Theory Binomial
section ‹Binomial Coefficients and Binomial Theorem›
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}"
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")
by (simp add: card_image)
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
by (simp add: binomial_def)
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
by (simp add: binomial_def)
qed
lemma binomial_0_Suc [simp]: "0 choose Suc k = 0"
by (simp add: binomial_def)
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)
(auto simp add: card.insert_remove subset_eq_atLeast0_lessThan_finite
Diff_subset_conv K_finite Suc_card_K)
qed
also have "{K∈?Q. n ∉ K} = ?P n (Suc k)"
by (auto simp add: atLeast0_lessThan_Suc)
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_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
by (auto simp add: add_mult_distrib add_mult_distrib2 le_Suc_eq binomial_eq_0)
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
using Suc.IH by (simp add: add_le_mono mult_2)
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 binomial_ring: "(a + b :: 'a::comm_semiring_1)^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}"
by auto
have 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)"]
by (simp add: sum.atLeast_Suc_atMost atMost_atLeast0)
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))"
by (simp add: field_simps)
also have "… = (k + (m - h)) * fact m"
using H[rule_format, OF mn hm'] H[rule_format, OF mn km]
by (simp add: field_simps)
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 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
by (simp add: binomial_fact')
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))"
by (simp add: sum.distrib)
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))"
by (simp add: sum_subtractf)
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
by (simp add: atMost_atLeast0)
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)"
by (simp add: gbinomial_prod_rev atLeastLessThanSuc_atLeastAtMost)
lemma gbinomial_1 [simp]: "a gchoose 1 = a"
by (simp add: gbinomial_prod_rev lessThan_Suc)
lemma gbinomial_Suc0 [simp]: "a gchoose Suc 0 = a"
by (simp add: gbinomial_prod_rev lessThan_Suc)
lemma gbinomial_mult_fact: "fact k * (a gchoose k) = (∏i = 0..<k. a - of_nat i)"
for a :: "'a::field_char_0"
by (simp_all add: gbinomial_prod_rev field_simps)
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
by (simp add: pochhammer_minus Suc)
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"
by (subst power_add [symmetric]) simp
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"
by (simp add: not_le)
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)"
by (simp add: fact_prod_rev)
also have "… = ((∏i∈{0..<k} ∪ {k..<m + k}. of_nat (m + k - i)) :: 'a)"
by (simp add: ivl_disj_un)
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)"
by (simp add: n)
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"
by (simp add: field_simps gbinomial_pochhammer)
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"
by (simp add: mult.commute gbinomial_mult_1)
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))"
by (simp add: field_simps)
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"
by (metis Suc_pred' diff_add_cancel gbinomial_Suc_Suc)
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))"
by (simp add: Suc)
also have "… = Suc m * 2 ^ m"
unfolding sum.atMost_Suc_shift Suc_times_binomial sum_distrib_left[symmetric]
by (simp add: choose_row_sum)
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))"
by (simp add: Suc)
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)
(simp add: algebra_simps)
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)"
by (simp add: pochhammer_rec algebra_simps)
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"
by (simp add: x zero_le_divide_iff)
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)"
by (subst gbinomial_negated_upper) (simp add: add_ac)
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)"
by (simp_all add: gbinomial_prod_rev atLeastLessThanSuc_atLeastAtMost)
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
by (simp add: Suc)
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]
by (subst gbinomial_Suc_Suc) (simp add: field_simps del: of_nat_Suc, simp add: algebra_simps)
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)
lemma gbinomial_addition_formula:
"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)
lemma binomial_addition_formula:
"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]
by (simp add: add_ac)
qed
subsubsection ‹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]
by (simp add: add_ac)
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"
by (simp add: field_simps)
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)"
by (simp add: G_def)
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)"
by (simp add: ring_distribs)
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)"
by (simp add: algebra_simps)
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"
by (simp add: power_minus[of x])
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))"
by (simp add: gbinomial_partial_sum_poly)
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"]
by (simp add: mult_2)
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)"
by (simp add: binomial_gbinomial add_ac)
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"]
by (simp add: add_ac)
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)"
by (metis add.commute add_diff_cancel_left' choose_dvd le_add2)
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))"
by (simp add: binomial_altdef_nat)
also have "... = fact (m + r + k) * fact (m + k) div
(fact (m + k) * fact (m + r - m) * (fact k * fact m))"
by (metis add_implies_diff add_le_mono1 choose_dvd diff_cancel2 div_mult_div_if_dvd le_add1 le_add2)
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
by (simp add: binomial_altdef_nat mult.commute)
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
by (simp add: binomial_altdef_of_nat prod_dividef)
qed
subsection ‹More on Binomial Coefficients›
lemma choose_one: "n choose 1 = n" for n :: nat
by simp
text ‹The famous inclusion-exclusion formula for the cardinality of a union›
lemma int_card_UNION:
assumes "finite A"
and "∀k ∈ A. finite k"
shows "int (card (⋃A)) = (∑I | I ⊆ A ∧ I ≠ {}. (- 1) ^ (card I + 1) * int (card (⋂I)))"
(is "?lhs = ?rhs")
proof -
have "?rhs = (∑I | I ⊆ A ∧ I ≠ {}. (- 1) ^ (card I + 1) * (∑_∈⋂I. 1))"
by simp
also have "… = (∑I | I ⊆ A ∧ I ≠ {}. (∑_∈⋂I. (- 1) ^ (card I + 1)))"
by (subst sum_distrib_left) simp
also have "… = (∑(I, _)∈Sigma {I. I ⊆ A ∧ I ≠ {}} Inter. (- 1) ^ (card I + 1))"
using assms by (subst sum.Sigma) auto
also have "… = (∑(x, I)∈(SIGMA x:UNIV. {I. I ⊆ A ∧ I ≠ {} ∧ x ∈ ⋂I}). (- 1) ^ (card I + 1))"
by (rule sum.reindex_cong [where l = "λ(x, y). (y, x)"]) (auto intro: inj_onI)
also have "… = (∑(x, I)∈(SIGMA x:⋃A. {I. I ⊆ A ∧ I ≠ {} ∧ x ∈ ⋂I}). (- 1) ^ (card I + 1))"
using assms
by (auto intro!: sum.mono_neutral_cong_right finite_SigmaI2 intro: finite_subset[where B="⋃A"])
also have "… = (∑x∈⋃A. (∑I|I ⊆ A ∧ I ≠ {} ∧ x ∈ ⋂I. (- 1) ^ (card I + 1)))"
using assms by (subst sum.Sigma) auto
also have "… = (∑_∈⋃A. 1)" (is "sum ?lhs _ = _")
proof (rule sum.cong[OF refl])
fix x
assume x: "x ∈ ⋃A"
define K where "K = {X ∈ A. x ∈ X}"
with ‹finite A› have K: "finite K"
by auto
let ?I = "λi. {I. I ⊆ A ∧ card I = i ∧ x ∈ ⋂I}"
have "inj_on snd (SIGMA i:{1..card A}. ?I i)"
using assms by (auto intro!: inj_onI)
moreover have [symmetric]: "snd ` (SIGMA i:{1..card A}. ?I i) = {I. I ⊆ A ∧ I ≠ {} ∧ x ∈ ⋂I}"
using assms
by (auto intro!: rev_image_eqI[where x="(card a, a)" for a]
simp add: card_gt_0_iff[folded Suc_le_eq]
dest: finite_subset intro: card_mono)
ultimately have "?lhs x = (∑(i, I)∈(SIGMA i:{1..card A}. ?I i). (- 1) ^ (i + 1))"
by (rule sum.reindex_cong [where l = snd]) fastforce
also have "… = (∑i=1..card A. (∑I|I ⊆ A ∧ card I = i ∧ x ∈ ⋂I. (- 1) ^ (i + 1)))"
using assms by (subst sum.Sigma) auto
also have "… = (∑i=1..card A. (- 1) ^ (i + 1) * (∑I|I ⊆ A ∧ card I = i ∧ x ∈ ⋂I. 1))"
by (subst sum_distrib_left) simp
also have "… = (∑i=1..card K. (- 1) ^ (i + 1) * (∑I|I ⊆ K ∧ card I = i. 1))"
(is "_ = ?rhs")
proof (rule sum.mono_neutral_cong_right[rule_format])
show "finite {1..card A}"
by simp
show "{1..card K} ⊆ {1..card A}"
using ‹finite A› by (auto simp add: K_def intro: card_mono)
next
fix i
assume "i ∈ {1..card A} - {1..card K}"
then have i: "i ≤ card A" "card K < i"
by auto
have "{I. I ⊆ A ∧ card I = i ∧ x ∈ ⋂I} = {I. I ⊆ K ∧ card I = i}"
by (auto simp add: K_def)
also have "… = {}"
using ‹finite A› i by (auto simp add: K_def dest: card_mono[rotated 1])
finally show "(- 1) ^ (i + 1) * (∑I | I ⊆ A ∧ card I = i ∧ x ∈ ⋂I. 1 :: int) = 0"
by (metis mult_zero_right sum.empty)
next
fix i
have "(∑I | I ⊆ A ∧ card I = i ∧ x ∈ ⋂I. 1) = (∑I | I ⊆ K ∧ card I = i. 1 :: int)"
(is "?lhs = ?rhs")
by (rule sum.cong) (auto simp add: K_def)
then show "(- 1) ^ (i + 1) * ?lhs = (- 1) ^ (i + 1) * ?rhs"
by simp
qed
also have "{I. I ⊆ K ∧ card I = 0} = {{}}"
using assms by (auto simp add: card_eq_0_iff K_def dest: finite_subset)
then have "?rhs = (∑i = 0..card K. (- 1) ^ (i + 1) * (∑I | I ⊆ K ∧ card I = i. 1 :: int)) + 1"
by (subst (2) sum.atLeast_Suc_atMost) simp_all
also have "… = (∑i = 0..card K. (- 1) * ((- 1) ^ i * int (card K choose i))) + 1"
using K by (subst n_subsets[symmetric]) simp_all
also have "… = - (∑i = 0..card K. (- 1) ^ i * int (card K choose i)) + 1"
by (subst sum_distrib_left[symmetric]) simp
also have "… = - ((-1 + 1) ^ card K) + 1"
by (subst binomial_ring) (simp add: ac_simps atMost_atLeast0)
also have "… = 1"
using x K by (auto simp add: K_def card_gt_0_iff)
finally show "?lhs x = 1" .
qed
also have "… = int (card (⋃A))"
by simp
finally show ?thesis ..
qed
lemma card_UNION:
assumes "finite A"
and "∀k ∈ A. finite k"
shows "card (⋃A) = nat (∑I | I ⊆ A ∧ I ≠ {}. (- 1) ^ (card I + 1) * int (card (⋂I)))"
by (simp only: flip: int_card_UNION [OF assms])
lemma card_UNION_nonneg:
assumes "finite A"
and "∀k ∈ A. finite k"
shows "(∑I | I ⊆ A ∧ I ≠ {}. (- 1) ^ (card I + 1) * int (card (⋂I))) ≥ 0"
using int_card_UNION [OF assms] by