 Solution

Suppose two distinct reversible adiabatic processes intersect at an equilibrium state $C$. Choose an isothermal process that meets the two branches at distinct states $A$ and $B$, forming a nonzero-area three-sided thermodynamic cycle. There is no heat exchange on $AC$ or $BC$, so the only net heat comes from the one heat reservoir on $AB$. Since $\Delta U=0$ around a cycle, $W_{\mathrm{cycle}}=Q_{AB}$.

Choose the direction in which this nonzero cycle delivers positive work. It would then convert heat from a single reservoir entirely into work, violating the Kelvin-Planck statement of the second law. Distinct reversible adiabats cannot intersect. Equivalently, their constant entropy labels cannot differ at one equilibrium state. The argument concerns an ordinary equilibrium family of reversible adiabats, not arbitrary irreversible insulated paths. The usual local uniqueness of an equilibrium adiabat excludes coincident segments described as distinct curves.