P. 157 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 15601-15700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	geoisum1c 15601*	The infinite sum of 𝐴 · (𝑅↑1) + 𝐴 · (𝑅↑2)... is (𝐴 · 𝑅) / (1 − 𝑅). (Contributed by NM, 2-Nov-2007.) (Revised by Mario Carneiro, 26-Apr-2014.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑅 ∈ ℂ ∧ (abs‘𝑅) < 1) → Σ𝑘 ∈ ℕ (𝐴 · (𝑅↑𝑘)) = ((𝐴 · 𝑅) / (1 − 𝑅)))
Theorem	0.999... 15602	The recurring decimal 0.999..., which is defined as the infinite sum 0.9 + 0.09 + 0.009 + ... i.e. 9 / 10↑1 + 9 / 10↑2 + 9 / 10↑3 + ..., is exactly equal to 1, according to ZF set theory. Interestingly, about 40% of the people responding to a poll at http://forum.physorg.com/index.php?showtopic=13177 disagree. (Contributed by NM, 2-Nov-2007.) (Revised by AV, 8-Sep-2021.)
⊢ Σ𝑘 ∈ ℕ (9 / (;10↑𝑘)) = 1
Theorem	geoihalfsum 15603	Prove that the infinite geometric series of 1/2, 1/2 + 1/4 + 1/8 + ... = 1. Uses geoisum1 15600. This is a representation of .111... in binary with an infinite number of 1's. Theorem 0.999... 15602 proves a similar claim for .999... in base 10. (Contributed by David A. Wheeler, 4-Jan-2017.) (Proof shortened by AV, 9-Jul-2022.)
⊢ Σ𝑘 ∈ ℕ (1 / (2↑𝑘)) = 1
5.10.10  Ratio test for infinite series convergence
Theorem	cvgrat 15604*	Ratio test for convergence of a complex infinite series. If the ratio 𝐴 of the absolute values of successive terms in an infinite sequence 𝐹 is less than 1 for all terms beyond some index 𝐵, then the infinite sum of the terms of 𝐹 converges to a complex number. Equivalent to first part of Exercise 4 of [Gleason] p. 182. (Contributed by NM, 26-Apr-2005.) (Proof shortened by Mario Carneiro, 27-Apr-2014.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ 𝑊 = (ℤ≥‘𝑁)    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐴 < 1)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑍)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑍) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝑊) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹‘𝑘))))    ⇒   ⊢ (𝜑 → seq𝑀( + , 𝐹) ∈ dom ⇝ )
5.10.11  Mertens' theorem
Theorem	mertenslem1 15605*	Lemma for mertens 15607. (Contributed by Mario Carneiro, 29-Apr-2014.)
⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → (𝐹‘𝑗) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → (𝐾‘𝑗) = (abs‘𝐴))    &   ⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → 𝐴 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → (𝐺‘𝑘) = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → (𝐻‘𝑘) = Σ𝑗 ∈ (0...𝑘)(𝐴 · (𝐺‘(𝑘 − 𝑗))))    &   ⊢ (𝜑 → seq0( + , 𝐾) ∈ dom ⇝ )    &   ⊢ (𝜑 → seq0( + , 𝐺) ∈ dom ⇝ )    &   ⊢ (𝜑 → 𝐸 ∈ ℝ+)    &   ⊢ 𝑇 = {𝑧 ∣ ∃𝑛 ∈ (0...(𝑠 − 1))𝑧 = (abs‘Σ𝑘 ∈ (ℤ≥‘(𝑛 + 1))(𝐺‘𝑘))}    &   ⊢ (𝜓 ↔ (𝑠 ∈ ℕ ∧ ∀𝑛 ∈ (ℤ≥‘𝑠)(abs‘Σ𝑘 ∈ (ℤ≥‘(𝑛 + 1))(𝐺‘𝑘)) < ((𝐸 / 2) / (Σ𝑗 ∈ ℕ0 (𝐾‘𝑗) + 1))))    &   ⊢ (𝜑 → (𝜓 ∧ (𝑡 ∈ ℕ0 ∧ ∀𝑚 ∈ (ℤ≥‘𝑡)(𝐾‘𝑚) < (((𝐸 / 2) / 𝑠) / (sup(𝑇, ℝ, < ) + 1)))))    &   ⊢ (𝜑 → (0 ≤ sup(𝑇, ℝ, < ) ∧ (𝑇 ⊆ ℝ ∧ 𝑇 ≠ ∅ ∧ ∃𝑧 ∈ ℝ ∀𝑤 ∈ 𝑇 𝑤 ≤ 𝑧)))    ⇒   ⊢ (𝜑 → ∃𝑦 ∈ ℕ0 ∀𝑚 ∈ (ℤ≥‘𝑦)(abs‘Σ𝑗 ∈ (0...𝑚)(𝐴 · Σ𝑘 ∈ (ℤ≥‘((𝑚 − 𝑗) + 1))𝐵)) < 𝐸)
Theorem	mertenslem2 15606*	Lemma for mertens 15607. (Contributed by Mario Carneiro, 28-Apr-2014.)
⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → (𝐹‘𝑗) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → (𝐾‘𝑗) = (abs‘𝐴))    &   ⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → 𝐴 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → (𝐺‘𝑘) = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → (𝐻‘𝑘) = Σ𝑗 ∈ (0...𝑘)(𝐴 · (𝐺‘(𝑘 − 𝑗))))    &   ⊢ (𝜑 → seq0( + , 𝐾) ∈ dom ⇝ )    &   ⊢ (𝜑 → seq0( + , 𝐺) ∈ dom ⇝ )    &   ⊢ (𝜑 → 𝐸 ∈ ℝ+)    &   ⊢ 𝑇 = {𝑧 ∣ ∃𝑛 ∈ (0...(𝑠 − 1))𝑧 = (abs‘Σ𝑘 ∈ (ℤ≥‘(𝑛 + 1))(𝐺‘𝑘))}    &   ⊢ (𝜓 ↔ (𝑠 ∈ ℕ ∧ ∀𝑛 ∈ (ℤ≥‘𝑠)(abs‘Σ𝑘 ∈ (ℤ≥‘(𝑛 + 1))(𝐺‘𝑘)) < ((𝐸 / 2) / (Σ𝑗 ∈ ℕ0 (𝐾‘𝑗) + 1))))    ⇒   ⊢ (𝜑 → ∃𝑦 ∈ ℕ0 ∀𝑚 ∈ (ℤ≥‘𝑦)(abs‘Σ𝑗 ∈ (0...𝑚)(𝐴 · Σ𝑘 ∈ (ℤ≥‘((𝑚 − 𝑗) + 1))𝐵)) < 𝐸)
Theorem	mertens 15607*	Mertens' theorem. If 𝐴(𝑗) is an absolutely convergent series and 𝐵(𝑘) is convergent, then (Σ𝑗 ∈ ℕ0𝐴(𝑗) · Σ𝑘 ∈ ℕ0𝐵(𝑘)) = Σ𝑘 ∈ ℕ0Σ𝑗 ∈ (0...𝑘)(𝐴(𝑗) · 𝐵(𝑘 − 𝑗)) (and this latter series is convergent). This latter sum is commonly known as the Cauchy product of the sequences. The proof follows the outline at http://en.wikipedia.org/wiki/Cauchy_product#Proof_of_Mertens.27_theorem. (Contributed by Mario Carneiro, 29-Apr-2014.)
⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → (𝐹‘𝑗) = 𝐴)    &   ⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → (𝐾‘𝑗) = (abs‘𝐴))    &   ⊢ ((𝜑 ∧ 𝑗 ∈ ℕ0) → 𝐴 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → (𝐺‘𝑘) = 𝐵)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → 𝐵 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → (𝐻‘𝑘) = Σ𝑗 ∈ (0...𝑘)(𝐴 · (𝐺‘(𝑘 − 𝑗))))    &   ⊢ (𝜑 → seq0( + , 𝐾) ∈ dom ⇝ )    &   ⊢ (𝜑 → seq0( + , 𝐺) ∈ dom ⇝ )    ⇒   ⊢ (𝜑 → seq0( + , 𝐻) ⇝ (Σ𝑗 ∈ ℕ0 𝐴 · Σ𝑘 ∈ ℕ0 𝐵))
5.10.12  Finite and infinite products
5.10.12.1  Product sequences
Theorem	prodf 15608*	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 15609*	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	clim2div 15610*	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	prodfmul 15611*	The product of two infinite products. (Contributed by Scott Fenton, 18-Dec-2017.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐺‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐻‘𝑘) = ((𝐹‘𝑘) · (𝐺‘𝑘)))    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐻)‘𝑁) = ((seq𝑀( · , 𝐹)‘𝑁) · (seq𝑀( · , 𝐺)‘𝑁)))
Theorem	prodf1 15612	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 15613	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 15614	The constant one product converges to one. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ 𝑍 = (ℤ≥‘𝑀)    ⇒   ⊢ (𝑀 ∈ ℤ → seq𝑀( · , (𝑍 × {1})) ⇝ 1)
Theorem	prodfn0 15615*	No term of a nonzero infinite product is zero. (Contributed by Scott Fenton, 14-Jan-2018.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) ≠ 0)    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐹)‘𝑁) ≠ 0)
Theorem	prodfrec 15616*	The reciprocal of an infinite product. (Contributed by Scott Fenton, 15-Jan-2018.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) ≠ 0)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐺‘𝑘) = (1 / (𝐹‘𝑘)))    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐺)‘𝑁) = (1 / (seq𝑀( · , 𝐹)‘𝑁)))
Theorem	prodfdiv 15617*	The quotient of two infinite products. (Contributed by Scott Fenton, 15-Jan-2018.)
⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐹‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐺‘𝑘) ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐺‘𝑘) ≠ 0)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑀...𝑁)) → (𝐻‘𝑘) = ((𝐹‘𝑘) / (𝐺‘𝑘)))    ⇒   ⊢ (𝜑 → (seq𝑀( · , 𝐻)‘𝑁) = ((seq𝑀( · , 𝐹)‘𝑁) / (seq𝑀( · , 𝐺)‘𝑁)))
5.10.12.2  Non-trivial convergence
Theorem	ntrivcvg 15618*	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 15619*	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 15620*	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 15621*	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 15622*	Lemma for ntrivcvgmul 15623. (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 15623*	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 15624	Extend class notation to include complex products.
class ∏𝑘 ∈ 𝐴 𝐵
Definition	df-prod 15625*	Define the product of a series with an index set of integers 𝐴. This definition takes most of the aspects of df-sum 15407 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 15626	A product is a set. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ ∏𝑘 ∈ 𝐴 𝐵 ∈ V
Theorem	prodeq1f 15627	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ Ⅎ𝑘𝐴    &   ⊢ Ⅎ𝑘𝐵    ⇒   ⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq1 15628*	Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	nfcprod1 15629*	Bound-variable hypothesis builder for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ Ⅎ𝑘𝐴    ⇒   ⊢ Ⅎ𝑘∏𝑘 ∈ 𝐴 𝐵
Theorem	nfcprod 15630*	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 15631*	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 15632*	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 15633*	Equality theorem for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (∀𝑘 ∈ 𝐴 𝐵 = 𝐶 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	cbvprod 15634*	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 15635*	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 15636*	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 15637*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶
Theorem	prodeq2i 15638*	Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝑘 ∈ 𝐴 → 𝐵 = 𝐶)    ⇒   ⊢ ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶
Theorem	prodeq12i 15639*	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 15640*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)
Theorem	prodeq2d 15641*	Equality deduction for product. Note that unlike prodeq2dv 15642, 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 15642*	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 15643*	Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
⊢ (𝜑 → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → ∏𝑘 ∈ 𝐴 𝐵 = ∏𝑘 ∈ 𝐴 𝐶)
Theorem	2cprodeq2dv 15644*	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 15645*	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 15646*	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 15647*	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 15648*	Lemma for prodrb 15651. (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 15649*	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 15650*	Lemma for prodrb 15651. (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 15651*	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 15652*	Lemma for prodmo 15655. (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 15653*	Lemma for prodmo 15655. (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 15654*	Lemma for prodmo 15655. (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 15655*	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 15656*	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 15657*	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 15658*	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 15659*	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 15660*	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 15661*	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 15662	A product over the empty set is one. (Contributed by Scott Fenton, 5-Dec-2017.)
⊢ ∏𝑘 ∈ ∅ 𝐴 = 1
Theorem	prod1 15663*	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 15664*	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 15665*	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 15666*	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 15667*	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 15668*	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 15669*	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 15670*	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 15671*	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 15672*	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 15673*	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 15674*	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 15675*	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 15676*	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 15677*	Finite product closure lemma. A version of fprodcllem 15670 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 15678*	Closure of a finite product of real numbers. A version of fprodrecl 15672 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 15679*	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 15680*	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 15681*	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 15682*	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 15683*	A product of a singleton is the term. A version of prodsn 15681 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 15684	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 15685*	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 15686*	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 15687*	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 15688*	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 15689*	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 15690*	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 15691*	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 15692*	Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
⊢ (𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
Theorem	fprodabs 15693*	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 15694*	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 15695*	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 15696*	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 15697*	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 15698*	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 15699*	Lemma for fprod2d 15700- 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 15700*	Write a double product as a product over a two-dimensional region. Compare fsum2d 15492. (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)