Theorem axregndlem1 10289

Description: Lemma for the Axiom of Regularity with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 2372. (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 2176	. 2 ⊢ (𝑥 ∈ 𝑦 → ∃𝑥 𝑥 ∈ 𝑦)
2		nfae 2433	. . 3 ⊢ Ⅎ𝑥∀𝑥 𝑥 = 𝑧
3		nfae 2433	. . . . . 6 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
4		elirrv 9285	. . . . . . . . 9 ⊢ ¬ 𝑥 ∈ 𝑥
5		elequ1 2115	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝑥 ↔ 𝑧 ∈ 𝑥))
6	4, 5	mtbii 325	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
7	6	sps 2180	. . . . . . 7 ⊢ (∀𝑥 𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
8	7	pm2.21d 121	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
9	3, 8	alrimi 2209	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑧 → ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
10	9	anim2i 616	. . . 4 ⊢ ((𝑥 ∈ 𝑦 ∧ ∀𝑥 𝑥 = 𝑧) → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦)))
11	10	expcom 413	. . 3 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑥 ∈ 𝑦 → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))
12	2, 11	eximd 2212	. 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 395  ∀wal 1537  ∃wex 1783

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-13 2372  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347  ax-reg 9281

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-v 3424  df-dif 3886  df-un 3888  df-nul 4254  df-sn 4559  df-pr 4561

This theorem is referenced by:  axregndlem2  10290  axregnd  10291