proving $X(s+t)|X(s) = x \sim \mathcal{N}(x,\sigma^2 t)$

Question

Let $\{X(t),t\geq 0\}$ be a Brownian motion so that $X(t)\sim \mathcal{N}(0,\sigma^2 t)$.
Then for any $s,t>0$, $$X(s+t)|(X(s)= x) \sim \mathcal{N}(x,\sigma^2 t) $$ I get this intuitively (after the time unit $s$, we go another $t$ seconds. We pretend we started at $x$ at time unit $0$ and go another $t$ units which must be Brownian motion by independent and stationary increments).
But how do I prove this mathematically?
I tried this but it doesn't make sense:
Let $Y = X(s+t) - X(s) \sim X(t) \sim \mathcal{N}(0,\sigma^2 t).$
Then $Y + X(s) = X(s+t) - X(s)+X(s) \sim X(s+t) - x + x | X(s) = x \sim X(s+t) | X(s) = x \sim ???$

Answer 1

You can write $X(s+t)=Y+X(s)$ where $Y:=X(s+t)-X(s)$.

Here $Y$ and $X(s)$ are independent and $Y\sim\mathsf{Norm}(0,\sigma^2t)$.

Then under condition $X(s)=x$ the distribution of $Y+X(s)$ is the same as the distribution of $Y+x$ which is $\mathsf{Norm}(x,\sigma^2t)$.