reserve a,b,c,h for Integer;
reserve k,m,n for Nat;
reserve i,j,z for Integer;
reserve p for Prime;

theorem Th100:
  2|^2 <= n & 2|^n divides i|^3 implies 2|^2 divides i
  proof
    per cases;
    suppose i = 0;
      hence thesis by INT_2:12;
    end;
    suppose
A1:   i <> 0;
      assume
A2:   2|^2 <= n;
      assume 2|^n divides i|^3;
      then 2|^n divides |.i|^3.| by Th4;
      then
A3:   2|^n divides |.i.||^3 by TAYLOR_2:1;
      consider a being Nat, b being odd Nat such that
A4:   |.i.| = b * 2|^a by A1,NAT_6:3;
A5:   |.i.||^3 = b|^3 * 2|^a|^3 by A4,NEWTON:7;
A6:   b|^3,2|^n are_coprime by NAT_5:3;
      2|^a|^3 = 2|^(3*a) by NEWTON:9;
      then n <= 3*a by A3,A5,A6,INT_2:25,PEPIN:31;
      then
A7:   2|^2 <= 3*a by A2,XXREAL_0:2;
      now
        assume a < 1+1;
        then a <= 1 by NAT_1:13;
        then 3*a <= 3*1 by XREAL_1:64;
        hence contradiction by A7,Lm1,XXREAL_0:2;
      end;
      then 2|^2 divides |.i.| by A4,INT_2:2,PEPIN:31;
      hence thesis by Th4;
    end;
  end;
