depends on the cut surfaces and consequently on the whole history of creating dislocations. Likewise, the gradient of plastic strain tensor $\mathbf{C}^p$ cannot be used as the state variable by the same reason. In contrary, the dislocation densities depend only on the characteristics of dislocations in the current state (Burgers vector and positions of dislocation lines) and not on how they are created, so $\rho_\perp $ and $\rho_\parallel $, in addition to $\mathbf{F}^e$, are the proper state variables. Thus, if we consider isothermal processes of deformation, then the free energy per unit volume of crystal (assumed as macroscopically homogeneous) must be a function of $\mathbf{F}^e$, $\rho_\perp $, and $\rho_\parallel $ $$\psi = \psi (\mathbf{F}^e,\rho_\perp ,\rho_\parallel ).$$ Now, if we superimpose an elastic rotation $\mathbf{R}$ onto the actual deformation of the body, then the total and elastic deformation change according to $$\mathbf{F}^{\ast}=\mathbf{R}\cdot \mathbf{F},\quad \mathbf{F}^{e\ast}=\mathbf{R}\cdot \mathbf{F}^e.$$ At the same time, the dislocation densities $\rho_\perp $ and $\rho_\parallel $ remain unchanged. As such superimposed elastic rotation does not change the elastic strain and the dislocation densities, we expect that the energy remains unchanged. The standard argument (see, e.g., [@Gurtin1981]) leads then to $$\psi = \psi (\mathbf{C}^e,\rho_\perp ,\rho_\parallel ),$$ where $\mathbf{C}^e$ is the elastic 