Theorem bj-nalset 10015

Description: nalset 3887 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 1539	. 2 ⊢ (∀𝑥∃𝑦 ¬ 𝑦 ∈ 𝑥 → ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥)
2		ax-bdel 9941	. . . . 5 ⊢ BOUNDED 𝑧 ∈ 𝑧
3	2	ax-bdn 9937	. . . 4 ⊢ BOUNDED ¬ 𝑧 ∈ 𝑧
4	3	bdsep1 10005	. . 3 ⊢ ∃𝑦∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧))
5		elequ1 1600	. . . . . 6 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑦 ↔ 𝑦 ∈ 𝑦))
6		elequ1 1600	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑦 ∈ 𝑥))
7		elequ1 1600	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑧))
8		elequ2 1601	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑦 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
9	7, 8	bitrd 177	. . . . . . . 8 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
10	9	notbid 592	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (¬ 𝑧 ∈ 𝑧 ↔ ¬ 𝑦 ∈ 𝑦))
11	6, 10	anbi12d 442	. . . . . 6 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧) ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
12	5, 11	bibi12d 224	. . . . 5 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) ↔ (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦))))
13	12	spv 1740	. . . 4 ⊢ (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
14		pclem6 1265	. . . 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 1493	. 2 ⊢ ∃𝑦 ¬ 𝑦 ∈ 𝑥
17	1, 16	mpg 1340	1 ⊢ ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥

Syntax hints:  ¬ wn 3   ∧ wa 97   ↔ wb 98  ∀wal 1241  ∃wex 1381

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-4 1400  ax-13 1404  ax-14 1405  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-bdn 9937  ax-bdel 9941  ax-bdsep 10004

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-fal 1249  df-nf 1350

This theorem is referenced by:  bj-vprc  10016