Theorem ruOLD 3789

Description: Obsolete version of ru 3788 as of 20-Jun-2025. (Contributed by NM, 7-Aug-1994.) Remove use of ax-13 2374. (Revised by BJ, 12-Oct-2019.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
ruOLD	⊢ {𝑥 ∣ 𝑥 ∉ 𝑥} ∉ V

Proof of Theorem ruOLD

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		pm5.19 386	. . . . . 6 ⊢ ¬ (𝑦 ∈ 𝑦 ↔ ¬ 𝑦 ∈ 𝑦)
2		eleq1w 2821	. . . . . . . 8 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝑦 ↔ 𝑦 ∈ 𝑦))
3		df-nel 3044	. . . . . . . . 9 ⊢ (𝑥 ∉ 𝑥 ↔ ¬ 𝑥 ∈ 𝑥)
4		id 22	. . . . . . . . . . 11 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑦)
5	4, 4	eleq12d 2832	. . . . . . . . . 10 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝑥 ↔ 𝑦 ∈ 𝑦))
6	5	notbid 318	. . . . . . . . 9 ⊢ (𝑥 = 𝑦 → (¬ 𝑥 ∈ 𝑥 ↔ ¬ 𝑦 ∈ 𝑦))
7	3, 6	bitrid 283	. . . . . . . 8 ⊢ (𝑥 = 𝑦 → (𝑥 ∉ 𝑥 ↔ ¬ 𝑦 ∈ 𝑦))
8	2, 7	bibi12d 345	. . . . . . 7 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝑦 ↔ 𝑥 ∉ 𝑥) ↔ (𝑦 ∈ 𝑦 ↔ ¬ 𝑦 ∈ 𝑦)))
9	8	spvv 1993	. . . . . 6 ⊢ (∀𝑥(𝑥 ∈ 𝑦 ↔ 𝑥 ∉ 𝑥) → (𝑦 ∈ 𝑦 ↔ ¬ 𝑦 ∈ 𝑦))
10	1, 9	mto 197	. . . . 5 ⊢ ¬ ∀𝑥(𝑥 ∈ 𝑦 ↔ 𝑥 ∉ 𝑥)
11		eqabb 2878	. . . . 5 ⊢ (𝑦 = {𝑥 ∣ 𝑥 ∉ 𝑥} ↔ ∀𝑥(𝑥 ∈ 𝑦 ↔ 𝑥 ∉ 𝑥))
12	10, 11	mtbir 323	. . . 4 ⊢ ¬ 𝑦 = {𝑥 ∣ 𝑥 ∉ 𝑥}
13	12	nex 1796	. . 3 ⊢ ¬ ∃𝑦 𝑦 = {𝑥 ∣ 𝑥 ∉ 𝑥}
14		isset 3491	. . 3 ⊢ ({𝑥 ∣ 𝑥 ∉ 𝑥} ∈ V ↔ ∃𝑦 𝑦 = {𝑥 ∣ 𝑥 ∉ 𝑥})
15	13, 14	mtbir 323	. 2 ⊢ ¬ {𝑥 ∣ 𝑥 ∉ 𝑥} ∈ V
16	15	nelir 3046	1 ⊢ {𝑥 ∣ 𝑥 ∉ 𝑥} ∉ V

Syntax hints:  ¬ wn 3   ↔ wb 206  ∀wal 1534   = wceq 1536  ∃wex 1775   ∈ wcel 2105  {cab 2711   ∉ wnel 3043  Vcvv 3477

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-10 2138  ax-11 2154  ax-12 2174  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1539  df-ex 1776  df-nf 1780  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-nel 3044  df-v 3479

This theorem is referenced by: (None)