P. 159 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 15801-15900   *Has distinct variable group(s)

Type	Label	Description
Statement
5.10.12.2  Non-trivial convergence
Theorem	ntrivcvg 15801*	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	ntrivcvgn0 15802*	A product that converges to a nonzero value 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))
Theorem	ntrivcvgfvn0 15803*	Any value of a product sequence that converges to a nonzero value is itself nonzero. (Contributed by Scott Fenton, 20-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐹)‘𝑁) ≠ 0)
Theorem	ntrivcvgtail 15804*	A tail of a non-trivially convergent sequence converges non-trivially. (Contributed by Scott Fenton, 20-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    ⇒   ⊢ (𝜑 → (( ⇝ ‘seq𝑁( · , 𝐹)) ≠ 0 ∧ seq𝑁( · , 𝐹) ⇝ ( ⇝ ‘seq𝑁( · , 𝐹))))
Theorem	ntrivcvgmullem 15805*	Lemma for ntrivcvgmul 15806. (Contributed by Scott Fenton, 19-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ (𝜑 → 𝑃 ∈ 𝑍)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    &   ⊢ (𝜑 → seq𝑁( · , 𝐹) ⇝ 𝑋)    &   ⊢ (𝜑 → seq𝑃( · , 𝐺) ⇝ 𝑌)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐺‘𝑘) ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ≤ 𝑃)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐻‘𝑘) = ((𝐹‘𝑘) · (𝐺‘𝑘)))    ⇒   ⊢ (𝜑 → ∃𝑞 ∈ 𝑍 ∃𝑤(𝑤 ≠ 0 ∧ seq𝑞( · , 𝐻) ⇝ 𝑤))
Theorem	ntrivcvgmul 15806*	The product of two non-trivially converging products converges non-trivially. (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 → ∃𝑚 ∈ 𝑍 ∃𝑧(𝑧 ≠ 0 ∧ seq𝑚( · , 𝐺) ⇝ 𝑧))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐺‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐻‘𝑘) = ((𝐹‘𝑘) · (𝐺‘𝑘)))    ⇒   ⊢ (𝜑 → ∃𝑝 ∈ 𝑍 ∃𝑤(𝑤 ≠ 0 ∧ seq𝑝( · , 𝐻) ⇝ 𝑤))
5.10.12.3  Complex products
Syntax	cprod 15807	Extend class notation to include complex products.
class ∏𝑘 ∈ 𝐴 𝐵
Definition	df-prod 15808*	Define the product of a series with an index set of integers 𝐴. This definition takes most of the aspects of df-sum 15591 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.)
⊢ ∏𝑘 ∈ 𝐴 𝐵 = (℩𝑥(∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃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 ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐵))‘𝑚))))
Theorem	prodex 15809	A product is a set. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ ∏𝑘 ∈ 𝐴 𝐵 ∈ V
Theorem	prodeq1f 15810	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ Ⅎ𝑘𝐴    &   ⊢ Ⅎ𝑘𝐵    ⇒   ⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq1 15811*	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	nfcprod1 15812*	Bound-variable hypothesis builder for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ Ⅎ𝑘𝐴    ⇒   ⊢ Ⅎ𝑘∏𝑘 ∈ 𝐴 𝐵
Theorem	nfcprod 15813*	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 15814*	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	prodeq2ii 15815*	Equality theorem for product, with the class expressions 𝐵 and 𝐶 guarded by I to be always sets. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (∀𝑘 ∈ 𝐴 ( I ‘𝐵) = ( I ‘𝐶) → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	prodeq2 15816*	Equality theorem for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (∀𝑘 ∈ 𝐴 𝐵 = 𝐶 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	cbvprod 15817*	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 15818*	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 15819*	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 15820	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.) Remove DV conditions. (Revised by GG, 1-Sep-2025.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶
Theorem	prodeq1iOLD 15821*	Obsolete version of prodeq1i 15820 as of 1-Sep-2025. (Contributed by Scott Fenton, 4-Dec-2017.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶
Theorem	prodeq2i 15822*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝑘 ∈ 𝐴 → 𝐵 = 𝐶)    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶
Theorem	prodeq12i 15823*	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 15824*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq2d 15825*	Equality deduction for product. Note that unlike prodeq2dv 15826, 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 15826*	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 15827*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.) Avoid axioms. (Revised by GG, 1-Sep-2025.)
⊢ (𝜑 → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	prodeq2sdvOLD 15828*	Obsolete version of prodeq2sdv 15827 as of 1-Sep-2025. (Contributed by Scott Fenton, 4-Dec-2017.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝜑 → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	2cprodeq2dv 15829*	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 15830*	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 15831*	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	prod2id 15832*	The second class argument to a product can be chosen so that it is always a set. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 ( I ‘𝐵)
Theorem	prodrblem 15833*	Lemma for prodrb 15836. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    ⇒   ⊢ ((𝜑 ∧ 𝐴 ⊆ (ℤ≥‘𝑁)) → (seq𝑀( · , 𝐹) ↾ (ℤ≥‘𝑁)) = seq𝑁( · , 𝐹))
Theorem	fprodcvg 15834*	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 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → 𝐴 ⊆ (𝑀...𝑁))    ⇒   ⊢ (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq𝑀( · , 𝐹)‘𝑁))
Theorem	prodrblem2 15835*	Lemma for prodrb 15836. (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 ∧ 𝑁 ∈ (ℤ≥‘𝑀)) → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
Theorem	prodrb 15836*	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 → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
Theorem	prodmolem3 15837*	Lemma for prodmo 15840. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ ⦋(𝑓‘𝑗) / 𝑘⦌𝐵)    &   ⊢ 𝐻 = (𝑗 ∈ ℕ ↦ ⦋(𝐾‘𝑗) / 𝑘⦌𝐵)    &   ⊢ (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	prodmolem2a 15838*	Lemma for prodmo 15840. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ ⦋(𝑓‘𝑗) / 𝑘⦌𝐵)    &   ⊢ 𝐻 = (𝑗 ∈ ℕ ↦ ⦋(𝐾‘𝑗) / 𝑘⦌𝐵)    &   ⊢ (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	prodmolem2 15839*	Lemma for prodmo 15840. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ ⦋(𝑓‘𝑗) / 𝑘⦌𝐵)    ⇒   ⊢ ((𝜑 ∧ ∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃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	prodmo 15840*	A product has at most one limit. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    &   ⊢ 𝐺 = (𝑗 ∈ ℕ ↦ ⦋(𝑓‘𝑗) / 𝑘⦌𝐵)    ⇒   ⊢ (𝜑 → ∃*𝑥(∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃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	zprod 15841*	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 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))
Theorem	iprod 15842*	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	zprodn0 15843*	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 → 𝐴 ⊆ 𝑍)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) = if(𝑘 ∈ 𝐴, 𝐵, 1))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = 𝑋)
Theorem	iprodn0 15844*	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)
5.10.12.4  Finite products
Theorem	fprod 15845*	The value of a product over a nonempty finite set. (Contributed by Scott Fenton, 6-Dec-2017.)
⊢ (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( · , 𝐺)‘𝑀))
Theorem	fprodntriv 15846*	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 15847	A product over the empty set is one. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ ∏𝑘 ∈ ∅ 𝐴 = 1
Theorem	prod1 15848*	Any product of one over a valid set is one. (Contributed by Scott Fenton, 7-Dec-2017.)
⊢ ((𝐴 ⊆ (ℤ≥‘𝑀) ∨ 𝐴 ∈ Fin) → ∏𝑘 ∈ 𝐴 1 = 1)
Theorem	prodfc 15849*	A lemma to facilitate conversions from the function form to the class-variable form of a product. (Contributed by Scott Fenton, 7-Dec-2017.)
⊢ ∏𝑗 ∈ 𝐴 ((𝑘 ∈ 𝐴 ↦ 𝐵)‘𝑗) = ∏𝑘 ∈ 𝐴 𝐵
Theorem	fprodf1o 15850*	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	prodss 15851*	Change the index set to a subset in an upper integer product. (Contributed by Scott Fenton, 11-Dec-2017.)
⊢ (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 ∧ 𝑘 ∈ (𝐵 ∖ 𝐴)) → 𝐶 = 1)    &   ⊢ (𝜑 → 𝐵 ⊆ (ℤ≥‘𝑀))    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	fprodss 15852*	Change the index set to a subset in a finite product. (Contributed by Scott Fenton, 16-Dec-2017.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝐵 ∖ 𝐴)) → 𝐶 = 1)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	fprodser 15853*	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.)
⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) = 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (seq𝑀( · , 𝐹)‘𝑁))
Theorem	fprodcl2lem 15854*	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 → 𝐴 ≠ ∅)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ∈ 𝑆)
Theorem	fprodcllem 15855*	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 15856*	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 15857*	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 15858*	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 15859*	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 15860*	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 15861*	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 15862*	Finite product closure lemma. A version of fprodcllem 15855 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 15863*	Closure of a finite product of real numbers. A version of fprodrecl 15857 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 15864*	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 15865*	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 15866*	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 15867*	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 15868*	A product of a singleton is the term. A version of prodsn 15866 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 15869	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 15870*	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 15871*	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 15872*	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 15873*	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 15874*	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 15875*	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 15876*	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 15877*	Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
⊢ (𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
Theorem	fprodabs 15878*	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 15879*	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 15880*	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 15881*	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 15882*	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 15883*	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 15884*	Lemma for fprod2d 15885- 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 15885*	Write a double product as a product over a two-dimensional region. Compare fsum2d 15675. (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 15886*	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 15887*	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 15888*	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 15889*	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 15890*	Two ways to express "the product of 𝐴(𝑗, 𝑘) over the triangular region 𝑀 ≤ 𝑗, 𝑀 ≤ 𝑘, 𝑗 + 𝑘 ≤ 𝑁. Compare fsum0diag 15681. (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 15891*	The quotient of two finite products. A version of fproddiv 15865 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 15892*	Split a finite product into two parts. A version of fprodsplit 15870 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 15893*	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 15894*	Separate out a term in a finite product. A version of fprodsplit1 45632 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 15895*	A finite product of nonzero terms is nonzero. A version of fprodn0 15883 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 15896*	Closure of a finite product of complex numbers. A version of fprodcl 15856 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 15897*	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 15898*	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 15899*	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 15900*	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)