I am trying to learn systems with stochastic inputs, especially deterministic systems.
$L[\cdot]$ represents a time-invariant system if $$ Y(t) = L\{X(t) \} \implies Y(t - t_0) = L\{X(t - t_0) \}. $$
Now I was trying to understand the following proof for a linear time-invariant system, where I cannot get the starred step ($\delta(\cdot)$ is Dirac's delta): \begin{align} \begin{array}{lll} Y(t) & = L\{X(t) \} & \textit{linear system}\\ & = L \{ \int\limits_{-\infty}^{\infty} X(\tau) \delta(t - \tau) d\tau\} & \textit{Dirac representation}\\ & = \int\limits_{-\infty}^{\infty} L \{ X(\tau) \delta(t - \tau) d\tau\} & \textit{linearity}\\ & = \int\limits_{-\infty}^{\infty} X(\tau) L \{ \delta(t - \tau) \}d\tau & \textit{linearity}\\ & = \int\limits_{-\infty}^{\infty} X(\tau) h(t - \tau)d\tau & \textit{}\\ & = \int\limits_{-\infty}^{\infty} X(t - \tau) h(\tau)d\tau & \textit{*****} \end{array} \end{align} What is the relationship between the last step and time-invariance? Maybe it follows since in the second last step $h(t-\tau)$ is $0$ everywhere except $\tau = t$ then it has $X(\tau)$ next to itself, and in the last step $h(\tau) = 0$ when $\tau = 0$ so $X(t - \tau)$ becomes $X(t)$ again. Still, I cannot get where we use time-invariance.