[proofplan]
Fix $x,y,z\in L$ with $x\le z$. The universal properties of meet and join first give the easy inequality $x\vee(y\wedge z)\le (x\vee y)\wedge z$. For the reverse inequality, name the two sides $a$ and $b$, compute $\rho(a)$ and $\rho(b)$ using the rank modularity identity on four carefully chosen pairs, and show the two ranks agree. Since $a\le b$ and rank strictly increases along proper comparable elements in a finite ranked lattice, equality of ranks forces $a=b$.
[/proofplan]
[step:Compare the two sides by the order properties of meet and join]
Fix $x,y,z\in L$ and assume $x\le z$. Define
\begin{align*}
a:=x\vee (y\wedge z)
\end{align*}
and
\begin{align*}
b:=(x\vee y)\wedge z.
\end{align*}
Since $y\wedge z\le y$, the join $x\vee(y\wedge z)$ is bounded above by $x\vee y$. Since $x\le z$ and $y\wedge z\le z$, the element $z$ is an upper bound for $\{x,y\wedge z\}$, so $a=x\vee(y\wedge z)\le z$. Thus $a\le x\vee y$ and $a\le z$, hence by the defining greatest-lower-bound property of meet,
\begin{align*}
a\le (x\vee y)\wedge z=b.
\end{align*}
[/step]
[step:Compute the rank of $x\vee(y\wedge z)$]
Because $x\le z$, the two lower bounds $x\wedge(y\wedge z)$ and $x\wedge y$ are equal. Indeed, $x\wedge(y\wedge z)\le x$ and $x\wedge(y\wedge z)\le y$, so $x\wedge(y\wedge z)\le x\wedge y$; conversely, $x\wedge y\le x\le z$ and $x\wedge y\le y$, so $x\wedge y\le y\wedge z$, hence $x\wedge y\le x\wedge(y\wedge z)$.
Apply the assumed rank identity to the pair $x$ and $y\wedge z$. Since $a=x\vee(y\wedge z)$ and $x\wedge(y\wedge z)=x\wedge y$, we obtain
\begin{align*}
\rho(a)=\rho(x)+\rho(y\wedge z)-\rho(x\wedge y).
\end{align*}
Apply the same rank identity to the pair $y,z$. This gives
\begin{align*}
\rho(y\wedge z)=\rho(y)+\rho(z)-\rho(y\vee z).
\end{align*}
Substituting this value into the previous display gives
\begin{align*}
\rho(a)=\rho(x)+\rho(y)+\rho(z)-\rho(y\vee z)-\rho(x\wedge y).
\end{align*}
[guided]
The purpose of this step is to express the rank of the left-hand side $a=x\vee(y\wedge z)$ only in terms of $x,y,z$, $x\wedge y$, and $y\vee z$. The first point to verify is the meet simplification
\begin{align*}
x\wedge(y\wedge z)=x\wedge y.
\end{align*}
The inequality $x\wedge(y\wedge z)\le x\wedge y$ holds because $x\wedge(y\wedge z)$ is below both $x$ and $y$. For the reverse inequality, use the hypothesis $x\le z$: the element $x\wedge y$ is below $x$, hence below $z$, and it is also below $y$. Therefore $x\wedge y$ is a lower bound of $y$ and $z$, so $x\wedge y\le y\wedge z$. Since it is also below $x$, it follows that $x\wedge y\le x\wedge(y\wedge z)$.
Now apply the assumed rank modularity identity
\begin{align*}
\rho(u)+\rho(v)=\rho(u\wedge v)+\rho(u\vee v)
\end{align*}
with $u=x$ and $v=y\wedge z$. The hypotheses allow this because $x$ and $y\wedge z$ are elements of $L$. Rearranging the resulting equality and using the definitions just established gives
\begin{align*}
\rho(a)=\rho(x)+\rho(y\wedge z)-\rho(x\wedge y).
\end{align*}
To eliminate $\rho(y\wedge z)$, apply the same rank identity with $u=y$ and $v=z$. This yields
\begin{align*}
\rho(y\wedge z)=\rho(y)+\rho(z)-\rho(y\vee z).
\end{align*}
Substitution gives the required expression:
\begin{align*}
\rho(a)=\rho(x)+\rho(y)+\rho(z)-\rho(y\vee z)-\rho(x\wedge y).
\end{align*}
[/guided]
[/step]
[step:Compute the rank of $(x\vee y)\wedge z$]
Because $x\le z$, the joins $(x\vee y)\vee z$ and $y\vee z$ are equal. Indeed, both are the least upper bound of $\{y,z\}$, since requiring an upper bound for $z$ already requires an upper bound for $x$.
Apply the rank identity to the pair $x\vee y$ and $z$. Since $b=(x\vee y)\wedge z$ and $(x\vee y)\vee z=y\vee z$, we get
\begin{align*}
\rho(x\vee y)+\rho(z)=\rho(b)+\rho(y\vee z).
\end{align*}
Hence
\begin{align*}
\rho(b)=\rho(x\vee y)+\rho(z)-\rho(y\vee z).
\end{align*}
Applying the rank identity to the pair $x,y$ gives
\begin{align*}
\rho(x\vee y)=\rho(x)+\rho(y)-\rho(x\wedge y).
\end{align*}
Substitution yields
\begin{align*}
\rho(b)=\rho(x)+\rho(y)+\rho(z)-\rho(x\wedge y)-\rho(y\vee z).
\end{align*}
Thus $\rho(a)=\rho(b)$.
[/step]
[step:Use strict rank increase to force equality]
We have shown that $a\le b$ and $\rho(a)=\rho(b)$. In a finite ranked lattice, if $u<v$, then every saturated chain from $u$ to $v$ has positive length, so $\rho(u)<\rho(v)$. Therefore $a<b$ is impossible. Hence $a=b$, namely
\begin{align*}
x\vee(y\wedge z)=(x\vee y)\wedge z.
\end{align*}
Since $x,y,z\in L$ with $x\le z$ were arbitrary, $L$ satisfies the modular identity.
[/step]
