Theorem prodeq1f 16107

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

Hypotheses

Ref	Expression
prodeq1f.1	⊢ Ⅎ𝑘𝐴
prodeq1f.2	⊢ Ⅎ𝑘𝐵

Assertion

Ref	Expression
prodeq1f	⊢ (𝐴 = 𝐵 → ∏𝑘 ∈ 𝐴 𝐶 = ∏𝑘 ∈ 𝐵 𝐶)

Proof of Theorem prodeq1f

Dummy variables 𝑓 𝑚 𝑛 𝑥 𝑦 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		prodeq1f.1	. . . . . . . . . . . . 13 ⊢ Ⅎ𝑘𝐴
3		prodeq1f.2	. . . . . . . . . . . . 13 ⊢ Ⅎ𝑘𝐵
4	2, 3	nfeq 2939	. . . . . . . . . . . 12 ⊢ Ⅎ𝑘 𝐴 = 𝐵
5		eleq2 2853	. . . . . . . . . . . . . 14 ⊢ (𝐴 = 𝐵 → (𝑘 ∈ 𝐴 ↔ 𝑘 ∈ 𝐵))
6	5	ifbid 4509	. . . . . . . . . . . . 13 ⊢ (𝐴 = 𝐵 → if(𝑘 ∈ 𝐴, 𝐶, 1) = if(𝑘 ∈ 𝐵, 𝐶, 1))
7	6	adantr 486	. . . . . . . . . . . 12 ⊢ ((𝐴 = 𝐵 ∧ 𝑘 ∈ ℤ) → if(𝑘 ∈ 𝐴, 𝐶, 1) = if(𝑘 ∈ 𝐵, 𝐶, 1))
8	4, 7	mpteq2da 5201	. . . . . . . . . . 11 ⊢ (𝐴 = 𝐵 → (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1)) = (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1)))
9	8	seqeq3d 14182	. . . . . . . . . 10 ⊢ (𝐴 = 𝐵 → seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) = seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))))
10	9	breq1d 5116	. . . . . . . . 9 ⊢ (𝐴 = 𝐵 → (seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑦 ↔ seq𝑛( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑦))
11	10	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)))
12	11	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)))
13	12	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)))
14	8	seqeq3d 14182	. . . . . . 7 ⊢ (𝐴 = 𝐵 → seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) = seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))))
15	14	breq1d 5116	. . . . . 6 ⊢ (𝐴 = 𝐵 → (seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐴, 𝐶, 1))) ⇝ 𝑥 ↔ seq𝑚( · , (𝑘 ∈ ℤ ↦ if(𝑘 ∈ 𝐵, 𝐶, 1))) ⇝ 𝑥))
16	1, 13, 15	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)))
17	16	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)))
18		f1oeq3 6839	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝑓:(1...𝑚)⟶1-1onto𝐴 ↔ 𝑓:(1...𝑚)⟶1-1onto𝐵))
19	18	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))))
20	19	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))))
21	20	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))))
22	17, 21	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)))))
23	22	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)))))
24		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))))
25		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))))
26	23, 24, 25	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  Ⅎwnfc 2911   ≠ 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-10 2180  ax-11 2197  ax-12 2216  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-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  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: (None)