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Theorem reldisjun 6050
Description: Split a relation into two disjoint parts based on its domain. (Contributed by Thierry Arnoux, 9-Oct-2023.)
Assertion
Ref Expression
reldisjun ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → 𝑅 = ((𝑅𝐴) ∪ (𝑅𝐵)))

Proof of Theorem reldisjun
StepHypRef Expression
1 reseq2 5992 . . 3 (dom 𝑅 = (𝐴𝐵) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴𝐵)))
213ad2ant2 1135 . 2 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴𝐵)))
3 resdm 6044 . . 3 (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
433ad2ant1 1134 . 2 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = 𝑅)
5 resundi 6011 . . 3 (𝑅 ↾ (𝐴𝐵)) = ((𝑅𝐴) ∪ (𝑅𝐵))
65a1i 11 . 2 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → (𝑅 ↾ (𝐴𝐵)) = ((𝑅𝐴) ∪ (𝑅𝐵)))
72, 4, 63eqtr3d 2785 1 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → 𝑅 = ((𝑅𝐴) ∪ (𝑅𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  w3a 1087   = wceq 1540  cun 3949  cin 3950  c0 4333  dom cdm 5685  cres 5687  Rel wrel 5690
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-xp 5691  df-rel 5692  df-dm 5695  df-res 5697
This theorem is referenced by:  fressupp  32697  cycpmconjslem2  33175  evlselvlem  42596
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