Theorem axacndlem1 10409

Description: Lemma for the Axiom of Choice with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 2370. (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 2431	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑦
2		nfae 2431	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑦
3		simpl 484	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑦 ∈ 𝑧)
4	3	alimi 1811	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑦 ∈ 𝑧)
5		nd1 10389	. . . . . 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 2204	. . 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 2204	. 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 2173	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 397  ∀wal 1537  ∃wex 1779

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-11 2152  ax-12 2169  ax-13 2370  ax-ext 2707  ax-sep 5232  ax-nul 5239  ax-pr 5361  ax-reg 9395

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 846  df-tru 1542  df-fal 1552  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2714  df-cleq 2728  df-clel 2814  df-ral 3063  df-rex 3072  df-v 3439  df-dif 3895  df-un 3897  df-nul 4263  df-sn 4566  df-pr 4568

This theorem is referenced by:  axacndlem4  10412  axacndlem5  10413