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Theorem List for Metamath Proof Explorer - 14501-14600   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremprodrb 14501* Rebase the starting point of a product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝐴 ⊆ (ℤ𝑀))    &   (𝜑𝐴 ⊆ (ℤ𝑁))       (𝜑 → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))

Theoremprodmolem3 14502* Lemma for prodmo 14505. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ (𝑓𝑗) / 𝑘𝐵)    &   𝐻 = (𝑗 ∈ ℕ ↦ (𝐾𝑗) / 𝑘𝐵)    &   (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ))    &   (𝜑𝑓:(1...𝑀)–1-1-onto𝐴)    &   (𝜑𝐾:(1...𝑁)–1-1-onto𝐴)       (𝜑 → (seq1( · , 𝐺)‘𝑀) = (seq1( · , 𝐻)‘𝑁))

Theoremprodmolem2a 14503* Lemma for prodmo 14505. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ (𝑓𝑗) / 𝑘𝐵)    &   𝐻 = (𝑗 ∈ ℕ ↦ (𝐾𝑗) / 𝑘𝐵)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴 ⊆ (ℤ𝑀))    &   (𝜑𝑓:(1...𝑁)–1-1-onto𝐴)    &   (𝜑𝐾 Isom < , < ((1...(#‘𝐴)), 𝐴))       (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq1( · , 𝐺)‘𝑁))

Theoremprodmolem2 14504* Lemma for prodmo 14505. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ (𝑓𝑗) / 𝑘𝐵)       ((𝜑 ∧ ∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦) ∧ seq𝑚( · , 𝐹) ⇝ 𝑥)) → (∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑧 = (seq1( · , 𝐺)‘𝑚)) → 𝑥 = 𝑧))

Theoremprodmo 14505* A product has at most one limit. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ (𝑓𝑗) / 𝑘𝐵)       (𝜑 → ∃*𝑥(∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦) ∧ seq𝑚( · , 𝐹) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , 𝐺)‘𝑚))))

Theoremzprod 14506* Series product with index set a subset of the upper integers. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   (𝜑𝐴𝑍)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))

Theoremiprod 14507* Series product with an upper integer index set (i.e. an infinite product.) (Contributed by Scott Fenton, 5-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐵)    &   ((𝜑𝑘𝑍) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))

Theoremzprodn0 14508* Nonzero series product with index set a subset of the upper integers. (Contributed by Scott Fenton, 6-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   (𝜑𝐴𝑍)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = 𝑋)

Theoremiprodn0 14509* Nonzero series product with an upper integer index set (i.e. an infinite product.) (Contributed by Scott Fenton, 6-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐵)    &   ((𝜑𝑘𝑍) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 𝐵 = 𝑋)

5.10.12.4  Finite products

Theoremfprod 14510* The value of a product over a nonempty finite set. (Contributed by Scott Fenton, 6-Dec-2017.)
(𝑘 = (𝐹𝑛) → 𝐵 = 𝐶)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝐹:(1...𝑀)–1-1-onto𝐴)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑛 ∈ (1...𝑀)) → (𝐺𝑛) = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = (seq1( · , 𝐺)‘𝑀))

Theoremfprodntriv 14511* A non-triviality lemma for finite sequences. (Contributed by Scott Fenton, 16-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   (𝜑𝐴 ⊆ (𝑀...𝑁))       (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘𝑍 ↦ if(𝑘𝐴, 𝐵, 1))) ⇝ 𝑦))

Theoremprod0 14512 A product over the empty set is one. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑘 ∈ ∅ 𝐴 = 1

Theoremprod1 14513* Any product of one over a valid set is one. (Contributed by Scott Fenton, 7-Dec-2017.)
((𝐴 ⊆ (ℤ𝑀) ∨ 𝐴 ∈ Fin) → ∏𝑘𝐴 1 = 1)

Theoremprodfc 14514* A lemma to facilitate conversions from the function form to the class-variable form of a product. (Contributed by Scott Fenton, 7-Dec-2017.)
𝑗𝐴 ((𝑘𝐴𝐵)‘𝑗) = ∏𝑘𝐴 𝐵

Theoremfprodf1o 14515* Re-index a finite product using a bijection. (Contributed by Scott Fenton, 7-Dec-2017.)
(𝑘 = 𝐺𝐵 = 𝐷)    &   (𝜑𝐶 ∈ Fin)    &   (𝜑𝐹:𝐶1-1-onto𝐴)    &   ((𝜑𝑛𝐶) → (𝐹𝑛) = 𝐺)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑛𝐶 𝐷)

Theoremprodss 14516* Change the index set to a subset in an upper integer product. (Contributed by Scott Fenton, 11-Dec-2017.)
(𝜑𝐴𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝜑 → ∃𝑛 ∈ (ℤ𝑀)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ (ℤ𝑀) ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦))    &   ((𝜑𝑘 ∈ (𝐵𝐴)) → 𝐶 = 1)    &   (𝜑𝐵 ⊆ (ℤ𝑀))       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)

Theoremfprodss 14517* Change the index set to a subset in a finite sum. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑𝐴𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   ((𝜑𝑘 ∈ (𝐵𝐴)) → 𝐶 = 1)    &   (𝜑𝐵 ∈ Fin)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)

Theoremfprodser 14518* A finite product expressed in terms of a partial product of an infinite sequence. The recursive definition of a finite product follows from here. (Contributed by Scott Fenton, 14-Dec-2017.)
((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐹𝑘) = 𝐴)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (seq𝑀( · , 𝐹)‘𝑁))

Theoremfprodcl2lem 14519* Finite product closure lemma. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝑆 ⊆ ℂ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵𝑆)    &   (𝜑𝐴 ≠ ∅)       (𝜑 → ∏𝑘𝐴 𝐵𝑆)

Theoremfprodcllem 14520* Finite product closure lemma. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝑆 ⊆ ℂ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵𝑆)    &   (𝜑 → 1 ∈ 𝑆)       (𝜑 → ∏𝑘𝐴 𝐵𝑆)

Theoremfprodcl 14521* Closure of a finite product of complex numbers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℂ)

Theoremfprodrecl 14522* Closure of a finite product of real numbers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℝ)

Theoremfprodzcl 14523* Closure of a finite product of integers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℤ)

Theoremfprodnncl 14524* Closure of a finite product of positive integers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℕ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℕ)

Theoremfprodrpcl 14525* Closure of a finite product of positive reals. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ+)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℝ+)

Theoremfprodnn0cl 14526* Closure of a finite product of nonnegative integers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℕ0)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℕ0)

Theoremfprodcllemf 14527* Finite product closure lemma. A version of fprodcllem 14520 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝑆 ⊆ ℂ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵𝑆)    &   (𝜑 → 1 ∈ 𝑆)       (𝜑 → ∏𝑘𝐴 𝐵𝑆)

Theoremfprodreclf 14528* Closure of a finite product of real numbers. A version of fprodrecl 14522 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℝ)

Theoremfprodmul 14529* The product of two finite products. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 (𝐵 · 𝐶) = (∏𝑘𝐴 𝐵 · ∏𝑘𝐴 𝐶))

Theoremfproddiv 14530* The quotient of two finite products. (Contributed by Scott Fenton, 15-Jan-2018.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ≠ 0)       (𝜑 → ∏𝑘𝐴 (𝐵 / 𝐶) = (∏𝑘𝐴 𝐵 / ∏𝑘𝐴 𝐶))

Theoremprodsn 14531* A product of a singleton is the term. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀𝑉𝐵 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝐵)

Theoremfprod1 14532* A finite product of only one term is the term itself. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀 ∈ ℤ ∧ 𝐵 ∈ ℂ) → ∏𝑘 ∈ (𝑀...𝑀)𝐴 = 𝐵)

Theoremprodsnf 14533* A product of a singleton is the term. A version of prodsn 14531 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝐵    &   (𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀𝑉𝐵 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝐵)

Theoremclimprod1 14534 The limit of a product over one. (Contributed by Scott Fenton, 15-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)       (𝜑 → seq𝑀( · , (𝑍 × {1})) ⇝ 1)

Theoremfprodsplit 14535* Split a finite product into two parts. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑 → (𝐴𝐵) = ∅)    &   (𝜑𝑈 = (𝐴𝐵))    &   (𝜑𝑈 ∈ Fin)    &   ((𝜑𝑘𝑈) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝑈 𝐶 = (∏𝑘𝐴 𝐶 · ∏𝑘𝐵 𝐶))

Theoremfprodm1 14536* Separate out the last term in a finite product. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑘 = 𝑁𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · 𝐵))

Theoremfprod1p 14537* Separate out the first term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑘 = 𝑀𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (𝐵 · ∏𝑘 ∈ ((𝑀 + 1)...𝑁)𝐴))

Theoremfprodp1 14538* Multiply in the last term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)    &   (𝑘 = (𝑁 + 1) → 𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · 𝐵))

Theoremfprodm1s 14539* Separate out the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · 𝑁 / 𝑘𝐴))

Theoremfprodp1s 14540* Multiply in the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · (𝑁 + 1) / 𝑘𝐴))

Theoremprodsns 14541* A product of the singleton is the term. (Contributed by Scott Fenton, 25-Dec-2017.)
((𝑀𝑉𝑀 / 𝑘𝐴 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝑀 / 𝑘𝐴)

Theoremfprodfac 14542* Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
(𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)

Theoremfprodabs 14543* The absolute value of a finite product. (Contributed by Scott Fenton, 25-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)       (𝜑 → (abs‘∏𝑘 ∈ (𝑀...𝑁)𝐴) = ∏𝑘 ∈ (𝑀...𝑁)(abs‘𝐴))

Theoremfprodeq0 14544* Anything finite product containing a zero term is itself zero. (Contributed by Scott Fenton, 27-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   ((𝜑𝑘 = 𝑁) → 𝐴 = 0)       ((𝜑𝐾 ∈ (ℤ𝑁)) → ∏𝑘 ∈ (𝑀...𝐾)𝐴 = 0)

Theoremfprodshft 14545* Shift the index of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝜑𝐾 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   ((𝜑𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑗 = (𝑘𝐾) → 𝐴 = 𝐵)       (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝑀 + 𝐾)...(𝑁 + 𝐾))𝐵)

Theoremfprodrev 14546* Reversal of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝜑𝐾 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   ((𝜑𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑗 = (𝐾𝑘) → 𝐴 = 𝐵)       (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝐾𝑁)...(𝐾𝑀))𝐵)

Theoremfprodconst 14547* The product of constant terms (𝑘 is not free in 𝐵.) (Contributed by Scott Fenton, 12-Jan-2018.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ ℂ) → ∏𝑘𝐴 𝐵 = (𝐵↑(#‘𝐴)))

Theoremfprodn0 14548* A finite product of nonzero terms is nonzero. (Contributed by Scott Fenton, 15-Jan-2018.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐵 ≠ 0)       (𝜑 → ∏𝑘𝐴 𝐵 ≠ 0)

Theoremfprod2dlem 14549* Lemma for fprod2d 14550- induction step. (Contributed by Scott Fenton, 30-Jan-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)    &   (𝜑 → ¬ 𝑦𝑥)    &   (𝜑 → (𝑥 ∪ {𝑦}) ⊆ 𝐴)    &   (𝜓 ↔ ∏𝑗𝑥𝑘𝐵 𝐶 = ∏𝑧 𝑗𝑥 ({𝑗} × 𝐵)𝐷)       ((𝜑𝜓) → ∏𝑗 ∈ (𝑥 ∪ {𝑦})∏𝑘𝐵 𝐶 = ∏𝑧 𝑗 ∈ (𝑥 ∪ {𝑦})({𝑗} × 𝐵)𝐷)

Theoremfprod2d 14550* Write a double product as a product over a two-dimensional region. Compare fsum2d 14344. (Contributed by Scott Fenton, 30-Jan-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑧 𝑗𝐴 ({𝑗} × 𝐵)𝐷)

Theoremfprodxp 14551* Combine two products into a single product over the cartesian product. (Contributed by Scott Fenton, 1-Feb-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑧 ∈ (𝐴 × 𝐵)𝐷)

Theoremfprodcnv 14552* 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)    &   (𝜑𝐴 ∈ Fin)    &   (𝜑 → Rel 𝐴)    &   ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑥𝐴 𝐵 = ∏𝑦 𝐴𝐶)

Theoremfprodcom2 14553* 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.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐶 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   (𝜑 → ((𝑗𝐴𝑘𝐵) ↔ (𝑘𝐶𝑗𝐷)))    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐸 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐸 = ∏𝑘𝐶𝑗𝐷 𝐸)

Theoremfprodcom2OLD 14554* Obsolete proof of fprodcom2 14553 as of 2-Aug-2021. (Contributed by Scott Fenton, 1-Feb-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐶 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   (𝜑 → ((𝑗𝐴𝑘𝐵) ↔ (𝑘𝐶𝑗𝐷)))    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐸 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐸 = ∏𝑘𝐶𝑗𝐷 𝐸)

Theoremfprodcom 14555* Interchange product order. (Contributed by Scott Fenton, 2-Feb-2018.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑘𝐵𝑗𝐴 𝐶)

Theoremfprod0diag 14556* Two ways to express "the product of 𝐴(𝑗, 𝑘) over the triangular region 𝑀𝑗, 𝑀𝑘, 𝑗 + 𝑘𝑁. Compare fsum0diag 14351. (Contributed by Scott Fenton, 2-Feb-2018.)
((𝜑 ∧ (𝑗 ∈ (0...𝑁) ∧ 𝑘 ∈ (0...(𝑁𝑗)))) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑗 ∈ (0...𝑁)∏𝑘 ∈ (0...(𝑁𝑗))𝐴 = ∏𝑘 ∈ (0...𝑁)∏𝑗 ∈ (0...(𝑁𝑘))𝐴)

Theoremfproddivf 14557* The quotient of two finite products. A version of fproddiv 14530 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ≠ 0)       (𝜑 → ∏𝑘𝐴 (𝐵 / 𝐶) = (∏𝑘𝐴 𝐵 / ∏𝑘𝐴 𝐶))

Theoremfprodsplitf 14558* Split a finite product into two parts. A version of fprodsplit 14535 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑 → (𝐴𝐵) = ∅)    &   (𝜑𝑈 = (𝐴𝐵))    &   (𝜑𝑈 ∈ Fin)    &   ((𝜑𝑘𝑈) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝑈 𝐶 = (∏𝑘𝐴 𝐶 · ∏𝑘𝐵 𝐶))

Theoremfprodsplitsn 14559* Separate out a term in a finite product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   𝑘𝐷    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐵𝑉)    &   (𝜑 → ¬ 𝐵𝐴)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝑘 = 𝐵𝐶 = 𝐷)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝐴 ∪ {𝐵})𝐶 = (∏𝑘𝐴 𝐶 · 𝐷))

Theoremfprodsplit1f 14560* Separate out a term in a finite product. A version of fprodsplit1 38660 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝑘𝐷)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   (𝜑𝐶𝐴)    &   ((𝜑𝑘 = 𝐶) → 𝐵 = 𝐷)       (𝜑 → ∏𝑘𝐴 𝐵 = (𝐷 · ∏𝑘 ∈ (𝐴 ∖ {𝐶})𝐵))

Theoremfprodn0f 14561* A finite product of nonzero terms is nonzero. A version of fprodn0 14548 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐵 ≠ 0)       (𝜑 → ∏𝑘𝐴 𝐵 ≠ 0)

Theoremfprodclf 14562* Closure of a finite product of complex numbers. A version of fprodcl 14521 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℂ)

Theoremfprodge0 14563* If all the terms of a finite product are nonnegative, so is the product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 0 ≤ 𝐵)       (𝜑 → 0 ≤ ∏𝑘𝐴 𝐵)

Theoremfprodeq0g 14564* Any finite product containing a zero term is itself zero. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   (𝜑𝐶𝐴)    &   ((𝜑𝑘 = 𝐶) → 𝐵 = 0)       (𝜑 → ∏𝑘𝐴 𝐵 = 0)

Theoremfprodge1 14565* If all of the terms of a finite product are larger or equal to 1, so is the product. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 1 ≤ 𝐵)       (𝜑 → 1 ≤ ∏𝑘𝐴 𝐵)

Theoremfprodle 14566* If all the terms of two finite products are nonnegative and compare, so do the two products. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 0 ≤ 𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 𝐵𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 ≤ ∏𝑘𝐴 𝐶)

Theoremfprodmodd 14567* If all factors of two finite products are equal modulo 𝑀, the products are equal modulo 𝑀. (Contributed by AV, 7-Jul-2021.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   ((𝜑𝑘𝐴) → (𝐵 mod 𝑀) = (𝐶 mod 𝑀))       (𝜑 → (∏𝑘𝐴 𝐵 mod 𝑀) = (∏𝑘𝐴 𝐶 mod 𝑀))

5.10.12.5  Infinite products

Theoremiprodclim 14568* An infinite product equals the value its sequence converges to. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝐵)       (𝜑 → ∏𝑘𝑍 𝐴 = 𝐵)

Theoremiprodclim2 14569* A converging product converges to its infinite product. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)       (𝜑 → seq𝑀( · , 𝐹) ⇝ ∏𝑘𝑍 𝐴)

Theoremiprodclim3 14570* 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 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘𝑍𝐴)) ⇝ 𝑦))    &   (𝜑𝐹 ∈ dom ⇝ )    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   ((𝜑𝑗𝑍) → (𝐹𝑗) = ∏𝑘 ∈ (𝑀...𝑗)𝐴)       (𝜑𝐹 ⇝ ∏𝑘𝑍 𝐴)

Theoremiprodcl 14571* The product of a non-trivially converging infinite sequence is a complex number. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 𝐴 ∈ ℂ)

Theoremiprodrecl 14572* The product of a non-trivially converging infinite real sequence is a real number. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℝ)       (𝜑 → ∏𝑘𝑍 𝐴 ∈ ℝ)

Theoremiprodmul 14573* Multiplication of infinite sums. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐴)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   (𝜑 → ∃𝑚𝑍𝑧(𝑧 ≠ 0 ∧ seq𝑚( · , 𝐺) ⇝ 𝑧))    &   ((𝜑𝑘𝑍) → (𝐺𝑘) = 𝐵)    &   ((𝜑𝑘𝑍) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 (𝐴 · 𝐵) = (∏𝑘𝑍 𝐴 · ∏𝑘𝑍 𝐵))

5.10.13  Falling and Rising Factorial

Syntaxcfallfac 14574 Declare the syntax for the falling factorial.
class FallFac

Syntaxcrisefac 14575 Declare the syntax for the rising factorial.
class RiseFac

Definitiondf-risefac 14576* 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))

Definitiondf-fallfac 14577* Define the falling factorial function. This is the function (𝐴 · (𝐴 − 1) · ...(𝐴𝑁)) for complex 𝐴 and nonnegative integers 𝑁. (Contributed by Scott Fenton, 5-Jan-2018.)
FallFac = (𝑥 ∈ ℂ, 𝑛 ∈ ℕ0 ↦ ∏𝑘 ∈ (0...(𝑛 − 1))(𝑥𝑘))

Theoremrisefacval 14578* 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))

Theoremfallfacval 14579* 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))(𝐴𝑘))

Theoremrisefacval2 14580* 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)))

Theoremfallfacval2 14581* 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)))

Theoremfallfacval3 14582* 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)

Theoremrisefaccllem 14583* Lemma for rising factorial closure laws. (Contributed by Scott Fenton, 5-Jan-2018.)
𝑆 ⊆ ℂ    &   1 ∈ 𝑆    &   ((𝑥𝑆𝑦𝑆) → (𝑥 · 𝑦) ∈ 𝑆)    &   ((𝐴𝑆𝑘 ∈ ℕ0) → (𝐴 + 𝑘) ∈ 𝑆)       ((𝐴𝑆𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ 𝑆)

Theoremfallfaccllem 14584* Lemma for falling factorial closure laws. (Contributed by Scott Fenton, 5-Jan-2018.)
𝑆 ⊆ ℂ    &   1 ∈ 𝑆    &   ((𝑥𝑆𝑦𝑆) → (𝑥 · 𝑦) ∈ 𝑆)    &   ((𝐴𝑆𝑘 ∈ ℕ0) → (𝐴𝑘) ∈ 𝑆)       ((𝐴𝑆𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ 𝑆)

Theoremrisefaccl 14585 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℂ)

Theoremfallfaccl 14586 Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℂ)

Theoremrerisefaccl 14587 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℝ)

Theoremrefallfaccl 14588 Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℝ)

Theoremnnrisefaccl 14589 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℕ)

Theoremzrisefaccl 14590 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℤ)

Theoremzfallfaccl 14591 Closure law for falling factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 FallFac 𝑁) ∈ ℤ)

Theoremnn0risefaccl 14592 Closure law for rising factorial. (Contributed by Scott Fenton, 5-Jan-2018.)
((𝐴 ∈ ℕ0𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℕ0)

Theoremrprisefaccl 14593 Closure law for rising factorial. (Contributed by Scott Fenton, 9-Jan-2018.)
((𝐴 ∈ ℝ+𝑁 ∈ ℕ0) → (𝐴 RiseFac 𝑁) ∈ ℝ+)

Theoremrisefallfac 14594 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)))

Theoremfallrisefac 14595 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)))

Theoremrisefall0lem 14596 Lemma for risefac0 14597 and fallfac0 14598. Show a particular set of finite integers is empty. (Contributed by Scott Fenton, 5-Jan-2018.)
(0...(0 − 1)) = ∅

Theoremrisefac0 14597 The value of the rising factorial when 𝑁 = 0. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝐴 ∈ ℂ → (𝐴 RiseFac 0) = 1)

Theoremfallfac0 14598 The value of the falling factorial when 𝑁 = 0. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝐴 ∈ ℂ → (𝐴 FallFac 0) = 1)

Theoremrisefacp1 14599 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)))

Theoremfallfacp1 14600 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) · (𝐴𝑁)))

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