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

theorem
  dom f misses dom g implies f \/ g is Function
proof
  assume
A1: dom f /\ dom g = {};
  now
    thus p in f \/ g implies ex x,y being object st [x,y] = p
by RELAT_1:def 1;
    let x,y1,y2;
    assume [x,y1] in f \/ g;
    then
A2: [x,y1] in f or [x,y1] in g by XBOOLE_0:def 3;
    assume [x,y2] in f \/ g;
    then
A3: [x,y2] in f or [x,y2] in g by XBOOLE_0:def 3;
    not (x in dom f & x in dom g) by A1,XBOOLE_0:def 4;
    hence y1 = y2 by A2,A3,FUNCT_1:def 1,XTUPLE_0:def 12;
  end;
  hence thesis by FUNCT_1:def 1;
end;
