probability - Use of the tower property of conditional expectation

Let

• $(\Omega,\mathcal A)$ and $(E,\mathcal E)$ be measurable spaces


• $I\subseteq[0,\infty)$ be at most countable and closed under addition with $0\in I$



• $X=(X_t)_{t\in I}$ be a stochastic process on $(\Omega,\mathcal A)$ with values in $(E,\mathcal E)$


• $\mathbb F=(\mathcal F_t)_{t\in I}$ be the filtration generated by $X$


• $\tau$ be a $\mathbb F$-stopping time


• $f:E^I\to\mathbb R$ be bounded and $\mathcal E^{\otimes I}$-measurable

Clearly, $$Y_s:=1_{\left\{\tau=s\right\}}\operatorname E\left[f\circ\left(X_{s+t}\right)_{t\in I}\mid\mathcal F_\tau\right]$$ is $\mathcal F_s$-measurable. Thus,

\begin{equation}
\begin{split}

\operatorname E\left[f\circ\left(X_{\tau+t}\right)_{t\in I}\mid\mathcal F_\tau\right]&=&\sum_{s\in I}Y_s\\&=&\sum_{s\in I}\operatorname E\left[Y_s\mid\mathcal F_s\right]\\&\color{red}=&\color{red}{\sum_{s\in I}\operatorname E\left[1_{\left\{\tau=s\right\}}\operatorname E\left[f\circ\left(X_{\tau+t}\right)_{t\in I}\mid\mathcal F_s\right]\mid\mathcal F_\tau\right]}\;,
\end{split}
\end{equation}

but I don't understand why the $\color{red}{\text{red}}$ part is true. It looks like the tower property, but we shouldn't be able to use it unless $\mathcal F_\tau\subseteq\mathcal F_s$, which is obviously wrong. So, how do we need to argue?