Theorem axacndlem2 10364

Description: Lemma for the Axiom of Choice with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 2372. (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 2433	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑧
2		nfae 2433	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
3		simpr 485	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑧 ∈ 𝑤)
4	3	alimi 1814	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑧 ∈ 𝑤)
5		nd1 10343	. . . . . 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 2206	. . 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 2206	. 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 2175	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 205   ∧ wa 396  ∀wal 1537  ∃wex 1782

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-13 2372  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352  ax-reg 9351

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-v 3434  df-dif 3890  df-un 3892  df-nul 4257  df-sn 4562  df-pr 4564

This theorem is referenced by:  axacndlem4  10366  axacnd  10368