Theorem reldisjun 6052

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 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

Proof of Theorem reldisjun

Step	Hyp	Ref	Expression
1		reseq2 5995	. . 3 ⊢ (dom 𝑅 = (𝐴 ∪ 𝐵) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
2	1	3ad2ant2 1133	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
3		resdm 6046	. . 3 ⊢ (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
4	3	3ad2ant1 1132	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = 𝑅)
5		resundi 6014	. . 3 ⊢ (𝑅 ↾ (𝐴 ∪ 𝐵)) = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵))
6	5	a1i 11	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝑅 ↾ (𝐴 ∪ 𝐵)) = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))
7	2, 4, 6	3eqtr3d 2783	1 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

Syntax hints:   → wi 4   ∧ w3a 1086   = wceq 1537   ∪ cun 3961   ∩ cin 3962  ∅c0 4339  dom cdm 5689   ↾ cres 5691  Rel wrel 5694

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 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-br 5149  df-opab 5211  df-xp 5695  df-rel 5696  df-dm 5699  df-res 5701

This theorem is referenced by:  fressupp  32703  cycpmconjslem2  33158  evlselvlem  42573