theorem Th5:
  cods(I-->f) = I-->(cod f)
proof
  set F = I-->f, F9 = I-->(cod f);
  now
    let x;
    assume
A1: x in I;
    then F/.x = f & F9/.x = cod f by Th2;
    hence (cods F)/.x = F9/.x by A1,Def2;
  end;
  hence thesis by Th1;
end;
