reserve x, y, z, v for set,
  n, m, k for Nat;
reserve G, G1, G2, G3 for Graph;
reserve x, y for Element of (the carrier of G);

theorem Th16:
  for G1,G2 being strict Graph holds G1 c= G2 & G2 c= G1 implies G1 = G2
proof
  let G1,G2 be strict Graph;
  assume that
A1: G1 c= G2 and
A2: G2 c= G1;
A3: G1 is Subgraph of G2 by A1;
A4: G2 is Subgraph of G1 by A2;
A5: (the carrier of G1) c= (the carrier of G2) by A3,Def18;
 (the carrier of G2) c= (the carrier of G1) by A4,Def18;
then A6: the carrier of G1 = the carrier of G2 by A5,XBOOLE_0:def 10;
A7: (the carrier' of G1) c= (the carrier' of G2) by A3,Def18;
 (the carrier' of G2) c= (the carrier' of G1) by A4,Def18;
then A8: the carrier' of G1 = the carrier' of G2 by A7,XBOOLE_0:def 10;
then A9: dom(the Source of G1) = the carrier' of G2 by FUNCT_2:def 1
    .= dom(the Source of G2) by FUNCT_2:def 1;
 for v being object st v in dom (the Source of G1) holds
  (the Source of G1).v = (the Source of G2).v by A3,Def18;
then A10: the Source of G1 = the Source of G2 by A9;
A11: dom (the Target of G1) = the carrier' of G2 by A8,FUNCT_2:def 1
    .= dom (the Target of G2) by FUNCT_2:def 1;
 for v being object st v in dom (the Target of G1) holds
  (the Target of G1).v = (the Target of G2).v by A3,Def18;
  hence thesis by A6,A8,A10,A11,FUNCT_1:2;
end;
