \<open>m\<close> summing to \<open>N\<close> is

  1446 @{term "(N + m - 1) choose N"}:\<close>

  1447

  1448 lemma card_length_listsum_rec:

  1449   assumes "m\<ge>1"

  1450   shows "card {l::nat list. length l = m \<and> listsum l = N} =

  1451     (card {l. length l = (m - 1) \<and> listsum l = N} +

  1452     card {l. length l = m \<and> listsum l + 1 =  N})"

  1453     (is "card ?C = (card ?A + card ?B)")

  1454 proof -

  1455   let ?A'="{l. length l = m \<and> listsum l = N \<and> hd l = 0}"

  1456   let ?B'="{l. length l = m \<and> listsum l = N \<and> hd l \<noteq> 0}"

  1457   let ?f ="\<lambda> l. 0#l"

  1458   let ?g ="\<lambda> l. (hd l + 1) # tl l"

  1459   have 1: "\<And>xs x. xs \<noteq> [] \<Longrightarrow> x = hd xs \<Longrightarrow> x # tl xs = xs" by simp

  1460   have 2: "\<And>xs. (xs::nat list) \<noteq> [] \<Longrightarrow> listsum(tl xs) = listsum xs - hd xs"

  1461     by(auto simp add: neq_Nil_conv)

  1462   have f: "bij_betw ?f ?A ?A'"

  1463     apply(rule bij_betw_byWitness[where f' = tl])

  1464     using assms

  1465     by (auto simp: 2 length_0_conv[symmetric] 1 simp del: length_0_conv)

  1466   have 3: "\<And>xs:: nat list. xs \<noteq> [] \<Longrightarrow> hd xs + (listsum xs - hd xs) = listsum xs"

  1467     by (metis 1 listsum_simps(2) 2)

  1468   have g: "bij_betw ?g ?B ?B'"

  1469     apply(rule bij_betw_byWitness[where f' = "\<lambda> l. (hd l - 1) # tl l"])

  1470     using assms

  1471     by (auto simp: 2 length_0_conv[symmetric] intro!: 3

  1472       simp del: length_greater_0_conv length_0_conv)

  1473   { fix M N :: nat have "finite {xs. size xs = M \<and> set xs \<subseteq> {0..<N}}"

  1474     using finite_lists_length_eq[OF finite_atLeastLessThan] conj_commute by auto }

  1475     note fin = this

  1476   have fin_A: "finite ?A" using fin[of _ "N+1"]

  1477     by (intro finite_subset[where ?A = "?A" and ?B = "{xs. size xs = m - 1 \<and> set xs \<subseteq> {0..<N+1}}"],

  1478       auto simp: member_le_listsum_nat less_Suc_eq_le)

  1479   have fin_B: "finite ?B"

  1480     by (intro finite_subset[where ?A = "?B" and ?B = "{xs. size xs = m \<and> set xs \<subseteq> {0..<N}}"],

  1481       auto simp: member_le_listsum_nat less_Suc_eq_le fin)

  1482   have uni: "?C = ?A' \<union> ?B'" by auto

  1483   have disj: "?A' \<inter> ?B' = {}" by auto

  1484   have "card ?C = card(?A' \<union> ?B')" using uni by simp

  1485   also have "\<dots> = card ?A + card ?B"

  1486     using card_Un_disjoint[OF _ _ disj] bij_betw_finite[OF f] bij_betw_finite[OF g]

  1487       bij_betw_same_card[OF f] bij_betw_same_card[OF g] fin_A fin_B

  1488     by presburger

  1489   finally show ?thesis .

  1490 qed

  1491

  1492 lemma card_length_listsum: \<comment>"By Holden Lee, tidied by Tobias Nipkow"

  1493   "card {l::nat list. size l = m \<and> listsum l = N} = (N + m - 1) choose N"

  1494 proof (cases m)

  1495   case 0 then show ?thesis

  1496     by (cases N) (auto simp: cong: conj_cong)

  1497 next

  1498   case (Suc m')

  1499     have m: "m\<ge>1" by (simp add: Suc)

  1500     then show ?thesis

  1501     proof (induct "N + m - 1" arbitrary: N m)

  1502       case 0   \<comment> "In the base case, the only solution is [0]."

  1503       have [simp]: "{l::nat list. length l = Suc 0 \<and> (\<forall>n\<in>set l. n = 0)} = {[0]}"

  1504         by (auto simp: length_Suc_conv)

  1505       have "m=1 \<and> N=0" using 0 by linarith

  1506       then show ?case by simp

  1507     next

  1508       case (Suc k)

  1509

  1510       have c1: "card {l::nat list. size l = (m - 1) \<and> listsum l =  N} =

  1511         (N + (m - 1) - 1) choose N"

  1512       proof cases

  1513         assume "m = 1"

  1514         with Suc.hyps have "N\<ge>1" by auto

  1515         with \<open>m = 1\<close> show ?thesis by (simp add: binomial_eq_0)

  1516       next

  1517         assume "m \<noteq> 1" thus ?thesis using Suc by fastforce

  1518       qed

  1519

  1520       from Suc have c2: "card {l::nat list. size l = m \<and> listsum l + 1 = N} =

  1521         (if N>0 then ((N - 1) + m - 1) choose (N - 1) else 0)"

  1522       proof -

  1523         have aux: "\<And>m n. n > 0 \<Longrightarrow> Suc m = n \<longleftrightarrow> m = n - 1" by arith

  1524         from Suc have "N>0 \<Longrightarrow>

  1525           card {l::nat list. size l = m \<and> listsum l + 1 = N} =

  1526           ((N - 1) + m - 1) choose (N - 1)" by (simp add: aux)

  1527         thus ?thesis by auto

  1528       qed

  1529

  1530       from Suc.prems have "(card {l::nat list. size l = (m - 1) \<and> listsum l = N} +

  1531           card {l::nat list. size l = m \<and> listsum l + 1 = N}) = (N + m - 1) choose N"

  1532         by (auto simp: c1 c2 choose_reduce_nat[of "N + m - 1" N] simp del: One_nat_def)

  1533       thus ?case using card_length_listsum_rec[OF Suc.prems] by auto

  1534     qed

  1535 qed

  1536

  1537

  1538 lemma Suc_times_binomial_add: \<comment> \<open>by Lukas Bulwahn\<close>

  1539   "Suc a * (Suc (a + b) choose Suc a) = Suc b * (Suc (a + b) choose a)"

  1540 proof -

  1541   have dvd: "Suc a * (fact a * fact b) dvd fact (Suc (a + b))" for a b

  1542     using fact_fact_dvd_fact[of "Suc a" "b", where 'a=nat]

  1543     by (simp only: fact_Suc add_Suc[symmetric] of_nat_id mult.assoc)

  1544

  1545   have "Suc a * (fact (Suc (a + b)) div (Suc a * fact a * fact b)) =

  1546       Suc a * fact (Suc (a + b)) div (Suc a * (fact a * fact b))"

  1547     by (subst div_mult_swap[symmetric]; simp only: mult.assoc dvd)

  1548   also have "\<dots> = Suc b * fact (Suc (a + b)) div (Suc b * (fact a * fact b))"

  1549     by (simp only: div_mult_mult1)

  1550   also have "\<dots> = Suc b * (fact (Suc (a + b)) div (Suc b * (fact a * fact b)))"

  1551     using dvd[of b a] by (subst div_mult_swap[symmetric]; simp only: ac_simps dvd)

  1552   finally show ?thesis

  1553     by (subst (1 2) binomial_altdef_nat)

  1554        (simp_all only: ac_simps diff_Suc_Suc Suc_diff_le diff_add_inverse fact_Suc of_nat_id)

  1555 qed

  1556

  1557

  1558

  1559 lemma fact_code [code]:

  1560   "fact n = (of_nat (fold_atLeastAtMost_nat (op *) 2 n 1) :: 'a :: semiring_char_0)"

  1561 proof -

  1562   have "fact n = (of_nat (\<Prod>{1..n}) :: 'a)" by (simp add: fact_altdef')

  1563   also have "\<Prod>{1..n} = \<Prod>{2..n}"

  1564     by (intro setprod.mono_neutral_right) auto

  1565   also have "\<dots> = fold_atLeastAtMost_nat (op *) 2 n 1"

  1566     by (simp add: setprod_atLeastAtMost_code)

  1567   finally show ?thesis .

  1568 qed

  1569

  1570 lemma pochhammer_code [code]:

  1571   "pochhammer a n = (if n = 0 then 1 else

  1572        fold_atLeastAtMost_nat (\<lambda>n acc. (a + of_nat n) * acc) 0 (n - 1) 1)"

  1573   by (simp add: setprod_atLeastAtMost_code pochhammer_def)

  1574

  1575 lemma gbinomial_code [code]:

  1576   "a gchoose n = (if n = 0 then 1 else

  1577      fold_atLeastAtMost_nat (\<lambda>n acc. (a - of_nat n) * acc) 0 (n - 1) 1 / fact n)"

  1578   by (simp add: setprod_atLeastAtMost_code gbinomial_def)

  1579

  1580 (*TODO: This code equation breaks Scala code generation in HOL-Codegenerator_Test. We have to figure out why and how to prevent that. *)

  1581

  1582 (*

  1583 lemma binomial_code [code]:

  1584   "(n choose k) =

  1585       (if k > n then 0

  1586        else if 2 * k > n then (n choose (n - k))

  1587        else (fold_atLeastAtMost_nat (op * ) (n-k+1) n 1 div fact k))"

  1588 proof -

  1589   {

  1590     assume "k \<le> n"

  1591     hence "{1..n} = {1..n-k} \<union> {n-k+1..n}" by auto

  1592     hence "(fact n :: nat) = fact (n-k) * \<Prod>{n-k+1..n}"

  1593       by (simp add: setprod.union_disjoint fact_altdef_nat)

  1594   }

  1595   thus ?thesis by (auto simp: binomial_altdef_nat mult_ac setprod_atLeastAtMost_code)

  1596 qed

  1597 *)

  1598

  1599 end
`