theorem Th6:
  dom (f^^) = dom (f|(dom (f^)))
proof
A1: dom (f^) = dom f \ f"{0} by Def2;
  thus dom (f^^) = dom (f^) \(f^)"{0} by Def2
    .= dom (f^) \ {} by Th4
    .= dom f /\ dom (f^) by A1,XBOOLE_1:28
    .= dom (f|(dom (f^))) by RELAT_1:61;
end;
