theorem
  f is translation & g is translation implies (f*g) is translation
proof
  assume that
A1: f is translation and
A2: g is translation;
  f is dilatation & g is dilatation by A1,A2;
  then
A3: (f*g) is dilatation by Th71;
  now
    assume
A4: (f*g)<>(id the carrier of AFS);
    for x holds (f*g).x<>x
    proof
      given x such that
A5:   (f*g).x=x;
      f.(g.x)=x by A5,FUNCT_2:15;
      then
A6:   g.x=f".x by Th2;
      f" is translation by A1,Th85;
      then (f*g) = f*f" by A2,A6,Th84;
      hence contradiction by A4,FUNCT_2:61;
    end;
    hence thesis by A3;
  end;
  hence thesis by A3;
end;
