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Theorem onfrALTlem1 40889
Description: Lemma for onfrALT 40890. (Contributed by Alan Sare, 22-Jul-2012.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
onfrALTlem1 ((𝑎 ⊆ On ∧ 𝑎 ≠ ∅) → ((𝑥𝑎 ∧ (𝑎𝑥) = ∅) → ∃𝑦𝑎 (𝑎𝑦) = ∅))
Distinct variable group:   𝑥,𝑎,𝑦

Proof of Theorem onfrALTlem1
StepHypRef Expression
1 19.8a 2180 . . . . 5 ((𝑥𝑎 ∧ (𝑎𝑥) = ∅) → ∃𝑥(𝑥𝑎 ∧ (𝑎𝑥) = ∅))
21a1i 11 . . . 4 ((𝑎 ⊆ On ∧ 𝑎 ≠ ∅) → ((𝑥𝑎 ∧ (𝑎𝑥) = ∅) → ∃𝑥(𝑥𝑎 ∧ (𝑎𝑥) = ∅)))
3 cbvexsv 40888 . . . 4 (∃𝑥(𝑥𝑎 ∧ (𝑎𝑥) = ∅) ↔ ∃𝑦[𝑦 / 𝑥](𝑥𝑎 ∧ (𝑎𝑥) = ∅))
42, 3syl6ib 253 . . 3 ((𝑎 ⊆ On ∧ 𝑎 ≠ ∅) → ((𝑥𝑎 ∧ (𝑎𝑥) = ∅) → ∃𝑦[𝑦 / 𝑥](𝑥𝑎 ∧ (𝑎𝑥) = ∅)))
5 sbsbc 3778 . . . . 5 ([𝑦 / 𝑥](𝑥𝑎 ∧ (𝑎𝑥) = ∅) ↔ [𝑦 / 𝑥](𝑥𝑎 ∧ (𝑎𝑥) = ∅))
6 onfrALTlem4 40884 . . . . 5 ([𝑦 / 𝑥](𝑥𝑎 ∧ (𝑎𝑥) = ∅) ↔ (𝑦𝑎 ∧ (𝑎𝑦) = ∅))
75, 6bitri 277 . . . 4 ([𝑦 / 𝑥](𝑥𝑎 ∧ (𝑎𝑥) = ∅) ↔ (𝑦𝑎 ∧ (𝑎𝑦) = ∅))
87exbii 1848 . . 3 (∃𝑦[𝑦 / 𝑥](𝑥𝑎 ∧ (𝑎𝑥) = ∅) ↔ ∃𝑦(𝑦𝑎 ∧ (𝑎𝑦) = ∅))
94, 8syl6ib 253 . 2 ((𝑎 ⊆ On ∧ 𝑎 ≠ ∅) → ((𝑥𝑎 ∧ (𝑎𝑥) = ∅) → ∃𝑦(𝑦𝑎 ∧ (𝑎𝑦) = ∅)))
10 df-rex 3146 . 2 (∃𝑦𝑎 (𝑎𝑦) = ∅ ↔ ∃𝑦(𝑦𝑎 ∧ (𝑎𝑦) = ∅))
119, 10syl6ibr 254 1 ((𝑎 ⊆ On ∧ 𝑎 ≠ ∅) → ((𝑥𝑎 ∧ (𝑎𝑥) = ∅) → ∃𝑦𝑎 (𝑎𝑦) = ∅))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 398   = wceq 1537  wex 1780  [wsb 2069  wne 3018  wrex 3141  [wsbc 3774  cin 3937  wss 3938  c0 4293  Oncon0 6193
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-13 2390  ax-ext 2795
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-fal 1550  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ral 3145  df-rex 3146  df-rab 3149  df-v 3498  df-sbc 3775  df-csb 3886  df-dif 3941  df-in 3945  df-nul 4294
This theorem is referenced by:  onfrALT  40890
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