Metamath Proof Explorer


Theorem releldmqs

Description: Elementhood in the domain quotient of a relation. (Contributed by Peter Mazsa, 24-Apr-2021)

Ref Expression
Assertion releldmqs ( 𝐴𝑉 → ( Rel 𝑅 → ( 𝐴 ∈ ( dom 𝑅 / 𝑅 ) ↔ ∃ 𝑢 ∈ dom 𝑅𝑥 ∈ [ 𝑢 ] 𝑅 𝐴 = [ 𝑢 ] 𝑅 ) ) )

Proof

Step Hyp Ref Expression
1 resdm ( Rel 𝑅 → ( 𝑅 ↾ dom 𝑅 ) = 𝑅 )
2 1 dmqseqd ( Rel 𝑅 → ( dom ( 𝑅 ↾ dom 𝑅 ) / ( 𝑅 ↾ dom 𝑅 ) ) = ( dom 𝑅 / 𝑅 ) )
3 2 eleq2d ( Rel 𝑅 → ( 𝐴 ∈ ( dom ( 𝑅 ↾ dom 𝑅 ) / ( 𝑅 ↾ dom 𝑅 ) ) ↔ 𝐴 ∈ ( dom 𝑅 / 𝑅 ) ) )
4 3 adantl ( ( 𝐴𝑉 ∧ Rel 𝑅 ) → ( 𝐴 ∈ ( dom ( 𝑅 ↾ dom 𝑅 ) / ( 𝑅 ↾ dom 𝑅 ) ) ↔ 𝐴 ∈ ( dom 𝑅 / 𝑅 ) ) )
5 eldmqsres2 ( 𝐴𝑉 → ( 𝐴 ∈ ( dom ( 𝑅 ↾ dom 𝑅 ) / ( 𝑅 ↾ dom 𝑅 ) ) ↔ ∃ 𝑢 ∈ dom 𝑅𝑥 ∈ [ 𝑢 ] 𝑅 𝐴 = [ 𝑢 ] 𝑅 ) )
6 5 adantr ( ( 𝐴𝑉 ∧ Rel 𝑅 ) → ( 𝐴 ∈ ( dom ( 𝑅 ↾ dom 𝑅 ) / ( 𝑅 ↾ dom 𝑅 ) ) ↔ ∃ 𝑢 ∈ dom 𝑅𝑥 ∈ [ 𝑢 ] 𝑅 𝐴 = [ 𝑢 ] 𝑅 ) )
7 4 6 bitr3d ( ( 𝐴𝑉 ∧ Rel 𝑅 ) → ( 𝐴 ∈ ( dom 𝑅 / 𝑅 ) ↔ ∃ 𝑢 ∈ dom 𝑅𝑥 ∈ [ 𝑢 ] 𝑅 𝐴 = [ 𝑢 ] 𝑅 ) )
8 7 ex ( 𝐴𝑉 → ( Rel 𝑅 → ( 𝐴 ∈ ( dom 𝑅 / 𝑅 ) ↔ ∃ 𝑢 ∈ dom 𝑅𝑥 ∈ [ 𝑢 ] 𝑅 𝐴 = [ 𝑢 ] 𝑅 ) ) )