[proofplan] Use the independent-set polytope description as input. The base polytope is the convex hull of the incidence vectors of bases, while the independent-set polytope is the convex hull of the incidence vectors of all independent sets. Bases are exactly the independent sets of size $r(E)$, so the base polytope should be the face of the independent-set polytope cut out by the equation $x(E)=r(E)$. [/proofplan] [step:Show that every point of the base polytope satisfies the displayed conditions] Let $B$ be a basis of $M$. Its incidence vector $\mathbf 1_B$ is nonnegative. For every $A\subset E$, \begin{align*} \mathbf 1_B(A)=|A\cap B|\le r(A), \end{align*} because $A\cap B$ is an independent subset of $A$. Also \begin{align*} \mathbf 1_B(E)=|B|=r(E). \end{align*} Therefore every basis incidence vector satisfies the displayed conditions. Since the displayed conditions are linear inequalities and one linear equality, their solution set is convex. Hence every convex combination of basis incidence vectors, and therefore every point of $P_B(M)$, satisfies the displayed conditions. [guided] The base polytope is built from basis incidence vectors. A basis has exactly $r(E)$ elements, so its total coordinate sum is $r(E)$. On any subset $A$, it can use at most $r(A)$ elements, because $A\cap B$ is independent in the restriction to $A$. Convex combinations preserve all these linear inequalities and the equality. [/guided] [/step] [step:Use the independent-set polytope theorem for the converse] Suppose $x\in\mathbb R^E$ satisfies \begin{align*} x_e\ge 0\text{ for all }e\in E,\qquad x(A)\le r(A)\text{ for all }A\subset E,\qquad x(E)=r(E). \end{align*} By the [Matroid Independence Polytope Theorem](/theorems/5830), the first two families of inequalities imply that $x$ lies in the independent-set polytope \begin{align*} P_I(M)=\operatorname{conv}\{\mathbf 1_I:I\text{ is independent in }M\}. \end{align*} Thus there are independent sets $I_1,\dots,I_m$ and coefficients $\lambda_1,\dots,\lambda_m\ge 0$ with $\sum_i\lambda_i=1$ such that \begin{align*} x=\sum_{i=1}^m \lambda_i \mathbf 1_{I_i}. \end{align*} Taking total coordinate sums gives \begin{align*} r(E)=x(E)=\sum_{i=1}^m\lambda_i |I_i|. \end{align*} Each $I_i$ is independent, so $|I_i|\le r(E)$. The weighted average of the numbers $|I_i|$ is equal to their common upper bound $r(E)$, hence every $I_i$ with $\lambda_i>0$ has $|I_i|=r(E)$. Such an independent set is a basis of $M$. Therefore $x$ is a convex combination of incidence vectors of bases, so $x\in P_B(M)$. [/step] [step:Conclude equality] The first step proves that $P_B(M)$ is contained in the displayed solution set. The second step proves that every point of the displayed solution set lies in $P_B(M)$. The two sets are therefore equal. [/step]