theorem
  (f")\g = (f\g)"
proof
  thus (f\g)" = (g")"*(g*f)" by FUNCT_1:44
    .= (g")"*(f"*g") by FUNCT_1:44
    .= g*(f"*g") by FUNCT_1:43
    .= f"\g by RELAT_1:36;
end;
