Theorem reldisjun 6007

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 5949	. . 3 ⊢ (dom 𝑅 = (𝐴 ∪ 𝐵) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
2	1	3ad2ant2 1143	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
3		resdm 6001	. . 3 ⊢ (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
4	3	3ad2ant1 1142	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → (𝑅 ↾ dom 𝑅) = 𝑅)
5		resundi 5968	. . 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 2795	1 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

Syntax hints:   → wi 4   ∧ w3a 1095   = wceq 1550   ∪ cun 3893   ∩ cin 3894  ∅c0 4276  dom cdm 5636   ↾ cres 5638  Rel wrel 5641

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1805  ax-4 1819  ax-5 1920  ax-6 1977  ax-7 2018  ax-8 2134  ax-9 2142  ax-ext 2724  ax-sep 5236  ax-pr 5380

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 857  df-3an 1097  df-tru 1553  df-fal 1563  df-ex 1790  df-sb 2081  df-clab 2731  df-cleq 2744  df-clel 2827  df-ral 3067  df-rex 3077  df-rab 3405  df-v 3446  df-dif 3898  df-un 3900  df-in 3902  df-ss 3912  df-nul 4277  df-if 4471  df-sn 4573  df-pr 4575  df-op 4579  df-br 5091  df-opab 5153  df-xp 5642  df-rel 5643  df-dm 5646  df-res 5648

This theorem is referenced by:  fressupp  32829  cycpmconjslem2  33285  esplyind  33816  evlselvlem  43108