
theorem LM002:
  for x be Nat,
  n1,n2 be Nat st
  2 to_power n1 <= x & x < 2 to_power (n1 + 1)
  &
  2 to_power n2 <= x & x < 2 to_power (n2 + 1)
  holds n1=n2
  proof
    let x be Nat, n1,n2 be Nat;
    assume that
    AS1:
    2 to_power n1 <= x & x < 2 to_power (n1 + 1)
    and
    AS2:
    2 to_power n2 <= x & x < 2 to_power (n2 + 1);
    assume n1<>n2;
    then
    per cases by XXREAL_0:1;
    suppose n1 < n2;
      then
      P1: n1 + 1 <= n2 by NAT_1:13;
      2 to_power (n1 + 1) <= 2 to_power n2
      proof
        per cases by P1,XXREAL_0:1;
        suppose n1+1 = n2;
          hence 2 to_power (n1 + 1) <= 2 to_power n2;
        end;
        suppose n1+1 < n2;
          hence 2 to_power (n1 + 1) <= 2 to_power n2 by POWER:39;
        end;
      end;
      hence contradiction by AS2,AS1,XXREAL_0:2;
    end;
    suppose n2 < n1; then
      P1: n2 + 1 <= n1 by NAT_1:13;
      2 to_power (n2 + 1) <= 2 to_power n1
      proof
        per cases by P1,XXREAL_0:1;
        suppose n2+1 = n1;
          hence 2 to_power (n2 + 1) <= 2 to_power n1;
        end;
        suppose n2+1 < n1;
          hence 2 to_power (n2 + 1) <= 2 to_power n1 by POWER:39;
        end;
      end;
      hence contradiction by AS1,AS2,XXREAL_0:2;
    end;
  end;
