theorem Th9:
  rng f = {x} implies f = (dom f) --> x
proof
  assume
A1: rng f = {x};
  then dom f <> {} by RELAT_1:42;
  then dom((dom f) --> x) = dom f & rng((dom f) -->x) = {x} by RELAT_1:160;
  hence thesis by A1,FUNCT_1:7;
end;
