Why is the quadratic covariation of two independent brownian motions zero?

Question

In my notes there is a remark saying that if $B$ and $B'$ are independent Brownian motions, then $\langle B,B'\rangle = 0$.

I know the following properties of quadratic covariation:

1. $\langle M, N \rangle = \frac{1}{2}\Bigl(\langle M+N, M+N \rangle - \langle M, M\rangle - \langle N,N \rangle \Bigr)$

2. The process $\langle M, N \rangle$ is the unique finite variation processes such that $(M_tN_t - \langle M,N \rangle_t)$ is a continuous local martingale.

3. $\langle M,N \rangle_t = \lim_{||\pi_n|| \rightarrow 0} \sum_{t_i \in \pi_n}(M_{t_{i+1}} - M_{t_i})(N_{t_{i+1}} - N_{t_i})$, where the limit is in probability.

I also know that the quadratic variation of a Brownian motion $\langle B, B \rangle_t = t$.

I feel like the result is a very basic combination of these properties but I cannot figure it out.

Thanks