theorem Th31:
  S1`2 = S2`2 implies CQC_Sub(Sub_&(S1,S2)) = (CQC_Sub(S1)) '&' ( CQC_Sub(S2))
proof
  set S = Sub_&(S1,S2);
  assume
A1: S1`2 = S2`2;
  then Sub_the_left_argument_of S = S1 & Sub_the_right_argument_of S = S2 by
Th18,Th19;
  hence thesis by A1,Th13,Th30;
end;
