P. 160 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 15901-16000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	fprodcl 15901*	Closure of a finite product of complex numbers. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℂ)
Theorem	fprodrecl 15902*	Closure of a finite product of real numbers. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℝ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℝ)
Theorem	fprodzcl 15903*	Closure of a finite product of integers. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℤ)
Theorem	fprodnncl 15904*	Closure of a finite product of positive integers. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℕ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℕ)
Theorem	fprodrpcl 15905*	Closure of a finite product of positive reals. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℝ+)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℝ+)
Theorem	fprodnn0cl 15906*	Closure of a finite product of nonnegative integers. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℕ0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℕ0)
Theorem	fprodcllemf 15907*	Finite product closure lemma. A version of fprodcllem 15900 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝑆 ⊆ ℂ)    &   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ 𝑆)    &   ⊢ (𝜑 → 1 ∈ 𝑆)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ 𝑆)
Theorem	fprodreclf 15908*	Closure of a finite product of real numbers. A version of fprodrecl 15902 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℝ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℝ)
Theorem	fprodmul 15909*	The product of two finite products. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 (𝐵 · 𝐶) = (∏𝑘 ∈ 𝐴 𝐵 · ∏𝑘 ∈ 𝐴 𝐶))
Theorem	fproddiv 15910*	The quotient of two finite products. (Contributed by Scott Fenton, 15-Jan-2018.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ≠ 0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 (𝐵 / 𝐶) = (∏𝑘 ∈ 𝐴 𝐵 / ∏𝑘 ∈ 𝐴 𝐶))
Theorem	prodsn 15911*	A product of a singleton is the term. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝑘 = 𝑀 → 𝐴 = 𝐵)    ⇒   ⊢ ((𝑀 ∈ 𝑉 ∧ 𝐵 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝐵)
Theorem	fprod1 15912*	A finite product of only one term is the term itself. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (𝑘 = 𝑀 → 𝐴 = 𝐵)    ⇒   ⊢ ((𝑀 ∈ ℤ ∧ 𝐵 ∈ ℂ) → ∏𝑘 ∈ (𝑀...𝑀)𝐴 = 𝐵)
Theorem	prodsnf 15913*	A product of a singleton is the term. A version of prodsn 15911 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝐵    &   ⊢ (𝑘 = 𝑀 → 𝐴 = 𝐵)    ⇒   ⊢ ((𝑀 ∈ 𝑉 ∧ 𝐵 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝐵)
Theorem	climprod1 15914	The limit of a product over one. (Contributed by Scott Fenton, 15-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    ⇒   ⊢ (𝜑 → seq𝑀( · , (𝑍 × {1})) ⇝ 1)
Theorem	fprodsplit 15915*	Split a finite product into two parts. (Contributed by Scott Fenton, 16-Dec-2017.)
⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)    &   ⊢ (𝜑 → 𝑈 = (𝐴 ∪ 𝐵))    &   ⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑈) → 𝐶 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑈 𝐶 = (∏𝑘 ∈ 𝐴 𝐶 · ∏𝑘 ∈ 𝐵 𝐶))
Theorem	fprodm1 15916*	Separate out the last term in a finite product. (Contributed by Scott Fenton, 16-Dec-2017.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   ⊢ (𝑘 = 𝑁 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · 𝐵))
Theorem	fprod1p 15917*	Separate out the first term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   ⊢ (𝑘 = 𝑀 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (𝐵 · ∏𝑘 ∈ ((𝑀 + 1)...𝑁)𝐴))
Theorem	fprodp1 15918*	Multiply in the last term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)    &   ⊢ (𝑘 = (𝑁 + 1) → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · 𝐵))
Theorem	fprodm1s 15919*	Separate out the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · ⦋𝑁 / 𝑘⦌𝐴))
Theorem	fprodp1s 15920*	Multiply in the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · ⦋(𝑁 + 1) / 𝑘⦌𝐴))
Theorem	prodsns 15921*	A product of the singleton is the term. (Contributed by Scott Fenton, 25-Dec-2017.)
⊢ ((𝑀 ∈ 𝑉 ∧ ⦋𝑀 / 𝑘⦌𝐴 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = ⦋𝑀 / 𝑘⦌𝐴)
Theorem	fprodfac 15922*	Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
⊢ (𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
Theorem	fprodabs 15923*	The absolute value of a finite product. (Contributed by Scott Fenton, 25-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → (abs‘∏𝑘 ∈ (𝑀...𝑁)𝐴) = ∏𝑘 ∈ (𝑀...𝑁)(abs‘𝐴))
Theorem	fprodeq0 15924*	Any finite product containing a zero term is itself zero. (Contributed by Scott Fenton, 27-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 = 𝑁) → 𝐴 = 0)    ⇒   ⊢ ((𝜑 ∧ 𝐾 ∈ (ℤ≥‘𝑁)) → ∏𝑘 ∈ (𝑀...𝐾)𝐴 = 0)
Theorem	fprodshft 15925*	Shift the index of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   ⊢ (𝑗 = (𝑘 − 𝐾) → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝑀 + 𝐾)...(𝑁 + 𝐾))𝐵)
Theorem	fprodrev 15926*	Reversal of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   ⊢ (𝑗 = (𝐾 − 𝑘) → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝐾 − 𝑁)...(𝐾 − 𝑀))𝐵)
Theorem	fprodconst 15927*	The product of constant terms (𝑘 is not free in 𝐵). (Contributed by Scott Fenton, 12-Jan-2018.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ ℂ) → ∏𝑘 ∈ 𝐴 𝐵 = (𝐵↑(♯‘𝐴)))
Theorem	fprodn0 15928*	A finite product of nonzero terms is nonzero. (Contributed by Scott Fenton, 15-Jan-2018.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ≠ 0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ≠ 0)
Theorem	fprod2dlem 15929*	Lemma for fprod2d 15930- induction step. (Contributed by Scott Fenton, 30-Jan-2018.)
⊢ (𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ 𝐴) → 𝐵 ∈ Fin)    &   ⊢ ((𝜑 ∧ (𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵)) → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → ¬ 𝑦 ∈ 𝑥)    &   ⊢ (𝜑 → (𝑥 ∪ {𝑦}) ⊆ 𝐴)    &   ⊢ (𝜓 ↔ ∏𝑗 ∈ 𝑥 ∏𝑘 ∈ 𝐵 𝐶 = ∏𝑧 ∈ ∪ 𝑗 ∈ 𝑥 ({𝑗} × 𝐵)𝐷)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → ∏𝑗 ∈ (𝑥 ∪ {𝑦})∏𝑘 ∈ 𝐵 𝐶 = ∏𝑧 ∈ ∪ 𝑗 ∈ (𝑥 ∪ {𝑦})({𝑗} × 𝐵)𝐷)
Theorem	fprod2d 15930*	Write a double product as a product over a two-dimensional region. Compare fsum2d 15722. (Contributed by Scott Fenton, 30-Jan-2018.)
⊢ (𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ 𝐴) → 𝐵 ∈ Fin)    &   ⊢ ((𝜑 ∧ (𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵)) → 𝐶 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ 𝐴 ∏𝑘 ∈ 𝐵 𝐶 = ∏𝑧 ∈ ∪ 𝑗 ∈ 𝐴 ({𝑗} × 𝐵)𝐷)
Theorem	fprodxp 15931*	Combine two products into a single product over the cartesian product. (Contributed by Scott Fenton, 1-Feb-2018.)
⊢ (𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    &   ⊢ ((𝜑 ∧ (𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵)) → 𝐶 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ 𝐴 ∏𝑘 ∈ 𝐵 𝐶 = ∏𝑧 ∈ (𝐴 × 𝐵)𝐷)
Theorem	fprodcnv 15932*	Transform a product region using the converse operation. (Contributed by Scott Fenton, 1-Feb-2018.)
⊢ (𝑥 = ⟨𝑗, 𝑘⟩ → 𝐵 = 𝐷)    &   ⊢ (𝑦 = ⟨𝑘, 𝑗⟩ → 𝐶 = 𝐷)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → Rel 𝐴)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑥 ∈ 𝐴 𝐵 = ∏𝑦 ∈ ◡ 𝐴𝐶)
Theorem	fprodcom2 15933*	Interchange order of multiplication. Note that 𝐵(𝑗) and 𝐷(𝑘) are not necessarily constant expressions. (Contributed by Scott Fenton, 1-Feb-2018.) (Proof shortened by JJ, 2-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐶 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ 𝐴) → 𝐵 ∈ Fin)    &   ⊢ (𝜑 → ((𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵) ↔ (𝑘 ∈ 𝐶 ∧ 𝑗 ∈ 𝐷)))    &   ⊢ ((𝜑 ∧ (𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵)) → 𝐸 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ 𝐴 ∏𝑘 ∈ 𝐵 𝐸 = ∏𝑘 ∈ 𝐶 ∏𝑗 ∈ 𝐷 𝐸)
Theorem	fprodcom 15934*	Interchange product order. (Contributed by Scott Fenton, 2-Feb-2018.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    &   ⊢ ((𝜑 ∧ (𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵)) → 𝐶 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ 𝐴 ∏𝑘 ∈ 𝐵 𝐶 = ∏𝑘 ∈ 𝐵 ∏𝑗 ∈ 𝐴 𝐶)
Theorem	fprod0diag 15935*	Two ways to express "the product of 𝐴(𝑗, 𝑘) over the triangular region 𝑀 ≤ 𝑗, 𝑀 ≤ 𝑘, 𝑗 + 𝑘 ≤ 𝑁. Compare fsum0diag 15728. (Contributed by Scott Fenton, 2-Feb-2018.)
⊢ ((𝜑 ∧ (𝑗 ∈ (0...𝑁) ∧ 𝑘 ∈ (0...(𝑁 − 𝑗)))) → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ (0...𝑁)∏𝑘 ∈ (0...(𝑁 − 𝑗))𝐴 = ∏𝑘 ∈ (0...𝑁)∏𝑗 ∈ (0...(𝑁 − 𝑘))𝐴)
Theorem	fproddivf 15936*	The quotient of two finite products. A version of fproddiv 15910 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ≠ 0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 (𝐵 / 𝐶) = (∏𝑘 ∈ 𝐴 𝐵 / ∏𝑘 ∈ 𝐴 𝐶))
Theorem	fprodsplitf 15937*	Split a finite product into two parts. A version of fprodsplit 15915 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)    &   ⊢ (𝜑 → 𝑈 = (𝐴 ∪ 𝐵))    &   ⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑈) → 𝐶 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑈 𝐶 = (∏𝑘 ∈ 𝐴 𝐶 · ∏𝑘 ∈ 𝐵 𝐶))
Theorem	fprodsplitsn 15938*	Separate out a term in a finite product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ Ⅎ𝑘𝐷    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐵 ∈ 𝑉)    &   ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    &   ⊢ (𝑘 = 𝐵 → 𝐶 = 𝐷)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝐴 ∪ {𝐵})𝐶 = (∏𝑘 ∈ 𝐴 𝐶 · 𝐷))
Theorem	fprodsplit1f 15939*	Separate out a term in a finite product. A version of fprodsplit1 44608 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → Ⅎ𝑘𝐷)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 = 𝐶) → 𝐵 = 𝐷)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = (𝐷 · ∏𝑘 ∈ (𝐴 ∖ {𝐶})𝐵))
Theorem	fprodn0f 15940*	A finite product of nonzero terms is nonzero. A version of fprodn0 15928 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ≠ 0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ≠ 0)
Theorem	fprodclf 15941*	Closure of a finite product of complex numbers. A version of fprodcl 15901 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ ℂ)
Theorem	fprodge0 15942*	If all the terms of a finite product are nonnegative, so is the product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℝ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 0 ≤ 𝐵)    ⇒   ⊢ (𝜑 → 0 ≤ ∏𝑘 ∈ 𝐴 𝐵)
Theorem	fprodeq0g 15943*	Any finite product containing a zero term is itself zero. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 = 𝐶) → 𝐵 = 0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = 0)
Theorem	fprodge1 15944*	If all of the terms of a finite product are greater than or equal to 1, so is the product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℝ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 1 ≤ 𝐵)    ⇒   ⊢ (𝜑 → 1 ≤ ∏𝑘 ∈ 𝐴 𝐵)
Theorem	fprodle 15945*	If all the terms of two finite products are nonnegative and compare, so do the two products. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℝ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 0 ≤ 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℝ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ≤ 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ≤ ∏𝑘 ∈ 𝐴 𝐶)
Theorem	fprodmodd 15946*	If all factors of two finite products are equal modulo 𝑀, the products are equal modulo 𝑀. (Contributed by AV, 7-Jul-2021.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → (𝐵 mod 𝑀) = (𝐶 mod 𝑀))    ⇒   ⊢ (𝜑 → (∏𝑘 ∈ 𝐴 𝐵 mod 𝑀) = (∏𝑘 ∈ 𝐴 𝐶 mod 𝑀))
5.10.12.5  Infinite products
Theorem	iprodclim 15947*	An infinite product equals the value its sequence converges to. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑍 𝐴 = 𝐵)
Theorem	iprodclim2 15948*	A converging product converges to its infinite product. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ ∏𝑘 ∈ 𝑍 𝐴)
Theorem	iprodclim3 15949*	The sequence of partial finite product of a converging infinite product converge to the infinite product of the series. Note that 𝑗 must not occur in 𝐴. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ 𝑍 ↦ 𝐴)) ⇝ 𝑦))    &   ⊢ (𝜑 → 𝐹 ∈ dom ⇝ )    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ 𝑍) → (𝐹‘𝑗) = ∏𝑘 ∈ (𝑀...𝑗)𝐴)    ⇒   ⊢ (𝜑 → 𝐹 ⇝ ∏𝑘 ∈ 𝑍 𝐴)
Theorem	iprodcl 15950*	The product of a non-trivially converging infinite sequence is a complex number. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑍 𝐴 ∈ ℂ)
Theorem	iprodrecl 15951*	The product of a non-trivially converging infinite real sequence is a real number. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℝ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑍 𝐴 ∈ ℝ)
Theorem	iprodmul 15952*	Multiplication of infinite sums. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → ∃𝑚 ∈ 𝑍 ∃𝑧(𝑧 ≠ 0 ∧ seq𝑚( · , 𝐺) ⇝ 𝑧))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐺‘𝑘) = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑍 (𝐴 · 𝐵) = (∏𝑘 ∈ 𝑍 𝐴 · ∏𝑘 ∈ 𝑍 𝐵))
5.10.13  Falling and Rising Factorial
Syntax	cfallfac 15953	Declare the syntax for the falling factorial.
class FallFac
Syntax	crisefac 15954	Declare the syntax for the rising factorial.
class RiseFac
Definition	df-risefac 15955*	Define the rising factorial function. This is the function (𝐴 · (𝐴 + 1) · ...(𝐴 + 𝑁)) for complex 𝐴 and nonnegative integers 𝑁. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ RiseFac = (𝑥 ∈ ℂ, 𝑛 ∈ ℕ0 ↦ ∏𝑘 ∈ (0...(𝑛 − 1))(𝑥 + 𝑘))
Definition	df-fallfac 15956*	Define the falling factorial function. This is the function (𝐴 · (𝐴 − 1) · ...(𝐴 − 𝑁)) for complex 𝐴 and nonnegative integers 𝑁. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ FallFac = (𝑥 ∈ ℂ, 𝑛 ∈ ℕ0 ↦ ∏𝑘 ∈ (0...(𝑛 − 1))(𝑥 − 𝑘))
Theorem	risefacval 15957*	The value of the rising factorial function. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) = ∏𝑘 ∈ (0...(𝑁 − 1))(𝐴 + 𝑘))
Theorem	fallfacval 15958*	The value of the falling factorial function. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) = ∏𝑘 ∈ (0...(𝑁 − 1))(𝐴 − 𝑘))
Theorem	risefacval2 15959*	One-based value of rising factorial. (Contributed by Scott Fenton, 15-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) = ∏𝑘 ∈ (1...𝑁)(𝐴 + (𝑘 − 1)))
Theorem	fallfacval2 15960*	One-based value of falling factorial. (Contributed by Scott Fenton, 15-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) = ∏𝑘 ∈ (1...𝑁)(𝐴 − (𝑘 − 1)))
Theorem	fallfacval3 15961*	A product representation of falling factorial when 𝐴 is a nonnegative integer. (Contributed by Scott Fenton, 20-Mar-2018.)
⊢ (𝑁 ∈ (0...𝐴) → (𝐴 FallFac 𝑁) = ∏𝑘 ∈ ((𝐴 − (𝑁 − 1))...𝐴)𝑘)
Theorem	risefaccllem 15962*	Lemma for rising factorial closure laws. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ 𝑆 ⊆ ℂ    &   ⊢ 1 ∈ 𝑆    &   ⊢ ((𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆) → (𝑥 · 𝑦) ∈ 𝑆)    &   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝑘 ∈ ℕ0) → (𝐴 + 𝑘) ∈ 𝑆)    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ 𝑆)
Theorem	fallfaccllem 15963*	Lemma for falling factorial closure laws. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ 𝑆 ⊆ ℂ    &   ⊢ 1 ∈ 𝑆    &   ⊢ ((𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆) → (𝑥 · 𝑦) ∈ 𝑆)    &   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝑘 ∈ ℕ0) → (𝐴 − 𝑘) ∈ 𝑆)    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ 𝑆)
Theorem	risefaccl 15964	Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℂ)
Theorem	fallfaccl 15965	Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℂ)
Theorem	rerisefaccl 15966	Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℝ)
Theorem	refallfaccl 15967	Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℝ)
Theorem	nnrisefaccl 15968	Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℕ)
Theorem	zrisefaccl 15969	Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℤ)
Theorem	zfallfaccl 15970	Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℤ)
Theorem	nn0risefaccl 15971	Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℕ0)
Theorem	rprisefaccl 15972	Closure law for rising factorial. (Contributed by Scott Fenton, 9-Jan-2018.)
⊢ ((𝐴 ∈ ℝ+ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℝ+)
Theorem	risefallfac 15973	A relationship between rising and falling factorials. (Contributed by Scott Fenton, 15-Jan-2018.)
⊢ ((𝑋 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝑋 RiseFac 𝑁) = ((-1↑𝑁) · (-𝑋 FallFac 𝑁)))
Theorem	fallrisefac 15974	A relationship between falling and rising factorials. (Contributed by Scott Fenton, 17-Jan-2018.)
⊢ ((𝑋 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝑋 FallFac 𝑁) = ((-1↑𝑁) · (-𝑋 RiseFac 𝑁)))
Theorem	risefall0lem 15975	Lemma for risefac0 15976 and fallfac0 15977. Show a particular set of finite integers is empty. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (0...(0 − 1)) = ∅
Theorem	risefac0 15976	The value of the rising factorial when 𝑁 = 0. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝐴 ∈ ℂ → (𝐴 RiseFac 0) = 1)
Theorem	fallfac0 15977	The value of the falling factorial when 𝑁 = 0. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝐴 ∈ ℂ → (𝐴 FallFac 0) = 1)
Theorem	risefacp1 15978	The value of the rising factorial at a successor. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac (𝑁 + 1)) = ((𝐴 RiseFac 𝑁) · (𝐴 + 𝑁)))
Theorem	fallfacp1 15979	The value of the falling factorial at a successor. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac (𝑁 + 1)) = ((𝐴 FallFac 𝑁) · (𝐴 − 𝑁)))
Theorem	risefacp1d 15980	The value of the rising factorial at a successor. (Contributed by Scott Fenton, 19-Mar-2018.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐴 RiseFac (𝑁 + 1)) = ((𝐴 RiseFac 𝑁) · (𝐴 + 𝑁)))
Theorem	fallfacp1d 15981	The value of the falling factorial at a successor. (Contributed by Scott Fenton, 19-Mar-2018.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐴 FallFac (𝑁 + 1)) = ((𝐴 FallFac 𝑁) · (𝐴 − 𝑁)))
Theorem	risefac1 15982	The value of rising factorial at one. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝐴 ∈ ℂ → (𝐴 RiseFac 1) = 𝐴)
Theorem	fallfac1 15983	The value of falling factorial at one. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝐴 ∈ ℂ → (𝐴 FallFac 1) = 𝐴)
Theorem	risefacfac 15984	Relate rising factorial to factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝑁 ∈ ℕ0 → (1 RiseFac 𝑁) = (!‘𝑁))
Theorem	fallfacfwd 15985	The forward difference of a falling factorial. (Contributed by Scott Fenton, 21-Jan-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ) → (((𝐴 + 1) FallFac 𝑁) − (𝐴 FallFac 𝑁)) = (𝑁 · (𝐴 FallFac (𝑁 − 1))))
Theorem	0fallfac 15986	The value of the zero falling factorial at natural 𝑁. (Contributed by Scott Fenton, 17-Feb-2018.)
⊢ (𝑁 ∈ ℕ → (0 FallFac 𝑁) = 0)
Theorem	0risefac 15987	The value of the zero rising factorial at natural 𝑁. (Contributed by Scott Fenton, 17-Feb-2018.)
⊢ (𝑁 ∈ ℕ → (0 RiseFac 𝑁) = 0)
Theorem	binomfallfaclem1 15988	Lemma for binomfallfac 15990. Closure law. (Contributed by Scott Fenton, 13-Mar-2018.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ ((𝜑 ∧ 𝐾 ∈ (0...𝑁)) → ((𝑁C𝐾) · ((𝐴 FallFac (𝑁 − 𝐾)) · (𝐵 FallFac (𝐾 + 1)))) ∈ ℂ)
Theorem	binomfallfaclem2 15989*	Lemma for binomfallfac 15990. Inductive step. (Contributed by Scott Fenton, 13-Mar-2018.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜓 → ((𝐴 + 𝐵) FallFac 𝑁) = Σ𝑘 ∈ (0...𝑁)((𝑁C𝑘) · ((𝐴 FallFac (𝑁 − 𝑘)) · (𝐵 FallFac 𝑘))))    ⇒   ⊢ ((𝜑 ∧ 𝜓) → ((𝐴 + 𝐵) FallFac (𝑁 + 1)) = Σ𝑘 ∈ (0...(𝑁 + 1))(((𝑁 + 1)C𝑘) · ((𝐴 FallFac ((𝑁 + 1) − 𝑘)) · (𝐵 FallFac 𝑘))))
Theorem	binomfallfac 15990*	A version of the binomial theorem using falling factorials instead of exponentials. (Contributed by Scott Fenton, 13-Mar-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → ((𝐴 + 𝐵) FallFac 𝑁) = Σ𝑘 ∈ (0...𝑁)((𝑁C𝑘) · ((𝐴 FallFac (𝑁 − 𝑘)) · (𝐵 FallFac 𝑘))))
Theorem	binomrisefac 15991*	A version of the binomial theorem using rising factorials instead of exponentials. (Contributed by Scott Fenton, 16-Mar-2018.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → ((𝐴 + 𝐵) RiseFac 𝑁) = Σ𝑘 ∈ (0...𝑁)((𝑁C𝑘) · ((𝐴 RiseFac (𝑁 − 𝑘)) · (𝐵 RiseFac 𝑘))))
Theorem	fallfacval4 15992	Represent the falling factorial via factorials when the first argument is a natural. (Contributed by Scott Fenton, 20-Mar-2018.)
⊢ (𝑁 ∈ (0...𝐴) → (𝐴 FallFac 𝑁) = ((!‘𝐴) / (!‘(𝐴 − 𝑁))))
Theorem	bcfallfac 15993	Binomial coefficient in terms of falling factorials. (Contributed by Scott Fenton, 20-Mar-2018.)
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) = ((𝑁 FallFac 𝐾) / (!‘𝐾)))
Theorem	fallfacfac 15994	Relate falling factorial to factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
⊢ (𝑁 ∈ ℕ0 → (𝑁 FallFac 𝑁) = (!‘𝑁))
5.10.14  Bernoulli polynomials and sums of k-th powers
Syntax	cbp 15995	Declare the constant for the Bernoulli polynomial operator.
class BernPoly
Definition	df-bpoly 15996*	Define the Bernoulli polynomials. Here we use well-founded recursion to define the Bernoulli polynomials. This agrees with most textbook definitions, although explicit formulas do exist. (Contributed by Scott Fenton, 22-May-2014.)
⊢ BernPoly = (𝑚 ∈ ℕ0, 𝑥 ∈ ℂ ↦ (wrecs( < , ℕ0, (𝑔 ∈ V ↦ ⦋(♯‘dom 𝑔) / 𝑛⦌((𝑥↑𝑛) − Σ𝑘 ∈ dom 𝑔((𝑛C𝑘) · ((𝑔‘𝑘) / ((𝑛 − 𝑘) + 1))))))‘𝑚))
Theorem	bpolylem 15997*	Lemma for bpolyval 15998. (Contributed by Scott Fenton, 22-May-2014.) (Revised by Mario Carneiro, 23-Aug-2014.)
⊢ 𝐺 = (𝑔 ∈ V ↦ ⦋(♯‘dom 𝑔) / 𝑛⦌((𝑋↑𝑛) − Σ𝑘 ∈ dom 𝑔((𝑛C𝑘) · ((𝑔‘𝑘) / ((𝑛 − 𝑘) + 1)))))    &   ⊢ 𝐹 = wrecs( < , ℕ0, 𝐺)    ⇒   ⊢ ((𝑁 ∈ ℕ0 ∧ 𝑋 ∈ ℂ) → (𝑁 BernPoly 𝑋) = ((𝑋↑𝑁) − Σ𝑘 ∈ (0...(𝑁 − 1))((𝑁C𝑘) · ((𝑘 BernPoly 𝑋) / ((𝑁 − 𝑘) + 1)))))
Theorem	bpolyval 15998*	The value of the Bernoulli polynomials. (Contributed by Scott Fenton, 16-May-2014.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑋 ∈ ℂ) → (𝑁 BernPoly 𝑋) = ((𝑋↑𝑁) − Σ𝑘 ∈ (0...(𝑁 − 1))((𝑁C𝑘) · ((𝑘 BernPoly 𝑋) / ((𝑁 − 𝑘) + 1)))))
Theorem	bpoly0 15999	The value of the Bernoulli polynomials at zero. (Contributed by Scott Fenton, 16-May-2014.)
⊢ (𝑋 ∈ ℂ → (0 BernPoly 𝑋) = 1)
Theorem	bpoly1 16000	The value of the Bernoulli polynomials at one. (Contributed by Scott Fenton, 16-May-2014.)
⊢ (𝑋 ∈ ℂ → (1 BernPoly 𝑋) = (𝑋 − (1 / 2)))