P. 116 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11501-11600   *Has distinct variable group(s)

Type	Label	Description
Statement
4.8.10  Finite and infinite products
4.8.10.1  Product sequences
Theorem	prodf 11501*	An infinite product of complex terms is a function from an upper set of integers to ℂ. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    ⇒   ⊢ (𝜑 → seq𝑀( · , 𝐹):𝑍⟶ℂ)
Theorem	clim2prod 11502*	The limit of an infinite product with an initial segment added. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ (𝜑 → seq(𝑁 + 1)( · , 𝐹) ⇝ 𝐴)    ⇒   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ ((seq𝑀( · , 𝐹)‘𝑁) · 𝐴))
Theorem	clim2divap 11503*	The limit of an infinite product with an initial segment removed. (Contributed by Scott Fenton, 20-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝐴)    &   ⊢ (𝜑 → (seq𝑀( · , 𝐹)‘𝑁) # 0)    ⇒   ⊢ (𝜑 → seq(𝑁 + 1)( · , 𝐹) ⇝ (𝐴 / (seq𝑀( · , 𝐹)‘𝑁)))
Theorem	prod3fmul 11504*	The product of two infinite products. (Contributed by Scott Fenton, 18-Dec-2017.) (Revised by Jim Kingdon, 22-Mar-2024.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐺‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐻‘𝑘) = ((𝐹‘𝑘) · (𝐺‘𝑘)))    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐻)‘𝑁) = ((seq𝑀( · , 𝐹)‘𝑁) · (seq𝑀( · , 𝐺)‘𝑁)))
Theorem	prodf1 11505	The value of the partial products in a one-valued infinite product. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ (𝑁 ∈ 𝑍 → (seq𝑀( · , (𝑍 × {1}))‘𝑁) = 1)
Theorem	prodf1f 11506	A one-valued infinite product is equal to the constant one function. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ (𝑀 ∈ ℤ → seq𝑀( · , (𝑍 × {1})) = (𝑍 × {1}))
Theorem	prodfclim1 11507	The constant one product converges to one. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ (𝑀 ∈ ℤ → seq𝑀( · , (𝑍 × {1})) ⇝ 1)
Theorem	prodfap0 11508*	The product of finitely many terms apart from zero is apart from zero. (Contributed by Scott Fenton, 14-Jan-2018.) (Revised by Jim Kingdon, 23-Mar-2024.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) # 0)    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐹)‘𝑁) # 0)
Theorem	prodfrecap 11509*	The reciprocal of a finite product. (Contributed by Scott Fenton, 15-Jan-2018.) (Revised by Jim Kingdon, 24-Mar-2024.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) # 0)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐺‘𝑘) = (1 / (𝐹‘𝑘)))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐺‘𝑘) ∈ ℂ)    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐺)‘𝑁) = (1 / (seq𝑀( · , 𝐹)‘𝑁)))
Theorem	prodfdivap 11510*	The quotient of two products. (Contributed by Scott Fenton, 15-Jan-2018.) (Revised by Jim Kingdon, 24-Mar-2024.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐺‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐺‘𝑘) # 0)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → (𝐻‘𝑘) = ((𝐹‘𝑘) / (𝐺‘𝑘)))    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐻)‘𝑁) = ((seq𝑀( · , 𝐹)‘𝑁) / (seq𝑀( · , 𝐺)‘𝑁)))
4.8.10.2  Non-trivial convergence
Theorem	ntrivcvgap 11511*	A non-trivially converging infinite product converges. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    ⇒   ⊢ (𝜑 → seq𝑀( · , 𝐹) ∈ dom ⇝ )
Theorem	ntrivcvgap0 11512*	A product that converges to a value apart from zero converges non-trivially. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   ⊢ (𝜑 → 𝑋 # 0)    ⇒   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))
4.8.10.3  Complex products
Syntax	cprod 11513	Extend class notation to include complex products.
class ∏𝑘 ∈ 𝐴 𝐵
Definition	df-proddc 11514*	Define the product of a series with an index set of integers 𝐴. This definition takes most of the aspects of df-sumdc 11317 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.)
⊢ ∏𝑘 ∈ 𝐴 𝐵 = (℩𝑥(∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∀𝑗 ∈ (ℤ≥‘𝑚)DECID 𝑗 ∈ 𝐴) ∧ (∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))) ⇝ 𝑥)) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto→𝐴 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ if(𝑛 ≤ 𝑚, ⦋(𝑓‘𝑛) / 𝑘⦌𝐵, 1)))‘𝑚))))
Theorem	prodeq1f 11515	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ Ⅎ𝑘𝐴    &   ⊢ Ⅎ𝑘𝐵    ⇒   ⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq1 11516*	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	nfcprod1 11517*	Bound-variable hypothesis builder for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ Ⅎ𝑘𝐴    ⇒   ⊢ Ⅎ𝑘∏𝑘 ∈ 𝐴 𝐵
Theorem	nfcprod 11518*	Bound-variable hypothesis builder for product: if 𝑥 is (effectively) not free in 𝐴 and 𝐵, it is not free in ∏𝑘 ∈ 𝐴𝐵. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐵    ⇒   ⊢ Ⅎ𝑥∏𝑘 ∈ 𝐴 𝐵
Theorem	prodeq2w 11519*	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 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	prodeq2 11520*	Equality theorem for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (∀𝑘 ∈ 𝐴 𝐵 = 𝐶 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	cbvprod 11521*	Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝑗 = 𝑘 → 𝐵 = 𝐶)    &   ⊢ Ⅎ𝑘𝐴    &   ⊢ Ⅎ𝑗𝐴    &   ⊢ Ⅎ𝑘𝐵    &   ⊢ Ⅎ𝑗𝐶    ⇒   ⊢ ∏𝑗 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶
Theorem	cbvprodv 11522*	Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝑗 = 𝑘 → 𝐵 = 𝐶)    ⇒   ⊢ ∏𝑗 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶
Theorem	cbvprodi 11523*	Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ Ⅎ𝑘𝐵    &   ⊢ Ⅎ𝑗𝐶    &   ⊢ (𝑗 = 𝑘 → 𝐵 = 𝐶)    ⇒   ⊢ ∏𝑗 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶
Theorem	prodeq1i 11524*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶
Theorem	prodeq2i 11525*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝑘 ∈ 𝐴 → 𝐵 = 𝐶)    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶
Theorem	prodeq12i 11526*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐴 = 𝐵    &   ⊢ (𝑘 ∈ 𝐴 → 𝐶 = 𝐷)    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐷
Theorem	prodeq1d 11527*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq2d 11528*	Equality deduction for product. Note that unlike prodeq2dv 11529, 𝑘 may occur in 𝜑. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → ∀𝑘 ∈ 𝐴 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	prodeq2dv 11529*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	prodeq2sdv 11530*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	2cprodeq2dv 11531*	Equality deduction for double product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ ((𝜑 ∧ 𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵) → 𝐶 = 𝐷)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ 𝐴 ∏𝑘 ∈ 𝐵 𝐶 = ∏𝑗 ∈ 𝐴 ∏𝑘 ∈ 𝐵 𝐷)
Theorem	prodeq12dv 11532*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐴 = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 = 𝐷)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐷)
Theorem	prodeq12rdv 11533*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐴 = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐵) → 𝐶 = 𝐷)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐷)
Theorem	prodrbdclem 11534*	Lemma for prodrbdc 11537. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 4-Apr-2024.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → DECID 𝑘 ∈ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    ⇒   ⊢ ((𝜑 ∧ 𝐴 ⊆ (ℤ≥‘𝑁)) → (seq𝑀( · , 𝐹) ↾ (ℤ≥‘𝑁)) = seq𝑁( · , 𝐹))
Theorem	fproddccvg 11535*	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(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → DECID 𝑘 ∈ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → 𝐴 ⊆ (𝑀...𝑁))    ⇒   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq𝑀( · , 𝐹)‘𝑁))
Theorem	prodrbdclem2 11536*	Lemma for prodrbdc 11537. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐴 ⊆ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → 𝐴 ⊆ (ℤ≥‘𝑁))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → DECID 𝑘 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑁)) → DECID 𝑘 ∈ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ (ℤ≥‘𝑀)) → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
Theorem	prodrbdc 11537*	Rebase the starting point of a product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐴 ⊆ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → 𝐴 ⊆ (ℤ≥‘𝑁))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → DECID 𝑘 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑁)) → DECID 𝑘 ∈ 𝐴)    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
Theorem	prodmodclem3 11538*	Lemma for prodmodc 11541. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 11-Apr-2024.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), ⦋(𝑓‘𝑗) / 𝑘⦌𝐵, 1))    &   ⊢ 𝐻 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), ⦋(𝐾‘𝑗) / 𝑘⦌𝐵, 1))    &   ⊢ (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ))    &   ⊢ (𝜑 → 𝑓:(1...𝑀)–1-1-onto→𝐴)    &   ⊢ (𝜑 → 𝐾:(1...𝑁)–1-1-onto→𝐴)    ⇒   ⊢ (𝜑 → (seq1( · , 𝐺)‘𝑀) = (seq1( · , 𝐻)‘𝑁))
Theorem	prodmodclem2a 11539*	Lemma for prodmodc 11541. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 11-Apr-2024.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), ⦋(𝑓‘𝑗) / 𝑘⦌𝐵, 1))    &   ⊢ 𝐻 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), ⦋(𝐾‘𝑗) / 𝑘⦌𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (ℤ≥‘𝑀)) → DECID 𝑘 ∈ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝐴 ⊆ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → 𝑓:(1...𝑁)–1-1-onto→𝐴)    &   ⊢ (𝜑 → 𝐾 Isom < , < ((1...(♯‘𝐴)), 𝐴))    ⇒   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq1( · , 𝐺)‘𝑁))
Theorem	prodmodclem2 11540*	Lemma for prodmodc 11541. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 13-Apr-2024.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), ⦋(𝑓‘𝑗) / 𝑘⦌𝐵, 1))    ⇒   ⊢ ((𝜑 ∧ ∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∀𝑗 ∈ (ℤ≥‘𝑚)DECID 𝑗 ∈ 𝐴) ∧ (∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦) ∧ seq𝑚( · , 𝐹) ⇝ 𝑥))) → (∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto→𝐴 ∧ 𝑧 = (seq1( · , 𝐺)‘𝑚)) → 𝑥 = 𝑧))
Theorem	prodmodc 11541*	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(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), ⦋(𝑓‘𝑗) / 𝑘⦌𝐵, 1))    ⇒   ⊢ (𝜑 → ∃*𝑥(∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∀𝑗 ∈ (ℤ≥‘𝑚)DECID 𝑗 ∈ 𝐴) ∧ (∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦) ∧ seq𝑚( · , 𝐹) ⇝ 𝑥)) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto→𝐴 ∧ 𝑥 = (seq1( · , 𝐺)‘𝑚))))
Theorem	zproddc 11542*	Series product with index set a subset of the upper integers. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ (𝜑 → 𝐴 ⊆ 𝑍)    &   ⊢ (𝜑 → ∀𝑗 ∈ 𝑍 DECID 𝑗 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))
Theorem	iprodap 11543*	Series product with an upper integer index set (i.e. an infinite product.) (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑍 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))
Theorem	zprodap0 11544*	Nonzero series product with index set a subset of the upper integers. (Contributed by Scott Fenton, 6-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑋 # 0)    &   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   ⊢ (𝜑 → ∀𝑗 ∈ 𝑍 DECID 𝑗 ∈ 𝐴)    &   ⊢ (𝜑 → 𝐴 ⊆ 𝑍)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = 𝑋)
Theorem	iprodap0 11545*	Nonzero series product with an upper integer index set (i.e. an infinite product.) (Contributed by Scott Fenton, 6-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑋 # 0)    &   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑍 𝐵 = 𝑋)
4.8.10.4  Finite products
Theorem	fprodseq 11546*	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 = (𝐹‘𝑛) → 𝐵 = 𝐶)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝐹:(1...𝑀)–1-1-onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ (1...𝑀)) → (𝐺‘𝑛) = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = (seq1( · , (𝑛 ∈ ℕ ↦ if(𝑛 ≤ 𝑀, (𝐺‘𝑛), 1)))‘𝑀))
Theorem	fprodntrivap 11547*	A non-triviality lemma for finite sequences. (Contributed by Scott Fenton, 16-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ (𝜑 → 𝐴 ⊆ (𝑀...𝑁))    ⇒   ⊢ (𝜑 → ∃𝑛 ∈ 𝑍 ∃𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘 ∈ 𝑍 ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))) ⇝ 𝑦))
Theorem	prod0 11548	A product over the empty set is one. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ ∏𝑘 ∈ ∅ 𝐴 = 1
Theorem	prod1dc 11549*	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 ⊆ (ℤ≥‘𝑀) ∧ ∀𝑗 ∈ (ℤ≥‘𝑀)DECID 𝑗 ∈ 𝐴) ∨ 𝐴 ∈ Fin) → ∏𝑘 ∈ 𝐴 1 = 1)
Theorem	prodfct 11550*	A lemma to facilitate conversions from the function form to the class-variable form of a product. (Contributed by Scott Fenton, 7-Dec-2017.)
⊢ (∀𝑘 ∈ 𝐴 𝐵 ∈ ℂ → ∏𝑗 ∈ 𝐴 ((𝑘 ∈ 𝐴 ↦ 𝐵)‘𝑗) = ∏𝑘 ∈ 𝐴 𝐵)
Theorem	fprodf1o 11551*	Re-index a finite product using a bijection. (Contributed by Scott Fenton, 7-Dec-2017.)
⊢ (𝑘 = 𝐺 → 𝐵 = 𝐷)    &   ⊢ (𝜑 → 𝐶 ∈ Fin)    &   ⊢ (𝜑 → 𝐹:𝐶–1-1-onto→𝐴)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ 𝐶) → (𝐹‘𝑛) = 𝐺)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑛 ∈ 𝐶 𝐷)
Theorem	prodssdc 11552*	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.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → ∃𝑛 ∈ (ℤ≥‘𝑀)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘 ∈ (ℤ≥‘𝑀) ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦))    &   ⊢ (𝜑 → ∀𝑗 ∈ (ℤ≥‘𝑀)DECID 𝑗 ∈ 𝐴)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝐵 ∖ 𝐴)) → 𝐶 = 1)    &   ⊢ (𝜑 → 𝐵 ⊆ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → ∀𝑗 ∈ (ℤ≥‘𝑀)DECID 𝑗 ∈ 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	fprodssdc 11553*	Change the index set to a subset in a finite sum. (Contributed by Scott Fenton, 16-Dec-2017.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → ∀𝑗 ∈ 𝐵 DECID 𝑗 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝐵 ∖ 𝐴)) → 𝐶 = 1)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	fprodmul 11554*	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	prodsnf 11555*	A product of a singleton is the term. A version of prodsn 11556 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	prodsn 11556*	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 11557*	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	climprod1 11558	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	fprodsplitdc 11559*	Split a finite product into two parts. New proofs should use fprodsplit 11560 which is the same but with one fewer hypothesis. (Contributed by Scott Fenton, 16-Dec-2017.) (New usage is discouraged.)
⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)    &   ⊢ (𝜑 → 𝑈 = (𝐴 ∪ 𝐵))    &   ⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → ∀𝑗 ∈ 𝑈 DECID 𝑗 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑈) → 𝐶 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝑈 𝐶 = (∏𝑘 ∈ 𝐴 𝐶 · ∏𝑘 ∈ 𝐵 𝐶))
Theorem	fprodsplit 11560*	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 11561*	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 11562*	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 11563*	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 11564*	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 11565*	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 11566*	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	fprodunsn 11567*	Multiply in an additional term in a finite product. See also fprodsplitsn 11596 which is the same but with a Ⅎ𝑘𝜑 hypothesis in place of the distinct variable condition between 𝜑 and 𝑘. (Contributed by Jim Kingdon, 16-Aug-2024.)
⊢ Ⅎ𝑘𝐷    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐵 ∈ 𝑉)    &   ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝑘 = 𝐵 → 𝐶 = 𝐷)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝐴 ∪ {𝐵})𝐶 = (∏𝑘 ∈ 𝐴 𝐶 · 𝐷))
Theorem	fprodcl2lem 11568*	Finite product closure lemma. (Contributed by Scott Fenton, 14-Dec-2017.) (Revised by Jim Kingdon, 17-Aug-2024.)
⊢ (𝜑 → 𝑆 ⊆ ℂ)    &   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐴 ≠ ∅)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ 𝑆)
Theorem	fprodcllem 11569*	Finite product closure lemma. (Contributed by Scott Fenton, 14-Dec-2017.)
⊢ (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	fprodcl 11570*	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 11571*	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 11572*	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 11573*	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 11574*	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 11575*	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 11576*	Finite product closure lemma. A version of fprodcllem 11569 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 11577*	Closure of a finite product of real numbers. A version of fprodrecl 11571 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	fprodfac 11578*	Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
⊢ (𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
Theorem	fprodabs 11579*	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 11580*	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 11581*	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 11582*	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 11583*	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	fprodap0 11584*	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	fprod2dlemstep 11585*	Lemma for fprod2d 11586- 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)    &   ⊢ (𝜑 → 𝑥 ∈ Fin)    &   ⊢ (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 11586*	Write a double product as a product over a two-dimensional region. Compare fsum2d 11398. (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 11587*	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 11588*	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	fprodcom2fi 11589*	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 ∧ 𝑘 ∈ 𝐶) → 𝐷 ∈ 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 11590*	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	fprod0diagfz 11591*	Two ways to express "the product of 𝐴(𝑗, 𝑘) over the triangular region 𝑀 ≤ 𝑗, 𝑀 ≤ 𝑘, 𝑗 + 𝑘 ≤ 𝑁. Compare fisum0diag 11404. (Contributed by Scott Fenton, 2-Feb-2018.)
⊢ ((𝜑 ∧ (𝑗 ∈ (0...𝑁) ∧ 𝑘 ∈ (0...(𝑁 − 𝑗)))) → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    ⇒   ⊢ (𝜑 → ∏𝑗 ∈ (0...𝑁)∏𝑘 ∈ (0...(𝑁 − 𝑗))𝐴 = ∏𝑘 ∈ (0...𝑁)∏𝑗 ∈ (0...(𝑁 − 𝑘))𝐴)
Theorem	fprodrec 11592*	The finite product of reciprocals is the reciprocal of the product. (Contributed by Jim Kingdon, 28-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 # 0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 (1 / 𝐵) = (1 / ∏𝑘 ∈ 𝐴 𝐵))
Theorem	fproddivap 11593*	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	fproddivapf 11594*	The quotient of two finite products. A version of fproddivap 11593 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 11595*	Split a finite product into two parts. A version of fprodsplit 11560 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 11596*	Separate out a term in a finite product. See also fprodunsn 11567 which is the same but with a distinct variable condition in place of Ⅎ𝑘𝜑. (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 11597*	Separate out a term in a finite product. (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	fprodclf 11598*	Closure of a finite product of complex numbers. A version of fprodcl 11570 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	fprodap0f 11599*	A finite product of terms apart from zero is apart from zero. A version of fprodap0 11584 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.) (Revised by Jim Kingdon, 30-Aug-2024.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 # 0)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 # 0)
Theorem	fprodge0 11600*	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)