Let $(X_{n})_{n \geq 0}$ be a sequence of random variables and $\tau,t$ stopping times with respect to the sequence $(X_{n})_{n \geq 0}$
$$\begin{align*}\{\tau+t =n\} = \{\tau+t = n\} \cap \{t \leq n\} &= \bigcup_{k=0}^n \{\tau+t = n\} \cap \{t = k\} \\ &= \bigcup_{k=0}^n \{\tau=n-k\} \cap \{t=k\}. \end{align*}$$
As $\{\tau =n-k\} \in \mathcal{F}_{n-k} \subseteq \mathcal{F}_n$ and $\{t = k\} \in \mathcal{F}_k \subseteq \mathcal{F}_n$ for any $k \leq n$, this implies that $\{\tau+t=n\} \in \mathcal{F}_n$, and so $\tau+t$ is a stopping time.
Now, my question is, let assume that I am considering $\tau-t$. I know that in general, $\tau-t$ is not a stopping time. However, if I were to consider my birthday this year (a stopping time), which is a deterministic stopping time. At any time, I know exactly when my birthday occurs. Also, I know two days before my birthday i.e, $\tau-2$. What kind of a formulated counterexample will show that $\tau-2$ is indeed a stopping time in this setting.
If $\tau$ is deterministic, $\tau-2$ is deterministic. As long as $\tau-2 \ge 0$, it is a stopping time because $\{\tau-2 = n\} = \emptyset$ or $\{\tau-2 = n\} = \Omega$ for all $n$.