
theorem Th7:
  for n,i being Nat st 0 < i & i <= 2|^n holds (i*2-1)/
  (2|^(n+1)) in dyadic(n+1) \ dyadic(n)
proof
  let n,i be Nat;
  assume that
A1: 0 < i and
A2: i <= 2|^n;
 0*2 < i*2 by A1,XREAL_1:68;
  then consider k being Nat such that
A3: i*2 = k + 1 by NAT_1:6;
A4: not (i*2-1)/(2|^(n+1)) in dyadic(n)
  proof
    assume (i*2-1)/(2|^(n+1)) in dyadic(n);
    then consider s being Nat such that
    s <= 2|^n and
A5: (i*2-1)/(2|^(n+1)) = s/(2|^n) by Def1;
A6: 2|^n <> 0 by NEWTON:83;
    2|^(n+1) <> 0 by NEWTON:83;
    then (i*2-1)*(2|^n) = s*(2|^(n+1)) by A5,A6,XCMPLX_1:95;
    then (i*2-1)*(2|^n) = s*((2|^n)*2) by NEWTON:6;
    then (i*2-1)*(2|^n) = s*2*(2|^n);
    then (i*2-1)/(2|^n) = s*2/(2|^n) by A6,XCMPLX_1:94;
    then i*2+(-1) = s*2 by A6,XCMPLX_1:53;
    then 2*i + 0 = 2*s + 1;
    then 0 = 1 by NAT_1:18;
    hence thesis;
  end;
  i*2 <= (2|^n)*2 by A2,XREAL_1:64;
  then i*2 <= 2|^(n+1) by NEWTON:6;
  then
A7: k <= 2|^(n+1) by A3,NAT_1:13;
  (i*2-1)/(2|^(n+1)) in dyadic(n+1) by A3,A7,Def1;
  hence thesis by A4,XBOOLE_0:def 5;
end;
