Suppose we have random variables $X$ and $Y$ with $X-Y \sim N(\mu,\sigma^2)$. I want to show $X|Y \sim N(Y+\mu, \sigma^2)$ in the sense that $$ E(g(X)|Y) = \int g(x) \frac{1}{\sqrt{2\pi \sigma^2}}\exp(-(x-(Y+\mu))^2/2\sigma^2)dx $$ for every measurable function $g:\mathbb{R} \to \mathbb{R}$.
Attempt
I thought I could proceed as follows. Set $h(z)=g(z)+Y$. \begin{align} E(g(X)|Y) &= E(g(X-Y+Y)|Y) \\ &= E(h(X-Y)|Y) \\ &= \label{1}\tag{1} \int h(z) \frac{1}{\sqrt{2\pi \sigma^2}}\exp(-(z-\mu)^2/2\sigma^2)dz \\ &= \int g(z+Y) \frac{1}{\sqrt{2\pi \sigma^2}}\exp(-(z-\mu)^2/2\sigma^2)dz \\ &= \int g(x) \frac{1}{\sqrt{2\pi \sigma^2}}\exp(-(x-(\mu+Y))^2/2\sigma^2)dx \end{align} where the last line is a change of variable.
However, \eqref{1} doesn't seem correct unless $X-Y$ is independent from $Y$.
Context
Brownian motion with $W_0 = 0$, $W_{t} - W_{s} \sim N(0,t-s)$ when $0 \leq s \leq t$, $W_{t} - W_{s}$ independent from $W_{t'} - W_{s'}$ when $0 \leq s < t \leq s' < t'$. I'd like to argue that $W_t | W_s \sim N(W_s,\sigma^2)$ with the meaning above.
Do I need to use independent increments somehow? If so, how?
Edit: I came up with this before seeing @Marcus M 's answer; it's the same I think:
Would it be okay to argue as follows? $W_t - W_s$ is independent from $W_s-W_0 = W_s$, so $E(h(W_t-W_s)|W_s) = E(h(W_t-W_s)|W_s-W_0) = E(h(W_t-W_s))$. I think this even works if $W_0$ is a non-zero constant because conditioning on $W_s-W_0$ would still be equivalent to conditioning on $W_s$?
As you note, line $(1)$ is only true if $X - Y$ is independent of $Y$. In the Brownian motion case, you're all good since you know $W_t - W_s$ is independent of $W_s$.
To see that your first statement isn't true in general, note that if you take $Y \sim N(0,1)$ and $X = 2Y$ then $X - Y = Y$ so $E[g(X)|Y] = g(X)$ since $X$ is measurable with respect to $\sigma(Y)$ (where $\sigma(Y)$ is the smallest $\sigma$-algebra for which $Y$ is measurable.)