The Entropy of the Universe Does Not Decrease
Theorem
The entropy of the universe does not decrease.
Explanation
The first thing that comes to mind upon seeing the statement above is ‘wow, that’s cool’. However, the truly cool people are those who can understand this equationally, and let’s strive to be one of those people.
Proof
This universe is unique, and thus it requires the assumption that there is no ‘outside’ of this universe.
Consider the cycles as shown above; $A \to B$ being irreversible, and $B \to A$ being reversible. These processes include irreversible processes, so overall, it is an irreversible process.
The following holds for a cyclic process:
$$ \oint {{\delta Q} \over {T}} \le 0 $$
According to the Clausius theorem, the following holds.
$$ \oint {{\delta Q} \over {T}} = \int_{A}^{B} { { \delta Q } \over { T }} + \int_{B}^{A} { {{ \delta Q_{\text{rev} } } \over { T }} } \le 0 $$
Summing up the top and bottom ends, we get the following.
$$ \int_{A}^{B} { { \delta Q } \over { T }} \le \int_{A}^{B} { {{ \delta Q_{\text{rev} } } \over { T }} } $$
The following equation defines entropy:
$$ dS = {{ \delta Q_{\text{rev} } } \over { T }} $$
By the definition of entropy, we obtain the following.
$$ \int_{A}^{B} { { \delta Q } \over { T }} \le d S $$
If the universe is unique and there’s nothing beyond it, it would not be able to exchange thermal energy with the outside. Therefore, looking at the universe as a whole system, all processes must be adiabatic, and equationally, this must be $\delta Q = 0$. Summarizing, this leads to $dS \ge 0$, therefore, the entropy of the universe cannot decrease.
■