theorem Th19:
  n <= m implies DYADIC(n) c= DYADIC(m)
proof
  assume n <= m;
  then reconsider mn = m-n as Nat by NAT_1:21;
  let o;
  assume
A1:o in DYADIC(n);
  then reconsider o as Dyadic;
  consider i be Integer such that
A2:o = i / (2|^n) by A1,Def4;
A3: (2|^n) * (2|^mn) = 2 |^ (n+mn) by NEWTON:8;
  o = (i * 2|^mn) / ((2|^n) * (2|^mn)) by A2,XCMPLX_1:91;
  hence thesis by A3,Def4;
end;
