
theorem LM001:
  for x be Nat
  st x <> 0
  ex n be Nat st 2 to_power n <= x & x < 2 to_power (n + 1)
  proof
    let x be Nat;
    assume x <> 0; then
    P1:1<=x by NAT_1:25;
    defpred P1[Nat] means x < 2 to_power $1;
    A2:ex m be Nat st P1[m] by LM000;
    consider k being Nat such that
    A3: P1[k] & ( for n being Nat st P1[n] holds
    k <= n ) from NAT_1:sch 5(A2);
    k <> 0 by P1,POWER:24,A3;
    then
    1 <= k by NAT_1:25;
    then
    0 <= k-1 by XREAL_1:48;
    then
    reconsider k1=k-1 as Nat by INT_1:3,ORDINAL1:def 12;
    take k1;
    thus 2 to_power k1 <= x
    proof
      assume ASP1:not 2 to_power k1 <= x;
      k -1 < k - 0 by XREAL_1:15;
      hence contradiction by A3,ASP1;
    end;
    thus x < 2 to_power (k1 + 1) by A3;
  end;
