Theorem axacndlem3 10634

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
axacndlem3	⊢ (∀𝑦 𝑦 = 𝑧 → ∃𝑥∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))

Proof of Theorem axacndlem3

Step	Hyp	Ref	Expression
1		nfae 2426	. . . 4 ⊢ Ⅎ𝑧∀𝑦 𝑦 = 𝑧
2		simpl 481	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑦 ∈ 𝑧)
3	2	alimi 1805	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑦 ∈ 𝑧)
4		nd3 10614	. . . . . 6 ⊢ (∀𝑦 𝑦 = 𝑧 → ¬ ∀𝑥 𝑦 ∈ 𝑧)
5	4	pm2.21d 121	. . . . 5 ⊢ (∀𝑦 𝑦 = 𝑧 → (∀𝑥 𝑦 ∈ 𝑧 → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
6	3, 5	syl5 34	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑧 → (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
7	1, 6	alrimi 2201	. . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → ∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
8	7	axc4i 2310	. 2 ⊢ (∀𝑦 𝑦 = 𝑧 → ∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
9	8	19.8ad 2170	1 ⊢ (∀𝑦 𝑦 = 𝑧 → ∃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 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 5300  ax-pr 5429  ax-reg 9617

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 3051  df-rex 3060  df-v 3463  df-un 3949  df-sn 4631  df-pr 4633

This theorem is referenced by:  axacndlem5  10636  axacnd  10637