Theorem opabn0 5532

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 4300	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ≠ ∅ ↔ ∃𝑧 𝑧 ∈ {⟨𝑥, 𝑦⟩ ∣ 𝜑})
2		elopab 5505	. . . 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 2202	. . . 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 3468	. . . . . 6 ⊢ ∃𝑧 𝑧 = ⟨𝑥, 𝑦⟩
7		19.41v 1982	. . . . . 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 2145   ≠ wne 2955  ∅c0 4279  ⟨cop 4590  {copab 5167

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-11 2194  ax-ext 2732  ax-sep 5251  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 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-sn 4585  df-pr 4587  df-op 4591  df-opab 5168

This theorem is used by:  opab0  5533  csbopab  5534  dvdsrval  20502  thlle  21910  bcthlem5  25556  lgsquadlem3  27618  disjecxrn  39160  br1cosscnvxrn  39312