Theorem bj-nalset 13082

Description: nalset 4053 from bounded separation. (Contributed by BJ, 18-Nov-2019.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-nalset	⊢ ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥

Distinct variable group:   𝑥,𝑦

Proof of Theorem bj-nalset

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		alexnim 1627	. 2 ⊢ (∀𝑥∃𝑦 ¬ 𝑦 ∈ 𝑥 → ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥)
2		ax-bdel 13008	. . . . 5 ⊢ BOUNDED 𝑧 ∈ 𝑧
3	2	ax-bdn 13004	. . . 4 ⊢ BOUNDED ¬ 𝑧 ∈ 𝑧
4	3	bdsep1 13072	. . 3 ⊢ ∃𝑦∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧))
5		elequ1 1690	. . . . . 6 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑦 ↔ 𝑦 ∈ 𝑦))
6		elequ1 1690	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑦 ∈ 𝑥))
7		elequ1 1690	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑧))
8		elequ2 1691	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑦 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
9	7, 8	bitrd 187	. . . . . . . 8 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
10	9	notbid 656	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (¬ 𝑧 ∈ 𝑧 ↔ ¬ 𝑦 ∈ 𝑦))
11	6, 10	anbi12d 464	. . . . . 6 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧) ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
12	5, 11	bibi12d 234	. . . . 5 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) ↔ (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦))))
13	12	spv 1832	. . . 4 ⊢ (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
14		pclem6 1352	. . . 4 ⊢ ((𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)) → ¬ 𝑦 ∈ 𝑥)
15	13, 14	syl 14	. . 3 ⊢ (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → ¬ 𝑦 ∈ 𝑥)
16	4, 15	eximii 1581	. 2 ⊢ ∃𝑦 ¬ 𝑦 ∈ 𝑥
17	1, 16	mpg 1427	1 ⊢ ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥

Syntax hints:  ¬ wn 3   ∧ wa 103   ↔ wb 104  ∀wal 1329  ∃wex 1468

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-4 1487  ax-13 1491  ax-14 1492  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-bdn 13004  ax-bdel 13008  ax-bdsep 13071

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-fal 1337  df-nf 1437

This theorem is referenced by:  bj-vprc  13083