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Theorem finxpeq1 38143
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 2849 . . . . . . . . . 10 (𝑈 = 𝑉 → (𝑥𝑈𝑥𝑉))
21anbi2d 642 . . . . . . . . 9 (𝑈 = 𝑉 → ((𝑛 = 1o𝑥𝑈) ↔ (𝑛 = 1o𝑥𝑉)))
3 xpeq2 5676 . . . . . . . . . . 11 (𝑈 = 𝑉 → (V × 𝑈) = (V × 𝑉))
43eleq2d 2846 . . . . . . . . . 10 (𝑈 = 𝑉 → (𝑥 ∈ (V × 𝑈) ↔ 𝑥 ∈ (V × 𝑉)))
54ifbid 4506 . . . . . . . . 9 (𝑈 = 𝑉 → if(𝑥 ∈ (V × 𝑈), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩) = if(𝑥 ∈ (V × 𝑉), ⟨ 𝑛, (1st𝑥)⟩, ⟨𝑛, 𝑥⟩))
62, 5ifbieq2d 4509 . . . . . . . 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 7493 . . . . . . 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 8401 . . . . . . 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 18 . . . . . 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 6881 . . . . 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 2771 . . . 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 642 . . 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 2826 . 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 38141 . 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 38141 . 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 2820 1 (𝑈 = 𝑉 → (𝑈↑↑𝑁) = (𝑉↑↑𝑁))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401   = wceq 1570  wcel 2145  {cab 2738  Vcvv 3450  c0 4279  ifcif 4482  cop 4590   cuni 4867   × cxp 5653  cfv 6533  cmpo 7416  ωcom 7863  1st c1st 7985  reccrdg 8399  1oc1o 8451  ↑↑cfinxp 38140
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 2147  ax-9 2155  ax-ext 2732
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 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-mpt 5187  df-xp 5661  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-pred 6299  df-iota 6489  df-fv 6541  df-ov 7417  df-oprab 7418  df-mpo 7419  df-frecs 8281  df-wrecs 8312  df-recs 8361  df-rdg 8400  df-finxp 38141
This theorem is used by: (None)
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