Theorem rab0OLD 4339

Description: Obsolete version of rab0 4338 as of 10-Jun-2026. (Contributed by NM, 15-Oct-2003.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof shortened by JJ, 14-Jul-2021.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
rab0OLD	⊢ {𝑥 ∈ ∅ ∣ 𝜑} = ∅

Proof of Theorem rab0OLD

Step	Hyp	Ref	Expression
1		df-rab 3417	. 2 ⊢ {𝑥 ∈ ∅ ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ ∅ ∧ 𝜑)}
2		ab0 4332	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ ∅ ∧ 𝜑)} = ∅ ↔ ∀𝑥 ¬ (𝑥 ∈ ∅ ∧ 𝜑))
3		noel 4286	. . . 4 ⊢ ¬ 𝑥 ∈ ∅
4	3	intnanr 493	. . 3 ⊢ ¬ (𝑥 ∈ ∅ ∧ 𝜑)
5	2, 4	mpgbir 1832	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ ∅ ∧ 𝜑)} = ∅
6	1, 5	eqtri 2786	1 ⊢ {𝑥 ∈ ∅ ∣ 𝜑} = ∅

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401   = wceq 1570   ∈ wcel 2147  {cab 2741  {crab 3416  ∅c0 4281

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-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-dif 4050  df-nul 4282

This theorem is used by: (None)