Theorem axacndlem1 10604

Description: Lemma for the Axiom of Choice 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
axacndlem1	⊢ (∀𝑥 𝑥 = 𝑦 → ∃𝑥∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))

Proof of Theorem axacndlem1

Step	Hyp	Ref	Expression
1		nfae 2426	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑦
2		nfae 2426	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑦
3		simpl 482	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑦 ∈ 𝑧)
4	3	alimi 1805	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑦 ∈ 𝑧)
5		nd1 10584	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑦 → ¬ ∀𝑥 𝑦 ∈ 𝑧)
6	5	pm2.21d 121	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥 𝑦 ∈ 𝑧 → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
7	4, 6	syl5 34	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
8	2, 7	alrimi 2198	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → ∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
9	1, 8	alrimi 2198	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → ∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
10	9	19.8ad 2167	1 ⊢ (∀𝑥 𝑥 = 𝑦 → ∃𝑥∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395  ∀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 2163  ax-13 2365  ax-ext 2697  ax-sep 5292  ax-pr 5420  ax-reg 9589

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-ral 3056  df-rex 3065  df-v 3470  df-un 3948  df-sn 4624  df-pr 4626

This theorem is referenced by:  axacndlem4  10607  axacndlem5  10608