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Theorem oprabidlem 5561
Description: Slight elaboration of exdistrfor 1722. A lemma for oprabid 5562. (Contributed by Jim Kingdon, 15-Jan-2019.)
Assertion
Ref Expression
oprabidlem (∃𝑥𝑦(𝑥 = 𝑧𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))
Distinct variable groups:   𝑥,𝑧   𝑦,𝑧
Allowed substitution hints:   𝜓(𝑥,𝑦,𝑧)

Proof of Theorem oprabidlem
StepHypRef Expression
1 ax-bndl 1440 . . 3 (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2 ax-10 1437 . . . 4 (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3 dtru 4305 . . . . . 6 ¬ ∀𝑦 𝑦 = 𝑧
4 pm2.53 674 . . . . . 6 ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → (¬ ∀𝑦 𝑦 = 𝑧 → ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
53, 4mpi 15 . . . . 5 ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
6 df-nf 1391 . . . . . 6 (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
76albii 1400 . . . . 5 (∀𝑥𝑦 𝑥 = 𝑧 ↔ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
85, 7sylibr 132 . . . 4 ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥𝑦 𝑥 = 𝑧)
92, 8orim12i 709 . . 3 ((∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))) → (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥𝑦 𝑥 = 𝑧))
101, 9ax-mp 7 . 2 (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥𝑦 𝑥 = 𝑧)
1110exdistrfor 1722 1 (∃𝑥𝑦(𝑥 = 𝑧𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102  wo 662  wal 1283  wnf 1390  wex 1422
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-14 1446  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064  ax-sep 3898  ax-pow 3950  ax-setind 4282
This theorem depends on definitions:  df-bi 115  df-3an 922  df-tru 1288  df-fal 1291  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-ne 2247  df-ral 2354  df-v 2604  df-dif 2976  df-in 2980  df-ss 2987  df-pw 3386  df-sn 3406
This theorem is referenced by:  oprabid  5562
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