This is a very difficult problem, I try to explain why.
In statistical mechanics, tending to the most probable distribution is a probability event, and for <PERSON>' entropy, $dS\ge 0$ is also a probability event but not an inevitable result. So you can’t prove $dS\ge 0$ as an inevitable result from statistical mechanics.
If we want to obtain the mathematical proof of the second law from thermodynamics, we must consider the mathematical proof of the entropy first, as a state function. <PERSON>’ definition $dS=δQ/T$ cannot be proven in mathematics, as an exact differential, so the definition $dS=δQ/T$ can only depend on imaginary reversible cycles. On the other hand, in <PERSON> or <PERSON> approaches, the expressions of the entropy are the mathematical equations but not the definition of a physical concept because the equation contains the difference of functions (please see bellow, <PERSON>’s equation), the physical image of the entropy and both the second law are not clear, that is why we cannot explain the physical meaning of entropy according to these approaches. In such case, to prove the second law in mathematics will be very difficult, it is not an isolated problem.
The following are the some steps of the link paper, the paper introduce a new approach, and the new statement on the second law can be considered as an axiom.
1) According to the fundamental equation of thermodynamics (<PERSON>’s equation).
\begin{align} dS=\frac{dU}{T}-\frac{Ydx}{T}-\sum_j\frac{\mu_jdN_j}{T}+\frac{pdV}{T}.\end{align}
2) Define the function that
\begin{align}dq=dU-Ydx-\sum_j\mu_jdN_j.\end{align}
Here $dq$ can be proven as an exact differential in mathematics, the physical meaning of $q$ is the heat energy within the system. 273
(but not the heat in transfer $Q$)
3) Such that we get
\begin{align}dS=\frac{dq}{T}+\frac{pdV}{T}.\end{align}
Here $dS$ can be proven as an exact differential in mathematics.
4) Consider an interaction between the two locals, then we can get the total differential of the entropy production.
\begin{align}d_iS=\nabla \left(\frac{1}{T}\right)dq+\frac{1}{T}\nabla Ydx+\sum_j\frac{1}{T}\nabla \mu_jdN_j+\nabla \left(\frac{p}{T}\right)dV.\end{align}
This is a non- equilibrium thermodynamic equation.
5) Prove the total differential of the entropy production $d_iS\ge0$.
The new statement of the second law: "irreversibility root in a fundamental principle: the gradients of the four thermodynamic forces spontaneously tend to zero".
The four gradients of thermodynamic forces are \begin{align} \nabla \left(\frac{1}{T}\right),\,\,\,\, \nabla Y,\,\,\,\, \nabla \mu_j,\,\,\,\, \nabla \left(\frac{p}{T}\right). \end{align}
The conditions of thermodynamic equilibrium are these four gradients equal to zero.
Please compare the different statements about the second law, and see which statement can be considered as an axiom. 1)-4) are quoted from http://en.wikipedia.org/wiki/Second_law_of_thermodynamics
1) <PERSON> statement
* Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.
2) <PERSON> statement
* It is impossible, by means of inanimate material agency, to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects.
3) <PERSON>'s statement
* Every process occurring in nature proceeds in the sense in which the sum of the entropies of all bodies taking part in the process is increased. In the limit, i.e. for reversible processes, the sum of the entropies remains unchanged. 749