theorem Th91:
  G1 == G2 & G3 is Subgraph of G1 implies G3 is Subgraph of G2
proof
  assume that
A1: G1 == G2 and
A2: G3 is Subgraph of G1;
  the_Vertices_of G3 c= the_Vertices_of G1 by A2,Def32;
  hence the_Vertices_of G3 c= the_Vertices_of G2 by A1;
  the_Edges_of G3 c= the_Edges_of G1 by A2,Def32;
  hence the_Edges_of G3 c= the_Edges_of G2 by A1;
  let e be set;
  assume
A3: e in the_Edges_of G3;
  hence (the_Source_of G3).e = (the_Source_of G1).e by A2,Def32
    .= (the_Source_of G2).e by A1;
  thus (the_Target_of G3).e = (the_Target_of G1).e by A2,A3,Def32
    .= (the_Target_of G2).e by A1;
end;
