Theory Binomial

(*  Title:      HOL/Binomial.thy
Author:     Jacques D. Fleuriot
Author:     Lawrence C Paulson
Author:     Chaitanya Mangla
Author:     Manuel Eberl
*)

section ‹Binomial Coefficients, Binomial Theorem, Inclusion-exclusion Principle›

theory Binomial
imports Presburger Factorial
begin

subsection ‹Binomial coefficients›

text ‹This development is based on the work of Andy Gordon and Florian Kammueller.›

text ‹Combinatorial definition›

definition binomial :: "nat ⇒ nat ⇒ nat"  (infixl "choose" 65)
where "n choose k = card {K∈Pow {0..<n}. card K = k}"

lemma binomial_mono:
assumes "m ≤ n" shows "m choose k ≤ n choose k"
proof -
have "{K. K ⊆ {0..<m} ∧ card K = k} ⊆ {K. K ⊆ {0..<n} ∧ card K = k}"
using assms by auto
then show ?thesis
qed

theorem n_subsets:
assumes "finite A"
shows "card {B. B ⊆ A ∧ card B = k} = card A choose k"
proof -
from assms obtain f where bij: "bij_betw f {0..<card A} A"
by (blast dest: ex_bij_betw_nat_finite)
then have [simp]: "card (f  C) = card C" if "C ⊆ {0..<card A}" for C
by (meson bij_betw_imp_inj_on bij_betw_subset card_image that)
from bij have "bij_betw (image f) (Pow {0..<card A}) (Pow A)"
by (rule bij_betw_Pow)
then have "inj_on (image f) (Pow {0..<card A})"
by (rule bij_betw_imp_inj_on)
moreover have "{K. K ⊆ {0..<card A} ∧ card K = k} ⊆ Pow {0..<card A}"
by auto
ultimately have "inj_on (image f) {K. K ⊆ {0..<card A} ∧ card K = k}"
by (rule inj_on_subset)
then have "card {K. K ⊆ {0..<card A} ∧ card K = k} =
card (image f  {K. K ⊆ {0..<card A} ∧ card K = k})" (is "_ = card ?C")
also have "?C = {K. K ⊆ f  {0..<card A} ∧ card K = k}"
by (auto elim!: subset_imageE)
also have "f  {0..<card A} = A"
by (meson bij bij_betw_def)
finally show ?thesis
qed

text ‹Recursive characterization›

lemma binomial_n_0 [simp]: "n choose 0 = 1"
proof -
have "{K ∈ Pow {0..<n}. card K = 0} = {{}}"
by (auto dest: finite_subset)
then show ?thesis
qed

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

lemma binomial_Suc_Suc [simp]: "Suc n choose Suc k = (n choose k) + (n choose Suc k)"
proof -
let ?P = "λn k. {K. K ⊆ {0..<n} ∧ card K = k}"
let ?Q = "?P (Suc n) (Suc k)"
have inj: "inj_on (insert n) (?P n k)"
by rule (auto; metis atLeastLessThan_iff insert_iff less_irrefl subsetCE)
have disjoint: "insert n  ?P n k ∩ ?P n (Suc k) = {}"
by auto
have "?Q = {K∈?Q. n ∈ K} ∪ {K∈?Q. n ∉ K}"
by auto
also have "{K∈?Q. n ∈ K} = insert n  ?P n k" (is "?A = ?B")
proof (rule set_eqI)
fix K
have K_finite: "finite K" if "K ⊆ insert n {0..<n}"
using that by (rule finite_subset) simp_all
have Suc_card_K: "Suc (card K - Suc 0) = card K" if "n ∈ K"
and "finite K"
proof -
from ‹n ∈ K› obtain L where "K = insert n L" and "n ∉ L"
by (blast elim: Set.set_insert)
with that show ?thesis by (simp add: card.insert_remove)
qed
show "K ∈ ?A ⟷ K ∈ ?B"
by (subst in_image_insert_iff)
Diff_subset_conv K_finite Suc_card_K)
qed
also have "{K∈?Q. n ∉ K} = ?P n (Suc k)"
finally show ?thesis using inj disjoint
by (simp add: binomial_def card_Un_disjoint card_image)
qed

lemma binomial_eq_0: "n < k ⟹ n choose k = 0"
by (auto simp add: binomial_def dest: subset_eq_atLeast0_lessThan_card)

lemma zero_less_binomial: "k ≤ n ⟹ n choose k > 0"
by (induct n k rule: diff_induct) simp_all

lemma binomial_eq_0_iff [simp]: "n choose k = 0 ⟷ n < k"
by (metis binomial_eq_0 less_numeral_extra(3) not_less zero_less_binomial)

lemma zero_less_binomial_iff [simp]: "n choose k > 0 ⟷ k ≤ n"
by (metis binomial_eq_0_iff not_less0 not_less zero_less_binomial)

lemma binomial_n_n [simp]: "n choose n = 1"
by (induct n) (simp_all add: binomial_eq_0)

lemma binomial_Suc_n [simp]: "Suc n choose n = Suc n"
by (induct n) simp_all

lemma binomial_1 [simp]: "n choose Suc 0 = n"
by (induct n) simp_all

lemma choose_one: "n choose 1 = n" for n :: nat
by simp

lemma choose_reduce_nat:
"0 < n ⟹ 0 < k ⟹
n choose k = ((n - 1) choose (k - 1)) + ((n - 1) choose k)"
using binomial_Suc_Suc [of "n - 1" "k - 1"] by simp

lemma Suc_times_binomial_eq: "Suc n * (n choose k) = (Suc n choose Suc k) * Suc k"
proof (induction n arbitrary: k)
case 0
then show ?case
by auto
next
case (Suc n)
show ?case
proof (cases k)
case (Suc k')
then show ?thesis
using Suc.IH
qed auto
qed

lemma binomial_le_pow2: "n choose k ≤ 2^n"
proof (induction n arbitrary: k)
case 0
then show ?case
using le_less less_le_trans by fastforce
next
case (Suc n)
show ?case
proof (cases k)
case (Suc k')
then show ?thesis
qed auto
qed

text ‹The absorption property.›
lemma Suc_times_binomial: "Suc k * (Suc n choose Suc k) = Suc n * (n choose k)"
using Suc_times_binomial_eq by auto

text ‹This is the well-known version of absorption, but it's harder to use
because of the need to reason about division.›
lemma binomial_Suc_Suc_eq_times: "(Suc n choose Suc k) = (Suc n * (n choose k)) div Suc k"
by (simp add: Suc_times_binomial_eq del: mult_Suc mult_Suc_right)

text ‹Another version of absorption, with ‹-1› instead of ‹Suc›.›
lemma times_binomial_minus1_eq: "0 < k ⟹ k * (n choose k) = n * ((n - 1) choose (k - 1))"
using Suc_times_binomial_eq [where n = "n - 1" and k = "k - 1"]
by (auto split: nat_diff_split)

subsection ‹The binomial theorem (courtesy of Tobias Nipkow):›

text ‹Avigad's version, generalized to any commutative ring›
theorem (in comm_semiring_1) binomial_ring:
"(a + b :: 'a)^n = (∑k≤n. (of_nat (n choose k)) * a^k * b^(n-k))"
proof (induct n)
case 0
then show ?case by simp
next
case (Suc n)
have decomp: "{0..n+1} = {0} ∪ {n + 1} ∪ {1..n}" and decomp2: "{0..n} = {0} ∪ {1..n}"
by auto
have "(a + b)^(n+1) = (a + b) * (∑k≤n. of_nat (n choose k) * a^k * b^(n - k))"
using Suc.hyps by simp
also have "… = a * (∑k≤n. of_nat (n choose k) * a^k * b^(n-k)) +
b * (∑k≤n. of_nat (n choose k) * a^k * b^(n-k))"
by (rule distrib_right)
also have "… = (∑k≤n. of_nat (n choose k) * a^(k+1) * b^(n-k)) +
(∑k≤n. of_nat (n choose k) * a^k * b^(n - k + 1))"
by (auto simp add: sum_distrib_left ac_simps)
also have "… = (∑k≤n. of_nat (n choose k) * a^k * b^(n + 1 - k)) +
(∑k=1..n+1. of_nat (n choose (k - 1)) * a^k * b^(n + 1 - k))"
by (simp add: atMost_atLeast0 sum.shift_bounds_cl_Suc_ivl Suc_diff_le field_simps del: sum.cl_ivl_Suc)
also have "… = b^(n + 1) +
(∑k=1..n. of_nat (n choose k) * a^k * b^(n + 1 - k)) + (a^(n + 1) +
(∑k=1..n. of_nat (n choose (k - 1)) * a^k * b^(n + 1 - k)))"
using sum.nat_ivl_Suc' [of 1 n "λk. of_nat (n choose (k-1)) * a ^ k * b ^ (n + 1 - k)"]
also have "… = a^(n + 1) + b^(n + 1) +
(∑k=1..n. of_nat (n + 1 choose k) * a^k * b^(n + 1 - k))"
by (auto simp add: field_simps sum.distrib [symmetric] choose_reduce_nat)
also have "… = (∑k≤n+1. of_nat (n + 1 choose k) * a^k * b^(n + 1 - k))"
using decomp by (simp add: atMost_atLeast0 field_simps)
finally show ?case
by simp
qed

text ‹Original version for the naturals.›
corollary binomial: "(a + b :: nat)^n = (∑k≤n. (of_nat (n choose k)) * a^k * b^(n - k))"
using binomial_ring [of "int a" "int b" n]
by (simp only: of_nat_add [symmetric] of_nat_mult [symmetric] of_nat_power [symmetric]
of_nat_sum [symmetric] of_nat_eq_iff of_nat_id)

lemma binomial_fact_lemma: "k ≤ n ⟹ fact k * fact (n - k) * (n choose k) = fact n"
proof (induct n arbitrary: k rule: nat_less_induct)
fix n k
assume H: "∀m<n. ∀x≤m. fact x * fact (m - x) * (m choose x) = fact m"
assume kn: "k ≤ n"
let ?ths = "fact k * fact (n - k) * (n choose k) = fact n"
consider "n = 0 ∨ k = 0 ∨ n = k" | m h where "n = Suc m" "k = Suc h" "h < m"
using kn by atomize_elim presburger
then show "fact k * fact (n - k) * (n choose k) = fact n"
proof cases
case 1
with kn show ?thesis by auto
next
case 2
note n = ‹n = Suc m›
note k = ‹k = Suc h›
note hm = ‹h < m›
have mn: "m < n"
using n by arith
have hm': "h ≤ m"
using hm by arith
have km: "k ≤ m"
using hm k n kn by arith
have "m - h = Suc (m - Suc h)"
using  k km hm by arith
with km k have "fact (m - h) = (m - h) * fact (m - k)"
by simp
with n k have "fact k * fact (n - k) * (n choose k) =
k * (fact h * fact (m - h) * (m choose h)) +
(m - h) * (fact k * fact (m - k) * (m choose k))"
also have "… = (k + (m - h)) * fact m"
using H[rule_format, OF mn hm'] H[rule_format, OF mn km]
finally show ?thesis
using k n km by simp
qed
qed

lemma binomial_fact':
assumes "k ≤ n"
shows "n choose k = fact n div (fact k * fact (n - k))"
using binomial_fact_lemma [OF assms]
by (metis fact_nonzero mult_eq_0_iff nonzero_mult_div_cancel_left)

lemma binomial_fact:
assumes kn: "k ≤ n"
shows "(of_nat (n choose k) :: 'a::field_char_0) = fact n / (fact k * fact (n - k))"
using binomial_fact_lemma[OF kn]
by (metis (mono_tags, lifting) fact_nonzero mult_eq_0_iff nonzero_mult_div_cancel_left of_nat_fact of_nat_mult)

lemma fact_binomial:
assumes "k ≤ n"
shows "fact k * of_nat (n choose k) = (fact n / fact (n - k) :: 'a::field_char_0)"
unfolding binomial_fact [OF assms] by (simp add: field_simps)

lemma binomial_fact_pow: "(n choose s) * fact s ≤ n^s"
proof (cases "s ≤ n")
case True
then show ?thesis
by (smt (verit) binomial_fact_lemma mult.assoc mult.commute fact_div_fact_le_pow fact_nonzero nonzero_mult_div_cancel_right)

lemma choose_two: "n choose 2 = n * (n - 1) div 2"
proof (cases "n ≥ 2")
case False
then have "n = 0 ∨ n = 1"
by auto
then show ?thesis by auto
next
case True
define m where "m = n - 2"
with True have "n = m + 2"
by simp
then have "fact n = n * (n - 1) * fact (n - 2)"
by (simp add: fact_prod_Suc atLeast0_lessThan_Suc algebra_simps)
with True show ?thesis
qed

lemma choose_row_sum: "(∑k≤n. n choose k) = 2^n"
using binomial [of 1 "1" n] by (simp add: numeral_2_eq_2)

lemma sum_choose_lower: "(∑k≤n. (r+k) choose k) = Suc (r+n) choose n"
by (induct n) auto

lemma sum_choose_upper: "(∑k≤n. k choose m) = Suc n choose Suc m"
by (induct n) auto

lemma choose_alternating_sum:
"n > 0 ⟹ (∑i≤n. (-1)^i * of_nat (n choose i)) = (0 :: 'a::comm_ring_1)"
using binomial_ring[of "-1 :: 'a" 1 n]
by (simp add: atLeast0AtMost mult_of_nat_commute zero_power)

lemma choose_even_sum:
assumes "n > 0"
shows "2 * (∑i≤n. if even i then of_nat (n choose i) else 0) = (2 ^ n :: 'a::comm_ring_1)"
proof -
have "2 ^ n = (∑i≤n. of_nat (n choose i)) + (∑i≤n. (-1) ^ i * of_nat (n choose i) :: 'a)"
using choose_row_sum[of n]
by (simp add: choose_alternating_sum assms atLeast0AtMost of_nat_sum[symmetric])
also have "… = (∑i≤n. of_nat (n choose i) + (-1) ^ i * of_nat (n choose i))"
also have "… = 2 * (∑i≤n. if even i then of_nat (n choose i) else 0)"
by (subst sum_distrib_left, intro sum.cong) simp_all
finally show ?thesis ..
qed

lemma choose_odd_sum:
assumes "n > 0"
shows "2 * (∑i≤n. if odd i then of_nat (n choose i) else 0) = (2 ^ n :: 'a::comm_ring_1)"
proof -
have "2 ^ n = (∑i≤n. of_nat (n choose i)) - (∑i≤n. (-1) ^ i * of_nat (n choose i) :: 'a)"
using choose_row_sum[of n]
by (simp add: choose_alternating_sum assms atLeast0AtMost of_nat_sum[symmetric])
also have "… = (∑i≤n. of_nat (n choose i) - (-1) ^ i * of_nat (n choose i))"
also have "… = 2 * (∑i≤n. if odd i then of_nat (n choose i) else 0)"
by (subst sum_distrib_left, intro sum.cong) simp_all
finally show ?thesis ..
qed

text‹NW diagonal sum property›
lemma sum_choose_diagonal:
assumes "m ≤ n"
shows "(∑k≤m. (n - k) choose (m - k)) = Suc n choose m"
proof -
have "(∑k≤m. (n-k) choose (m - k)) = (∑k≤m. (n - m + k) choose k)"
using sum.atLeastAtMost_rev [of "λk. (n - k) choose (m - k)" 0 m] assms
also have "… = Suc (n - m + m) choose m"
by (rule sum_choose_lower)
also have "… = Suc n choose m"
using assms by simp
finally show ?thesis .
qed

subsection ‹Generalized binomial coefficients›

definition gbinomial :: "'a::{semidom_divide,semiring_char_0} ⇒ nat ⇒ 'a"  (infixl "gchoose" 65)
where gbinomial_prod_rev: "a gchoose k = prod (λi. a - of_nat i) {0..<k} div fact k"

lemma gbinomial_0 [simp]:
"a gchoose 0 = 1"
"0 gchoose (Suc k) = 0"
by (simp_all add: gbinomial_prod_rev prod.atLeast0_lessThan_Suc_shift del: prod.op_ivl_Suc)

lemma gbinomial_Suc: "a gchoose (Suc k) = prod (λi. a - of_nat i) {0..k} div fact (Suc k)"

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

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

lemma gbinomial_mult_fact: "fact k * (a gchoose k) = (∏i = 0..<k. a - of_nat i)"
for a :: "'a::field_char_0"

lemma gbinomial_mult_fact': "(a gchoose k) * fact k = (∏i = 0..<k. a - of_nat i)"
for a :: "'a::field_char_0"
using gbinomial_mult_fact [of k a] by (simp add: ac_simps)

lemma gbinomial_pochhammer: "a gchoose k = (- 1) ^ k * pochhammer (- a) k / fact k"
for a :: "'a::field_char_0"
proof (cases k)
case (Suc k')
then have "a gchoose k = pochhammer (a - of_nat k') (Suc k') / ((1 + of_nat k') * fact k')"
by (simp add: gbinomial_prod_rev pochhammer_prod_rev atLeastLessThanSuc_atLeastAtMost
prod.atLeast_Suc_atMost_Suc_shift of_nat_diff flip: power_mult_distrib prod.cl_ivl_Suc)
then show ?thesis
qed auto

lemma gbinomial_pochhammer': "a gchoose k = pochhammer (a - of_nat k + 1) k / fact k"
for a :: "'a::field_char_0"
proof -
have "a gchoose k = ((-1)^k * (-1)^k) * pochhammer (a - of_nat k + 1) k / fact k"
by (simp add: gbinomial_pochhammer pochhammer_minus mult_ac)
also have "(-1 :: 'a)^k * (-1)^k = 1"
finally show ?thesis
by simp
qed

lemma gbinomial_binomial: "n gchoose k = n choose k"
proof (cases "k ≤ n")
case False
then have "n < k"
then have "0 ∈ ((-) n)  {0..<k}"
by auto
then have "prod ((-) n) {0..<k} = 0"
by (auto intro: prod_zero)
with ‹n < k› show ?thesis
by (simp add: binomial_eq_0 gbinomial_prod_rev prod_zero)
next
case True
from True have *: "prod ((-) n) {0..<k} = ∏{Suc (n - k)..n}"
by (intro prod.reindex_bij_witness[of _ "λi. n - i" "λi. n - i"]) auto
from True have "n choose k = fact n div (fact k * fact (n - k))"
by (rule binomial_fact')
with * show ?thesis
by (simp add: gbinomial_prod_rev mult.commute [of "fact k"] div_mult2_eq fact_div_fact)
qed

lemma of_nat_gbinomial: "of_nat (n gchoose k) = (of_nat n gchoose k :: 'a::field_char_0)"
proof (cases "k ≤ n")
case False
then show ?thesis
by (simp add: not_le gbinomial_binomial binomial_eq_0 gbinomial_prod_rev)
next
case True
define m where "m = n - k"
with True have n: "n = m + k"
by arith
from n have "fact n = ((∏i = 0..<m + k. of_nat (m + k - i) ):: 'a)"
also have "… = ((∏i∈{0..<k} ∪ {k..<m + k}. of_nat (m + k - i)) :: 'a)"
finally have "fact n = (fact m * (∏i = 0..<k. of_nat m + of_nat k - of_nat i) :: 'a)"
using prod.shift_bounds_nat_ivl [of "λi. of_nat (m + k - i) :: 'a" 0 k m]
by (simp add: fact_prod_rev [of m] prod.union_disjoint of_nat_diff)
then have "fact n / fact (n - k) = ((∏i = 0..<k. of_nat n - of_nat i) :: 'a)"
with True have "fact k * of_nat (n gchoose k) = (fact k * (of_nat n gchoose k) :: 'a)"
by (simp only: gbinomial_mult_fact [of k "of_nat n"] gbinomial_binomial [of n k] fact_binomial)
then show ?thesis
by simp
qed

lemma binomial_gbinomial: "of_nat (n choose k) = (of_nat n gchoose k :: 'a::field_char_0)"
by (simp add: gbinomial_binomial [symmetric] of_nat_gbinomial)

setup
‹Sign.add_const_constraint (\<^const_name>‹gbinomial›, SOME \<^typ>‹'a::field_char_0 ⇒ nat ⇒ 'a›)›

lemma gbinomial_mult_1:
fixes a :: "'a::field_char_0"
shows "a * (a gchoose k) = of_nat k * (a gchoose k) + of_nat (Suc k) * (a gchoose (Suc k))"
(is "?l = ?r")
proof -
have "?r = ((- 1) ^k * pochhammer (- a) k / fact k) * (of_nat k - (- a + of_nat k))"
unfolding gbinomial_pochhammer pochhammer_Suc right_diff_distrib power_Suc
by (auto simp add: field_simps simp del: of_nat_Suc)
also have "… = ?l"
finally show ?thesis ..
qed

lemma gbinomial_mult_1':
"(a gchoose k) * a = of_nat k * (a gchoose k) + of_nat (Suc k) * (a gchoose (Suc k))"
for a :: "'a::field_char_0"

lemma gbinomial_Suc_Suc: "(a + 1) gchoose (Suc k) = a gchoose k + (a gchoose (Suc k))"
for a :: "'a::field_char_0"
proof (cases k)
case 0
then show ?thesis by simp
next
case (Suc h)
have eq0: "(∏i∈{1..k}. (a + 1) - of_nat i) = (∏i∈{0..h}. a - of_nat i)"
proof (rule prod.reindex_cong)
show "{1..k} = Suc  {0..h}"
using Suc by (auto simp add: image_Suc_atMost)
qed auto
have "fact (Suc k) * (a gchoose k + (a gchoose (Suc k))) =
(a gchoose Suc h) * (fact (Suc (Suc h))) +
(a gchoose Suc (Suc h)) * (fact (Suc (Suc h)))"
by (simp add: Suc field_simps del: fact_Suc)
also have "… =
(a gchoose Suc h) * of_nat (Suc (Suc h) * fact (Suc h)) + (∏i=0..Suc h. a - of_nat i)"
apply (simp only: gbinomial_mult_fact field_simps mult.left_commute [of _ "2"])
apply (simp del: fact_Suc add: fact_Suc [of "Suc h"] field_simps gbinomial_mult_fact
mult.left_commute [of _ "2"] atLeastLessThanSuc_atLeastAtMost)
done
also have "… =
(fact (Suc h) * (a gchoose Suc h)) * of_nat (Suc (Suc h)) + (∏i=0..Suc h. a - of_nat i)"
by (simp only: fact_Suc mult.commute mult.left_commute of_nat_fact of_nat_id of_nat_mult)
also have "… =
of_nat (Suc (Suc h)) * (∏i=0..h. a - of_nat i) + (∏i=0..Suc h. a - of_nat i)"
unfolding gbinomial_mult_fact atLeastLessThanSuc_atLeastAtMost by auto
also have "… =
(∏i=0..Suc h. a - of_nat i) + (of_nat h * (∏i=0..h. a - of_nat i) + 2 * (∏i=0..h. a - of_nat i))"
also have "… =
((a gchoose Suc h) * (fact (Suc h)) * of_nat (Suc k)) + (∏i∈{0..Suc h}. a - of_nat i)"
unfolding gbinomial_mult_fact'
by (simp add: comm_semiring_class.distrib field_simps Suc atLeastLessThanSuc_atLeastAtMost)
also have "… = (∏i∈{0..h}. a - of_nat i) * (a + 1)"
unfolding gbinomial_mult_fact' atLeast0_atMost_Suc
by (simp add: field_simps Suc atLeastLessThanSuc_atLeastAtMost)
also have "… = (∏i∈{0..k}. (a + 1) - of_nat i)"
using eq0
by (simp add: Suc prod.atLeast0_atMost_Suc_shift del: prod.cl_ivl_Suc)
also have "… = (fact (Suc k)) * ((a + 1) gchoose (Suc k))"
by (simp only: gbinomial_mult_fact atLeastLessThanSuc_atLeastAtMost)
finally show ?thesis
using fact_nonzero [of "Suc k"] by auto
qed

lemma gbinomial_reduce_nat: "0 < k ⟹ a gchoose k = (a - 1) gchoose (k - 1) + ((a - 1) gchoose k)"
for a :: "'a::field_char_0"

lemma gchoose_row_sum_weighted:
"(∑k = 0..m. (r gchoose k) * (r/2 - of_nat k)) = of_nat(Suc m) / 2 * (r gchoose (Suc m))"
for r :: "'a::field_char_0"
by (induct m) (simp_all add: field_simps distrib gbinomial_mult_1)

lemma binomial_symmetric:
assumes kn: "k ≤ n"
shows "n choose k = n choose (n - k)"
proof -
have kn': "n - k ≤ n"
using kn by arith
from binomial_fact_lemma[OF kn] binomial_fact_lemma[OF kn']
have "fact k * fact (n - k) * (n choose k) = fact (n - k) * fact (n - (n - k)) * (n choose (n - k))"
by simp
then show ?thesis
using kn by simp
qed

lemma choose_rising_sum:
"(∑j≤m. ((n + j) choose n)) = ((n + m + 1) choose (n + 1))"
"(∑j≤m. ((n + j) choose n)) = ((n + m + 1) choose m)"
proof -
show "(∑j≤m. ((n + j) choose n)) = ((n + m + 1) choose (n + 1))"
by (induct m) simp_all
also have "… = (n + m + 1) choose m"
by (subst binomial_symmetric) simp_all
finally show "(∑j≤m. ((n + j) choose n)) = (n + m + 1) choose m" .
qed

lemma choose_linear_sum: "(∑i≤n. i * (n choose i)) = n * 2 ^ (n - 1)"
proof (cases n)
case 0
then show ?thesis by simp
next
case (Suc m)
have "(∑i≤n. i * (n choose i)) = (∑i≤Suc m. i * (Suc m choose i))"
also have "… = Suc m * 2 ^ m"
unfolding sum.atMost_Suc_shift Suc_times_binomial sum_distrib_left[symmetric]
finally show ?thesis
using Suc by simp
qed

lemma choose_alternating_linear_sum:
assumes "n ≠ 1"
shows "(∑i≤n. (-1)^i * of_nat i * of_nat (n choose i) :: 'a::comm_ring_1) = 0"
proof (cases n)
case 0
then show ?thesis by simp
next
case (Suc m)
with assms have "m > 0"
by simp
have "(∑i≤n. (-1) ^ i * of_nat i * of_nat (n choose i) :: 'a) =
(∑i≤Suc m. (-1) ^ i * of_nat i * of_nat (Suc m choose i))"
also have "… = (∑i≤m. (-1) ^ (Suc i) * of_nat (Suc i * (Suc m choose Suc i)))"
by (simp only: sum.atMost_Suc_shift sum_distrib_left[symmetric] mult_ac of_nat_mult) simp
also have "… = - of_nat (Suc m) * (∑i≤m. (-1) ^ i * of_nat (m choose i))"
by (subst sum_distrib_left, rule sum.cong[OF refl], subst Suc_times_binomial)
also have "(∑i≤m. (-1 :: 'a) ^ i * of_nat ((m choose i))) = 0"
using choose_alternating_sum[OF ‹m > 0›] by simp
finally show ?thesis
by simp
qed

lemma vandermonde: "(∑k≤r. (m choose k) * (n choose (r - k))) = (m + n) choose r"
proof (induct n arbitrary: r)
case 0
have "(∑k≤r. (m choose k) * (0 choose (r - k))) = (∑k≤r. if k = r then (m choose k) else 0)"
by (intro sum.cong) simp_all
also have "… = m choose r"
by simp
finally show ?case
by simp
next
case (Suc n r)
show ?case
by (cases r) (simp_all add: Suc [symmetric] algebra_simps sum.distrib Suc_diff_le)
qed

lemma choose_square_sum: "(∑k≤n. (n choose k)^2) = ((2*n) choose n)"
using vandermonde[of n n n]
by (simp add: power2_eq_square mult_2 binomial_symmetric [symmetric])

lemma pochhammer_binomial_sum:
fixes a b :: "'a::comm_ring_1"
shows "pochhammer (a + b) n = (∑k≤n. of_nat (n choose k) * pochhammer a k * pochhammer b (n - k))"
proof (induction n arbitrary: a b)
case 0
then show ?case by simp
next
case (Suc n a b)
have "(∑k≤Suc n. of_nat (Suc n choose k) * pochhammer a k * pochhammer b (Suc n - k)) =
(∑i≤n. of_nat (n choose i) * pochhammer a (Suc i) * pochhammer b (n - i)) +
((∑i≤n. of_nat (n choose Suc i) * pochhammer a (Suc i) * pochhammer b (n - i)) +
pochhammer b (Suc n))"
by (subst sum.atMost_Suc_shift) (simp add: ring_distribs sum.distrib)
also have "(∑i≤n. of_nat (n choose i) * pochhammer a (Suc i) * pochhammer b (n - i)) =
a * pochhammer ((a + 1) + b) n"
by (subst Suc) (simp add: sum_distrib_left pochhammer_rec mult_ac)
also have "(∑i≤n. of_nat (n choose Suc i) * pochhammer a (Suc i) * pochhammer b (n - i)) +
pochhammer b (Suc n) =
(∑i=0..Suc n. of_nat (n choose i) * pochhammer a i * pochhammer b (Suc n - i))"
apply (subst sum.atLeast_Suc_atMost, simp)
apply (simp add: sum.shift_bounds_cl_Suc_ivl atLeast0AtMost del: sum.cl_ivl_Suc)
done
also have "… = (∑i≤n. of_nat (n choose i) * pochhammer a i * pochhammer b (Suc n - i))"
using Suc by (intro sum.mono_neutral_right) (auto simp: not_le binomial_eq_0)
also have "… = (∑i≤n. of_nat (n choose i) * pochhammer a i * pochhammer b (Suc (n - i)))"
by (intro sum.cong) (simp_all add: Suc_diff_le)
also have "… = b * pochhammer (a + (b + 1)) n"
by (subst Suc) (simp add: sum_distrib_left mult_ac pochhammer_rec)
also have "a * pochhammer ((a + 1) + b) n + b * pochhammer (a + (b + 1)) n =
pochhammer (a + b) (Suc n)"
finally show ?case ..
qed

text ‹Contributed by Manuel Eberl, generalised by LCP.
Alternative definition of the binomial coefficient as \<^term>‹∏i<k. (n - i) / (k - i)›.›
lemma gbinomial_altdef_of_nat: "a gchoose k = (∏i = 0..<k. (a - of_nat i) / of_nat (k - i) :: 'a)"
for k :: nat and a :: "'a::field_char_0"
by (simp add: prod_dividef gbinomial_prod_rev fact_prod_rev)

lemma gbinomial_ge_n_over_k_pow_k:
fixes k :: nat
and a :: "'a::linordered_field"
assumes "of_nat k ≤ a"
shows "(a / of_nat k :: 'a) ^ k ≤ a gchoose k"
proof -
have x: "0 ≤ a"
using assms of_nat_0_le_iff order_trans by blast
have "(a / of_nat k :: 'a) ^ k = (∏i = 0..<k. a / of_nat k :: 'a)"
by simp
also have "… ≤ a gchoose k"
proof -
have "⋀i. i < k ⟹ 0 ≤ a / of_nat k"
moreover have "a / of_nat k ≤ (a - of_nat i) / of_nat (k - i)" if "i < k" for i
proof -
from assms have "a * of_nat i ≥ of_nat (i * k)"
by (metis mult.commute mult_le_cancel_right of_nat_less_0_iff of_nat_mult)
then have "a * of_nat k - a * of_nat i ≤ a * of_nat k - of_nat (i * k)"
by arith
then have "a * of_nat (k - i) ≤ (a - of_nat i) * of_nat k"
using ‹i < k› by (simp add: algebra_simps zero_less_mult_iff of_nat_diff)
then have "a * of_nat (k - i) ≤ (a - of_nat i) * (of_nat k :: 'a)"
by blast
with assms show ?thesis
using ‹i < k› by (simp add: field_simps)
qed
ultimately show ?thesis
unfolding gbinomial_altdef_of_nat
by (intro prod_mono) auto
qed
finally show ?thesis .
qed

lemma gbinomial_negated_upper: "(a gchoose k) = (-1) ^ k * ((of_nat k - a - 1) gchoose k)"
by (simp add: gbinomial_pochhammer pochhammer_minus algebra_simps)

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

lemma Suc_times_gbinomial: "of_nat (Suc k) * ((a + 1) gchoose (Suc k)) = (a + 1) * (a gchoose k)"
proof (cases k)
case 0
then show ?thesis by simp
next
case (Suc b)
then have "((a + 1) gchoose (Suc (Suc b))) = (∏i = 0..Suc b. a + (1 - of_nat i)) / fact (b + 2)"
by (simp add: field_simps gbinomial_prod_rev atLeastLessThanSuc_atLeastAtMost)
also have "(∏i = 0..Suc b. a + (1 - of_nat i)) = (a + 1) * (∏i = 0..b. a - of_nat i)"
by (simp add: prod.atLeast0_atMost_Suc_shift del: prod.cl_ivl_Suc)
also have "… / fact (b + 2) = (a + 1) / of_nat (Suc (Suc b)) * (a gchoose Suc b)"
finally show ?thesis by (simp add: Suc field_simps del: of_nat_Suc)
qed

lemma gbinomial_factors: "((a + 1) gchoose (Suc k)) = (a + 1) / of_nat (Suc k) * (a gchoose k)"
proof (cases k)
case 0
then show ?thesis by simp
next
case (Suc b)
then have "((a + 1) gchoose (Suc (Suc b))) = (∏i = 0 .. Suc b. a + (1 - of_nat i)) / fact (b + 2)"
by (simp add: field_simps gbinomial_prod_rev atLeastLessThanSuc_atLeastAtMost)
also have "(∏i = 0 .. Suc b. a + (1 - of_nat i)) = (a + 1) * (∏i = 0..b. a - of_nat i)"
by (simp add: prod.atLeast0_atMost_Suc_shift del: prod.cl_ivl_Suc)
also have "… / fact (b + 2) = (a + 1) / of_nat (Suc (Suc b)) * (a gchoose Suc b)"
by (simp_all add: gbinomial_prod_rev atLeastLessThanSuc_atLeastAtMost atLeast0AtMost)
finally show ?thesis
qed

lemma gbinomial_rec: "((a + 1) gchoose (Suc k)) = (a gchoose k) * ((a + 1) / of_nat (Suc k))"
using gbinomial_mult_1[of a k]

lemma gbinomial_of_nat_symmetric: "k ≤ n ⟹ (of_nat n) gchoose k = (of_nat n) gchoose (n - k)"
using binomial_symmetric[of k n] by (simp add: binomial_gbinomial [symmetric])

text ‹The absorption identity (equation 5.5 \<^cite>‹‹p.~157› in GKP_CM›):
${r \choose k} = \frac{r}{k}{r - 1 \choose k - 1},\quad \textnormal{integer } k \neq 0.$›
lemma gbinomial_absorption': "k > 0 ⟹ a gchoose k = (a / of_nat k) * (a - 1 gchoose (k - 1))"
using gbinomial_rec[of "a - 1" "k - 1"]
by (simp_all add: gbinomial_rec field_simps del: of_nat_Suc)

text ‹The absorption identity is written in the following form to avoid
division by $k$ (the lower index) and therefore remove the $k \neq 0$
restriction \<^cite>‹‹p.~157› in GKP_CM›:
$k{r \choose k} = r{r - 1 \choose k - 1}, \quad \textnormal{integer } k.$›
lemma gbinomial_absorption: "of_nat (Suc k) * (a gchoose Suc k) = a * ((a - 1) gchoose k)"
using gbinomial_absorption'[of "Suc k" a] by (simp add: field_simps del: of_nat_Suc)

text ‹The absorption identity for natural number binomial coefficients:›
lemma binomial_absorption: "Suc k * (n choose Suc k) = n * ((n - 1) choose k)"
by (cases n) (simp_all add: binomial_eq_0 Suc_times_binomial del: binomial_Suc_Suc mult_Suc)

text ‹The absorption companion identity for natural number coefficients,
following the proof by GKP \<^cite>‹‹p.~157› in GKP_CM›:›
lemma binomial_absorb_comp: "(n - k) * (n choose k) = n * ((n - 1) choose k)"
(is "?lhs = ?rhs")
proof (cases "n ≤ k")
case True
then show ?thesis by auto
next
case False
then have "?rhs = Suc ((n - 1) - k) * (n choose Suc ((n - 1) - k))"
using binomial_symmetric[of k "n - 1"] binomial_absorption[of "(n - 1) - k" n]
by simp
also have "Suc ((n - 1) - k) = n - k"
using False by simp
also have "n choose … = n choose k"
using False by (intro binomial_symmetric [symmetric]) simp_all
finally show ?thesis ..
qed

text ‹The generalised absorption companion identity:›
lemma gbinomial_absorb_comp: "(a - of_nat k) * (a gchoose k) = a * ((a - 1) gchoose k)"
using pochhammer_absorb_comp[of a k] by (simp add: gbinomial_pochhammer)

"a gchoose (Suc k) = ((a - 1) gchoose (Suc k)) + ((a - 1) gchoose k)"
using gbinomial_Suc_Suc[of "a - 1" k] by (simp add: algebra_simps)

"0 < n ⟹ n choose (Suc k) = ((n - 1) choose (Suc k)) + ((n - 1) choose k)"
by (subst choose_reduce_nat) simp_all

text ‹
Equation 5.9 of the reference material \<^cite>‹‹p.~159› in GKP_CM› is a useful
summation formula, operating on both indices:
$\sum\limits_{k \leq n}{r + k \choose k} = {r + n + 1 \choose n}, \quad \textnormal{integer } n.$
›
lemma gbinomial_parallel_sum: "(∑k≤n. (a + of_nat k) gchoose k) = (a + of_nat n + 1) gchoose n"
proof (induct n)
case 0
then show ?case by simp
next
case (Suc m)
then show ?case
using gbinomial_Suc_Suc[of "(a + of_nat m + 1)" m]
qed

subsection ‹Summation on the upper index›

text ‹
Another summation formula is equation 5.10 of the reference material \<^cite>‹‹p.~160› in GKP_CM›,
aptly named \emph{summation on the upper index}:$\sum_{0 \leq k \leq n} {k \choose m} = {n + 1 \choose m + 1}, \quad \textnormal{integers } m, n \geq 0.$
›
lemma gbinomial_sum_up_index:
"(∑j = 0..n. (of_nat j gchoose k) :: 'a::field_char_0) = (of_nat n + 1) gchoose (k + 1)"
proof (induct n)
case 0
show ?case
using gbinomial_Suc_Suc[of 0 k]
by (cases k) auto
next
case (Suc n)
then show ?case
using gbinomial_Suc_Suc[of "of_nat (Suc n) :: 'a" k]
qed

lemma gbinomial_index_swap:
"((-1) ^ k) * ((- (of_nat n) - 1) gchoose k) = ((-1) ^ n) * ((- (of_nat k) - 1) gchoose n)"
(is "?lhs = ?rhs")
proof -
have "?lhs = (of_nat (k + n) gchoose k)"
by (subst gbinomial_negated_upper) (simp add: power_mult_distrib [symmetric])
also have "… = (of_nat (k + n) gchoose n)"
by (subst gbinomial_of_nat_symmetric) simp_all
also have "… = ?rhs"
by (subst gbinomial_negated_upper) simp
finally show ?thesis .
qed

lemma gbinomial_sum_lower_neg: "(∑k≤m. (a gchoose k) * (- 1) ^ k) = (- 1) ^ m * (a - 1 gchoose m)"
(is "?lhs = ?rhs")
proof -
have "?lhs = (∑k≤m. -(a + 1) + of_nat k gchoose k)"
by (intro sum.cong[OF refl]) (subst gbinomial_negated_upper, simp add: power_mult_distrib)
also have "…  = - a + of_nat m gchoose m"
by (subst gbinomial_parallel_sum) simp
also have "… = ?rhs"
by (subst gbinomial_negated_upper) (simp add: power_mult_distrib)
finally show ?thesis .
qed

lemma gbinomial_partial_row_sum:
"(∑k≤m. (a gchoose k) * ((a / 2) - of_nat k)) = ((of_nat m + 1)/2) * (a gchoose (m + 1))"
proof (induct m)
case 0
then show ?case by simp
next
case (Suc mm)
then have "(∑k≤Suc mm. (a gchoose k) * (a / 2 - of_nat k)) =
(a - of_nat (Suc mm)) * (a gchoose Suc mm) / 2"
also have "… = a * (a - 1 gchoose Suc mm) / 2"
by (subst gbinomial_absorb_comp) (rule refl)
also have "… = (of_nat (Suc mm) + 1) / 2 * (a gchoose (Suc mm + 1))"
by (subst gbinomial_absorption [symmetric]) simp
finally show ?case .
qed

lemma sum_bounds_lt_plus1: "(∑k<mm. f (Suc k)) = (∑k=1..mm. f k)"
by (induct mm) simp_all

lemma gbinomial_partial_sum_poly:
"(∑k≤m. (of_nat m + a gchoose k) * x^k * y^(m-k)) =
(∑k≤m. (-a gchoose k) * (-x)^k * (x + y)^(m-k))"
(is "?lhs m = ?rhs m")
proof (induction m)
case 0
then show ?case by simp
next
case (Suc mm)
define G where "G i k = (of_nat i + a gchoose k) * x^k * y^(i - k)" for i k
define S where "S = ?lhs"
have SG_def: "S = (λi. (∑k≤i. (G i k)))"
unfolding S_def G_def ..

have "S (Suc mm) = G (Suc mm) 0 + (∑k=Suc 0..Suc mm. G (Suc mm) k)"
using SG_def by (simp add: sum.atLeast_Suc_atMost atLeast0AtMost [symmetric])
also have "(∑k=Suc 0..Suc mm. G (Suc mm) k) = (∑k=0..mm. G (Suc mm) (Suc k))"
by (subst sum.shift_bounds_cl_Suc_ivl) simp
also have "… = (∑k=0..mm. ((of_nat mm + a gchoose (Suc k)) +
(of_nat mm + a gchoose k)) * x^(Suc k) * y^(mm - k))"
unfolding G_def by (subst gbinomial_addition_formula) simp
also have "… = (∑k=0..mm. (of_nat mm + a gchoose (Suc k)) * x^(Suc k) * y^(mm - k)) +
(∑k=0..mm. (of_nat mm + a gchoose k) * x^(Suc k) * y^(mm - k))"
by (subst sum.distrib [symmetric]) (simp add: algebra_simps)
also have "(∑k=0..mm. (of_nat mm + a gchoose (Suc k)) * x^(Suc k) * y^(mm - k)) =
(∑k<Suc mm. (of_nat mm + a gchoose (Suc k)) * x^(Suc k) * y^(mm - k))"
by (simp only: atLeast0AtMost lessThan_Suc_atMost)
also have "… = (∑k<mm. (of_nat mm + a gchoose Suc k) * x^(Suc k) * y^(mm-k)) +
(of_nat mm + a gchoose (Suc mm)) * x^(Suc mm)"
(is "_ = ?A + ?B")
by (subst sum.lessThan_Suc) simp
also have "?A = (∑k=1..mm. (of_nat mm + a gchoose k) * x^k * y^(mm - k + 1))"
proof (subst sum_bounds_lt_plus1 [symmetric], intro sum.cong[OF refl], clarify)
fix k
assume "k < mm"
then have "mm - k = mm - Suc k + 1"
by linarith
then show "(of_nat mm + a gchoose Suc k) * x ^ Suc k * y ^ (mm - k) =
(of_nat mm + a gchoose Suc k) * x ^ Suc k * y ^ (mm - Suc k + 1)"
by (simp only:)
qed
also have "… + ?B = y * (∑k=1..mm. (G mm k)) + (of_nat mm + a gchoose (Suc mm)) * x^(Suc mm)"
unfolding G_def by (subst sum_distrib_left) (simp add: algebra_simps)
also have "(∑k=0..mm. (of_nat mm + a gchoose k) * x^(Suc k) * y^(mm - k)) = x * (S mm)"
unfolding S_def by (subst sum_distrib_left) (simp add: atLeast0AtMost algebra_simps)
also have "(G (Suc mm) 0) = y * (G mm 0)"
finally have "S (Suc mm) =
y * (G mm 0 + (∑k=1..mm. (G mm k))) + (of_nat mm + a gchoose (Suc mm)) * x^(Suc mm) + x * (S mm)"
also have "G mm 0 + (∑k=1..mm. (G mm k)) = S mm"
by (simp add: sum.atLeast_Suc_atMost[symmetric] SG_def atLeast0AtMost)
finally have "S (Suc mm) = (x + y) * (S mm) + (of_nat mm + a gchoose (Suc mm)) * x^(Suc mm)"
also have "(of_nat mm + a gchoose (Suc mm)) = (-1) ^ (Suc mm) * (- a gchoose (Suc mm))"
by (subst gbinomial_negated_upper) simp
also have "(-1) ^ Suc mm * (- a gchoose Suc mm) * x ^ Suc mm =
(- a gchoose (Suc mm)) * (-x) ^ Suc mm"
also have "(x + y) * S mm + … = (x + y) * ?rhs mm + (- a gchoose (Suc mm)) * (- x)^Suc mm"
unfolding S_def by (subst Suc.IH) simp
also have "(x + y) * ?rhs mm = (∑n≤mm. ((- a gchoose n) * (- x) ^ n * (x + y) ^ (Suc mm - n)))"
by (subst sum_distrib_left, rule sum.cong) (simp_all add: Suc_diff_le)
also have "… + (-a gchoose (Suc mm)) * (-x)^Suc mm =
(∑n≤Suc mm. (- a gchoose n) * (- x) ^ n * (x + y) ^ (Suc mm - n))"
by simp
finally show ?case
by (simp only: S_def)
qed

lemma gbinomial_partial_sum_poly_xpos:
"(∑k≤m. (of_nat m + a gchoose k) * x^k * y^(m-k)) =
(∑k≤m. (of_nat k + a - 1 gchoose k) * x^k * (x + y)^(m-k))" (is "?lhs = ?rhs")
proof -
have "?lhs = (∑k≤m. (- a gchoose k) * (- x) ^ k * (x + y) ^ (m - k))"
also have "... = (∑k≤m. (-1) ^ k * (of_nat k - - a - 1 gchoose k) * (- x) ^ k * (x + y) ^ (m - k))"
by (metis (no_types, opaque_lifting) gbinomial_negated_upper)
also have "... = ?rhs"
by (intro sum.cong) (auto simp flip: power_mult_distrib)
finally show ?thesis .
qed

lemma binomial_r_part_sum: "(∑k≤m. (2 * m + 1 choose k)) = 2 ^ (2 * m)"
proof -
have "2 * 2^(2*m) = (∑k = 0..(2 * m + 1). (2 * m + 1 choose k))"
using choose_row_sum[where n="2 * m + 1"]  by (simp add: atMost_atLeast0)
also have "(∑k = 0..(2 * m + 1). (2 * m + 1 choose k)) =
(∑k = 0..m. (2 * m + 1 choose k)) +
(∑k = m+1..2*m+1. (2 * m + 1 choose k))"
using sum.ub_add_nat[of 0 m "λk. 2 * m + 1 choose k" "m+1"]
also have "(∑k = m+1..2*m+1. (2 * m + 1 choose k)) =
(∑k = 0..m. (2 * m + 1 choose (k + (m + 1))))"
by (subst sum.shift_bounds_cl_nat_ivl [symmetric]) (simp add: mult_2)
also have "… = (∑k = 0..m. (2 * m + 1 choose (m - k)))"
by (intro sum.cong[OF refl], subst binomial_symmetric) simp_all
also have "… = (∑k = 0..m. (2 * m + 1 choose k))"
using sum.atLeastAtMost_rev [of "λk. 2 * m + 1 choose (m - k)" 0 m]
by simp
also have "… + … = 2 * …"
by simp
finally show ?thesis
by (subst (asm) mult_cancel1) (simp add: atLeast0AtMost)
qed

lemma gbinomial_r_part_sum: "(∑k≤m. (2 * (of_nat m) + 1 gchoose k)) = 2 ^ (2 * m)"
(is "?lhs = ?rhs")
proof -
have "?lhs = of_nat (∑k≤m. (2 * m + 1) choose k)"
also have "… = of_nat (2 ^ (2 * m))"
by (subst binomial_r_part_sum) (rule refl)
finally show ?thesis by simp
qed

lemma gbinomial_sum_nat_pow2:
"(∑k≤m. (of_nat (m + k) gchoose k :: 'a::field_char_0) / 2 ^ k) = 2 ^ m"
(is "?lhs = ?rhs")
proof -
have "2 ^ m * 2 ^ m = (2 ^ (2*m) :: 'a)"
by (induct m) simp_all
also have "… = (∑k≤m. (2 * (of_nat m) + 1 gchoose k))"
using gbinomial_r_part_sum ..
also have "… = (∑k≤m. (of_nat (m + k) gchoose k) * 2 ^ (m - k))"
using gbinomial_partial_sum_poly_xpos[where x="1" and y="1" and a="of_nat m + 1" and m="m"]
also have "… = 2 ^ m * (∑k≤m. (of_nat (m + k) gchoose k) / 2 ^ k)"
by (subst sum_distrib_left) (simp add: algebra_simps power_diff)
finally show ?thesis
by (subst (asm) mult_left_cancel) simp_all
qed

lemma gbinomial_trinomial_revision:
assumes "k ≤ m"
shows "(a gchoose m) * (of_nat m gchoose k) = (a gchoose k) * (a - of_nat k gchoose (m - k))"
proof -
have "(a gchoose m) * (of_nat m gchoose k) = (a gchoose m) * fact m / (fact k * fact (m - k))"
using assms by (simp add: binomial_gbinomial [symmetric] binomial_fact)
also have "… = (a gchoose k) * (a - of_nat k gchoose (m - k))"
using assms by (simp add: gbinomial_pochhammer power_diff pochhammer_product)
finally show ?thesis .
qed

text ‹Versions of the theorems above for the natural-number version of "choose"›
lemma binomial_altdef_of_nat:
"k ≤ n ⟹ of_nat (n choose k) = (∏i = 0..<k. of_nat (n - i) / of_nat (k - i) :: 'a)"
for n k :: nat and x :: "'a::field_char_0"
by (simp add: gbinomial_altdef_of_nat binomial_gbinomial of_nat_diff)

lemma binomial_ge_n_over_k_pow_k: "k ≤ n ⟹ (of_nat n / of_nat k :: 'a) ^ k ≤ of_nat (n choose k)"
for k n :: nat and x :: "'a::linordered_field"
by (simp add: gbinomial_ge_n_over_k_pow_k binomial_gbinomial of_nat_diff)

lemma binomial_le_pow:
assumes "r ≤ n"
shows "n choose r ≤ n ^ r"
proof -
have "n choose r ≤ fact n div fact (n - r)"
using assms by (subst binomial_fact_lemma[symmetric]) auto
with fact_div_fact_le_pow [OF assms] show ?thesis
by auto
qed

lemma binomial_altdef_nat: "k ≤ n ⟹ n choose k = fact n div (fact k * fact (n - k))"
for k n :: nat
by (subst binomial_fact_lemma [symmetric]) auto

lemma choose_dvd:
assumes "k ≤ n" shows "fact k * fact (n - k) dvd (fact n :: 'a::linordered_semidom)"
unfolding dvd_def
proof
show "fact n = fact k * fact (n - k) * of_nat (n choose k)"
by (metis assms binomial_fact_lemma of_nat_fact of_nat_mult)
qed

lemma fact_fact_dvd_fact:
"fact k * fact n dvd (fact (k + n) :: 'a::linordered_semidom)"

lemma choose_mult_lemma:
"((m + r + k) choose (m + k)) * ((m + k) choose k) = ((m + r + k) choose k) * ((m + r) choose m)"
(is "?lhs = _")
proof -
have "?lhs =
fact (m + r + k) div (fact (m + k) * fact (m + r - m)) * (fact (m + k) div (fact k * fact m))"
also have "... = fact (m + r + k) * fact (m + k) div
(fact (m + k) * fact (m + r - m) * (fact k * fact m))"
also have "… = fact (m + r + k) div (fact r * (fact k * fact m))"
by (auto simp: algebra_simps fact_fact_dvd_fact)
also have "… = (fact (m + r + k) * fact (m + r)) div (fact r * (fact k * fact m) * fact (m + r))"
by simp
also have "… =
(fact (m + r + k) div (fact k * fact (m + r)) * (fact (m + r) div (fact r * fact m)))"
by (auto simp: div_mult_div_if_dvd fact_fact_dvd_fact algebra_simps)
finally show ?thesis
qed

text ‹The "Subset of a Subset" identity.›
lemma choose_mult:
"k ≤ m ⟹ m ≤ n ⟹ (n choose m) * (m choose k) = (n choose k) * ((n - k) choose (m - k))"
using choose_mult_lemma [of "m-k" "n-m" k] by simp

lemma of_nat_binomial_eq_mult_binomial_Suc:
assumes "k ≤ n"
shows "(of_nat :: (nat ⇒ ('a :: field_char_0))) (n choose k) = of_nat (n + 1 - k) / of_nat (n + 1) * of_nat (Suc n choose k)"
proof (cases k)
case 0 then show ?thesis
using of_nat_neq_0 by auto
next
case (Suc l)
have "of_nat (n + 1) * (∏i=0..<k. of_nat (n - i)) = (of_nat :: (nat ⇒ 'a)) (n + 1 - k) * (∏i=0..<k. of_nat (Suc n - i))"
using prod.atLeast0_lessThan_Suc [where ?'a = 'a, symmetric, of "λi. of_nat (Suc n - i)" k]
by (simp add: ac_simps prod.atLeast0_lessThan_Suc_shift del: prod.op_ivl_Suc)
also have "... = (of_nat :: (nat ⇒ 'a)) (Suc n - k) * (∏i=0..<k. of_nat (Suc n - i))"
by (simp add: Suc atLeast0_atMost_Suc atLeastLessThanSuc_atLeastAtMost)
also have "... = (of_nat :: (nat ⇒ 'a)) (n + 1 - k) * (∏i=0..<k. of_nat (Suc n - i))"
by (simp only: Suc_eq_plus1)
finally have "(∏i=0..<k. of_nat (n - i)) = (of_nat :: (nat ⇒ 'a)) (n + 1 - k) / of_nat (n + 1) * (∏i=0..<k. of_nat (Suc n - i))"
using of_nat_neq_0 by (auto simp: mult.commute divide_simps)
with assms show ?thesis
qed

subsection ‹More on Binomial Coefficients›

text ‹The number of nat lists of length ‹m› summing to ‹N› is \<^term>‹(N + m - 1) choose N›:›
lemma card_length_sum_list_rec:
assumes "m ≥ 1"
shows "card {l::nat list. length l = m ∧ sum_list l = N} =
card {l. length l = (m - 1) ∧ sum_list l = N} +
card {l. length l = m ∧ sum_list l + 1 = N}"
(is "card ?C = card ?A + card ?B")
proof -
let ?A' = "{l. length l = m ∧ sum_list l = N ∧ hd l = 0}"
let ?B' = "{l. length l = m ∧ sum_list l = N ∧ hd l ≠ 0}"
let ?f = "λl. 0 # l"
let ?g = "λl. (hd l + 1) # tl l"
have 1: "xs ≠ [] ⟹ x = hd xs ⟹ x # tl xs = xs" for x :: nat and xs
by simp
have 2: "xs ≠ [] ⟹ sum_list(tl xs) = sum_list xs - hd xs" for xs :: "nat list"
have f: "bij_betw ?f ?A ?A'"
by (rule bij_betw_byWitness[where f' = tl]) (use assms in ‹auto simp: 2 1 simp flip: length_0_conv›)
have 3: "xs ≠ [] ⟹ hd xs + (sum_list xs - hd xs) = sum_list xs" for xs :: "nat list"
by (metis 1 sum_list_simps(2) 2)
have g: "bij_betw ?g ?B ?B'"
apply (rule bij_betw_byWitness[where f' = "λl. (hd l - 1) # tl l"])
using assms
by (auto simp: 2 simp flip: length_0_conv intro!: 3)
have fin: "finite {xs. size xs = M ∧ set xs ⊆ {0..<N}}" for M N :: nat
using finite_lists_length_eq[OF finite_atLeastLessThan] conj_commute by auto
have fin_A: "finite ?A" using fin[of _ "N+1"]
by (intro finite_subset[where ?A = "?A" and ?B = "{xs. size xs = m - 1 ∧ set xs ⊆ {0..<N+1}}"])
(auto simp: member_le_sum_list less_Suc_eq_le)
have fin_B: "finite ?B"
by (intro finite_subset[where ?A = "?B" and ?B = "{xs. size xs = m ∧ set xs ⊆ {0..<N}}"])
(auto simp: member_le_sum_list less_Suc_eq_le fin)
have uni: "?C = ?A' ∪ ?B'"
by auto
have disj: "?A' ∩ ?B' = {}" by blast
have "card ?C = card(?A' ∪ ?B')"
using uni by simp
also have "… = card ?A + card ?B"
using card_Un_disjoint[OF _ _ disj] bij_betw_finite[OF f] bij_betw_finite[OF g]
bij_betw_same_card[OF f] bij_betw_same_card[OF g] fin_A fin_B
by presburger
finally show ?thesis .
qed

lemma card_length_sum_list: "card {l::nat list. size l = m ∧ sum_list l = N} = (N + m - 1) choose N"
― ‹by Holden Lee, tidied by Tobias Nipkow›
proof (cases m)
case 0
then show ?thesis
by (cases N) (auto cong: conj_cong)
next
case (Suc m')
have m: "m ≥ 1"
then show ?thesis
proof (induct "N + m - 1" arbitrary: N m)
case 0  ― ‹In the base case, the only solution is [0].›
have [simp]: "{l::nat list. length l = Suc 0 ∧ (∀n∈set l. n = 0)} = {[0]}"
by (auto simp: length_Suc_conv)
have "m = 1 ∧ N = 0"
using 0 by linarith
then show ?case
by simp
next
case (Suc k)
have c1: "card {l::nat list. size l = (m - 1) ∧ sum_list l =  N} = (N + (m - 1) - 1) choose N"
proof (cases "m = 1")
case True
with Suc.hyps have "N ≥ 1"
by auto
with True show ?thesis
next
case False
then show ?thesis
using Suc by fastforce
qed
from Suc have c2: "card {l::nat list. size l = m ∧ sum_list l + 1 = N} =
(if N > 0 then ((N - 1) + m - 1) choose (N - 1) else 0)"
proof -
have *: "n > 0 ⟹ Suc m = n ⟷ m = n - 1" for m n
by arith
from Suc have "N > 0 ⟹
card {l::nat list. size l = m ∧ sum_list l + 1 = N} =
((N - 1) + m - 1) choose (N - 1)"
then show ?thesis
by auto
qed
from Suc.prems have "(card {l::nat list. size l = (m - 1) ∧ sum_list l = N} +
card {l::nat list. size l = m ∧ sum_list l + 1 = N}) = (N + m - 1) choose N"
by (auto simp: c1 c2 choose_reduce_nat[of "N + m - 1" N] simp del: One_nat_def)
then show ?case
using card_length_sum_list_rec[OF Suc.prems] by auto
qed
qed

lemma card_disjoint_shuffles:
assumes "set xs ∩ set ys = {}"
shows   "card (shuffles xs ys) = (length xs + length ys) choose length xs"
using assms
proof (induction xs ys rule: shuffles.induct)
case (3 x xs y ys)
have "shuffles (x # xs) (y # ys) = (#) x  shuffles xs (y # ys) ∪ (#) y  shuffles (x # xs) ys"
by (rule shuffles.simps)
also have "card … = card ((#) x  shuffles xs (y # ys)) + card ((#) y  shuffles (x # xs) ys)"
by (rule card_Un_disjoint) (insert "3.prems", auto)
also have "card ((#) x  shuffles xs (y # ys)) = card (shuffles xs (y # ys))"
by (rule card_image) auto
also have "… = (length xs + length (y # ys)) choose length xs"
using "3.prems" by (intro "3.IH") auto
also have "card ((#) y  shuffles (x # xs) ys