Reduce HALT to $\mathrm{L}_2$ by constructing $\mathrm{M}^{\prime}$ that first simulates $\mathrm{M}$ on $\mathrm{x}$ and, only if that halting occurs, executes a carefully designed reversible halting computation; otherwise it never produces a reversible halting trace. Hence deciding $\mathrm{L}_2$ would decide HALT, so $\mathrm{L}_2$ is undecidable.