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Theorem oprabidlem 5802
 Description: Slight elaboration of exdistrfor 1772. A lemma for oprabid 5803. (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 1486 . . 3 (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2 ax-10 1483 . . . 4 (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3 dtru 4475 . . . . . 6 ¬ ∀𝑦 𝑦 = 𝑧
4 pm2.53 711 . . . . . 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 1437 . . . . . 6 (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
76albii 1446 . . . . 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 748 . . 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 1772 1 (∃𝑥𝑦(𝑥 = 𝑧𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ∨ wo 697  ∀wal 1329  Ⅎwnf 1436  ∃wex 1468 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 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-14 1492  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121  ax-sep 4046  ax-pow 4098  ax-setind 4452 This theorem depends on definitions:  df-bi 116  df-3an 964  df-tru 1334  df-fal 1337  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-ne 2309  df-ral 2421  df-v 2688  df-dif 3073  df-in 3077  df-ss 3084  df-pw 3512  df-sn 3533 This theorem is referenced by:  oprabid  5803
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