theorem
  Sub_not(S) = Sub_not(S9) implies S = S9
proof
  assume Sub_not(S) = Sub_not(S9);
  then
A1: 'not' S`1 = 'not' (S9)`1 & S`2 = (S9)`2 by XTUPLE_0:1;
  S = [S`1,S`2] & S9 = [(S9)`1,(S9)`2] by Th10;
  hence thesis by A1,FINSEQ_1:33;
end;
