Theorem axregndlem1 10623

Description: Lemma for the Axiom of Regularity with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 2365. (Contributed by NM, 3-Jan-2002.) (New usage is discouraged.)

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 2169	. 2 ⊢ (𝑥 ∈ 𝑦 → ∃𝑥 𝑥 ∈ 𝑦)
2		nfae 2426	. . 3 ⊢ Ⅎ𝑥∀𝑥 𝑥 = 𝑧
3		nfae 2426	. . . . . 6 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
4		elirrv 9617	. . . . . . . . 9 ⊢ ¬ 𝑥 ∈ 𝑥
5		elequ1 2105	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝑥 ↔ 𝑧 ∈ 𝑥))
6	4, 5	mtbii 325	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
7	6	sps 2173	. . . . . . 7 ⊢ (∀𝑥 𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
8	7	pm2.21d 121	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
9	3, 8	alrimi 2201	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑧 → ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
10	9	anim2i 615	. . . 4 ⊢ ((𝑥 ∈ 𝑦 ∧ ∀𝑥 𝑥 = 𝑧) → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦)))
11	10	expcom 412	. . 3 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑥 ∈ 𝑦 → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))
12	2, 11	eximd 2204	. 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 394  ∀wal 1531  ∃wex 1773

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2166  ax-13 2365  ax-ext 2696  ax-sep 5294  ax-pr 5423  ax-reg 9613

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-ral 3052  df-rex 3061  df-v 3465  df-un 3945  df-sn 4625  df-pr 4627

This theorem is referenced by:  axregndlem2  10624  axregnd  10625