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
Definition	df-proddc 11501*	Define the product of a series with an index set of integers 𝐴. This definition takes most of the aspects of df-sumdc 11304 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 11502	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ Ⅎ𝑘𝐴    &   ⊢ Ⅎ𝑘𝐵    ⇒   ⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq1 11503*	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	nfcprod1 11504*	Bound-variable hypothesis builder for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ Ⅎ𝑘𝐴    ⇒   ⊢ Ⅎ𝑘∏𝑘 ∈ 𝐴 𝐵
Theorem	nfcprod 11505*	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 11506*	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 11507*	Equality theorem for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (∀𝑘 ∈ 𝐴 𝐵 = 𝐶 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	cbvprod 11508*	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 11509*	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 11510*	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 11511*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶
Theorem	prodeq2i 11512*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝑘 ∈ 𝐴 → 𝐵 = 𝐶)    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶
Theorem	prodeq12i 11513*	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 11514*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq2d 11515*	Equality deduction for product. Note that unlike prodeq2dv 11516, 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 11516*	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 11517*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	2cprodeq2dv 11518*	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 11519*	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 11520*	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 11521*	Lemma for prodrbdc 11524. (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 11522*	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 11523*	Lemma for prodrbdc 11524. (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 11524*	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 11525*	Lemma for prodmodc 11528. (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 11526*	Lemma for prodmodc 11528. (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 11527*	Lemma for prodmodc 11528. (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 11528*	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 11529*	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 11530*	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 11531*	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 11532*	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 11533*	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 11534*	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 11535	A product over the empty set is one. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ ∏𝑘 ∈ ∅ 𝐴 = 1
Theorem	prod1dc 11536*	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 11537*	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 11538*	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 11539*	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 11540*	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 11541*	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 11542*	A product of a singleton is the term. A version of prodsn 11543 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 11543*	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 11544*	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 11545	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 11546*	Split a finite product into two parts. New proofs should use fprodsplit 11547 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 11547*	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 11548*	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 11549*	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 11550*	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 11551*	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 11552*	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 11553*	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 11554*	Multiply in an additional term in a finite product. See also fprodsplitsn 11583 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 11555*	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 11556*	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 11557*	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 11558*	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 11559*	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 11560*	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 11561*	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 11562*	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 11563*	Finite product closure lemma. A version of fprodcllem 11556 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 11564*	Closure of a finite product of real numbers. A version of fprodrecl 11558 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 11565*	Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
⊢ (𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
Theorem	fprodabs 11566*	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 11567*	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 11568*	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 11569*	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 11570*	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 11571*	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 11572*	Lemma for fprod2d 11573- 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 11573*	Write a double product as a product over a two-dimensional region. Compare fsum2d 11385. (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 11574*	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 11575*	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 11576*	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 11577*	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 11578*	Two ways to express "the product of 𝐴(𝑗, 𝑘) over the triangular region 𝑀 ≤ 𝑗, 𝑀 ≤ 𝑘, 𝑗 + 𝑘 ≤ 𝑁. Compare fisum0diag 11391. (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 11579*	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 11580*	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 11581*	The quotient of two finite products. A version of fproddivap 11580 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 11582*	Split a finite product into two parts. A version of fprodsplit 11547 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 11583*	Separate out a term in a finite product. See also fprodunsn 11554 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 11584*	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 11585*	Closure of a finite product of complex numbers. A version of fprodcl 11557 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 11586*	A finite product of terms apart from zero is apart from zero. A version of fprodap0 11571 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 11587*	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 11588*	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 11589*	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 11590*	If all the terms of two finite products are nonnegative and compare, so do the two products. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ Ⅎ𝑘𝜑    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℝ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 0 ≤ 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℝ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ≤ 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 ≤ ∏𝑘 ∈ 𝐴 𝐶)
Theorem	fprodmodd 11591*	If all factors of two finite products are equal modulo 𝑀, the products are equal modulo 𝑀. (Contributed by AV, 7-Jul-2021.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → (𝐵 mod 𝑀) = (𝐶 mod 𝑀))    ⇒   ⊢ (𝜑 → (∏𝑘 ∈ 𝐴 𝐵 mod 𝑀) = (∏𝑘 ∈ 𝐴 𝐶 mod 𝑀))
4.9  Elementary trigonometry
4.9.1  The exponential, sine, and cosine functions
Syntax	ce 11592	Extend class notation to include the exponential function.
class exp
Syntax	ceu 11593	Extend class notation to include Euler's constant e = 2.71828....
class e
Syntax	csin 11594	Extend class notation to include the sine function.
class sin
Syntax	ccos 11595	Extend class notation to include the cosine function.
class cos
Syntax	ctan 11596	Extend class notation to include the tangent function.
class tan
Syntax	cpi 11597	Extend class notation to include the constant pi, π = 3.14159....
class π
Definition	df-ef 11598*	Define the exponential function. Its value at the complex number 𝐴 is (exp‘𝐴) and is called the "exponential of 𝐴"; see efval 11611. (Contributed by NM, 14-Mar-2005.)
⊢ exp = (𝑥 ∈ ℂ ↦ Σ𝑘 ∈ ℕ0 ((𝑥↑𝑘) / (!‘𝑘)))
Definition	df-e 11599	Define Euler's constant e = 2.71828.... (Contributed by NM, 14-Mar-2005.)
⊢ e = (exp‘1)
Definition	df-sin 11600	Define the sine function. (Contributed by NM, 14-Mar-2005.)
⊢ sin = (𝑥 ∈ ℂ ↦ (((exp‘(i · 𝑥)) − (exp‘(-i · 𝑥))) / (2 · i)))