Theorem axacndlem1 9825

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

Assertion

Ref	Expression
axacndlem1	⊢ (∀𝑥 𝑥 = 𝑦 → ∃𝑥∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))

Proof of Theorem axacndlem1

Step	Hyp	Ref	Expression
1		nfae 2370	. . 3 ⊢ Ⅎ𝑦∀𝑥 𝑥 = 𝑦
2		nfae 2370	. . . 4 ⊢ Ⅎ𝑧∀𝑥 𝑥 = 𝑦
3		simpl 475	. . . . . 6 ⊢ ((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → 𝑦 ∈ 𝑧)
4	3	alimi 1775	. . . . 5 ⊢ (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∀𝑥 𝑦 ∈ 𝑧)
5		nd1 9805	. . . . . 6 ⊢ (∀𝑥 𝑥 = 𝑦 → ¬ ∀𝑥 𝑦 ∈ 𝑧)
6	5	pm2.21d 119	. . . . 5 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥 𝑦 ∈ 𝑧 → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
7	4, 6	syl5 34	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
8	2, 7	alrimi 2144	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → ∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
9	1, 8	alrimi 2144	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → ∀𝑦∀𝑧(∀𝑥(𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) → ∃𝑤∀𝑦(∃𝑤((𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤) ∧ (𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥)) ↔ 𝑦 = 𝑤)))
10	9	19.8ad 2111	1 ⊢ (∀𝑥 𝑥 = 𝑦 → ∃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 198   ∧ wa 387  ∀wal 1506  ∃wex 1743

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1759  ax-4 1773  ax-5 1870  ax-6 1929  ax-7 1966  ax-8 2053  ax-9 2060  ax-10 2080  ax-11 2094  ax-12 2107  ax-13 2302  ax-ext 2743  ax-sep 5056  ax-nul 5063  ax-pr 5182  ax-reg 8849

This theorem depends on definitions:  df-bi 199  df-an 388  df-or 835  df-tru 1511  df-ex 1744  df-nf 1748  df-sb 2017  df-clab 2752  df-cleq 2764  df-clel 2839  df-nfc 2911  df-ral 3086  df-rex 3087  df-v 3410  df-dif 3825  df-un 3827  df-nul 4173  df-sn 4436  df-pr 4438

This theorem is referenced by:  axacndlem4  9828  axacndlem5  9829