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Metamath Proof Explorer

Theorem reldisjun

Description: Split a relation into two disjoint parts based on its domain. (Contributed by Thierry Arnoux, 9-Oct-2023)

Ref Expression
Assertion reldisjun ( ( Rel 𝑅 ∧ dom 𝑅 = ( 𝐴 ∪ 𝐵 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → 𝑅 = ( ( 𝑅 ↾ 𝐴 ) ∪ ( 𝑅 ↾ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 reseq2 ⊢ ( dom 𝑅 = ( 𝐴 ∪ 𝐵 ) → ( 𝑅 ↾ dom 𝑅 ) = ( 𝑅 ↾ ( 𝐴 ∪ 𝐵 ) ) )
2 1 3ad2ant2 ⊢ ( ( Rel 𝑅 ∧ dom 𝑅 = ( 𝐴 ∪ 𝐵 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → ( 𝑅 ↾ dom 𝑅 ) = ( 𝑅 ↾ ( 𝐴 ∪ 𝐵 ) ) )
3 resdm ⊢ ( Rel 𝑅 → ( 𝑅 ↾ dom 𝑅 ) = 𝑅 )
4 3 3ad2ant1 ⊢ ( ( Rel 𝑅 ∧ dom 𝑅 = ( 𝐴 ∪ 𝐵 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → ( 𝑅 ↾ dom 𝑅 ) = 𝑅 )
5 resundi ⊢ ( 𝑅 ↾ ( 𝐴 ∪ 𝐵 ) ) = ( ( 𝑅 ↾ 𝐴 ) ∪ ( 𝑅 ↾ 𝐵 ) )
6 5 a1i ⊢ ( ( Rel 𝑅 ∧ dom 𝑅 = ( 𝐴 ∪ 𝐵 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → ( 𝑅 ↾ ( 𝐴 ∪ 𝐵 ) ) = ( ( 𝑅 ↾ 𝐴 ) ∪ ( 𝑅 ↾ 𝐵 ) ) )
7 2 4 6 3eqtr3d ⊢ ( ( Rel 𝑅 ∧ dom 𝑅 = ( 𝐴 ∪ 𝐵 ) ∧ ( 𝐴 ∩ 𝐵 ) = ∅ ) → 𝑅 = ( ( 𝑅 ↾ 𝐴 ) ∪ ( 𝑅 ↾ 𝐵 ) ) )