How can we show that the total variation distance of $X_s$ and $Y_s$ is bounded by the distance of $(X_t)_{t\ge s}$ and $(Y_t)_{t\ge s}$?

Let $ (X_t)_{t\ge0}$ and $ (Y_t)_{t\ge0}$ be real-valued time-homogeneous Markov processes with a common transition semigroup $ (\kappa_t)_{t\ge0}$ . Let $ \mathcal L(Z)$ denote the distribution of a random variable $ Z$ and $ $ \left|\mu-\nu\right|:=\sup_{B\in\mathcal E}\left|\mu(B)-\nu(B)\right|$ $ denote the total variation distance of probability measures $ \mu,\nu$ defined on a common $ \sigma$ -algebra $ \mathcal E$ .

How can we show that $ $ \left|\mathcal L(X_s)-\mathcal (Y_s)\right|\le\left|\mathcal L\left(\left(X_{s+t}\right)_{t\ge0}\right)-\mathcal L\left(\left(Y_{s+t}\right)_{t\ge0}\right)\right|\tag1$ $ for all $ s\ge0$ ?

We know that $ $ \mathcal L\left(X_{t_0},\ldots,X_{t_n}\right)=\mathcal L\left(X_{t_0}\right)\otimes\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}\tag2$ $ for all $ n\in\mathbb N_0$ and $ 0\le t_0\le\cdots\le t_n$ , where the right-hand side of $ (2)$ denote the product of transition kernels. Thus, $ $ \mathcal L(X_t)=\mathcal L(X_0)\kappa_t\;\;\;\text{for all }t\ge0\tag3,$ $ where the right-hand side of $ (3)$ denotes the composition of transition kernels. Moreover, $ \mathcal L\left(\left(X_{s+t}\right)_{t\ge0}\right)$ is uniquely determined of the finite-dimensional distributions $ (2)$ . So, it intuitively seems to be reasonable that $ (1)$ holds. But how can we prove it?