HomeHome Intuitionistic Logic Explorer
Theorem List (p. 120 of 161)
< Previous  Next >
Bad symbols? Try the
GIF version.

Mirrors  >  Metamath Home Page  >  ILE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

Theorem List for Intuitionistic Logic Explorer - 11901-12000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremprodfrecap 11901* The reciprocal of a finite product. (Contributed by Scott Fenton, 15-Jan-2018.) (Revised by Jim Kingdon, 24-Mar-2024.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐹𝑘) # 0)    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐺𝑘) = (1 / (𝐹𝑘)))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℂ)       (𝜑 → (seq𝑀( · , 𝐺)‘𝑁) = (1 / (seq𝑀( · , 𝐹)‘𝑁)))
 
Theoremprodfdivap 11902* The quotient of two products. (Contributed by Scott Fenton, 15-Jan-2018.) (Revised by Jim Kingdon, 24-Mar-2024.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) # 0)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐻𝑘) = ((𝐹𝑘) / (𝐺𝑘)))       (𝜑 → (seq𝑀( · , 𝐻)‘𝑁) = ((seq𝑀( · , 𝐹)‘𝑁) / (seq𝑀( · , 𝐺)‘𝑁)))
 
4.9.10.2  Non-trivial convergence
 
Theoremntrivcvgap 11903* A non-trivially converging infinite product converges. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ)       (𝜑 → seq𝑀( · , 𝐹) ∈ dom ⇝ )
 
Theoremntrivcvgap0 11904* A product that converges to a value apart from zero converges non-trivially. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   (𝜑𝑋 # 0)       (𝜑 → ∃𝑛𝑍𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))
 
4.9.10.3  Complex products
 
Syntaxcprod 11905 Extend class notation to include complex products.
class 𝑘𝐴 𝐵
 
Definitiondf-proddc 11906* Define the product of a series with an index set of integers 𝐴. This definition takes most of the aspects of df-sumdc 11709 and adapts them for multiplication instead of addition. However, we insist that in the infinite case, there is a nonzero tail of the sequence. This ensures that the convergence criteria match those of infinite sums. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 21-Mar-2024.)
𝑘𝐴 𝐵 = (℩𝑥(∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ𝑚) ∧ ∀𝑗 ∈ (ℤ𝑚)DECID 𝑗𝐴) ∧ (∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))) ⇝ 𝑥)) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ if(𝑛𝑚, (𝑓𝑛) / 𝑘𝐵, 1)))‘𝑚))))
 
Theoremprodeq1f 11907 Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
𝑘𝐴    &   𝑘𝐵       (𝐴 = 𝐵 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremprodeq1 11908* Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
(𝐴 = 𝐵 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremnfcprod1 11909* Bound-variable hypothesis builder for product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝑘𝐴       𝑘𝑘𝐴 𝐵
 
Theoremnfcprod 11910* Bound-variable hypothesis builder for product: if 𝑥 is (effectively) not free in 𝐴 and 𝐵, it is not free in 𝑘𝐴𝐵. (Contributed by Scott Fenton, 1-Dec-2017.)
𝑥𝐴    &   𝑥𝐵       𝑥𝑘𝐴 𝐵
 
Theoremprodeq2w 11911* Equality theorem for product, when the class expressions 𝐵 and 𝐶 are equal everywhere. Proved using only Extensionality. (Contributed by Scott Fenton, 4-Dec-2017.)
(∀𝑘 𝐵 = 𝐶 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremprodeq2 11912* Equality theorem for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(∀𝑘𝐴 𝐵 = 𝐶 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremcbvprod 11913* Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝑗 = 𝑘𝐵 = 𝐶)    &   𝑘𝐴    &   𝑗𝐴    &   𝑘𝐵    &   𝑗𝐶       𝑗𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremcbvprodv 11914* Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝑗 = 𝑘𝐵 = 𝐶)       𝑗𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremcbvprodi 11915* Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝑘𝐵    &   𝑗𝐶    &   (𝑗 = 𝑘𝐵 = 𝐶)       𝑗𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremprodeq1i 11916* Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐴 = 𝐵       𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶
 
Theoremprodeq2i 11917* Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝑘𝐴𝐵 = 𝐶)       𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremprodeq12i 11918* Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐴 = 𝐵    &   (𝑘𝐴𝐶 = 𝐷)       𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐷
 
Theoremprodeq1d 11919* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐴 = 𝐵)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremprodeq2d 11920* Equality deduction for product. Note that unlike prodeq2dv 11921, 𝑘 may occur in 𝜑. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑 → ∀𝑘𝐴 𝐵 = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremprodeq2dv 11921* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
((𝜑𝑘𝐴) → 𝐵 = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremprodeq2sdv 11922* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐵 = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theorem2cprodeq2dv 11923* Equality deduction for double product. (Contributed by Scott Fenton, 4-Dec-2017.)
((𝜑𝑗𝐴𝑘𝐵) → 𝐶 = 𝐷)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑗𝐴𝑘𝐵 𝐷)
 
Theoremprodeq12dv 11924* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐴 = 𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 = 𝐷)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐷)
 
Theoremprodeq12rdv 11925* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐴 = 𝐵)    &   ((𝜑𝑘𝐵) → 𝐶 = 𝐷)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐷)
 
Theoremprodrbdclem 11926* Lemma for prodrbdc 11929. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 4-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   (𝜑𝑁 ∈ (ℤ𝑀))       ((𝜑𝐴 ⊆ (ℤ𝑁)) → (seq𝑀( · , 𝐹) ↾ (ℤ𝑁)) = seq𝑁( · , 𝐹))
 
Theoremfproddccvg 11927* The sequence of partial products of a finite product converges to the whole product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐴 ⊆ (𝑀...𝑁))       (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq𝑀( · , 𝐹)‘𝑁))
 
Theoremprodrbdclem2 11928* Lemma for prodrbdc 11929. (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))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   ((𝜑𝑘 ∈ (ℤ𝑁)) → DECID 𝑘𝐴)       ((𝜑𝑁 ∈ (ℤ𝑀)) → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
 
Theoremprodrbdc 11929* 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))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   ((𝜑𝑘 ∈ (ℤ𝑁)) → DECID 𝑘𝐴)       (𝜑 → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
 
Theoremprodmodclem3 11930* Lemma for prodmodc 11933. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 11-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))    &   𝐻 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝐾𝑗) / 𝑘𝐵, 1))    &   (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ))    &   (𝜑𝑓:(1...𝑀)–1-1-onto𝐴)    &   (𝜑𝐾:(1...𝑁)–1-1-onto𝐴)       (𝜑 → (seq1( · , 𝐺)‘𝑀) = (seq1( · , 𝐻)‘𝑁))
 
Theoremprodmodclem2a 11931* Lemma for prodmodc 11933. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 11-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))    &   𝐻 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝐾𝑗) / 𝑘𝐵, 1))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴 ⊆ (ℤ𝑀))    &   (𝜑𝑓:(1...𝑁)–1-1-onto𝐴)    &   (𝜑𝐾 Isom < , < ((1...(♯‘𝐴)), 𝐴))       (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq1( · , 𝐺)‘𝑁))
 
Theoremprodmodclem2 11932* Lemma for prodmodc 11933. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 13-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))       ((𝜑 ∧ ∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ𝑚) ∧ ∀𝑗 ∈ (ℤ𝑚)DECID 𝑗𝐴) ∧ (∃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))
 
Theoremprodmodc 11933* A product has at most one limit. (Contributed by Scott Fenton, 4-Dec-2017.) (Modified by Jim Kingdon, 14-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))       (𝜑 → ∃*𝑥(∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ𝑚) ∧ ∀𝑗 ∈ (ℤ𝑚)DECID 𝑗𝐴) ∧ (∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦) ∧ seq𝑚( · , 𝐹) ⇝ 𝑥)) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , 𝐺)‘𝑚))))
 
Theoremzproddc 11934* 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)    &   (𝜑 → ∀𝑗𝑍 DECID 𝑗𝐴)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))
 
Theoremiprodap 11935* 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)))
 
Theoremzprodap0 11936* 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)    &   (𝜑 → ∀𝑗𝑍 DECID 𝑗𝐴)    &   (𝜑𝐴𝑍)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = 𝑋)
 
Theoremiprodap0 11937* 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)
 
4.9.10.4  Finite products
 
Theoremfprodseq 11938* The value of a product over a nonempty finite set. (Contributed by Scott Fenton, 6-Dec-2017.) (Revised by Jim Kingdon, 15-Jul-2024.)
(𝑘 = (𝐹𝑛) → 𝐵 = 𝐶)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝐹:(1...𝑀)–1-1-onto𝐴)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑛 ∈ (1...𝑀)) → (𝐺𝑛) = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = (seq1( · , (𝑛 ∈ ℕ ↦ if(𝑛𝑀, (𝐺𝑛), 1)))‘𝑀))
 
Theoremfprodntrivap 11939* 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 11940 A product over the empty set is one. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑘 ∈ ∅ 𝐴 = 1
 
Theoremprod1dc 11941* Any product of one over a valid set is one. (Contributed by Scott Fenton, 7-Dec-2017.) (Revised by Jim Kingdon, 5-Aug-2024.)
(((𝑀 ∈ ℤ ∧ 𝐴 ⊆ (ℤ𝑀) ∧ ∀𝑗 ∈ (ℤ𝑀)DECID 𝑗𝐴) ∨ 𝐴 ∈ Fin) → ∏𝑘𝐴 1 = 1)
 
Theoremprodfct 11942* A lemma to facilitate conversions from the function form to the class-variable form of a product. (Contributed by Scott Fenton, 7-Dec-2017.)
(∀𝑘𝐴 𝐵 ∈ ℂ → ∏𝑗𝐴 ((𝑘𝐴𝐵)‘𝑗) = ∏𝑘𝐴 𝐵)
 
Theoremfprodf1o 11943* 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)
 
Theoremprodssdc 11944* Change the index set to a subset in an upper integer product. (Contributed by Scott Fenton, 11-Dec-2017.) (Revised by Jim Kingdon, 6-Aug-2024.)
(𝜑𝐴𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝜑 → ∃𝑛 ∈ (ℤ𝑀)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘 ∈ (ℤ𝑀) ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦))    &   (𝜑 → ∀𝑗 ∈ (ℤ𝑀)DECID 𝑗𝐴)    &   (𝜑𝑀 ∈ ℤ)    &   ((𝜑𝑘 ∈ (𝐵𝐴)) → 𝐶 = 1)    &   (𝜑𝐵 ⊆ (ℤ𝑀))    &   (𝜑 → ∀𝑗 ∈ (ℤ𝑀)DECID 𝑗𝐵)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremfprodssdc 11945* Change the index set to a subset in a finite sum. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑𝐴𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝜑 → ∀𝑗𝐵 DECID 𝑗𝐴)    &   ((𝜑𝑘 ∈ (𝐵𝐴)) → 𝐶 = 1)    &   (𝜑𝐵 ∈ Fin)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremfprodmul 11946* 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))
 
Theoremprodsnf 11947* A product of a singleton is the term. A version of prodsn 11948 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)
 
Theoremprodsn 11948* 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 11949* 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)
 
Theoremclimprod1 11950 The limit of a product over one. (Contributed by Scott Fenton, 15-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)       (𝜑 → seq𝑀( · , (𝑍 × {1})) ⇝ 1)
 
Theoremfprodsplitdc 11951* Split a finite product into two parts. New proofs should use fprodsplit 11952 which is the same but with one fewer hypothesis. (Contributed by Scott Fenton, 16-Dec-2017.) (New usage is discouraged.)
(𝜑 → (𝐴𝐵) = ∅)    &   (𝜑𝑈 = (𝐴𝐵))    &   (𝜑𝑈 ∈ Fin)    &   (𝜑 → ∀𝑗𝑈 DECID 𝑗𝐴)    &   ((𝜑𝑘𝑈) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝑈 𝐶 = (∏𝑘𝐴 𝐶 · ∏𝑘𝐵 𝐶))
 
Theoremfprodsplit 11952* 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 11953* 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 11954* 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 11955* 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 11956* 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 11957* 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 11958* A product of the singleton is the term. (Contributed by Scott Fenton, 25-Dec-2017.)
((𝑀𝑉𝑀 / 𝑘𝐴 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝑀 / 𝑘𝐴)
 
Theoremfprodunsn 11959* Multiply in an additional term in a finite product. See also fprodsplitsn 11988 which is the same but with a 𝑘𝜑 hypothesis in place of the distinct variable condition between 𝜑 and 𝑘. (Contributed by Jim Kingdon, 16-Aug-2024.)
𝑘𝐷    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐵𝑉)    &   (𝜑 → ¬ 𝐵𝐴)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝑘 = 𝐵𝐶 = 𝐷)       (𝜑 → ∏𝑘 ∈ (𝐴 ∪ {𝐵})𝐶 = (∏𝑘𝐴 𝐶 · 𝐷))
 
Theoremfprodcl2lem 11960* Finite product closure lemma. (Contributed by Scott Fenton, 14-Dec-2017.) (Revised by Jim Kingdon, 17-Aug-2024.)
(𝜑𝑆 ⊆ ℂ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵𝑆)    &   (𝜑𝐴 ≠ ∅)       (𝜑 → ∏𝑘𝐴 𝐵𝑆)
 
Theoremfprodcllem 11961* 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 11962* 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 11963* 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 11964* Closure of a finite product of integers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℤ)
 
Theoremfprodnncl 11965* 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 11966* 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 11967* 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 11968* Finite product closure lemma. A version of fprodcllem 11961 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 11969* Closure of a finite product of real numbers. A version of fprodrecl 11963 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)
 
Theoremfprodfac 11970* Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
(𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
 
Theoremfprodabs 11971* 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 11972* Any 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 11973* 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 11974* 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 11975* 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)))
 
Theoremfprodap0 11976* 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)
 
Theoremfprod2dlemstep 11977* Lemma for fprod2d 11978- 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)    &   (𝜑𝑥 ∈ Fin)    &   (𝜓 ↔ ∏𝑗𝑥𝑘𝐵 𝐶 = ∏𝑧 𝑗𝑥 ({𝑗} × 𝐵)𝐷)       ((𝜑𝜓) → ∏𝑗 ∈ (𝑥 ∪ {𝑦})∏𝑘𝐵 𝐶 = ∏𝑧 𝑗 ∈ (𝑥 ∪ {𝑦})({𝑗} × 𝐵)𝐷)
 
Theoremfprod2d 11978* Write a double product as a product over a two-dimensional region. Compare fsum2d 11790. (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 11979* 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 11980* 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)
 
Theoremfprodcom2fi 11981* 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)    &   ((𝜑𝑘𝐶) → 𝐷 ∈ Fin)    &   (𝜑 → ((𝑗𝐴𝑘𝐵) ↔ (𝑘𝐶𝑗𝐷)))    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐸 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐸 = ∏𝑘𝐶𝑗𝐷 𝐸)
 
Theoremfprodcom 11982* Interchange product order. (Contributed by Scott Fenton, 2-Feb-2018.)
(𝜑𝐴 ∈ Fin)    &   (𝜑𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑘𝐵𝑗𝐴 𝐶)
 
Theoremfprod0diagfz 11983* Two ways to express "the product of 𝐴(𝑗, 𝑘) over the triangular region 𝑀𝑗, 𝑀𝑘, 𝑗 + 𝑘𝑁. Compare fisum0diag 11796. (Contributed by Scott Fenton, 2-Feb-2018.)
((𝜑 ∧ (𝑗 ∈ (0...𝑁) ∧ 𝑘 ∈ (0...(𝑁𝑗)))) → 𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → ∏𝑗 ∈ (0...𝑁)∏𝑘 ∈ (0...(𝑁𝑗))𝐴 = ∏𝑘 ∈ (0...𝑁)∏𝑗 ∈ (0...(𝑁𝑘))𝐴)
 
Theoremfprodrec 11984* The finite product of reciprocals is the reciprocal of the product. (Contributed by Jim Kingdon, 28-Aug-2024.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐵 # 0)       (𝜑 → ∏𝑘𝐴 (1 / 𝐵) = (1 / ∏𝑘𝐴 𝐵))
 
Theoremfproddivap 11985* 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))
 
Theoremfproddivapf 11986* The quotient of two finite products. A version of fproddivap 11985 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 11987* Split a finite product into two parts. A version of fprodsplit 11952 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 11988* Separate out a term in a finite product. See also fprodunsn 11959 which is the same but with a distinct variable condition in place of 𝑘𝜑. (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 11989* Separate out a term in a finite product. (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))
 
Theoremfprodclf 11990* Closure of a finite product of complex numbers. A version of fprodcl 11962 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)
 
Theoremfprodap0f 11991* A finite product of terms apart from zero is apart from zero. A version of fprodap0 11976 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.) (Revised by Jim Kingdon, 30-Aug-2024.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐵 # 0)       (𝜑 → ∏𝑘𝐴 𝐵 # 0)
 
Theoremfprodge0 11992* 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 11993* 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 11994* 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.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)    &   ((𝜑𝑘𝐴) → 1 ≤ 𝐵)       (𝜑 → 1 ≤ ∏𝑘𝐴 𝐵)
 
Theoremfprodle 11995* 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 11996* 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))
 
4.10  Elementary trigonometry
 
4.10.1  The exponential, sine, and cosine functions
 
Syntaxce 11997 Extend class notation to include the exponential function.
class exp
 
Syntaxceu 11998 Extend class notation to include Euler's constant e = 2.71828....
class e
 
Syntaxcsin 11999 Extend class notation to include the sine function.
class sin
 
Syntaxccos 12000 Extend class notation to include the cosine function.
class cos
    < Previous  Next >

Page List
Jump to page: Contents  1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12300 124 12301-12400 125 12401-12500 126 12501-12600 127 12601-12700 128 12701-12800 129 12801-12900 130 12901-13000 131 13001-13100 132 13101-13200 133 13201-13300 134 13301-13400 135 13401-13500 136 13501-13600 137 13601-13700 138 13701-13800 139 13801-13900 140 13901-14000 141 14001-14100 142 14101-14200 143 14201-14300 144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16097
  Copyright terms: Public domain < Previous  Next >