theorem Th23:
  S is Sub_conjunctive implies len @((Sub_the_left_argument_of(S))
  `1) < len @(S`1) & len @((Sub_the_right_argument_of(S))`1) < len @(S`1)
proof
  assume S is Sub_conjunctive;
  then consider S1,S2 such that
A1: S = Sub_&(S1,S2) & S1`2 = S2`2;
  S = [(S1`1) '&' (S2`1),S1`2] by A1,Def21;
  then
A2: S`1 = (S1`1) '&' (S2`1);
  (S1`1) '&' (S2`1) is conjunctive;
  then
A3: len @the_left_argument_of (S1`1) '&' (S2`1) < len @(S`1) & len @
  the_right_argument_of (S1`1) '&' (S2`1) < len @(S`1) by A2,QC_LANG1:15;
  (Sub_the_right_argument_of(S))`1 = S2`1 & (Sub_the_left_argument_of(S))
  `1 = S1`1 by A1,Th18,Th19;
  hence thesis by A3,QC_LANG2:4;
end;
