Theorem prodeq1 16108

Description: Equality theorem for a product. (Contributed by Scott Fenton, 1-Dec-2017.)

Assertion

Ref	Expression
prodeq1	⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 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.

Step	Hyp	Ref	Expression
1		sseq1 3938	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ⊆ (ℤ≥‘𝑚) ↔ 𝐵 ⊆ (ℤ≥‘𝑚)))
2		eleq2 2853	. . . . . . . . . . . . 13 ⊢ (𝐴 = 𝐵 → (𝑘 ∈ 𝐴 ↔ 𝑘 ∈ 𝐵))
3	2	ifbid 4509	. . . . . . . . . . . 12 ⊢ (𝐴 = 𝐵 → if(𝑘 ∈ 𝐴, 𝐶, 1) = if(𝑘 ∈ 𝐵, 𝐶, 1))
4	3	mpteq2dv 5203	. . . . . . . . . . 11 ⊢ (𝐴 = 𝐵 → (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1)) = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1)))
5	4	seqeq3d 14182	. . . . . . . . . 10 ⊢ (𝐴 = 𝐵 → seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) = seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))))
6	5	breq1d 5116	. . . . . . . . 9 ⊢ (𝐴 = 𝐵 → (seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦 ↔ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦))
7	6	anbi2d 642	. . . . . . . 8 ⊢ (𝐴 = 𝐵 → ((𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ↔ (𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦)))
8	7	exbidv 1956	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ↔ ∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦)))
9	8	rexbidv 3190	. . . . . 6 ⊢ (𝐴 = 𝐵 → (∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ↔ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦)))
10	4	seqeq3d 14182	. . . . . . 7 ⊢ (𝐴 = 𝐵 → seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) = seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))))
11	10	breq1d 5116	. . . . . 6 ⊢ (𝐴 = 𝐵 → (seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑥 ↔ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑥))
12	1, 9, 11	3anbi123d 1465	. . . . 5 ⊢ (𝐴 = 𝐵 → ((𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑥) ↔ (𝐵 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑥)))
13	12	rexbidv 3190	. . . 4 ⊢ (𝐴 = 𝐵 → (∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑥) ↔ ∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑥)))
14		f1oeq3 6839	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝑓:(1...𝑚)⟶1-1onto𝐴 ↔ 𝑓:(1...𝑚)⟶1-1onto𝐵))
15	14	anbi1d 643	. . . . . 6 ⊢ (𝐴 = 𝐵 → ((𝑓:(1...𝑚)⟶1-1onto𝐴 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚)) ↔ (𝑓:(1...𝑚)⟶1-1onto𝐵 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚))))
16	15	exbidv 1956	. . . . 5 ⊢ (𝐴 = 𝐵 → (∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐴 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚)) ↔ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐵 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚))))
17	16	rexbidv 3190	. . . 4 ⊢ (𝐴 = 𝐵 → (∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐴 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚)) ↔ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐵 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚))))
18	13, 17	orbi12d 932	. . 3 ⊢ (𝐴 = 𝐵 → ((∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐴 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚))) ↔ (∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐵 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚)))))
19	18	iotabidv 6544	. 2 ⊢ (𝐴 = 𝐵 → (℩𝑥(∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐴 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚)))) = (℩𝑥(∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐵 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚)))))
20		df-prod 16105	. 2 ⊢ ∏𝑘 ∈ 𝐴 𝐶 = (℩𝑥(∃𝑚 ∈ ℤ (𝐴 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐴 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚))))
21		df-prod 16105	. 2 ⊢ ∏𝑘 ∈ 𝐵 𝐶 = (℩𝑥(∃𝑚 ∈ ℤ (𝐵 ⊆ (ℤ≥‘𝑚) ∧ ∃𝑛 ∈ (ℤ≥‘𝑚)∃𝑦(𝑦 ≠ 0 ∧ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦) ∧ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑥) ∨ ∃𝑚 ∈ ℕ ∃𝑓(𝑓:(1...𝑚)⟶1-1onto𝐵 ∧ 𝑥 = (seq1( · , (𝑛 ∈ ℕ ↦ ⦋(𝑓‘𝑛) / 𝑘⦌𝐶))‘𝑚))))
22	19, 20, 21	3eqtr4g 2823	1 ⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∨ wo 861   ∧ w3a 1103   = wceq 1570  ∃wex 1812   ∈ wcel 2147   ≠ wne 2959  ∃wrex 3090  ⦋csb 3851   ⊆ wss 3900  ifcif 4485   class class class wbr 5106   ↦ cmpt 5190  ℩cio 6513  ⟶1-1ontowf1o 6560  ‘cfv 6561  (class class class)co 7446  0cc0 11230  1c1 11231   · cmul 11235  ℕcn 12365  ℤcz 12723  ℤ_≥cuz 12995  ...cfz 13669  seqcseq 14174   ⇝ cli 15683  ∏cprod 16104

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-opab 5172  df-mpt 5191  df-xp 5662  df-cnv 5664  df-co 5665  df-dm 5666  df-rn 5667  df-res 5668  df-ima 5669  df-pred 6325  df-iota 6515  df-f 6565  df-f1 6566  df-fo 6567  df-f1o 6568  df-fv 6569  df-ov 7449  df-oprab 7450  df-mpo 7451  df-frecs 8330  df-wrecs 8361  df-recs 8410  df-rdg 8449  df-seq 14175  df-prod 16105

This theorem is used by:  prodeq1d  16120  prod1  16143  fprodf1o  16145  fprodss  16147  fprodcllem  16150  fprodmul  16159  fproddiv  16160  fprodconst  16177  fprodn0  16178  fprod2d  16180  fprodmodd  16196  coprmprod  16868  coprmproddvds  16870  fprodexp  46821  fprodabs2  46822  mccl  46825  fprodcn  46827  fprodcncf  47125  dvmptfprod  47170  dvnprodlem3  47173  hoidmvval  47802  ovnhoi  47828  hspmbllem2  47852  fmtnorec2  48843