Theorem axregndlem1 9409

Description: Lemma for the Axiom of Regularity with no distinct variable conditions. (Contributed by NM, 3-Jan-2002.)

Assertion

Ref	Expression
axregndlem1	⊢ (∀𝑥 𝑥 = 𝑧 → (𝑥 ∈ 𝑦 → ∃𝑥(𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))

Proof of Theorem axregndlem1

Step	Hyp	Ref	Expression
1		19.8a 2050	. 2 ⊢ (𝑥 ∈ 𝑦 → ∃𝑥 𝑥 ∈ 𝑦)
2		nfae 2314	. . 3 ⊢ Ⅎ𝑥∀𝑥 𝑥 = 𝑧
3		nfae 2314	. . . . . 6 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
4		elirrv 8489	. . . . . . . . 9 ⊢ ¬ 𝑥 ∈ 𝑥
5		elequ1 1995	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝑥 ↔ 𝑧 ∈ 𝑥))
6	4, 5	mtbii 316	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
7	6	sps 2053	. . . . . . 7 ⊢ (∀𝑥 𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
8	7	pm2.21d 118	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
9	3, 8	alrimi 2080	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑧 → ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
10	9	anim2i 592	. . . 4 ⊢ ((𝑥 ∈ 𝑦 ∧ ∀𝑥 𝑥 = 𝑧) → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦)))
11	10	expcom 451	. . 3 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑥 ∈ 𝑦 → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))
12	2, 11	eximd 2083	. 2 ⊢ (∀𝑥 𝑥 = 𝑧 → (∃𝑥 𝑥 ∈ 𝑦 → ∃𝑥(𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))
13	1, 12	syl5 34	1 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑥 ∈ 𝑦 → ∃𝑥(𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 384  ∀wal 1479  ∃wex 1702

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1720  ax-4 1735  ax-5 1837  ax-6 1886  ax-7 1933  ax-8 1990  ax-9 1997  ax-10 2017  ax-11 2032  ax-12 2045  ax-13 2244  ax-ext 2600  ax-sep 4772  ax-nul 4780  ax-pr 4897  ax-reg 8482

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1484  df-ex 1703  df-nf 1708  df-sb 1879  df-clab 2607  df-cleq 2613  df-clel 2616  df-nfc 2751  df-ral 2914  df-rex 2915  df-v 3197  df-dif 3570  df-un 3572  df-nul 3908  df-sn 4169  df-pr 4171

This theorem is referenced by:  axregndlem2  9410  axregnd  9411