reserve X,Y for set, p,x,x1,x2,y,y1,y2,z,z1,z2 for object;
reserve f,g,h for Function;

theorem
  f is one-to-one & g is one-to-one & h = f \/ g & rng f misses rng g
  implies h is one-to-one
proof
  assume that
A1: f is one-to-one & g is one-to-one and
A2: h = f \/ g and
A3: rng f /\ rng g = {};
  now
    let x1,x2;
    assume that
A4: x1 in dom h & x2 in dom h and
A5: h.x1 = h.x2;
A6: now
      assume x1 in dom f & x2 in dom g or x1 in dom g & x2 in dom f;
      then
      h.x1 = f.x1 & h.x2 = g.x2 & f.x1 <> g.x2 or h.x1 = g.x1 & h.x2 = f.
      x2 & f.x2 <> g.x1 by A2,A3,Lm1,Th15;
      hence x1 = x2 by A5;
    end;
A7: x1 in dom g & x2 in dom g implies h.x1 = g.x1 & h.x2 = g.x2 by A2,Th15;
    x1 in dom f & x2 in dom f implies h.x1 = f.x1 & h.x2 = f.x2 by A2,Th15;
    then x1 in dom f & x2 in dom f or x1 in dom g & x2 in dom g implies x1 =
    x2 by A1,A5,A7;
    hence x1 = x2 by A2,A4,A6,Lm3;
  end;
  hence thesis;
end;
