Theorem axacndlem2 9632

Description: Lemma for the Axiom of Choice with no distinct variable conditions. (Contributed by NM, 3-Jan-2002.)

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 2468	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑧
2		nfae 2468	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
3		simpr 471	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑧 ∈ 𝑤)
4	3	alimi 1887	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑧 ∈ 𝑤)
5		nd1 9611	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑧 → ¬ ∀𝑥 𝑧 ∈ 𝑤)
6	5	pm2.21d 119	. . . . 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 2238	. . 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 2238	. 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		19.8a 2206	. 2 ⊢ (∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)) → ∃𝑥∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
11	9, 10	syl 17	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 196   ∧ wa 382  ∀wal 1629  ∃wex 1852

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-8 2147  ax-9 2154  ax-10 2174  ax-11 2190  ax-12 2203  ax-13 2408  ax-ext 2751  ax-sep 4915  ax-nul 4923  ax-pr 5034  ax-reg 8653

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-tru 1634  df-ex 1853  df-nf 1858  df-sb 2050  df-clab 2758  df-cleq 2764  df-clel 2767  df-nfc 2902  df-ral 3066  df-rex 3067  df-v 3353  df-dif 3726  df-un 3728  df-nul 4064  df-sn 4317  df-pr 4319

This theorem is referenced by:  axacndlem4  9634  axacnd  9636