Metamath Proof Explorer


Theorem refsymrels3

Description: Elements of the class of reflexive relations which are elements of the class of symmetric relations as well (like the elements of the class of equivalence relations dfeqvrels3 ) can use the A. x e. dom r x r x version for their reflexive part, not just the A. x e. dom r A. y e. ran r ( x = y -> x r y ) version of dfrefrels3 , cf. the comment of dfrefrel3 . (Contributed by Peter Mazsa, 22-Jul-2019) (Proof modification is discouraged.)

Ref Expression
Assertion refsymrels3 ( RefRels ∩ SymRels ) = { 𝑟 ∈ Rels ∣ ( ∀ 𝑥 ∈ dom 𝑟 𝑥 𝑟 𝑥 ∧ ∀ 𝑥𝑦 ( 𝑥 𝑟 𝑦𝑦 𝑟 𝑥 ) ) }

Proof

Step Hyp Ref Expression
1 refsymrels2 ( RefRels ∩ SymRels ) = { 𝑟 ∈ Rels ∣ ( ( I ↾ dom 𝑟 ) ⊆ 𝑟 𝑟𝑟 ) }
2 idrefALT ( ( I ↾ dom 𝑟 ) ⊆ 𝑟 ↔ ∀ 𝑥 ∈ dom 𝑟 𝑥 𝑟 𝑥 )
3 cnvsym ( 𝑟𝑟 ↔ ∀ 𝑥𝑦 ( 𝑥 𝑟 𝑦𝑦 𝑟 𝑥 ) )
4 2 3 anbi12i ( ( ( I ↾ dom 𝑟 ) ⊆ 𝑟 𝑟𝑟 ) ↔ ( ∀ 𝑥 ∈ dom 𝑟 𝑥 𝑟 𝑥 ∧ ∀ 𝑥𝑦 ( 𝑥 𝑟 𝑦𝑦 𝑟 𝑥 ) ) )
5 1 4 rabbieq ( RefRels ∩ SymRels ) = { 𝑟 ∈ Rels ∣ ( ∀ 𝑥 ∈ dom 𝑟 𝑥 𝑟 𝑥 ∧ ∀ 𝑥𝑦 ( 𝑥 𝑟 𝑦𝑦 𝑟 𝑥 ) ) }