Metamath Proof Explorer


Theorem relsubgr

Description: The class of the subgraph relation is a relation. (Contributed by AV, 16-Nov-2020)

Ref Expression
Assertion relsubgr Rel SubGraph

Proof

Step Hyp Ref Expression
1 df-subgr SubGraph = { ⟨ 𝑠 , 𝑔 ⟩ ∣ ( ( Vtx ‘ 𝑠 ) ⊆ ( Vtx ‘ 𝑔 ) ∧ ( iEdg ‘ 𝑠 ) = ( ( iEdg ‘ 𝑔 ) ↾ dom ( iEdg ‘ 𝑠 ) ) ∧ ( Edg ‘ 𝑠 ) ⊆ 𝒫 ( Vtx ‘ 𝑠 ) ) }
2 1 relopabiv Rel SubGraph