Theorem reldmun 6016

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 5956	. . 3 ⊢ (dom 𝑅 = (𝐴 ∪ 𝐵) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
2	1	adantl 485	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → (𝑅 ↾ dom 𝑅) = (𝑅 ↾ (𝐴 ∪ 𝐵)))
3		resdm 6008	. . 3 ⊢ (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
4	3	adantr 484	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → (𝑅 ↾ dom 𝑅) = 𝑅)
5		resundi 5975	. . 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 2804	1 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → 𝑅 = ((𝑅 ↾ 𝐴) ∪ (𝑅 ↾ 𝐵)))

Syntax hints:   → wi 4   ∧ wa 399   = wceq 1559   ∪ cun 3900  dom cdm 5643   ↾ cres 5645  Rel wrel 5648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733  ax-sep 5243  ax-pr 5387

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3076  df-rex 3086  df-rab 3414  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4284  df-if 4478  df-sn 4580  df-pr 4582  df-op 4586  df-br 5098  df-opab 5160  df-xp 5649  df-rel 5650  df-dm 5653  df-res 5655

This theorem is referenced by:  fressupp  32851  cycpmconjslem2  33296  esplyind  33833  evlselvlem  43131