reserve X,X1,X2,Y,Y1,Y2 for set, p,x,x1,x2,y,y1,y2,z,z1,z2 for object;
reserve f,g,g1,g2,h for Function,
  R,S for Relation;

theorem
  rng(g*f) = rng g & g is one-to-one implies dom g c= rng f
proof
  assume that
A1: rng(g*f) = rng g and
A2: g is one-to-one;
  let y be object;
  assume
A3: y in dom g;
  then g.y in rng(g*f) by A1,Def3;
  then consider x being object such that
A4: x in dom(g*f) and
A5: g.y = (g*f).x by Def3;
  (g*f).x = g.(f.x) & f.x in dom g by A4,Th11,Th12;
  then
A6: y = f.x by A2,A3,A5;
  x in dom f by A4,Th11;
  hence thesis by A6,Def3;
end;
