Theorem nalset 4219

Description: No set contains all sets. Theorem 41 of [Suppes] p. 30. (Contributed by NM, 23-Aug-1993.)

Assertion

Ref	Expression
nalset	⊢ ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥

Distinct variable group:   𝑥,𝑦

Proof of Theorem nalset

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		alexnim 1696	. 2 ⊢ (∀𝑥∃𝑦 ¬ 𝑦 ∈ 𝑥 → ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥)
2		ax-sep 4207	. . 3 ⊢ ∃𝑦∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧))
3		elequ1 2206	. . . . . 6 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑦 ↔ 𝑦 ∈ 𝑦))
4		elequ1 2206	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑦 ∈ 𝑥))
5		elequ1 2206	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑧))
6		elequ2 2207	. . . . . . . . 9 ⊢ (𝑧 = 𝑦 → (𝑦 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
7	5, 6	bitrd 188	. . . . . . . 8 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
8	7	notbid 673	. . . . . . 7 ⊢ (𝑧 = 𝑦 → (¬ 𝑧 ∈ 𝑧 ↔ ¬ 𝑦 ∈ 𝑦))
9	4, 8	anbi12d 473	. . . . . 6 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧) ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
10	3, 9	bibi12d 235	. . . . 5 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) ↔ (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦))))
11	10	spv 1908	. . . 4 ⊢ (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
12		pclem6 1418	. . . 4 ⊢ ((𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)) → ¬ 𝑦 ∈ 𝑥)
13	11, 12	syl 14	. . 3 ⊢ (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → ¬ 𝑦 ∈ 𝑥)
14	2, 13	eximii 1650	. 2 ⊢ ∃𝑦 ¬ 𝑦 ∈ 𝑥
15	1, 14	mpg 1499	1 ⊢ ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥

Syntax hints:  ¬ wn 3   ∧ wa 104   ↔ wb 105  ∀wal 1395  ∃wex 1540

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 619  ax-in2 620  ax-5 1495  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-13 2204  ax-14 2205  ax-sep 4207

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-fal 1403  df-nf 1509

This theorem is referenced by:  vnex  4220