theorem Th20:
  S1`2 = S2`2 implies CQCSub_&(S1,S2)`1 = (S1`1) '&' (S2`1) &
  CQCSub_&(S1,S2)`2 = S1`2
proof
  assume
A1: S1`2 = S2`2;
  then Sub_&(S1,S2) = [(S1`1) '&' (S2`1),S1`2] by SUBSTUT1:def 21;
  then CQCSub_&(S1,S2) = [(S1`1) '&' (S2`1),S1`2] by A1,Def3;
  hence thesis;
end;
