Theorem reldisjun 6061

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 6004	. . 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 6055	. . 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 6023	. . 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 2788	1 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵) ∧ (𝐴 ∩ 𝐵) = ∅) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

Syntax hints:   → wi 4   ∧ w3a 1087   = wceq 1537   ∪ cun 3974   ∩ cin 3975  ∅c0 4352  dom cdm 5700   ↾ cres 5702  Rel wrel 5705

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pr 5447

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-br 5167  df-opab 5229  df-xp 5706  df-rel 5707  df-dm 5710  df-res 5712

This theorem is referenced by:  fressupp  32700  cycpmconjslem2  33148  evlselvlem  42541