Consider a standard Brownian motion $(\Omega,\mathcal{F}_t,\mathcal{F}, W_t,\mathbb{P})$. Let $dX_t = b(X_t)dt + \sigma(X_t)dW_t$ be a SDE. Assume the existence of a unique strong solution.
Let $X_t(x, s)$ denote the solution of the SDE starting from $x\in\mathbb{R}$ at time $s$.
For $u<s<t$, can the freezing lemma be applied to establish the equality:
$$ \mathbb{E}[X_t(X_s(x,u),s)\mid X_s(x,u)] = \Phi(X_s(x,u)), $$
where $$ \Phi(y) = \mathbb{E}[X_t(y,s)], $$ even in cases where $X_t$ and $ X_s $ are not independent?
This follows from the Markov property, as mentioned here The solution $X$ of the SDE $\mathrm d X_t = f(t, X_t) \mathrm d t + g(t, X_t) \mathrm d B_t$ is a Markov process one source is in : Schilling; Th.21.23. p. 403