theorem Th27:
  p <> p1 & p <> p2 & p in LSeg(p1,p2) implies not p1 in LSeg(p,p2 )
proof
  assume that
A1: p <> p1 & p <> p2 and
A2: p in LSeg(p1,p2);
A3: LSeg(p1,p) \/ LSeg(p,p2) = LSeg(p1,p2) by A2,TOPREAL1:5;
  now
    assume p1 in LSeg(p,p2);
    then
A4: LSeg(p,p1) \/ LSeg(p1,p2) = LSeg(p,p2) by TOPREAL1:5;
    LSeg(p,p1) \/ LSeg(p1,p2) = LSeg(p1,p2) by A3,XBOOLE_1:7,12;
    hence contradiction by A1,A4,SPPOL_1:8;
  end;
  hence thesis;
end;
