I am self-learning introductory stochastic calculus from A first course in stochastic Calculus, by Louis Pierre Arguin. I would like to ask, if the below proof attempt is technically correct and rigorous.
Proposition(Reflection at time $s$). Let $(B_t,t\geq 0)$ be a standard Brownian motion. The process $(-B_{t},t\geq0)$ is a Brownian motion. More generally, for any $s\geq0$, the process $(\tilde{B}(t),t\geq0)$ defined by:
\begin{align} \tilde{B}(t) & =\begin{cases} B_{t} & \text{if }t\leq s\\ B_{s}-(B_{t}-B_{s}) & \text{if }t>s \end{cases}\tag{1} \end{align}
is a Brownian motion.
My Proof Attempt.
(a) Consider the process $\tilde{B}(t)=(-B_{t},t\geq0)$.
(1) $\tilde{B}(0)=0$.
(2) If $X$ is a Gaussian random variable with mean $0$ and variance $t-s$, $-X$ is also Gaussian with mean $0$ and variance $t-s$. Thus, $\tilde{B}(t)-\tilde{B}(s)=-(B(t)-B(s))$ is also Gaussian with mean $0$ and variance $(t-s)$. Hence condition (2) is satisfied.
(3) Assume that $0\leq t_{0}\leq t_{1}\leq\ldots\leq t_{n}$. Then, the random variables $-(B(t_{k})-B(t_{k-1}))$ are independent for $k=1,2,3,\ldots,n$. Hence, condition (3) is satisfied.
Consequently, $(-B(t),t\geq0)$ is a Brownian Motion.
(b) Fix an $s\geq0$. Consider the process $\tilde{B}(t)$ as defined in (1).
(1) Let $t=0$. Then, $t\leq s$. $\tilde{B}(t)=\tilde{B}(0)=B(0)=0$.
(2) Let $t_{1}<t_{2}\leq s$. Then, $\tilde{B}(t_{2})-\tilde{B}(t_{1})=B(t_{2})-B(t_{1})$. This is a Gaussian random variable with mean $0$ and variance $t_{2}-t_{1}$.
Let $t_{1}<s<t_{2}$. Then, $\tilde{B}(t_{2})-\tilde{B}(t_{1})=B(s)-(B(t_{2})-B(s))-B(t_{1})=(B(s)-B(t_{1}))-(B(t_{2})-B(s))$. Since, $B(s)-B(t_{1})$ and $B(t_{2})-B(s)$ are independent Gaussian random variables, any linear combination of these is Gaussian. Moreover, its mean is zero. The variance is given by:
\begin{align*} Var[\tilde{B}(t_{2})-\tilde{B}(t_{1})] & =Var[B(s)-B(t_{1})]+Var[B(t_{2})-B(s)]\\ & =(s-t_{1})+(t_{2}-s)\\ & =t_{2}-t_{1} \end{align*}
Let $s<t_{1}<t_{2}$. Then, \begin{align*} \tilde{B}(t_{2})-\tilde{B}(t_{1}) & =B_{s}-(B_{t_{2}}-B_{s})-(B_{s}-(B_{t_{1}}-B_{s}))\\ & =-(B_{t_{2}}-B_{t_{1}}) \end{align*}
Hence, $\tilde{B}(t_{2})-\tilde{B}(t_{1})$ is again a Gaussian random variable with mean $0$ and variance $t_{2}-t_{1}$. Hence, condition (3) is satisfied.
(3) Assume that $0\leq t_{1}\leq\ldots\leq t_{k-1}\leq s\leq t_{k}\leq\ldots\leq t_{n}$.
From the above discussion, the increments $\tilde{B}(t_{2})-\tilde{B}(t_{1})$, $\ldots$, $\tilde{B}(s)-\tilde{B}(t_{k-1})$, $\tilde{B}(t_{k})-\tilde{B}(s)$, $\ldots$, $\tilde{B}(t_{n})-\tilde{B}(t_{n-1})$ are independent increments. The increment $\tilde{B}(t_{k})-\tilde{B}(t_{k-1})$ only depends on the random variables $\tilde{B}(s)-\tilde{B}(t_{k-1})$ and $\tilde{B}(t_{k})-\tilde{B}(s)$. Thus, $\tilde{B}(t_{2})-\tilde{B}(t_{1})$, $\ldots$, $\tilde{B}(t_{k})-\tilde{B}(t_{k-1})$, $\ldots$, $\tilde{B}(t_{n})-\tilde{B}(t_{n-1})$
are independent.