reserve k, l, m, n, i, j for Nat,
  K, N for non empty Subset of NAT,
  Ke, Ne, Me for Subset of NAT,
  X,Y for set;
reserve f for Function of Segm n,Segm k;
reserve x,y for set;

theorem
  n >= 3 implies n block 4 = 1/24 * ( 4 |^ n - 4 * 3 |^ n + 6* 2|^ n - 4 )
proof
  defpred P[Nat] means $1 block 4= 1/24 * ( 4 |^ $1 - 4 * 3 |^ $1 + 6* 2|^ $1
  - 4 );
A1: for k be Nat st k>=3 & P[k] holds P[k+1]
  proof
    let k be Nat such that
A2: k>=3 and
A3: P[k];
    k block 3 = 1/6 * (3|^k-3*2|^k+3) by A2,Th48,XXREAL_0:2;
    hence (k+1) block 4 = 4*(k block (3+1)) +(1/6*(3|^k-3*2|^k+3)) by Th46
      .=1/24*((4*4|^k)- 4*(3*3|^k)+ 6*(2*2|^k) - 4) by A3
      .=1/24*(4|^(k+1)- 4*(3*3|^k)+ 6*(2*2|^k) - 4) by NEWTON:6
      .=1/24*(4|^(k+1)- 4*3|^(k+1)+ 6*(2*2|^k) - 4) by NEWTON:6
      .=1/24*(4|^(k+1)- 4*(3|^(k+1))+ 6*2|^(k+1) - 4) by NEWTON:6;
  end;
  1/24*(4|^3-4*3|^3+6*2|^3-4)= 1/24*(4*4*4-4*3|^3+6*2|^3-4) by Lm3
    .=1/24*(64-4*(3*3*3)+6*2|^3-4) by Lm3
    .=1/24*(64-4*27+6*(2*2*2)-4) by Lm3
    .=3 block 4 by Th29;
  then
A4: P[3];
  for k be Nat st k>=3 holds P[k] from NAT_1:sch 8(A4,A1);
  hence thesis;
end;
