Theorem reldisjun 5987

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 5929	. . 3 ⊢ (dom 𝑅 = (𝐴 ∪ 𝐵) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
2	1	3ad2ant2 1134	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
3		resdm 5981	. . 3 ⊢ (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
4	3	3ad2ant1 1133	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = 𝑅)
5		resundi 5948	. . 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 2772	1 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

Syntax hints:   → wi 4   ∧ w3a 1086   = wceq 1540   ∪ cun 3903   ∩ cin 3904  ∅c0 4286  dom cdm 5623   ↾ cres 5625  Rel wrel 5628

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 2008  ax-8 2111  ax-9 2119  ax-ext 2701  ax-sep 5238  ax-nul 5248  ax-pr 5374

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-ral 3045  df-rex 3054  df-rab 3397  df-v 3440  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4479  df-sn 4580  df-pr 4582  df-op 4586  df-br 5096  df-opab 5158  df-xp 5629  df-rel 5630  df-dm 5633  df-res 5635

This theorem is referenced by:  fressupp  32644  cycpmconjslem2  33110  evlselvlem  42559