theorem Th60:
  X '||' Y & Y '||' Z & Y <> {} implies X '||' Z
proof
  assume that
A1: X '||' Y and
A2: Y '||' Z and
A3: Y <> {};
  set x = the Element of Y;
  reconsider p=x as Element of AS by A3,Lm1;
  now
    let a,A;
    assume that
A4: a in Z and
A5: A is being_line and
A6: A c= X;
    p*A c= Y & p*A is being_line by A1,A3,A5,A6,Th27;
    then a*(p*A) c= Z by A2,A4;
    hence a*A c= Z by A5,Th31;
  end;
  hence thesis;
end;
