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Theorem oprabidlem 5927
Description: Slight elaboration of exdistrfor 1811. A lemma for oprabid 5928. (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 1520 . . 3 (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2 ax-10 1516 . . . 4 (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3 dtru 4577 . . . . . 6 ¬ ∀𝑦 𝑦 = 𝑧
4 pm2.53 723 . . . . . 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 1472 . . . . . 6 (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
76albii 1481 . . . . 5 (∀𝑥𝑦 𝑥 = 𝑧 ↔ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
85, 7sylibr 134 . . . 4 ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥𝑦 𝑥 = 𝑧)
92, 8orim12i 760 . . 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 1811 1 (∃𝑥𝑦(𝑥 = 𝑧𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104  wo 709  wal 1362  wnf 1471  wex 1503
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-14 2163  ax-ext 2171  ax-sep 4136  ax-pow 4192  ax-setind 4554
This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-fal 1370  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-ne 2361  df-ral 2473  df-v 2754  df-dif 3146  df-in 3150  df-ss 3157  df-pw 3592  df-sn 3613
This theorem is referenced by:  oprabid  5928
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