Equivalent statements in the definition of Brownian Motion

Question

I'm reading Brownian Motion: An Introduction to Stochastic Processes by Schilling. In the first chapter they give the following definition for some random variable $X_t$ to be considered a Brownian motion:

1. $B_0(\omega)=0$ for almost all $\omega$
2. $B_{t_n}-B_{t_{n-1}},...,B_{t_1}-B_{t_0}$ are independent for $0 = t_0\leq t_1\leq ... \leq t_n$ for $n\geq 0$
3. $B_t-B_s\sim B_{t+h}-B_{s+h}$ for $0\leq s< t$ and $h\geq -s$
4. $B_t-B_s\sim \mathcal{N}(0,t-s)$
5. $t\mapsto B_t$ is continuous

My confusion is that it seems like condition 2 and 4 are redundant, in that:

$$\begin{align}Cov(B_t-B_s,B_a-B_b)&=\mathbb{E}((B_t-B_s)(B_a-B_b))\\ &=\mathbb{E}(B_tB_a-B_sB_a-B_tB_b+B_sB_b)\\ &=(t\wedge a -s\wedge a-t\wedge b+s\wedge b)\end{align}$$

For nonoverlapping intervals $[t,s]$ and $[a,b]$, such as those described in condition 2, the covariance will always evaluate to 0. Meaning that condition 4 implies condition 2.

I'm wondering if there's some flaw in my reasoning with this, or if I made some mistake up to this point.

Answer 1

Condition 2 ensures the increments $B_{t_n}-B_{t_{n-1}},...,B_{t_1}-B_{t_0}$ are jointly Gaussian, which is needed for you to use that having $0$ covariance implies independence. I think that Condition 3 is redundant with Condition 4, though, since Condition 4 clearly implies Condition 3.