theorem Th23:
  S1`2 = S2`2 implies CQC_Sub(CQCSub_&(S1,S2)) = (CQC_Sub(S1)) '&'
  (CQC_Sub(S2))
proof
  assume
A1: S1`2 = S2`2;
  then CQCSub_&(S1,S2) = Sub_&(S1,S2) by Def3;
  hence thesis by A1,SUBSTUT1:31;
end;
