Proof that $X_t=(t-T)^+$, where $T$ is exponentially distributed, defines a Markov process $(X_t)$

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Let $(\mathbb{R}^{+}, \mathbb{B}, P)$ be a probability space with probability $P$ given by its distribution function

$$F(y) = \left\{\begin{array}{ll} 1 - e^{-y} &, y\ge 0\\ 0 &,\text{otherwise} \end{array}\right.$$

Let $X_t(w) = \left\{\begin{array}{ll} 0 &, w\ge t\\ t-w &, 0<w<t \end{array}\right.\quad $ be a stochastic process, with $w\in \mathbb{R}^{+}$. Is it a Markov process?

My answer was no, because (with $0 < u < s < t$):

$$P[X_t = 0\mid X_s \in (-\infty, s]] = \frac{P([t,+\infty) \cap \mathbb{R}^{+})}{P[X_s \in (-\infty, s]]} = \frac{P([t,+\infty) \cap \mathbb{R}^{+})}{P(\mathbb{R}^{+})} = P([t, +\infty))$$

that is not the same as:

$$P[X_t = 0\mid X_u \in (-\infty, 0], X_s \in (-\infty, s]] = \frac{P([t,+\infty))}{P([u, +\infty))} = \frac{P([t, +\infty))}{1-F(u)}$$

The professor told me my argument was wrong, but I don't know why. In case it is indeed a Markov process, could you give a proof?