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Theorem opabn0 5528
Description: Nonempty ordered pair class abstraction. (Contributed by NM, 10-Oct-2007.)
Assertion
Ref Expression
opabn0 ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ≠ ∅ ↔ ∃𝑥∃𝑦𝜑)

Proof of Theorem opabn0
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 n0 4299 . 2 ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ≠ ∅ ↔ ∃𝑧 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑})
2 elopab 5501 . . . 4 (𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
32exbii 1880 . . 3 (∃𝑧 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ ∃𝑧∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
4 exrot3 2201 . . . 4 (∃𝑧∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ ∃𝑥∃𝑦∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
5 opex 5432 . . . . . . 7 ⟨𝑥, 𝑦⟩ ∈ V
65isseti 3468 . . . . . 6 ∃𝑧 𝑧 = ⟨𝑥, 𝑦⟩
7 19.41v 1981 . . . . . 6 (∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ (∃𝑧 𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
86, 7mpbiran 722 . . . . 5 (∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ 𝜑)
982exbii 1881 . . . 4 (∃𝑥∃𝑦∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ ∃𝑥∃𝑦𝜑)
104, 9bitri 278 . . 3 (∃𝑧∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ ∃𝑥∃𝑦𝜑)
113, 10bitri 278 . 2 (∃𝑧 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ ∃𝑥∃𝑦𝜑)
121, 11bitri 278 1 ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ≠ ∅ ↔ ∃𝑥∃𝑦𝜑)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1569  ∃wex 1811   ∈ wcel 2144   ≠ wne 2955  ∅c0 4277  ⟨cop 4834  {copab 5166
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-11 2193  ax-ext 2732  ax-sep 5248  ax-pr 5391
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-sn 4585  df-pr 4587  df-op 4835  df-opab 5167
This theorem is used by:  opab0  5529  csbopab  5530  dvdsrval  20600  thlle  22012  bcthlem5  25658  lgsquadlem3  27718  disjecxrn  39453  br1cosscnvxrn  39605
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