Asymptotic notation properties

Let $f(n)$ and $g(n)$ be asymptotically positive functions. Prove or disprove each of the following conjectures.

1. $f(n) = O(g(n))$ implies $g(n) = O(f(n))$.
2. $f(n) + g(n) = \Theta(min(f(n), g(n)))$.
3. $f(n) = O(g(n))$ implies $\lg (f(n)) = O(\lg(g(n)))$, where $\lg(g(n)) \ge 1$ and $f(n) \ge 1$ for all sufficiently large $n$.
4. $f(n) = O(g(n))$ implies $2^{f(n)} = O\left(2^{g(n)}\right)$.
5. $f(n) = O((f(n))^2)$.
6. $f(n) = O(g(n))$ implies $g(n) = \Omega(f(n))$.
7. $f(n) = \Theta(f(n/2))$.
8. $f(n) + o(f(n)) = \Theta(f(n))$.
1. Let $f(n) = n$ and $g(n) = n^2$. Hence, $n = O(n^2)$ but $n^2 \ne O(n)$.
2. Take the same example as above. $n^2 + n \ne \Theta(min(n^2, n))$.
3. $f(n) = O(g(n))$ implies $0 \le f(n) \le c \cdot g(n)$ for all $n \ge n_0$ such that the constants $c, n_0 > 0$. Hence, $0 \le \lg (f(n)) \le \lg c + \lg (g(n)) \le k \cdot \lg(g(n))$. Therefore, $\lg (f(n)) = O(\lg(g(n)))$.
4. Let $f(n) = 2n$ and $g(n) = n$. Hence, $f(n) = O(g(n))$ but $2^{2n} = 4^n \ne O\left(2^n\right)$.
5. If $f(n) \ge 1$ for sufficiently large values of $n$, $0 \le f(n) \le c \cdot (f(n))^2$, i.e. $f(n) = O((f(n))^2)$. However, if $% $, this conjecture does not hold.
6. $f(n) = O(g(n))$ implies $0 \le f(n) \le c \cdot g(n)$ for all $n \ge n_0$ such that the constants $c, n_0 > 0$. This inequality can be rearranged as $0 \le \frac 1 c \cdot f(n) \;e g(n)$, i.e. $g(n) = \Omega(f(n))$.
7. Let $f(n) = 4^n$. $4^n \ne \Theta(4^(n/2)) = \Theta(2^n))$.
8. From definition, $0 \le o(f(n)) \le f(n)$. Hence, $f(n) \le f(n) + o(f(n)) \le 2f(n)$. Therefore, $f(n) + o(f(n)) = \Theta(f(n))$.