theorem
  a,b // A & not a in A implies not b in A
proof
  assume that
A1: a,b // A and
A2: not a in A and
A3: b in A;
A4: b,a // A by A1,Th33;
  A is being_line by A1;
  hence contradiction by A2,A3,A4,Th22;
end;
