[proofplan]
We prove that the tropical Plucker repeated-minimum condition is equivalent to the valuated basis exchange inequality. In one direction, we encode a proposed exchange from $B$ to $B'$ by taking $S=B\setminus\{e\}$ and $T=B'\cup\{e\}$, so the tropical Plucker relation forces either a strictly smaller or an equally minimal exchanged term. In the converse direction, a uniquely minimal tropical Plucker term would give two finite bases to which the exchange inequality applies, producing a second term with the same minimum value and contradicting uniqueness.
[/proofplan]
[step:Derive the exchange inequality from the tropical Plucker relation]
Assume first that $p$ is a valuated matroid in the tropical Plucker sense. Then $\mathcal{B}\neq\varnothing$ by definition. Let $B,B'\in\mathcal{B}$, and let $e\in B\setminus B'$.
If $B\setminus B'=\{e\}$, then $|B|=|B'|=r$ implies $B'\setminus B=\{f\}$ for a unique element $f\in E$. In this case
\begin{align*}
(B\setminus\{e\})\cup\{f\}=B'
\end{align*}
and
\begin{align*}
(B'\setminus\{f\})\cup\{e\}=B.
\end{align*}
Therefore
\begin{align*}
p(B)+p(B')=p((B\setminus\{e\})\cup\{f\})+p((B'\setminus\{f\})\cup\{e\}),
\end{align*}
which is the desired inequality.
Now assume $B\setminus B'\neq\{e\}$. Define two subsets of $E$ by
\begin{align*}
S:=B\setminus\{e\}
\end{align*}
and
\begin{align*}
T:=B'\cup\{e\}.
\end{align*}
Then $S\in\binom{E}{r-1}$ and $T\in\binom{E}{r+1}$. Since $e\notin B'$ and $e\notin S$, while every element of $B'\setminus B$ lies in $T$ and not in $S$, we have
\begin{align*}
T\setminus S=\{e\}\cup(B'\setminus B).
\end{align*}
For each $i\in T\setminus S$, define the extended real number
\begin{align*}
a_i:=p(S\cup\{i\})+p(T\setminus\{i\}).
\end{align*}
The term indexed by $e$ is
\begin{align*}
a_e=p(B)+p(B').
\end{align*}
For each $f\in B'\setminus B$, the corresponding term is
\begin{align*}
a_f=p((B\setminus\{e\})\cup\{f\})+p((B'\setminus\{f\})\cup\{e\}).
\end{align*}
By the tropical Plucker relation for the pair $(S,T)$, the minimum of the finite family $(a_i)_{i\in T\setminus S}$ is attained at least twice. If $a_e$ is not minimal, then there exists $f\in B'\setminus B$ with $a_f<a_e$, hence $a_f\leq a_e$. If $a_e$ is minimal, then repeated attainment gives an index $j\in T\setminus S$ with $j\neq e$ and $a_j=a_e$. Since $T\setminus S=\{e\}\cup(B'\setminus B)$, this index has the form $j=f$ for some $f\in B'\setminus B$. In both cases there exists $f\in B'\setminus B$ such that
\begin{align*}
p(B)+p(B')\geq p((B\setminus\{e\})\cup\{f\})+p((B'\setminus\{f\})\cup\{e\}).
\end{align*}
[/step]
[step:Use the exchange inequality to force repeated minima]
Conversely, assume that $\mathcal{B}\neq\varnothing$ and that the stated valuated basis exchange condition holds for all $B,B'\in\mathcal{B}$ and all $e\in B\setminus B'$.
Fix $S\in\binom{E}{r-1}$ and $T\in\binom{E}{r+1}$. For each $i\in T\setminus S$, define
\begin{align*}
a_i:=p(S\cup\{i\})+p(T\setminus\{i\}).
\end{align*}
We prove that the minimum of the finite family $(a_i)_{i\in T\setminus S}$ is attained at least twice.
Suppose, toward a contradiction, that the minimum is attained at a unique index $i_0\in T\setminus S$. Since $|T|=r+1$ and $|S|=r-1$, the set $T\setminus S$ has at least two elements. If $a_{i_0}=\infty$, then every $a_i$ is equal to $\infty$, so the minimum is not unique. Thus $a_{i_0}<\infty$.
Define
\begin{align*}
B:=S\cup\{i_0\}
\end{align*}
and
\begin{align*}
B':=T\setminus\{i_0\}.
\end{align*}
The inequality $a_{i_0}<\infty$ implies $p(B)<\infty$ and $p(B')<\infty$, so $B,B'\in\mathcal{B}$. Also $i_0\in B\setminus B'$.
Applying the assumed exchange condition to $B$, $B'$, and $e=i_0$, there exists $f\in B'\setminus B$ such that
\begin{align*}
p(B)+p(B')\geq p((B\setminus\{i_0\})\cup\{f\})+p((B'\setminus\{f\})\cup\{i_0\}).
\end{align*}
Since $B\setminus\{i_0\}=S$ and $B'\setminus B=T\setminus(S\cup\{i_0\})$, this element $f$ belongs to $T\setminus S$ and satisfies $f\neq i_0$. Moreover
\begin{align*}
(B\setminus\{i_0\})\cup\{f\}=S\cup\{f\}
\end{align*}
and
\begin{align*}
(B'\setminus\{f\})\cup\{i_0\}=T\setminus\{f\}.
\end{align*}
Therefore
\begin{align*}
a_{i_0}\geq a_f.
\end{align*}
But $a_{i_0}$ was chosen as the minimum of the family, so $a_f\geq a_{i_0}$. Hence
\begin{align*}
a_f=a_{i_0}.
\end{align*}
This gives a second minimizer $f\neq i_0$, contradicting the assumed uniqueness. Thus the minimum is attained at least twice.
[guided]
We now prove the converse carefully. The goal is to verify the tropical Plucker repeated-minimum condition from the exchange inequality.
Fix $S\in\binom{E}{r-1}$ and $T\in\binom{E}{r+1}$. For every index $i\in T\setminus S$, define
\begin{align*}
a_i:=p(S\cup\{i\})+p(T\setminus\{i\}).
\end{align*}
These are exactly the tropical Plucker terms attached to the pair $(S,T)$. We must prove that the smallest value among the $a_i$ occurs for at least two different indices.
Assume, for contradiction, that there is a unique minimizer $i_0\in T\setminus S$. The set $T\setminus S$ has at least two elements, because $T$ has size $r+1$ and $S$ has size $r-1$. If the unique minimum were $\infty$, then every term would be $\infty$, and every index in $T\setminus S$ would be a minimizer. Therefore the unique minimum must be finite:
\begin{align*}
a_{i_0}<\infty.
\end{align*}
Now define two $r$-element subsets of $E$ by
\begin{align*}
B:=S\cup\{i_0\}
\end{align*}
and
\begin{align*}
B':=T\setminus\{i_0\}.
\end{align*}
The finiteness of $a_{i_0}$ means
\begin{align*}
p(B)+p(B')<\infty.
\end{align*}
Since both summands lie in $\mathbb{R}\cup\{\infty\}$, this forces $p(B)<\infty$ and $p(B')<\infty$. Hence $B,B'\in\mathcal{B}$. Also $i_0\in B$ by construction, while $i_0\notin B'$ because $B'=T\setminus\{i_0\}$, so $i_0\in B\setminus B'$.
We may therefore apply the assumed valuated basis exchange condition to the bases $B,B'$ and the element $e=i_0$. It gives an element $f\in B'\setminus B$ such that
\begin{align*}
p(B)+p(B')\geq p((B\setminus\{i_0\})\cup\{f\})+p((B'\setminus\{f\})\cup\{i_0\}).
\end{align*}
Because $B=S\cup\{i_0\}$, we have $B\setminus\{i_0\}=S$. Because $B'=T\setminus\{i_0\}$, removing $f$ from $B'$ and adding $i_0$ gives $T\setminus\{f\}$. Thus the exchange inequality becomes
\begin{align*}
p(S\cup\{i_0\})+p(T\setminus\{i_0\})\geq p(S\cup\{f\})+p(T\setminus\{f\}).
\end{align*}
In the notation of the Plucker terms, this is
\begin{align*}
a_{i_0}\geq a_f.
\end{align*}
Since $i_0$ was chosen to minimize all the $a_i$, we also have
\begin{align*}
a_f\geq a_{i_0}.
\end{align*}
Combining the two inequalities gives
\begin{align*}
a_f=a_{i_0}.
\end{align*}
Finally, $f\in B'\setminus B$, while $i_0\in B\setminus B'$, so $f\neq i_0$. Thus $f$ is a second minimizer, contradicting the assumption that $i_0$ was the unique minimizer. The tropical Plucker minimum is therefore attained at least twice.
[/guided]
[/step]
[step:Conclude the equivalence]
The first step proves that every valuated matroid satisfies the stated basis exchange inequality. The second step proves that the basis exchange inequality forces every tropical Plucker minimum to be attained at least twice. Together with the shared nonemptiness condition $\mathcal{B}\neq\varnothing$, this is exactly the equivalence claimed in the theorem.
[/step]
