Theorem axacndlem2 10646

Description: Lemma for the Axiom of Choice with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 2375. (Contributed by NM, 3-Jan-2002.) (New usage is discouraged.)

Assertion

Ref	Expression
axacndlem2	⊢ (∀𝑥 𝑥 = 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 axacndlem2

Step	Hyp	Ref	Expression
1		nfae 2436	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑧
2		nfae 2436	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
3		simpr 484	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑧 ∈ 𝑤)
4	3	alimi 1808	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑧 ∈ 𝑤)
5		nd1 10625	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑧 → ¬ ∀𝑥 𝑧 ∈ 𝑤)
6	5	pm2.21d 121	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑧 → (∀𝑥 𝑧 ∈ 𝑤 → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
7	4, 6	syl5 34	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑧 → (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
8	2, 7	alrimi 2211	. . 3 ⊢ (∀𝑥 𝑥 = 𝑧 → ∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
9	1, 8	alrimi 2211	. 2 ⊢ (∀𝑥 𝑥 = 𝑧 → ∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
10	9	19.8ad 2180	1 ⊢ (∀𝑥 𝑥 = 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 206   ∧ wa 395  ∀wal 1535  ∃wex 1776

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-13 2375  ax-ext 2706  ax-sep 5302  ax-pr 5438  ax-reg 9630

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-ex 1777  df-nf 1781  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ral 3060  df-rex 3069  df-v 3480  df-un 3968  df-sn 4632  df-pr 4634

This theorem is referenced by:  axacndlem4  10648  axacnd  10650