theorem Th21:
  (for Sub holds QuantNbr(p) = QuantNbr(CQC_Sub([p,Sub]))) & (for
  Sub holds QuantNbr(q) = QuantNbr(CQC_Sub([q,Sub]))) implies for Sub holds
  QuantNbr(p '&' q) = QuantNbr(CQC_Sub[p '&' q,Sub])
proof
  assume that
A1: for Sub holds QuantNbr(p) = QuantNbr(CQC_Sub([p,Sub])) and
A2: for Sub holds QuantNbr(q) = QuantNbr(CQC_Sub([q,Sub]));
  let Sub;
  set S = [p '&' q,Sub];
  set S1 = [p,Sub];
  set S2 = [q,Sub];
A3: S1`2 = Sub & S2`2 = Sub;
  S = CQCSub_&(S1,S2) by Th19;
  then S = Sub_&(S1,S2) by A3,SUBLEMMA:def 3;
  then CQC_Sub(S) = (CQC_Sub(S1)) '&' (CQC_Sub(S2)) by A3,SUBSTUT1:31;
  then
  QuantNbr(CQC_Sub(S)) = QuantNbr(CQC_Sub(S1)) + QuantNbr(CQC_Sub(S2)) by
CQC_SIM1:17
    .= QuantNbr(p) + QuantNbr(CQC_Sub(S2)) by A1
    .= QuantNbr(p) + QuantNbr(q) by A2;
  hence thesis by CQC_SIM1:17;
end;
