theorem Th8:
  p on P & q on P & p on Q & q on Q implies p = q or P = Q
proof
  reconsider p9= p, q9= q as Point of CPS;
  reconsider P9= P, Q9= Q as LINE of CPS by Th1;
  assume that
A1: p on P & q on P and
A2: p on Q & q on Q and
A3: p<>q;
A4: p9 in Q9 & q9 in Q9 by A2,Th5;
  p9 in P9 & q9 in P9 by A1,Th5;
  hence thesis by A3,A4,COLLSP:20;
end;
