
theorem Th11:
  for n being Nat st n > 0 holds
  [\ log (2, 2*n) /] +1 <> [\ log (2, 2*n + 1) /]
proof
  let n be Nat;
  set l22n = log (2, 2*n);
  set l22np1 = log (2, 2*n + 1);
  set k = [\ l22n + 1 /];
  l22n+1-1 < k by INT_1:def 6;
  then
A1: 2 to_power l22n < 2 to_power k by POWER:39;
  assume
A2: n > 0;
  then
A3: 2*n < 2 to_power k by A1,POWER:def 3;
  assume [\ log (2, 2*n) /] +1 = [\ log (2, 2*n + 1) /];
  then
A4: [\l22n+1/] = [\l22np1/] by INT_1:28;
  then k <= l22np1 by INT_1:def 6;
  then 2 to_power k <= 2 to_power l22np1 by Th8;
  then
A5: 2 to_power k <= 2*n + 1 by POWER:def 3;
  0+1 <= 2*n + 1 by XREAL_1:7;
  then log (2, 1) <= l22np1 by Th10;
  then 0 <= l22np1 by POWER:51;
  then [\ 0 /] <= k by A4,Th9;
  then 0 <= k;
  then reconsider k as Element of NAT by INT_1:3;
  reconsider T2tpk = 2 |^ k as Element of NAT;
  2*n < T2tpk by A3,POWER:41;
  then
A6: 2*n + 1 <= T2tpk by NAT_1:13;
  T2tpk <= 2*n + 1 by A5,POWER:41;
  then
A7: T2tpk = 2*n+1 by A6,XXREAL_0:1;
  per cases by NAT_1:6;
  suppose
    k = 0;
    then 1-1 = 2*n+1-1 by A7,NEWTON:4;
    hence contradiction by A2;
  end;
  suppose
    ex m being Nat st k = m + 1;
    then consider m being Nat such that
A8: k = m + 1;
    reconsider m as Element of NAT by ORDINAL1:def 12;
    2*2|^m + 0 = 2*n+1 by A7,A8,NEWTON:6;
    then 0 = (2*n+1) mod 2 by NAT_D:def 2;
    hence contradiction by NAT_D:def 2;
  end;
end;
