Theorem axacndlem2 10611

Description: Lemma for the Axiom of Choice with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 2407. (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 2468	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑧
2		nfae 2468	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑧
3		simpr 490	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑧 ∈ 𝑤)
4	3	alimi 1844	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑧 ∈ 𝑤)
5		nd1 10590	. . . . . 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 2252	. . 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 2252	. 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 2221	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)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-13 2407  ax-sep 5262  ax-reg 9564

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by:  axacndlem4  10613  axacnd  10615