Theorem axregndlem1 10602

Description: Lemma for the Axiom of Regularity with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 2406. (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 2220	. 2 ⊢ (𝑥 ∈ 𝑦 → ∃𝑥 𝑥 ∈ 𝑦)
2		nfae 2467	. . 3 ⊢ Ⅎ𝑥∀𝑥 𝑥 = 𝑧
3		nfae 2467	. . . . . 6 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
4		elirrv 9566	. . . . . . . . 9 ⊢ ¬ 𝑥 ∈ 𝑥
5		elequ1 2153	. . . . . . . . 9 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝑥 ↔ 𝑧 ∈ 𝑥))
6	4, 5	mtbii 329	. . . . . . . 8 ⊢ (𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
7	6	sps 2224	. . . . . . 7 ⊢ (∀𝑥 𝑥 = 𝑧 → ¬ 𝑧 ∈ 𝑥)
8	7	pm2.21d 122	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
9	3, 8	alrimi 2252	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑧 → ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))
10	9	anim2i 629	. . . 4 ⊢ ((𝑥 ∈ 𝑦 ∧ ∀𝑥 𝑥 = 𝑧) → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦)))
11	10	expcom 419	. . 3 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑥 ∈ 𝑦 → (𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))
12	2, 11	eximd 2255	. 2 ⊢ (∀𝑥 𝑥 = 𝑧 → (∃𝑥 𝑥 ∈ 𝑦 → ∃𝑥(𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))
13	1, 12	syl5 35	1 ⊢ (∀𝑥 𝑥 = 𝑧 → (𝑥 ∈ 𝑦 → ∃𝑥(𝑥 ∈ 𝑦 ∧ ∀𝑧(𝑧 ∈ 𝑥 → ¬ 𝑧 ∈ 𝑦))))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401  ∀wal 1568  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-13 2406  ax-sep 5259  ax-reg 9561

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817

This theorem is used by:  axregndlem2  10603  axregnd  10604