Theorem rru 3704

Description: Relative version of Russell's paradox ru 3705 (which corresponds to the case 𝐴 = V).

Originally a subproof in pwnss 5141. (Contributed by Stefan O'Rear, 22-Feb-2015.) Avoid df-nel 3091. (Revised by Steven Nguyen, 23-Nov-2022.)

Assertion

Ref	Expression
rru	⊢ ¬ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ 𝐴

Distinct variable group:   𝑥,𝐴

Proof of Theorem rru

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq12 2872	. . . . 5 ⊢ ((𝑦 = {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∧ 𝑦 = {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥}) → (𝑦 ∈ 𝑦 ↔ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥}))
2	1	anidms 567	. . . 4 ⊢ (𝑦 = {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} → (𝑦 ∈ 𝑦 ↔ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥}))
3	2	notbid 319	. . 3 ⊢ (𝑦 = {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} → (¬ 𝑦 ∈ 𝑦 ↔ ¬ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥}))
4		eleq12 2872	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝑥 = 𝑦) → (𝑥 ∈ 𝑥 ↔ 𝑦 ∈ 𝑦))
5	4	anidms 567	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝑥 ↔ 𝑦 ∈ 𝑦))
6	5	notbid 319	. . . 4 ⊢ (𝑥 = 𝑦 → (¬ 𝑥 ∈ 𝑥 ↔ ¬ 𝑦 ∈ 𝑦))
7	6	cbvrabv 3434	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} = {𝑦 ∈ 𝐴 ∣ ¬ 𝑦 ∈ 𝑦}
8	3, 7	elrab2 3621	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ↔ ({𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ 𝐴 ∧ ¬ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥}))
9		pclem6 1020	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ↔ ({𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ 𝐴 ∧ ¬ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥})) → ¬ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ 𝐴)
10	8, 9	ax-mp 5	1 ⊢ ¬ {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝑥} ∈ 𝐴

Syntax hints:  ¬ wn 3   ↔ wb 207   ∧ wa 396   = wceq 1522   ∈ wcel 2081  {crab 3109

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1777  ax-4 1791  ax-5 1888  ax-6 1947  ax-7 1992  ax-8 2083  ax-9 2091  ax-10 2112  ax-11 2126  ax-12 2141  ax-ext 2769

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-tru 1525  df-ex 1762  df-nf 1766  df-sb 2043  df-clab 2776  df-cleq 2788  df-clel 2863  df-nfc 2935  df-rab 3114  df-v 3439

This theorem is referenced by:  pwnss  5141