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 + 2) block n = 3*((n + 2) choose 4) + ((n + 2) choose 3)
proof
  defpred P[Nat] means ($1+2)block $1=3*(($1+2)choose 4)+(($1+2)
  choose 3);
A1: 2 choose 3=0 by NEWTON:def 3;
A2: for k st P[k] holds P[k+1]
  proof
    let k such that
A3: P[k];
    set k1=k+1;
    set k2=k+2;
    set k3=k2+1;
A4: k1*((k1+1) block k1)=k1*(k2 choose 2) by Th52
      .=k1*(k2*(k2-1)/2) by Th51
      .=k2*(k2-1)*(k1*12/24);
    k2 block k=3*(k2*(k2-1)*(k2-2)*(k2-3)/24)+(k2 choose 3) by A3,Th51
      .=3*(k2*(k2-1)*(k2-2)*(k2-3)/24)+(k2*(k2-1)*(k2-2)/6) by Th51
      .=k2*(k2-1)*(3*(k2-2)*(k2-3)/24 +4*(k2-2)/24);
    then k3 block k1= k2*k1*(k1*12/24)+k2*k1*(3*k*(k2-3)/24+4*k/24) by A4,Th46
      .=3*(k3*(k3-1)*(k3-2)*(k3-3)/24)+k3*k2*k1/6
      .=3*(k3 choose 4) +(k3*(k3-1)*(k3-2)/6) by Th51
      .=3*(k3 choose 4)+(k3 choose 3) by Th51;
    hence thesis;
  end;
  2 choose 4=0 by NEWTON:def 3;
  then
A5: P[0] by A1,Th31;
  for k holds P[k] from NAT_1:sch 2(A5,A2);
  hence thesis;
end;
