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Theorem reldisjun 30347
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 5843 . . 3 (dom 𝑅 = (𝐴𝐵) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴𝐵)))
213ad2ant2 1130 . 2 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴𝐵)))
3 resdm 5892 . . 3 (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
433ad2ant1 1129 . 2 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = 𝑅)
5 resundi 5862 . . 3 (𝑅 ↾ (𝐴𝐵)) = ((𝑅𝐴) ∪ (𝑅𝐵))
65a1i 11 . 2 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → (𝑅 ↾ (𝐴𝐵)) = ((𝑅𝐴) ∪ (𝑅𝐵)))
72, 4, 63eqtr3d 2864 1 ((Rel 𝑅 ∧ dom 𝑅 = (𝐴𝐵) ∧ (𝐴𝐵) = ∅) → 𝑅 = ((𝑅𝐴) ∪ (𝑅𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  w3a 1083   = wceq 1533  cun 3934  cin 3935  c0 4291  dom cdm 5550  cres 5552  Rel wrel 5555
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2156  ax-12 2172  ax-ext 2793  ax-sep 5196  ax-nul 5203  ax-pr 5322
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rex 3144  df-rab 3147  df-v 3497  df-dif 3939  df-un 3941  df-in 3943  df-ss 3952  df-nul 4292  df-if 4468  df-sn 4562  df-pr 4564  df-op 4568  df-br 5060  df-opab 5122  df-xp 5556  df-rel 5557  df-dm 5560  df-res 5562
This theorem is referenced by:  cycpmconjslem2  30792
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