
theorem th0n:
for n being Nat st n >= 3 holds n + n < 2|^n
proof
let n be Nat;
assume AS: n >= 3;
defpred P[Nat] means $1 + $1 < 2|^($1);
IA: P[3]
    proof
    2|^(2+1) = 2|^(1+1) * 2 by NEWTON:6
            .= (2|^1 * 2|^1) * 2 by NEWTON:6
            .= (2 * 2) * 2;
    hence 2|^3 > 3 + 3;
    end;
IS: now let k be Nat;
    assume A: k >= 3;
    assume B: P[k];
    C: 2|^(k+1) = 2|^k * 2 by NEWTON:6
               .= 2|^k + 2|^k;
    D: 2|^k + 2|^k > (k + k) + (k + k) by B,XREAL_1:8;
    k > 1 by A,XXREAL_0:2; then
    k + k > k + 1 by XREAL_1:8; then
    (k + k) + (k + k) > (k + 1) + (k + 1) by XREAL_1:8;
    hence P[k+1] by C,D,XXREAL_0:2;
    end;
for k being Nat st k >= 3 holds P[k] from NAT_1:sch 8(IA,IS);
hence thesis by AS;
end;
