Proposition: Let us consider a Brownian motion $W(t)$, $t\geq0$. For fixed $t_0\geq0$, the stochastic process $\widetilde{W}(t)=W(t+t_0)-W(t_0)$ is also a Brownian Motion.
Proof: Let us take properties $1.$, $3.$ and $4.$ in here for granted as to stochastic process $\widetilde{W}(t)$ and focus on the proof of property $2.$ for $\widetilde{W}(t)$. First, consider that for any $s<t$: $$\widetilde{W}(t)-\widetilde{W}(s)=W(t+t_0)-W(s+t_0)\tag{1}$$ To check property $2.$, we may assume that $t_0>0$.
Then, for any $0\leq t_1<t_2<\ldots<t_n$, we have $0<t_0\leq t_1+t_0<\ldots<t_n+t_0$. By property $2.$ for $W(t)$, $W(t_k+t_0)-W(t_{k-1}+t_0)$, $k=1,2,\ldots,n$ are independent random variables. Thus, by $(1)$, the random variables $\widetilde{W}(t_k)-\widetilde{W}(t_{k-1})$, $k=1,2,\ldots,n$ are independent and so $\widetilde{W}(t)$ satisfies property $2.$.
Starting from $0<t_0\leq t_1+t_0<\ldots<t_n+t_0$, I would say that property $2.$ does not necessary apply for $W_t$ in correspondence of $t=1$, since, given the above assumptions, I cannot be sure that, considering $W(t_1+t_0)-W(t_0+t_0)$, $(t_1+t_0)>(t_0+t_0)$, that is, in other words, that $W(t_1+t_0)-W(t_0+t_0)$ can be actually considered as an increment in time.
So, I would say that there is a missing assumption so as that $(1)$ holds true for all $k\geq1$ and not just for $k>1$, that is $$t_1>t_0\tag{2}$$ Would you agree with me as to the fact that assumption $(2)$ is necessary so as that $(1)$ holds true for $k=1$? If not, why am I mistaken?