Theorem reldmun 6039

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 5979	. . 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 6031	. . 3 ⊢ (Rel 𝑅 → (𝑅 ↾ dom 𝑅) = 𝑅)
4	3	adantr 486	. 2 ⊢ ((Rel 𝑅 ∧ dom 𝑅 = (𝐴 ∪ 𝐵)) → (𝑅 ↾ dom 𝑅) = 𝑅)
5		resundi 5998	. . 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 2806	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 4152  dom cdm 5656   ↾ cres 5658  Rel wrel 5661

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5253  ax-pr 5396

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-xp 5662  df-rel 5663  df-dm 5666  df-res 5668

This theorem is used by:  fressupp  33368  cycpmconjslem2  33805  esplyind  34311  evlselvlem  43876