theorem
  p => 'not' q in TAUT(A) implies q => 'not' p in TAUT(A)
proof
  assume
A1: p => 'not' q in TAUT(A);
  (p => 'not' q) => (q => 'not' p) in TAUT(A) by Th30;
  hence thesis by A1,CQC_THE1:46;
end;
