theorem Th36:
  CQC_Sub(S) is Element of CQC-WFF(A) implies CQC_Sub(Sub_not S) is
  Element of CQC-WFF(A)
proof
  set S9 = Sub_not S;
  assume
A1: CQC_Sub(S) is Element of CQC-WFF(A);
  CQC_Sub(S9) = 'not' CQC_Sub(S) by Th29;
  hence thesis by A1,CQC_LANG:8;
end;
