theorem Th18:
  (for Sub holds QuantNbr(p) = QuantNbr(CQC_Sub([p,Sub]))) implies
  for Sub holds QuantNbr('not' p) = QuantNbr(CQC_Sub(['not' p,Sub]))
proof
  assume
A1: for Sub holds QuantNbr(p) = QuantNbr(CQC_Sub([p,Sub]));
  let Sub;
  set S = ['not' p,Sub];
  S = Sub_not [p,Sub] by Th16;
  then QuantNbr(CQC_Sub(S)) = QuantNbr('not' CQC_Sub([p,Sub])) by SUBSTUT1:29
    .= QuantNbr(CQC_Sub([p,Sub])) by CQC_SIM1:16
    .= QuantNbr(p) by A1;
  hence thesis by CQC_SIM1:16;
end;
