Theorem opabn0 5535

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.

Step	Hyp	Ref	Expression
1		n0 4303	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ≠ ∅ ↔ ∃𝑧 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑})
2		elopab 5508	. . . 4 ⊢ (𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
3	2	exbii 1881	. . 3 ⊢ (∃𝑧 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ ∃𝑧∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
4		exrot3 2205	. . . 4 ⊢ (∃𝑧∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ ∃𝑥∃𝑦∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
5		opex 5439	. . . . . . 7 ⊢ ⟨𝑥, 𝑦⟩ ∈ V
6	5	isseti 3472	. . . . . 6 ⊢ ∃𝑧 𝑧 = ⟨𝑥, 𝑦⟩
7		19.41v 1984	. . . . . 6 ⊢ (∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ (∃𝑧 𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑))
8	6, 7	mpbiran 722	. . . . 5 ⊢ (∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ 𝜑)
9	8	2exbii 1882	. . . 4 ⊢ (∃𝑥∃𝑦∃𝑧(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ ∃𝑥∃𝑦𝜑)
10	4, 9	bitri 278	. . 3 ⊢ (∃𝑧∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ ∃𝑥∃𝑦𝜑)
11	3, 10	bitri 278	. 2 ⊢ (∃𝑧 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ↔ ∃𝑥∃𝑦𝜑)
12	1, 11	bitri 278	1 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ≠ ∅ ↔ ∃𝑥∃𝑦𝜑)

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2147   ≠ wne 2959  ∅c0 4281  ⟨cop 4840  {copab 5173

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-11 2197  ax-ext 2735  ax-sep 5255  ax-pr 5398

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-ne 2960  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-sn 4589  df-pr 4591  df-op 4841  df-opab 5174

This theorem is used by:  opab0  5536  csbopab  5537  dvdsrval  20642  thlle  22058  bcthlem5  25704  lgsquadlem3  27766  disjecxrn  39610  br1cosscnvxrn  39762