Metamath Proof Explorer


Theorem reldvdsr

Description: The divides relation is a relation. (Contributed by Mario Carneiro, 1-Dec-2014)

Ref Expression
Hypothesis reldvdsr.1 = ( ∥r𝑅 )
Assertion reldvdsr Rel

Proof

Step Hyp Ref Expression
1 reldvdsr.1 = ( ∥r𝑅 )
2 df-dvdsr r = ( 𝑤 ∈ V ↦ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ ( Base ‘ 𝑤 ) ∧ ∃ 𝑧 ∈ ( Base ‘ 𝑤 ) ( 𝑧 ( .r𝑤 ) 𝑥 ) = 𝑦 ) } )
3 2 relmptopab Rel ( ∥r𝑅 )
4 1 releqi ( Rel ↔ Rel ( ∥r𝑅 ) )
5 3 4 mpbir Rel