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Theorem finxpeq1 36774
Description: Equality theorem for Cartesian exponentiation. (Contributed by ML, 19-Oct-2020.)
Assertion
Ref Expression
finxpeq1 (𝑈 = 𝑉 → (𝑈↑↑𝑁) = (𝑉↑↑𝑁))

Proof of Theorem finxpeq1
Dummy variables 𝑛 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eleq2 2816 . . . . . . . . . 10 (𝑈 = 𝑉 → (𝑥𝑈𝑥𝑉))
21anbi2d 628 . . . . . . . . 9 (𝑈 = 𝑉 → ((𝑛 = 1o𝑥𝑈) ↔ (𝑛 = 1o𝑥𝑉)))
3 xpeq2 5690 . . . . . . . . . . 11 (𝑈 = 𝑉 → (V × 𝑈) = (V × 𝑉))
43eleq2d 2813 . . . . . . . . . 10 (𝑈 = 𝑉 → (𝑥 ∈ (V × 𝑈) ↔ 𝑥 ∈ (V × 𝑉)))
54ifbid 4546 . . . . . . . . 9 (𝑈 = 𝑉 → if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩) = if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))
62, 5ifbieq2d 4549 . . . . . . . 8 (𝑈 = 𝑉 → if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩)) = if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩)))
76mpoeq3dv 7484 . . . . . . 7 (𝑈 = 𝑉 → (𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))) = (𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))))
8 rdgeq1 8412 . . . . . . 7 ((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))) = (𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))) → rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩) = rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩))
97, 8syl 17 . . . . . 6 (𝑈 = 𝑉 → rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩) = rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩))
109fveq1d 6887 . . . . 5 (𝑈 = 𝑉 → (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁) = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁))
1110eqeq2d 2737 . . . 4 (𝑈 = 𝑉 → (∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁) ↔ ∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁)))
1211anbi2d 628 . . 3 (𝑈 = 𝑉 → ((𝑁 ∈ ω ∧ ∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁)) ↔ (𝑁 ∈ ω ∧ ∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁))))
1312abbidv 2795 . 2 (𝑈 = 𝑉 → {𝑦 ∣ (𝑁 ∈ ω ∧ ∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁))} = {𝑦 ∣ (𝑁 ∈ ω ∧ ∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁))})
14 df-finxp 36772 . 2 (𝑈↑↑𝑁) = {𝑦 ∣ (𝑁 ∈ ω ∧ ∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑈), ∅, if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁))}
15 df-finxp 36772 . 2 (𝑉↑↑𝑁) = {𝑦 ∣ (𝑁 ∈ ω ∧ ∅ = (rec((𝑛 ∈ ω, 𝑥 ∈ V ↦ if((𝑛 = 1o𝑥𝑉), ∅, if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))), ⟨𝑁, 𝑦⟩)‘𝑁))}
1613, 14, 153eqtr4g 2791 1 (𝑈 = 𝑉 → (𝑈↑↑𝑁) = (𝑉↑↑𝑁))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1533  wcel 2098  {cab 2703  Vcvv 3468  c0 4317  ifcif 4523  cop 4629   cuni 4902   × cxp 5667  cfv 6537  cmpo 7407  ωcom 7852  1st c1st 7972  reccrdg 8410  1oc1o 8460  ↑↑cfinxp 36771
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2697
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-ral 3056  df-rab 3427  df-v 3470  df-dif 3946  df-un 3948  df-in 3950  df-ss 3960  df-nul 4318  df-if 4524  df-sn 4624  df-pr 4626  df-op 4630  df-uni 4903  df-br 5142  df-opab 5204  df-mpt 5225  df-xp 5675  df-cnv 5677  df-co 5678  df-dm 5679  df-rn 5680  df-res 5681  df-ima 5682  df-pred 6294  df-iota 6489  df-fv 6545  df-ov 7408  df-oprab 7409  df-mpo 7410  df-frecs 8267  df-wrecs 8298  df-recs 8372  df-rdg 8411  df-finxp 36772
This theorem is referenced by: (None)
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