reserve d,i,j,k,m,n,p,q,x,k1,k2 for Nat,
  a,c,i1,i2,i3,i5 for Integer;

theorem
  2 |^ (n + 1) = (2 |^ n)+(2 |^ n)
proof
  defpred P[Nat] means 2 |^ ($1+1) = (2 |^ $1)+(2 |^ $1);
A1: for m st P[m] holds P[m+1]
  proof
    let m;
    assume
A2: 2 |^ (m+1) = (2 |^ m)+(2 |^ m);
    (2 |^ (m+1))+(2 |^ (m+1)) = 2*(2 |^ m)+(2 |^ (m+1)) by NEWTON:6
      .= 2*(2 |^ m)+2*(2 |^ m) by NEWTON:6
      .= 2*((2 |^ m)+(2 |^ m))
      .= 2 |^ ((m+1)+1) by A2,NEWTON:6;
    hence thesis;
  end;
  2 |^ (0+1) = 1 + 1
    .= (2 |^ 0)+1 by NEWTON:4
    .= (2 |^ 0)+(2 |^ 0) by NEWTON:4;
  then
A3: P[0];
  for n holds P[n] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;
