Theorem axacndlem2 10581

Description: Lemma for the Axiom of Choice 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
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 2467	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑧
2		nfae 2467	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
3		simpr 489	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑧 ∈ 𝑤)
4	3	alimi 1834	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑧 ∈ 𝑤)
5		nd1 10560	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑧 → ¬ ∀𝑥 𝑧 ∈ 𝑤)
6	5	pm2.21d 122	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑧 → (∀𝑥 𝑧 ∈ 𝑤 → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
7	4, 6	syl5 35	. . . 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 2251	. . 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 2251	. 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 2220	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 209   ∧ wa 400  ∀wal 1561  ∃wex 1802

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-13 2406  ax-sep 5251  ax-reg 9542

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1566  df-ex 1803  df-nf 1807  df-sb 2094

This theorem is referenced by:  axacndlem4  10583  axacnd  10585