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Theorem List for Intuitionistic Logic Explorer - 12001-12100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremcvgratnnlemnexp 12001* Lemma for cvgratnn 12008. (Contributed by Jim Kingdon, 15-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))    &   (𝜑𝑁 ∈ ℕ)       (𝜑 → (abs‘(𝐹𝑁)) ≤ ((abs‘(𝐹‘1)) · (𝐴↑(𝑁 − 1))))
 
Theoremcvgratnnlemmn 12002* Lemma for cvgratnn 12008. (Contributed by Jim Kingdon, 15-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → (abs‘(𝐹𝑁)) ≤ ((abs‘(𝐹𝑀)) · (𝐴↑(𝑁𝑀))))
 
Theoremcvgratnnlemseq 12003* Lemma for cvgratnn 12008. (Contributed by Jim Kingdon, 21-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → ((seq1( + , 𝐹)‘𝑁) − (seq1( + , 𝐹)‘𝑀)) = Σ𝑖 ∈ ((𝑀 + 1)...𝑁)(𝐹𝑖))
 
Theoremcvgratnnlemabsle 12004* Lemma for cvgratnn 12008. (Contributed by Jim Kingdon, 21-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → (abs‘Σ𝑖 ∈ ((𝑀 + 1)...𝑁)(𝐹𝑖)) ≤ ((abs‘(𝐹𝑀)) · Σ𝑖 ∈ ((𝑀 + 1)...𝑁)(𝐴↑(𝑖𝑀))))
 
Theoremcvgratnnlemsumlt 12005* Lemma for cvgratnn 12008. (Contributed by Jim Kingdon, 23-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → Σ𝑖 ∈ ((𝑀 + 1)...𝑁)(𝐴↑(𝑖𝑀)) < (𝐴 / (1 − 𝐴)))
 
Theoremcvgratnnlemfm 12006* Lemma for cvgratnn 12008. (Contributed by Jim Kingdon, 23-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))    &   (𝜑𝑀 ∈ ℕ)       (𝜑 → (abs‘(𝐹𝑀)) < ((((1 / ((1 / 𝐴) − 1)) / 𝐴) · ((abs‘(𝐹‘1)) + 1)) / 𝑀))
 
Theoremcvgratnnlemrate 12007* Lemma for cvgratnn 12008. (Contributed by Jim Kingdon, 21-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → (abs‘((seq1( + , 𝐹)‘𝑁) − (seq1( + , 𝐹)‘𝑀))) < (((((1 / ((1 / 𝐴) − 1)) / 𝐴) · ((abs‘(𝐹‘1)) + 1)) · (𝐴 / (1 − 𝐴))) / 𝑀))
 
Theoremcvgratnn 12008* 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, then the infinite sum of the terms of 𝐹 converges to a complex number. Although this theorem is similar to cvgratz 12009 and cvgratgt0 12010, the decision to index starting at one is not merely cosmetic, as proving convergence using climcvg1n 11827 is sensitive to how a sequence is indexed. (Contributed by NM, 26-Apr-2005.) (Revised by Jim Kingdon, 12-Nov-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘 ∈ ℕ) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))       (𝜑 → seq1( + , 𝐹) ∈ dom ⇝ )
 
Theoremcvgratz 12009* 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, then the infinite sum of the terms of 𝐹 converges to a complex number. (Contributed by NM, 26-Apr-2005.) (Revised by Jim Kingdon, 11-Nov-2022.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘𝑍) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))       (𝜑 → seq𝑀( + , 𝐹) ∈ dom ⇝ )
 
Theoremcvgratgt0 12010* 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. (Contributed by NM, 26-Apr-2005.) (Revised by Jim Kingdon, 11-Nov-2022.)
𝑍 = (ℤ𝑀)    &   𝑊 = (ℤ𝑁)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑 → 0 < 𝐴)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘𝑊) → (abs‘(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (abs‘(𝐹𝑘))))       (𝜑 → seq𝑀( + , 𝐹) ∈ dom ⇝ )
 
4.9.9  Mertens' theorem
 
Theoremmertenslemub 12011* Lemma for mertensabs 12014. An upper bound for 𝑇. (Contributed by Jim Kingdon, 3-Dec-2022.)
((𝜑𝑘 ∈ ℕ0) → (𝐺𝑘) = 𝐵)    &   ((𝜑𝑘 ∈ ℕ0) → 𝐵 ∈ ℂ)    &   (𝜑 → seq0( + , 𝐺) ∈ dom ⇝ )    &   𝑇 = {𝑧 ∣ ∃𝑛 ∈ (0...(𝑆 − 1))𝑧 = (abs‘Σ𝑘 ∈ (ℤ‘(𝑛 + 1))(𝐺𝑘))}    &   (𝜑𝑋𝑇)    &   (𝜑𝑆 ∈ ℕ)       (𝜑𝑋 ≤ Σ𝑛 ∈ (0...(𝑆 − 1))(abs‘Σ𝑘 ∈ (ℤ‘(𝑛 + 1))(𝐺𝑘)))
 
Theoremmertenslemi1 12012* Lemma for mertensabs 12014. (Contributed by Mario Carneiro, 29-Apr-2014.) (Revised by Jim Kingdon, 2-Dec-2022.)
((𝜑𝑗 ∈ ℕ0) → (𝐹𝑗) = 𝐴)    &   ((𝜑𝑗 ∈ ℕ0) → (𝐾𝑗) = (abs‘𝐴))    &   ((𝜑𝑗 ∈ ℕ0) → 𝐴 ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ0) → (𝐺𝑘) = 𝐵)    &   ((𝜑𝑘 ∈ ℕ0) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ0) → (𝐻𝑘) = Σ𝑗 ∈ (0...𝑘)(𝐴 · (𝐺‘(𝑘𝑗))))    &   (𝜑 → seq0( + , 𝐾) ∈ dom ⇝ )    &   (𝜑 → seq0( + , 𝐺) ∈ dom ⇝ )    &   (𝜑𝐸 ∈ ℝ+)    &   𝑇 = {𝑧 ∣ ∃𝑛 ∈ (0...(𝑠 − 1))𝑧 = (abs‘Σ𝑘 ∈ (ℤ‘(𝑛 + 1))(𝐺𝑘))}    &   (𝜓 ↔ (𝑠 ∈ ℕ ∧ ∀𝑛 ∈ (ℤ𝑠)(abs‘Σ𝑘 ∈ (ℤ‘(𝑛 + 1))(𝐺𝑘)) < ((𝐸 / 2) / (Σ𝑗 ∈ ℕ0 (𝐾𝑗) + 1))))    &   (𝜑𝑃 ∈ ℝ)    &   (𝜑 → (𝜓 ∧ (𝑡 ∈ ℕ0 ∧ ∀𝑚 ∈ (ℤ𝑡)(𝐾𝑚) < (((𝐸 / 2) / 𝑠) / (𝑃 + 1)))))    &   (𝜑 → 0 ≤ 𝑃)    &   (𝜑 → ∀𝑤𝑇 𝑤𝑃)       (𝜑 → ∃𝑦 ∈ ℕ0𝑚 ∈ (ℤ𝑦)(abs‘Σ𝑗 ∈ (0...𝑚)(𝐴 · Σ𝑘 ∈ (ℤ‘((𝑚𝑗) + 1))𝐵)) < 𝐸)
 
Theoremmertenslem2 12013* Lemma for mertensabs 12014. (Contributed by Mario Carneiro, 28-Apr-2014.)
((𝜑𝑗 ∈ ℕ0) → (𝐹𝑗) = 𝐴)    &   ((𝜑𝑗 ∈ ℕ0) → (𝐾𝑗) = (abs‘𝐴))    &   ((𝜑𝑗 ∈ ℕ0) → 𝐴 ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ0) → (𝐺𝑘) = 𝐵)    &   ((𝜑𝑘 ∈ ℕ0) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ0) → (𝐻𝑘) = Σ𝑗 ∈ (0...𝑘)(𝐴 · (𝐺‘(𝑘𝑗))))    &   (𝜑 → seq0( + , 𝐾) ∈ dom ⇝ )    &   (𝜑 → seq0( + , 𝐺) ∈ dom ⇝ )    &   (𝜑𝐸 ∈ ℝ+)    &   𝑇 = {𝑧 ∣ ∃𝑛 ∈ (0...(𝑠 − 1))𝑧 = (abs‘Σ𝑘 ∈ (ℤ‘(𝑛 + 1))(𝐺𝑘))}    &   (𝜓 ↔ (𝑠 ∈ ℕ ∧ ∀𝑛 ∈ (ℤ𝑠)(abs‘Σ𝑘 ∈ (ℤ‘(𝑛 + 1))(𝐺𝑘)) < ((𝐸 / 2) / (Σ𝑗 ∈ ℕ0 (𝐾𝑗) + 1))))       (𝜑 → ∃𝑦 ∈ ℕ0𝑚 ∈ (ℤ𝑦)(abs‘Σ𝑗 ∈ (0...𝑚)(𝐴 · Σ𝑘 ∈ (ℤ‘((𝑚𝑗) + 1))𝐵)) < 𝐸)
 
Theoremmertensabs 12014* Mertens' theorem. If 𝐴(𝑗) is an absolutely convergent series and 𝐵(𝑘) is convergent, then 𝑗 ∈ ℕ0𝐴(𝑗) · Σ𝑘 ∈ ℕ0𝐵(𝑘)) = Σ𝑘 ∈ ℕ0Σ𝑗 ∈ (0...𝑘)(𝐴(𝑗) · 𝐵(𝑘𝑗)) (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.) (Revised by Jim Kingdon, 8-Dec-2022.)
((𝜑𝑗 ∈ ℕ0) → (𝐹𝑗) = 𝐴)    &   ((𝜑𝑗 ∈ ℕ0) → (𝐾𝑗) = (abs‘𝐴))    &   ((𝜑𝑗 ∈ ℕ0) → 𝐴 ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ0) → (𝐺𝑘) = 𝐵)    &   ((𝜑𝑘 ∈ ℕ0) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘 ∈ ℕ0) → (𝐻𝑘) = Σ𝑗 ∈ (0...𝑘)(𝐴 · (𝐺‘(𝑘𝑗))))    &   (𝜑 → seq0( + , 𝐾) ∈ dom ⇝ )    &   (𝜑 → seq0( + , 𝐺) ∈ dom ⇝ )    &   (𝜑 → seq0( + , 𝐹) ∈ dom ⇝ )       (𝜑 → seq0( + , 𝐻) ⇝ (Σ𝑗 ∈ ℕ0 𝐴 · Σ𝑘 ∈ ℕ0 𝐵))
 
4.9.10  Finite and infinite products
 
4.9.10.1  Product sequences
 
Theoremprodf 12015* An infinite product of complex terms is a function from an upper set of integers to . (Contributed by Scott Fenton, 4-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ)       (𝜑 → seq𝑀( · , 𝐹):𝑍⟶ℂ)
 
Theoremclim2prod 12016* The limit of an infinite product with an initial segment added. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ)    &   (𝜑 → seq(𝑁 + 1)( · , 𝐹) ⇝ 𝐴)       (𝜑 → seq𝑀( · , 𝐹) ⇝ ((seq𝑀( · , 𝐹)‘𝑁) · 𝐴))
 
Theoremclim2divap 12017* The limit of an infinite product with an initial segment removed. (Contributed by Scott Fenton, 20-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝐴)    &   (𝜑 → (seq𝑀( · , 𝐹)‘𝑁) # 0)       (𝜑 → seq(𝑁 + 1)( · , 𝐹) ⇝ (𝐴 / (seq𝑀( · , 𝐹)‘𝑁)))
 
Theoremprod3fmul 12018* The product of two infinite products. (Contributed by Scott Fenton, 18-Dec-2017.) (Revised by Jim Kingdon, 22-Mar-2024.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐻𝑘) = ((𝐹𝑘) · (𝐺𝑘)))       (𝜑 → (seq𝑀( · , 𝐻)‘𝑁) = ((seq𝑀( · , 𝐹)‘𝑁) · (seq𝑀( · , 𝐺)‘𝑁)))
 
Theoremprodf1 12019 The value of the partial products in a one-valued infinite product. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑍 = (ℤ𝑀)       (𝑁𝑍 → (seq𝑀( · , (𝑍 × {1}))‘𝑁) = 1)
 
Theoremprodf1f 12020 A one-valued infinite product is equal to the constant one function. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑍 = (ℤ𝑀)       (𝑀 ∈ ℤ → seq𝑀( · , (𝑍 × {1})) = (𝑍 × {1}))
 
Theoremprodfclim1 12021 The constant one product converges to one. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑍 = (ℤ𝑀)       (𝑀 ∈ ℤ → seq𝑀( · , (𝑍 × {1})) ⇝ 1)
 
Theoremprodfap0 12022* 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.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐹𝑘) # 0)       (𝜑 → (seq𝑀( · , 𝐹)‘𝑁) # 0)
 
Theoremprodfrecap 12023* The reciprocal of a finite product. (Contributed by Scott Fenton, 15-Jan-2018.) (Revised by Jim Kingdon, 24-Mar-2024.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐹𝑘) # 0)    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐺𝑘) = (1 / (𝐹𝑘)))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℂ)       (𝜑 → (seq𝑀( · , 𝐺)‘𝑁) = (1 / (seq𝑀( · , 𝐹)‘𝑁)))
 
Theoremprodfdivap 12024* The quotient of two products. (Contributed by Scott Fenton, 15-Jan-2018.) (Revised by Jim Kingdon, 24-Mar-2024.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) # 0)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐻𝑘) = ((𝐹𝑘) / (𝐺𝑘)))       (𝜑 → (seq𝑀( · , 𝐻)‘𝑁) = ((seq𝑀( · , 𝐹)‘𝑁) / (seq𝑀( · , 𝐺)‘𝑁)))
 
4.9.10.2  Non-trivial convergence
 
Theoremntrivcvgap 12025* A non-trivially converging infinite product converges. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ)       (𝜑 → seq𝑀( · , 𝐹) ∈ dom ⇝ )
 
Theoremntrivcvgap0 12026* A product that converges to a value apart from zero converges non-trivially. (Contributed by Scott Fenton, 18-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   (𝜑𝑋 # 0)       (𝜑 → ∃𝑛𝑍𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))
 
4.9.10.3  Complex products
 
Syntaxcprod 12027 Extend class notation to include complex products.
class 𝑘𝐴 𝐵
 
Definitiondf-proddc 12028* Define the product of a series with an index set of integers 𝐴. This definition takes most of the aspects of df-sumdc 11831 and adapts them for multiplication instead of addition. However, we insist that in the infinite case, there is a nonzero tail of the sequence. This ensures that the convergence criteria match those of infinite sums. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 21-Mar-2024.)
𝑘𝐴 𝐵 = (℩𝑥(∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ𝑚) ∧ ∀𝑗 ∈ (ℤ𝑚)DECID 𝑗𝐴) ∧ (∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))) ⇝ 𝑥)) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ if(𝑛𝑚, (𝑓𝑛) / 𝑘𝐵, 1)))‘𝑚))))
 
Theoremprodeq1f 12029 Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
𝑘𝐴    &   𝑘𝐵       (𝐴 = 𝐵 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremprodeq1 12030* Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
(𝐴 = 𝐵 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremnfcprod1 12031* Bound-variable hypothesis builder for product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝑘𝐴       𝑘𝑘𝐴 𝐵
 
Theoremnfcprod 12032* Bound-variable hypothesis builder for product: if 𝑥 is (effectively) not free in 𝐴 and 𝐵, it is not free in 𝑘𝐴𝐵. (Contributed by Scott Fenton, 1-Dec-2017.)
𝑥𝐴    &   𝑥𝐵       𝑥𝑘𝐴 𝐵
 
Theoremprodeq2w 12033* Equality theorem for product, when the class expressions 𝐵 and 𝐶 are equal everywhere. Proved using only Extensionality. (Contributed by Scott Fenton, 4-Dec-2017.)
(∀𝑘 𝐵 = 𝐶 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremprodeq2 12034* Equality theorem for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(∀𝑘𝐴 𝐵 = 𝐶 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremcbvprod 12035* Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝑗 = 𝑘𝐵 = 𝐶)    &   𝑘𝐴    &   𝑗𝐴    &   𝑘𝐵    &   𝑗𝐶       𝑗𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremcbvprodv 12036* Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝑗 = 𝑘𝐵 = 𝐶)       𝑗𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremcbvprodi 12037* Change bound variable in a product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝑘𝐵    &   𝑗𝐶    &   (𝑗 = 𝑘𝐵 = 𝐶)       𝑗𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremprodeq1i 12038* Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐴 = 𝐵       𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶
 
Theoremprodeq2i 12039* Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝑘𝐴𝐵 = 𝐶)       𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶
 
Theoremprodeq12i 12040* Equality inference for product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐴 = 𝐵    &   (𝑘𝐴𝐶 = 𝐷)       𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐷
 
Theoremprodeq1d 12041* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐴 = 𝐵)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremprodeq2d 12042* Equality deduction for product. Note that unlike prodeq2dv 12043, 𝑘 may occur in 𝜑. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑 → ∀𝑘𝐴 𝐵 = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremprodeq2dv 12043* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
((𝜑𝑘𝐴) → 𝐵 = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theoremprodeq2sdv 12044* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐵 = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑘𝐴 𝐶)
 
Theorem2cprodeq2dv 12045* Equality deduction for double product. (Contributed by Scott Fenton, 4-Dec-2017.)
((𝜑𝑗𝐴𝑘𝐵) → 𝐶 = 𝐷)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑗𝐴𝑘𝐵 𝐷)
 
Theoremprodeq12dv 12046* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐴 = 𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 = 𝐷)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐷)
 
Theoremprodeq12rdv 12047* Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017.)
(𝜑𝐴 = 𝐵)    &   ((𝜑𝑘𝐵) → 𝐶 = 𝐷)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐷)
 
Theoremprodrbdclem 12048* Lemma for prodrbdc 12051. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 4-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   (𝜑𝑁 ∈ (ℤ𝑀))       ((𝜑𝐴 ⊆ (ℤ𝑁)) → (seq𝑀( · , 𝐹) ↾ (ℤ𝑁)) = seq𝑁( · , 𝐹))
 
Theoremfproddccvg 12049* The sequence of partial products of a finite product converges to the whole product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐴 ⊆ (𝑀...𝑁))       (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq𝑀( · , 𝐹)‘𝑁))
 
Theoremprodrbdclem2 12050* Lemma for prodrbdc 12051. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝐴 ⊆ (ℤ𝑀))    &   (𝜑𝐴 ⊆ (ℤ𝑁))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   ((𝜑𝑘 ∈ (ℤ𝑁)) → DECID 𝑘𝐴)       ((𝜑𝑁 ∈ (ℤ𝑀)) → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
 
Theoremprodrbdc 12051* Rebase the starting point of a product. (Contributed by Scott Fenton, 4-Dec-2017.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝐴 ⊆ (ℤ𝑀))    &   (𝜑𝐴 ⊆ (ℤ𝑁))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   ((𝜑𝑘 ∈ (ℤ𝑁)) → DECID 𝑘𝐴)       (𝜑 → (seq𝑀( · , 𝐹) ⇝ 𝐶 ↔ seq𝑁( · , 𝐹) ⇝ 𝐶))
 
Theoremprodmodclem3 12052* Lemma for prodmodc 12055. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 11-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))    &   𝐻 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝐾𝑗) / 𝑘𝐵, 1))    &   (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ))    &   (𝜑𝑓:(1...𝑀)–1-1-onto𝐴)    &   (𝜑𝐾:(1...𝑁)–1-1-onto𝐴)       (𝜑 → (seq1( · , 𝐺)‘𝑀) = (seq1( · , 𝐻)‘𝑁))
 
Theoremprodmodclem2a 12053* Lemma for prodmodc 12055. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 11-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))    &   𝐻 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝐾𝑗) / 𝑘𝐵, 1))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → DECID 𝑘𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴 ⊆ (ℤ𝑀))    &   (𝜑𝑓:(1...𝑁)–1-1-onto𝐴)    &   (𝜑𝐾 Isom < , < ((1...(♯‘𝐴)), 𝐴))       (𝜑 → seq𝑀( · , 𝐹) ⇝ (seq1( · , 𝐺)‘𝑁))
 
Theoremprodmodclem2 12054* Lemma for prodmodc 12055. (Contributed by Scott Fenton, 4-Dec-2017.) (Revised by Jim Kingdon, 13-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))       ((𝜑 ∧ ∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ𝑚) ∧ ∀𝑗 ∈ (ℤ𝑚)DECID 𝑗𝐴) ∧ (∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦) ∧ seq𝑚( · , 𝐹) ⇝ 𝑥))) → (∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑧 = (seq1( · , 𝐺)‘𝑚)) → 𝑥 = 𝑧))
 
Theoremprodmodc 12055* A product has at most one limit. (Contributed by Scott Fenton, 4-Dec-2017.) (Modified by Jim Kingdon, 14-Apr-2024.)
𝐹 = (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   𝐺 = (𝑗 ∈ ℕ ↦ if(𝑗 ≤ (♯‘𝐴), (𝑓𝑗) / 𝑘𝐵, 1))       (𝜑 → ∃*𝑥(∃𝑚 ∈ ℤ ((𝐴 ⊆ (ℤ𝑚) ∧ ∀𝑗 ∈ (ℤ𝑚)DECID 𝑗𝐴) ∧ (∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦) ∧ seq𝑚( · , 𝐹) ⇝ 𝑥)) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , 𝐺)‘𝑚))))
 
Theoremzproddc 12056* Series product with index set a subset of the upper integers. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   (𝜑𝐴𝑍)    &   (𝜑 → ∀𝑗𝑍 DECID 𝑗𝐴)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))
 
Theoremiprodap 12057* Series product with an upper integer index set (i.e. an infinite product.) (Contributed by Scott Fenton, 5-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑛𝑍𝑦(𝑦 # 0 ∧ seq𝑛( · , 𝐹) ⇝ 𝑦))    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐵)    &   ((𝜑𝑘𝑍) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 𝐵 = ( ⇝ ‘seq𝑀( · , 𝐹)))
 
Theoremzprodap0 12058* Nonzero series product with index set a subset of the upper integers. (Contributed by Scott Fenton, 6-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑋 # 0)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   (𝜑 → ∀𝑗𝑍 DECID 𝑗𝐴)    &   (𝜑𝐴𝑍)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = if(𝑘𝐴, 𝐵, 1))    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = 𝑋)
 
Theoremiprodap0 12059* Nonzero series product with an upper integer index set (i.e. an infinite product.) (Contributed by Scott Fenton, 6-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑋 # 0)    &   (𝜑 → seq𝑀( · , 𝐹) ⇝ 𝑋)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = 𝐵)    &   ((𝜑𝑘𝑍) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝑍 𝐵 = 𝑋)
 
4.9.10.4  Finite products
 
Theoremfprodseq 12060* The value of a product over a nonempty finite set. (Contributed by Scott Fenton, 6-Dec-2017.) (Revised by Jim Kingdon, 15-Jul-2024.)
(𝑘 = (𝐹𝑛) → 𝐵 = 𝐶)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝐹:(1...𝑀)–1-1-onto𝐴)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑛 ∈ (1...𝑀)) → (𝐺𝑛) = 𝐶)       (𝜑 → ∏𝑘𝐴 𝐵 = (seq1( · , (𝑛 ∈ ℕ ↦ if(𝑛𝑀, (𝐺𝑛), 1)))‘𝑀))
 
Theoremfprodntrivap 12061* A non-triviality lemma for finite sequences. (Contributed by Scott Fenton, 16-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   (𝜑𝐴 ⊆ (𝑀...𝑁))       (𝜑 → ∃𝑛𝑍𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘𝑍 ↦ if(𝑘𝐴, 𝐵, 1))) ⇝ 𝑦))
 
Theoremprod0 12062 A product over the empty set is one. (Contributed by Scott Fenton, 5-Dec-2017.)
𝑘 ∈ ∅ 𝐴 = 1
 
Theoremprod1dc 12063* Any product of one over a valid set is one. (Contributed by Scott Fenton, 7-Dec-2017.) (Revised by Jim Kingdon, 5-Aug-2024.)
(((𝑀 ∈ ℤ ∧ 𝐴 ⊆ (ℤ𝑀) ∧ ∀𝑗 ∈ (ℤ𝑀)DECID 𝑗𝐴) ∨ 𝐴 ∈ Fin) → ∏𝑘𝐴 1 = 1)
 
Theoremprodfct 12064* A lemma to facilitate conversions from the function form to the class-variable form of a product. (Contributed by Scott Fenton, 7-Dec-2017.)
(∀𝑘𝐴 𝐵 ∈ ℂ → ∏𝑗𝐴 ((𝑘𝐴𝐵)‘𝑗) = ∏𝑘𝐴 𝐵)
 
Theoremfprodf1o 12065* Re-index a finite product using a bijection. (Contributed by Scott Fenton, 7-Dec-2017.)
(𝑘 = 𝐺𝐵 = 𝐷)    &   (𝜑𝐶 ∈ Fin)    &   (𝜑𝐹:𝐶1-1-onto𝐴)    &   ((𝜑𝑛𝐶) → (𝐹𝑛) = 𝐺)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 = ∏𝑛𝐶 𝐷)
 
Theoremprodssdc 12066* Change the index set to a subset in an upper integer product. (Contributed by Scott Fenton, 11-Dec-2017.) (Revised by Jim Kingdon, 6-Aug-2024.)
(𝜑𝐴𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝜑 → ∃𝑛 ∈ (ℤ𝑀)∃𝑦(𝑦 # 0 ∧ seq𝑛( · , (𝑘 ∈ (ℤ𝑀) ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦))    &   (𝜑 → ∀𝑗 ∈ (ℤ𝑀)DECID 𝑗𝐴)    &   (𝜑𝑀 ∈ ℤ)    &   ((𝜑𝑘 ∈ (𝐵𝐴)) → 𝐶 = 1)    &   (𝜑𝐵 ⊆ (ℤ𝑀))    &   (𝜑 → ∀𝑗 ∈ (ℤ𝑀)DECID 𝑗𝐵)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremfprodssdc 12067* Change the index set to a subset in a finite sum. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑𝐴𝐵)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝜑 → ∀𝑗𝐵 DECID 𝑗𝐴)    &   ((𝜑𝑘 ∈ (𝐵𝐴)) → 𝐶 = 1)    &   (𝜑𝐵 ∈ Fin)       (𝜑 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
 
Theoremfprodmul 12068* The product of two finite products. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 (𝐵 · 𝐶) = (∏𝑘𝐴 𝐵 · ∏𝑘𝐴 𝐶))
 
Theoremprodsnf 12069* A product of a singleton is the term. A version of prodsn 12070 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝐵    &   (𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀𝑉𝐵 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝐵)
 
Theoremprodsn 12070* A product of a singleton is the term. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀𝑉𝐵 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝐵)
 
Theoremfprod1 12071* A finite product of only one term is the term itself. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝑘 = 𝑀𝐴 = 𝐵)       ((𝑀 ∈ ℤ ∧ 𝐵 ∈ ℂ) → ∏𝑘 ∈ (𝑀...𝑀)𝐴 = 𝐵)
 
Theoremclimprod1 12072 The limit of a product over one. (Contributed by Scott Fenton, 15-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)       (𝜑 → seq𝑀( · , (𝑍 × {1})) ⇝ 1)
 
Theoremfprodsplitdc 12073* Split a finite product into two parts. New proofs should use fprodsplit 12074 which is the same but with one fewer hypothesis. (Contributed by Scott Fenton, 16-Dec-2017.) (New usage is discouraged.)
(𝜑 → (𝐴𝐵) = ∅)    &   (𝜑𝑈 = (𝐴𝐵))    &   (𝜑𝑈 ∈ Fin)    &   (𝜑 → ∀𝑗𝑈 DECID 𝑗𝐴)    &   ((𝜑𝑘𝑈) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝑈 𝐶 = (∏𝑘𝐴 𝐶 · ∏𝑘𝐵 𝐶))
 
Theoremfprodsplit 12074* Split a finite product into two parts. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑 → (𝐴𝐵) = ∅)    &   (𝜑𝑈 = (𝐴𝐵))    &   (𝜑𝑈 ∈ Fin)    &   ((𝜑𝑘𝑈) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑘𝑈 𝐶 = (∏𝑘𝐴 𝐶 · ∏𝑘𝐵 𝐶))
 
Theoremfprodm1 12075* Separate out the last term in a finite product. (Contributed by Scott Fenton, 16-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑘 = 𝑁𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · 𝐵))
 
Theoremfprod1p 12076* Separate out the first term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑘 = 𝑀𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (𝐵 · ∏𝑘 ∈ ((𝑀 + 1)...𝑁)𝐴))
 
Theoremfprodp1 12077* Multiply in the last term in a finite product. (Contributed by Scott Fenton, 24-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)    &   (𝑘 = (𝑁 + 1) → 𝐴 = 𝐵)       (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · 𝐵))
 
Theoremfprodm1s 12078* Separate out the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝑀...𝑁)𝐴 = (∏𝑘 ∈ (𝑀...(𝑁 − 1))𝐴 · 𝑁 / 𝑘𝐴))
 
Theoremfprodp1s 12079* Multiply in the last term in a finite product. (Contributed by Scott Fenton, 27-Dec-2017.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...(𝑁 + 1))) → 𝐴 ∈ ℂ)       (𝜑 → ∏𝑘 ∈ (𝑀...(𝑁 + 1))𝐴 = (∏𝑘 ∈ (𝑀...𝑁)𝐴 · (𝑁 + 1) / 𝑘𝐴))
 
Theoremprodsns 12080* A product of the singleton is the term. (Contributed by Scott Fenton, 25-Dec-2017.)
((𝑀𝑉𝑀 / 𝑘𝐴 ∈ ℂ) → ∏𝑘 ∈ {𝑀}𝐴 = 𝑀 / 𝑘𝐴)
 
Theoremfprodunsn 12081* Multiply in an additional term in a finite product. See also fprodsplitsn 12110 which is the same but with a 𝑘𝜑 hypothesis in place of the distinct variable condition between 𝜑 and 𝑘. (Contributed by Jim Kingdon, 16-Aug-2024.)
𝑘𝐷    &   (𝜑𝐴 ∈ Fin)    &   (𝜑𝐵𝑉)    &   (𝜑 → ¬ 𝐵𝐴)    &   ((𝜑𝑘𝐴) → 𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝑘 = 𝐵𝐶 = 𝐷)       (𝜑 → ∏𝑘 ∈ (𝐴 ∪ {𝐵})𝐶 = (∏𝑘𝐴 𝐶 · 𝐷))
 
Theoremfprodcl2lem 12082* Finite product closure lemma. (Contributed by Scott Fenton, 14-Dec-2017.) (Revised by Jim Kingdon, 17-Aug-2024.)
(𝜑𝑆 ⊆ ℂ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵𝑆)    &   (𝜑𝐴 ≠ ∅)       (𝜑 → ∏𝑘𝐴 𝐵𝑆)
 
Theoremfprodcllem 12083* Finite product closure lemma. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝑆 ⊆ ℂ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵𝑆)    &   (𝜑 → 1 ∈ 𝑆)       (𝜑 → ∏𝑘𝐴 𝐵𝑆)
 
Theoremfprodcl 12084* Closure of a finite product of complex numbers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℂ)
 
Theoremfprodrecl 12085* Closure of a finite product of real numbers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℝ)
 
Theoremfprodzcl 12086* Closure of a finite product of integers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℤ)
 
Theoremfprodnncl 12087* Closure of a finite product of positive integers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℕ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℕ)
 
Theoremfprodrpcl 12088* Closure of a finite product of positive reals. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ+)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℝ+)
 
Theoremfprodnn0cl 12089* Closure of a finite product of nonnegative integers. (Contributed by Scott Fenton, 14-Dec-2017.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℕ0)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℕ0)
 
Theoremfprodcllemf 12090* Finite product closure lemma. A version of fprodcllem 12083 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝑆 ⊆ ℂ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 · 𝑦) ∈ 𝑆)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵𝑆)    &   (𝜑 → 1 ∈ 𝑆)       (𝜑 → ∏𝑘𝐴 𝐵𝑆)
 
Theoremfprodreclf 12091* Closure of a finite product of real numbers. A version of fprodrecl 12085 using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
𝑘𝜑    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℝ)       (𝜑 → ∏𝑘𝐴 𝐵 ∈ ℝ)
 
Theoremfprodfac 12092* Factorial using product notation. (Contributed by Scott Fenton, 15-Dec-2017.)
(𝐴 ∈ ℕ0 → (!‘𝐴) = ∏𝑘 ∈ (1...𝐴)𝑘)
 
Theoremfprodabs 12093* The absolute value of a finite product. (Contributed by Scott Fenton, 25-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)       (𝜑 → (abs‘∏𝑘 ∈ (𝑀...𝑁)𝐴) = ∏𝑘 ∈ (𝑀...𝑁)(abs‘𝐴))
 
Theoremfprodeq0 12094* Any finite product containing a zero term is itself zero. (Contributed by Scott Fenton, 27-Dec-2017.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑁𝑍)    &   ((𝜑𝑘𝑍) → 𝐴 ∈ ℂ)    &   ((𝜑𝑘 = 𝑁) → 𝐴 = 0)       ((𝜑𝐾 ∈ (ℤ𝑁)) → ∏𝑘 ∈ (𝑀...𝐾)𝐴 = 0)
 
Theoremfprodshft 12095* Shift the index of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝜑𝐾 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   ((𝜑𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑗 = (𝑘𝐾) → 𝐴 = 𝐵)       (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝑀 + 𝐾)...(𝑁 + 𝐾))𝐵)
 
Theoremfprodrev 12096* Reversal of a finite product. (Contributed by Scott Fenton, 5-Jan-2018.)
(𝜑𝐾 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   ((𝜑𝑗 ∈ (𝑀...𝑁)) → 𝐴 ∈ ℂ)    &   (𝑗 = (𝐾𝑘) → 𝐴 = 𝐵)       (𝜑 → ∏𝑗 ∈ (𝑀...𝑁)𝐴 = ∏𝑘 ∈ ((𝐾𝑁)...(𝐾𝑀))𝐵)
 
Theoremfprodconst 12097* The product of constant terms (𝑘 is not free in 𝐵). (Contributed by Scott Fenton, 12-Jan-2018.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ ℂ) → ∏𝑘𝐴 𝐵 = (𝐵↑(♯‘𝐴)))
 
Theoremfprodap0 12098* A finite product of nonzero terms is nonzero. (Contributed by Scott Fenton, 15-Jan-2018.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℂ)    &   ((𝜑𝑘𝐴) → 𝐵 # 0)       (𝜑 → ∏𝑘𝐴 𝐵 # 0)
 
Theoremfprod2dlemstep 12099* Lemma for fprod2d 12100- induction step. (Contributed by Scott Fenton, 30-Jan-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)    &   (𝜑 → ¬ 𝑦𝑥)    &   (𝜑 → (𝑥 ∪ {𝑦}) ⊆ 𝐴)    &   (𝜑𝑥 ∈ Fin)    &   (𝜓 ↔ ∏𝑗𝑥𝑘𝐵 𝐶 = ∏𝑧 𝑗𝑥 ({𝑗} × 𝐵)𝐷)       ((𝜑𝜓) → ∏𝑗 ∈ (𝑥 ∪ {𝑦})∏𝑘𝐵 𝐶 = ∏𝑧 𝑗 ∈ (𝑥 ∪ {𝑦})({𝑗} × 𝐵)𝐷)
 
Theoremfprod2d 12100* Write a double product as a product over a two-dimensional region. Compare fsum2d 11912. (Contributed by Scott Fenton, 30-Jan-2018.)
(𝑧 = ⟨𝑗, 𝑘⟩ → 𝐷 = 𝐶)    &   (𝜑𝐴 ∈ Fin)    &   ((𝜑𝑗𝐴) → 𝐵 ∈ Fin)    &   ((𝜑 ∧ (𝑗𝐴𝑘𝐵)) → 𝐶 ∈ ℂ)       (𝜑 → ∏𝑗𝐴𝑘𝐵 𝐶 = ∏𝑧 𝑗𝐴 ({𝑗} × 𝐵)𝐷)
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