theorem
  for H being non empty SubStr of G holds .:(H,X) is SubStr of .:(G,X)
proof
  let H be non empty SubStr of G;
A1: op(.:(H,X)) = (op(H), carr(H)).:X by Th17;
  op(H) c= op(G) & op(.:(G,X)) = (op(G), carr(G)).:X by Th17,MONOID_0:def 23;
  hence op(.:(H,X)) c= op(.:(G,X)) by A1,Th16;
end;
