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Theorem prodeq1 15987
Description: Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)
Assertion
Ref Expression
prodeq1 (𝐴 = 𝐵 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
Distinct variable groups:   𝐴,𝑘   𝐵,𝑘
Allowed substitution hint:   𝐶(𝑘)

Proof of Theorem prodeq1
Dummy variables 𝑥 𝑦 𝑚 𝑛 𝑓 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 sseq1 3965 . . . . . 6 (𝐴 = 𝐵 → (𝐴 ⊆ (ℤ𝑚) ↔ 𝐵 ⊆ (ℤ𝑚)))
2 eleq2 2855 . . . . . . . . . . . . 13 (𝐴 = 𝐵 → (𝑘𝐴𝑘𝐵))
32ifbid 4516 . . . . . . . . . . . 12 (𝐴 = 𝐵 → if(𝑘𝐴, 𝐶, 1) = if(𝑘𝐵, 𝐶, 1))
43mpteq2dv 5210 . . . . . . . . . . 11 (𝐴 = 𝐵 → (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1)) = (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1)))
54seqeq3d 14065 . . . . . . . . . 10 (𝐴 = 𝐵 → seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) = seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))))
65breq1d 5124 . . . . . . . . 9 (𝐴 = 𝐵 → (seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦 ↔ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦))
76anbi2d 642 . . . . . . . 8 (𝐴 = 𝐵 → ((𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ↔ (𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦)))
87exbidv 1954 . . . . . . 7 (𝐴 = 𝐵 → (∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ↔ ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦)))
98rexbidv 3192 . . . . . 6 (𝐴 = 𝐵 → (∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ↔ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦)))
104seqeq3d 14065 . . . . . . 7 (𝐴 = 𝐵 → seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) = seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))))
1110breq1d 5124 . . . . . 6 (𝐴 = 𝐵 → (seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑥 ↔ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑥))
121, 9, 113anbi123d 1464 . . . . 5 (𝐴 = 𝐵 → ((𝐴 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑥) ↔ (𝐵 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑥)))
1312rexbidv 3192 . . . 4 (𝐴 = 𝐵 → (∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑥) ↔ ∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑥)))
14 f1oeq3 6817 . . . . . . 7 (𝐴 = 𝐵 → (𝑓:(1...𝑚)–1-1-onto𝐴𝑓:(1...𝑚)–1-1-onto𝐵))
1514anbi1d 643 . . . . . 6 (𝐴 = 𝐵 → ((𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚)) ↔ (𝑓:(1...𝑚)–1-1-onto𝐵𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚))))
1615exbidv 1954 . . . . 5 (𝐴 = 𝐵 → (∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚)) ↔ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐵𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚))))
1716rexbidv 3192 . . . 4 (𝐴 = 𝐵 → (∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚)) ↔ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐵𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚))))
1813, 17orbi12d 932 . . 3 (𝐴 = 𝐵 → ((∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚))) ↔ (∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐵𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚)))))
1918iotabidv 6527 . 2 (𝐴 = 𝐵 → (℩𝑥(∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚)))) = (℩𝑥(∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐵𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚)))))
20 df-prod 15984 . 2 𝑘𝐴 𝐶 = (℩𝑥(∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐴, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐴𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚))))
21 df-prod 15984 . 2 𝑘𝐵 𝐶 = (℩𝑥(∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ𝑚) ∧ ∃𝑛 ∈ (ℤ𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘𝐵, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)–1-1-onto𝐵𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ (𝑓𝑛) / 𝑘𝐶))‘𝑚))))
2219, 20, 213eqtr4g 2826 1 (𝐴 = 𝐵 → ∏𝑘𝐴 𝐶 = ∏𝑘𝐵 𝐶)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401  wo 861  w3a 1103   = wceq 1570  wex 1812  wcel 2146  wne 2961  wrex 3092  csb 3856  wss 3908  ifcif 4492   class class class wbr 5114  cmpt 5197  cio 6497  1-1-ontowf1o 6542  cfv 6543  (class class class)co 7423  0cc0 11118  1c1 11119   · cmul 11123  cn 12251  cz 12609  cuz 12880  ...cfz 13553  seqcseq 14057  cli 15561  cprod 15983
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-uni 4878  df-br 5115  df-opab 5179  df-mpt 5198  df-xp 5672  df-cnv 5674  df-co 5675  df-dm 5676  df-rn 5677  df-res 5678  df-ima 5679  df-pred 6309  df-iota 6499  df-f 6547  df-f1 6548  df-fo 6549  df-f1o 6550  df-fv 6551  df-ov 7426  df-oprab 7427  df-mpo 7428  df-frecs 8287  df-wrecs 8318  df-recs 8367  df-rdg 8406  df-seq 14058  df-prod 15984
This theorem is used by:  prodeq1iOLD  15997  prodeq1d  16000  prod1  16024  fprodf1o  16026  fprodss  16028  fprodcllem  16031  fprodmul  16040  fproddiv  16041  fprodconst  16058  fprodn0  16059  fprod2d  16061  fprodmodd  16077  coprmprod  16744  coprmproddvds  16746  fprodexp  46351  fprodabs2  46352  mccl  46355  fprodcn  46357  fprodcncf  46655  dvmptfprod  46700  dvnprodlem3  46703  hoidmvval  47332  ovnhoi  47358  hspmbllem2  47382  fmtnorec2  48336
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