Metamath Proof Explorer


Theorem relfld

Description: The double union of a relation is its field. (Contributed by NM, 17-Sep-2006)

Ref Expression
Assertion relfld ( Rel 𝑅 𝑅 = ( dom 𝑅 ∪ ran 𝑅 ) )

Proof

Step Hyp Ref Expression
1 relssdmrn ( Rel 𝑅𝑅 ⊆ ( dom 𝑅 × ran 𝑅 ) )
2 uniss ( 𝑅 ⊆ ( dom 𝑅 × ran 𝑅 ) → 𝑅 ( dom 𝑅 × ran 𝑅 ) )
3 uniss ( 𝑅 ( dom 𝑅 × ran 𝑅 ) → 𝑅 ( dom 𝑅 × ran 𝑅 ) )
4 1 2 3 3syl ( Rel 𝑅 𝑅 ( dom 𝑅 × ran 𝑅 ) )
5 unixpss ( dom 𝑅 × ran 𝑅 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 )
6 4 5 sstrdi ( Rel 𝑅 𝑅 ⊆ ( dom 𝑅 ∪ ran 𝑅 ) )
7 dmrnssfld ( dom 𝑅 ∪ ran 𝑅 ) ⊆ 𝑅
8 7 a1i ( Rel 𝑅 → ( dom 𝑅 ∪ ran 𝑅 ) ⊆ 𝑅 )
9 6 8 eqssd ( Rel 𝑅 𝑅 = ( dom 𝑅 ∪ ran 𝑅 ) )