Theorem rabn0OLD 3959

Description: Obsolete proof of rabn0 3958 as of 16-Jul-2021. (Contributed by NM, 29-Aug-1999.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
rabn0OLD	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ≠ ∅ ↔ ∃𝑥 ∈ 𝐴 𝜑)

Proof of Theorem rabn0OLD

Step	Hyp	Ref	Expression
1		abn0 3954	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ≠ ∅ ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
2		df-rab 2921	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
3	2	neeq1i 2858	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ≠ ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ≠ ∅)
4		df-rex 2918	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
5	1, 3, 4	3bitr4i 292	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ≠ ∅ ↔ ∃𝑥 ∈ 𝐴 𝜑)

Syntax hints:   ↔ wb 196   ∧ wa 384  ∃wex 1704   ∈ wcel 1990  {cab 2608   ≠ wne 2794  ∃wrex 2913  {crab 2916  ∅c0 3915

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-rex 2918  df-rab 2921  df-v 3202  df-dif 3577  df-nul 3916

This theorem is referenced by:  rabeq0OLD  3960