This answer is too long for a comment, and it is for your question following @Themis's answer below.
Denote by $T^0,S^0$ the reservoir's parameters and by $\delta W^0$ the externally controlled work, then for any process connecting two infinitesimally close equilibrium states $$dU=TdS+\delta W=T^0dS^0+\delta W^0$$ For an isothermal process $T=T^0$ and then $$T(dS-dS^0)=\delta W^0-\delta W=T\sigma \ge 0$$ where $\sigma = dS-dS^0$ is the internally generated entropy that is never negative, and is zero only for a reversible process. Therefore $dF=d(U-TS)=dU-TdS-SdT=-SdT + \delta W$ that can be written for an isothermal process, $dT=0$, as $ dF|_T = \delta W = - T\sigma +\delta W^0 $ which in the special case of no work done, $\delta W^0=0$, is just $$ dF|_{T,\delta W^0=0} = - T\sigma \le 0 $$ with equality only in a reversible process.