[proofplan]
We compare the entropy growth of $T$ and $T^{-1}$ on an arbitrary finite measurable partition $\mathcal P$. The $n$-block partition for $T^{-1}$ is a translate of the $n$-block partition for $T$ by the measure-preserving automorphism $T^{n-1}$. Since applying a measure-preserving automorphism to every atom of a finite partition does not change the atom measures, the two block entropies are equal for every $n$. Taking limits and then suprema over finite partitions gives the equality of Kolmogorov-Sinai entropies.
[/proofplan]
[step:Fix a finite partition and define the two block partitions]
Let $\mathcal P=\{P_1,\dots,P_m\}$ be a finite measurable partition of $X$. For a finite measurable partition $\mathcal Q$ of $X$, define its entropy by
\begin{align*}
H_\mu(\mathcal Q):=-\sum_{Q\in\mathcal Q}\mu(Q)\log\mu(Q),
\end{align*}
with the convention $0\log 0=0$.
For $n\in\mathbb N$, define the $n$-block partition for $T$ by
\begin{align*}
\mathcal P_{T,n}:=\bigvee_{j=0}^{n-1}T^{-j}\mathcal P,
\end{align*}
where $T^{-j}\mathcal P:=\{T^{-j}(P_i):1\le i\le m\}$. Similarly, since $(T^{-1})^{-j}=T^j$, define the $n$-block partition for $T^{-1}$ by
\begin{align*}
\mathcal P_{T^{-1},n}:=\bigvee_{j=0}^{n-1}T^j\mathcal P,
\end{align*}
where $T^j\mathcal P:=\{T^j(P_i):1\le i\le m\}$.
[/step]
[step:Show that translating a finite partition by a measure-preserving automorphism preserves entropy]
Let $k\in\mathbb Z$, and let $T^k\mathcal Q:=\{T^k(Q):Q\in\mathcal Q\}$ for any finite measurable partition $\mathcal Q$ of $X$. Since $T$ is an invertible measure-preserving transformation, each integer iterate $T^k$ is measurable, invertible, and measure-preserving. Hence, for every $Q\in\mathcal Q$,
\begin{align*}
\mu(T^k(Q))=\mu(Q).
\end{align*}
Therefore the list of atom measures of $T^k\mathcal Q$ is exactly the list of atom measures of $\mathcal Q$, and so
\begin{align*}
H_\mu(T^k\mathcal Q)=H_\mu(\mathcal Q).
\end{align*}
[guided]
We need one elementary invariance fact: applying a measure-preserving automorphism to all atoms of a finite partition does not change partition entropy.
Let $k\in\mathbb Z$, and let $\mathcal Q$ be a finite measurable partition of $X$. Define
\begin{align*}
T^k\mathcal Q:=\{T^k(Q):Q\in\mathcal Q\}.
\end{align*}
Because $T$ is invertible and both $T$ and $T^{-1}$ are measurable, every integer iterate $T^k:X\to X$ is a measurable bijection whose inverse is $T^{-k}$. Because $T$ is measure-preserving, every iterate is also measure-preserving. Thus, for each atom $Q\in\mathcal Q$,
\begin{align*}
\mu(T^k(Q))=\mu(Q).
\end{align*}
The entropy of a finite partition depends only on the measures of its atoms:
\begin{align*}
H_\mu(\mathcal Q):=-\sum_{Q\in\mathcal Q}\mu(Q)\log\mu(Q).
\end{align*}
Since $Q\mapsto T^k(Q)$ is a bijection from the atoms of $\mathcal Q$ onto the atoms of $T^k\mathcal Q$, and since corresponding atoms have the same measure, the two sums are term-by-term equal. Hence
\begin{align*}
H_\mu(T^k\mathcal Q)=H_\mu(\mathcal Q).
\end{align*}
[/guided]
[/step]
[step:Identify the inverse block partition as a translate of the forward block partition]
For each $n\in\mathbb N$, we claim
\begin{align*}
\mathcal P_{T^{-1},n}=T^{n-1}\mathcal P_{T,n}
\end{align*}
up to reordering of the join factors.
Indeed,
\begin{align*}
T^{n-1}\mathcal P_{T,n}=T^{n-1}\left(\bigvee_{j=0}^{n-1}T^{-j}\mathcal P\right).
\end{align*}
Since $T^{n-1}$ is a bijection, it distributes over finite joins of partitions:
\begin{align*}
T^{n-1}\left(\bigvee_{j=0}^{n-1}T^{-j}\mathcal P\right)=\bigvee_{j=0}^{n-1}T^{n-1-j}\mathcal P.
\end{align*}
As $j$ ranges over $\{0,\dots,n-1\}$, the integer $n-1-j$ also ranges over $\{0,\dots,n-1\}$. Since joins are unchanged by reordering their factors,
\begin{align*}
\bigvee_{j=0}^{n-1}T^{n-1-j}\mathcal P=\bigvee_{r=0}^{n-1}T^r\mathcal P=\mathcal P_{T^{-1},n}.
\end{align*}
[/step]
[step:Compare entropy growth for the fixed partition]
Using entropy invariance under the measure-preserving automorphism $T^{n-1}$ and the partition identity above, for every $n\in\mathbb N$ we have
\begin{align*}
H_\mu(\mathcal P_{T^{-1},n})=H_\mu(T^{n-1}\mathcal P_{T,n}).
\end{align*}
Therefore
\begin{align*}
H_\mu(\mathcal P_{T^{-1},n})=H_\mu(\mathcal P_{T,n}).
\end{align*}
For the fixed finite partition $\mathcal P$, the standard subadditivity argument for finite partition entropy gives the existence of the limits defining the partition entropy rates:
\begin{align*}
h_\mu(T,\mathcal P):=\lim_{n\to\infty}\frac{1}{n}H_\mu(\mathcal P_{T,n}).
\end{align*}
and
\begin{align*}
h_\mu(T^{-1},\mathcal P):=\lim_{n\to\infty}\frac{1}{n}H_\mu(\mathcal P_{T^{-1},n}).
\end{align*}
Dividing the preceding equality by $n$ and passing to these existing limits gives
\begin{align*}
h_\mu(T^{-1},\mathcal P)=h_\mu(T,\mathcal P).
\end{align*}
[/step]
[step:Take suprema over finite partitions]
The finite measurable partition $\mathcal P$ was arbitrary. By definition,
\begin{align*}
h_\mu(T)=\sup_{\mathcal P}h_\mu(T,\mathcal P)
\end{align*}
and
\begin{align*}
h_\mu(T^{-1})=\sup_{\mathcal P}h_\mu(T^{-1},\mathcal P),
\end{align*}
where both suprema range over all finite measurable partitions of $X$. Since $h_\mu(T^{-1},\mathcal P)=h_\mu(T,\mathcal P)$ for every such $\mathcal P$, the two suprema are equal:
\begin{align*}
h_\mu(T^{-1})=h_\mu(T).
\end{align*}
This proves the theorem.
[/step]
