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Theorem axacndlem1 9635
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 → ∃𝑥𝑦𝑧(∀𝑥(𝑦𝑧𝑧𝑤) → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)))

Proof of Theorem axacndlem1
StepHypRef Expression
1 nfae 2468 . . 3 𝑦𝑥 𝑥 = 𝑦
2 nfae 2468 . . . 4 𝑧𝑥 𝑥 = 𝑦
3 simpl 468 . . . . . 6 ((𝑦𝑧𝑧𝑤) → 𝑦𝑧)
43alimi 1887 . . . . 5 (∀𝑥(𝑦𝑧𝑧𝑤) → ∀𝑥 𝑦𝑧)
5 nd1 9615 . . . . . 6 (∀𝑥 𝑥 = 𝑦 → ¬ ∀𝑥 𝑦𝑧)
65pm2.21d 119 . . . . 5 (∀𝑥 𝑥 = 𝑦 → (∀𝑥 𝑦𝑧 → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)))
74, 6syl5 34 . . . 4 (∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑦𝑧𝑧𝑤) → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)))
82, 7alrimi 2238 . . 3 (∀𝑥 𝑥 = 𝑦 → ∀𝑧(∀𝑥(𝑦𝑧𝑧𝑤) → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)))
91, 8alrimi 2238 . 2 (∀𝑥 𝑥 = 𝑦 → ∀𝑦𝑧(∀𝑥(𝑦𝑧𝑧𝑤) → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)))
10 19.8a 2206 . 2 (∀𝑦𝑧(∀𝑥(𝑦𝑧𝑧𝑤) → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)) → ∃𝑥𝑦𝑧(∀𝑥(𝑦𝑧𝑧𝑤) → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)))
119, 10syl 17 1 (∀𝑥 𝑥 = 𝑦 → ∃𝑥𝑦𝑧(∀𝑥(𝑦𝑧𝑧𝑤) → ∃𝑤𝑦(∃𝑤((𝑦𝑧𝑧𝑤) ∧ (𝑦𝑤𝑤𝑥)) ↔ 𝑦 = 𝑤)))
Colors of variables: wff setvar class
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 4916  ax-nul 4924  ax-pr 5035  ax-reg 8657
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 4318  df-pr 4320
This theorem is referenced by:  axacndlem4  9638  axacndlem5  9639
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