theorem Th23:
  (f|^i).v1 = 0.V1 implies (f|^(i+j)).v1=0.V1
proof
  assume
A1: (f|^i).v1 = 0.V1;
A2: dom (f|^i)=the carrier of V1 by FUNCT_2:def 1;
  thus (f|^(i+j)).v1 = ((f|^j)*(f|^i)).v1 by Th20
    .= (f|^j).(0.V1) by A1,A2,FUNCT_1:13
    .=0.V1 by RANKNULL:9;
end;
