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 ) C_ ( Vtx ` g ) /\ ( iEdg ` s ) = ( ( iEdg ` g ) |` dom ( iEdg ` s ) ) /\ ( Edg ` s ) C_ ~P ( Vtx ` s ) ) }
2 1 relopabiv
 |-  Rel SubGraph