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Theorem oprabidlem 5770
Description: Slight elaboration of exdistrfor 1756. A lemma for oprabid 5771. (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 1471 . . 3 (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2 ax-10 1468 . . . 4 (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3 dtru 4445 . . . . . 6 ¬ ∀𝑦 𝑦 = 𝑧
4 pm2.53 696 . . . . . 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 1422 . . . . . 6 (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
76albii 1431 . . . . 5 (∀𝑥𝑦 𝑥 = 𝑧 ↔ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
85, 7sylibr 133 . . . 4 ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥𝑦 𝑥 = 𝑧)
92, 8orim12i 733 . . 3 ((∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))) → (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥𝑦 𝑥 = 𝑧))
101, 9ax-mp 5 . 2 (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥𝑦 𝑥 = 𝑧)
1110exdistrfor 1756 1 (∃𝑥𝑦(𝑥 = 𝑧𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103  wo 682  wal 1314  wnf 1421  wex 1453
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-14 1477  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099  ax-sep 4016  ax-pow 4068  ax-setind 4422
This theorem depends on definitions:  df-bi 116  df-3an 949  df-tru 1319  df-fal 1322  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-ne 2286  df-ral 2398  df-v 2662  df-dif 3043  df-in 3047  df-ss 3054  df-pw 3482  df-sn 3503
This theorem is referenced by:  oprabid  5771
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