Theorem bj-nalset 15541

Description: nalset 4163 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 1662	. 2 ⊢ (∀𝑥∃𝑦 ¬ 𝑦 ∈ 𝑥 → ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥)
2		ax-bdel 15467	. . . . 5 ⊢ BOUNDED 𝑧 ∈ 𝑧
3	2	ax-bdn 15463	. . . 4 ⊢ BOUNDED ¬ 𝑧 ∈ 𝑧
4	3	bdsep1 15531	. . 3 ⊢ ∃𝑦∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧))
5		elequ1 2171	. . . . . 6 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑦 ↔ 𝑦 ∈ 𝑦))
6		elequ1 2171	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑦 ∈ 𝑥))
7		elequ1 2171	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑧))
8		elequ2 2172	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑦 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
9	7, 8	bitrd 188	. . . . . . . 8 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
10	9	notbid 668	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (¬ 𝑧 ∈ 𝑧 ↔ ¬ 𝑦 ∈ 𝑦))
11	6, 10	anbi12d 473	. . . . . 6 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧) ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
12	5, 11	bibi12d 235	. . . . 5 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) ↔ (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦))))
13	12	spv 1874	. . . 4 ⊢ (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
14		pclem6 1385	. . . 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 1616	. 2 ⊢ ∃𝑦 ¬ 𝑦 ∈ 𝑥
17	1, 16	mpg 1465	1 ⊢ ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥

Syntax hints:  ¬ wn 3   ∧ wa 104   ↔ wb 105  ∀wal 1362  ∃wex 1506

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-5 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-13 2169  ax-14 2170  ax-bdn 15463  ax-bdel 15467  ax-bdsep 15530

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-fal 1370  df-nf 1475

This theorem is referenced by:  bj-vprc  15542