Theorem reldmun 6035

Description: Split a relation into two parts based on its domain. (Contributed by Thierry Arnoux, 9-Oct-2023.) Remove requirement that 𝐴 and 𝐵 are disjoint. (Revised by Eric Schmidt, 20-Jun-2026.)

Assertion

Ref	Expression
reldmun	⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

Proof of Theorem reldmun

Step	Hyp	Ref	Expression
1		reseq2 5975	. . 3 ⊢ (dom 𝑅 = (𝐴 ∪ 𝐵) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
2	1	adantl 487	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
3		resdm 6027	. . 3 ⊢ (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
4	3	adantr 486	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → (𝑅 ↾ dom 𝑅) = 𝑅)
5		resundi 5994	. . 3 ⊢ (𝑅 ↾ (𝐴 ∪ 𝐵)) = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵))
6	5	a1i 11	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → (𝑅 ↾ (𝐴 ∪ 𝐵)) = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))
7	2, 4, 6	3eqtr3d 2808	1 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∪ cun 3904  dom cdm 5663   ↾ cres 5665  Rel wrel 5668

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-xp 5669  df-rel 5670  df-dm 5673  df-res 5675

This theorem is used by:  fressupp  33062  cycpmconjslem2  33498  esplyind  33988  evlselvlem  43353