[proofplan] We prove both identities by comparing rank functions on the common ground set $E \setminus A$. The rank formula for a dual matroid converts ranks in $(M \setminus A)^*$ and $(M / A)^*$ into ranks in $M$, while the rank formula for contraction converts ranks in $M^*/A$ into ranks in $M^*$. After substituting the dual rank formula for $M^*$, the two expressions agree for every subset $X \subset E \setminus A$. Since a matroid is determined by its rank function, the desired matroid identities follow. [/proofplan]

[step:Fix the common ground set and rank notation] Let $S := E \setminus A$. For any matroid $N$ on a finite ground set $F$, write $r_N : 2^F \to \mathbb{N} \cup \{0\}$ for its rank function. We use the standard rank formula for the dual matroid: for every $Y \subset F$, \begin{align*} r_{N^*}(Y) = |Y| - r_N(F) + r_N(F \setminus Y). \end{align*} We also use the standard rank formula for matroid contraction: for every $B \subset F$ and every $Y \subset F \setminus B$, \begin{align*} r_{N/B}(Y) = r_N(Y \cup B) - r_N(B). \end{align*} We use matroid deletion in the form that, for every $Y \subset F \setminus B$, the deletion $N \setminus B$ has rank $r_{N \setminus B}(Y) = r_N(Y)$. A matroid is determined by its rank function: a subset $I$ of its ground set is independent exactly when $r(I) = |I|$. Therefore, if two matroids on the same finite ground set have the same rank function, they have the same independent sets and hence are equal as matroids. [/step]

[step:Compare the rank functions of $(M \setminus A)^*$ and $M^*/A$]Let $X \subset S$. Since $M \setminus A$ is a matroid on $S$, the dual rank formula gives \begin{align*} r_{(M \setminus A)^*}(X) = |X| - r_{M \setminus A}(S) + r_{M \setminus A}(S \setminus X). \end{align*} By the definition of matroid deletion, ranks of subsets of $S$ are unchanged from $M$, so \begin{align*} r_{(M \setminus A)^*}(X) = |X| - r_M(S) + r_M(S \setminus X). \end{align*} On the other hand, by the rank formula for contraction applied to the matroid $M^*$ and the subset $A \subset E$, \begin{align*} r_{M^*/A}(X) = r_{M^*}(X \cup A) - r_{M^*}(A). \end{align*} Applying the dual rank formula for $M^*$ first to $X \cup A$ and then to $A$, we get \begin{align*} r_{M^*}(X \cup A) = |X \cup A| - r_M(E) + r_M(E \setminus (X \cup A)). \end{align*} Also, \begin{align*} r_{M^*}(A) = |A| - r_M(E) + r_M(E \setminus A). \end{align*} Since $X \subset E \setminus A$, the sets $X$ and $A$ are disjoint, so $|X \cup A| - |A| = |X|$. Moreover, \begin{align*} E \setminus (X \cup A) = (E \setminus A) \setminus X = S \setminus X. \end{align*} Substituting these identities gives \begin{align*} r_{M^*/A}(X) = |X| - r_M(S) + r_M(S \setminus X). \end{align*} Thus \begin{align*} r_{(M \setminus A)^*}(X) = r_{M^*/A}(X) \end{align*} for every $X \subset S$.[/step]

[guided]We want to prove equality of two matroids on the same ground set $S = E \setminus A$. The most efficient way is to compare their rank functions on every subset $X \subset S$. First consider $(M \setminus A)^*$. The matroid $M \setminus A$ has ground set $S$, so the dual rank formula gives \begin{align*} r_{(M \setminus A)^*}(X) = |X| - r_{M \setminus A}(S) + r_{M \setminus A}(S \setminus X). \end{align*} Matroid deletion does not change the rank of any subset of the remaining ground set. Since both $S$ and $S \setminus X$ are subsets of $E \setminus A$, we have \begin{align*} r_{M \setminus A}(S) = r_M(S). \end{align*} and \begin{align*} r_{M \setminus A}(S \setminus X) = r_M(S \setminus X). \end{align*} Therefore \begin{align*} r_{(M \setminus A)^*}(X) = |X| - r_M(S) + r_M(S \setminus X). \end{align*} Now consider $M^*/A$. Since matroid contraction by $A$ has ground set $E \setminus A = S$, the contraction rank formula gives \begin{align*} r_{M^*/A}(X) = r_{M^*}(X \cup A) - r_{M^*}(A). \end{align*} The point of using the dual rank formula is that it rewrites both terms entirely in ranks of the original matroid $M$. Applying the dual rank formula to $X \cup A \subset E$ gives \begin{align*} r_{M^*}(X \cup A) = |X \cup A| - r_M(E) + r_M(E \setminus (X \cup A)). \end{align*} Applying it to $A \subset E$ gives \begin{align*} r_{M^*}(A) = |A| - r_M(E) + r_M(E \setminus A). \end{align*} Subtracting these expressions, the two terms $r_M(E)$ cancel: \begin{align*} r_{M^*/A}(X) = |X \cup A| - |A| + r_M(E \setminus (X \cup A)) - r_M(E \setminus A). \end{align*} Because $X \subset E \setminus A$, the sets $X$ and $A$ are disjoint, so \begin{align*} |X \cup A| - |A| = |X|. \end{align*} Also $S = E \setminus A$, and removing both $X$ and $A$ from $E$ is the same as first passing to $S$ and then removing $X$: \begin{align*} E \setminus (X \cup A) = S \setminus X. \end{align*} Finally, $E \setminus A = S$, so the preceding expression becomes \begin{align*} r_{M^*/A}(X) = |X| - r_M(S) + r_M(S \setminus X). \end{align*} This is exactly the formula already obtained for $r_{(M \setminus A)^*}(X)$. Hence the two rank functions agree on every subset $X \subset S$.[/guided]

[step:Conclude the first matroid identity from equality of ranks] Both $(M \setminus A)^*$ and $M^*/A$ are matroids on the same ground set $S$. The preceding step proves that their rank functions agree on every subset $X \subset S$. Since a matroid is determined by its rank function, we obtain \begin{align*} (M \setminus A)^* = M^*/A. \end{align*} [/step]

[step:Compare the rank functions of $(M/A)^*$ and $M^* \setminus A$] Let $X \subset S$. Since $M/A$ is a matroid on $S$, the dual rank formula gives \begin{align*} r_{(M/A)^*}(X) = |X| - r_{M/A}(S) + r_{M/A}(S \setminus X). \end{align*} Using the contraction rank formula in $M$, we have \begin{align*} r_{M/A}(S) = r_M(S \cup A) - r_M(A). \end{align*} Since $S \cup A = E$, this is \begin{align*} r_{M/A}(S) = r_M(E) - r_M(A). \end{align*} Similarly, \begin{align*} r_{M/A}(S \setminus X) = r_M((S \setminus X) \cup A) - r_M(A). \end{align*} Because $X \subset S = E \setminus A$, we have \begin{align*} (S \setminus X) \cup A = E \setminus X. \end{align*} Therefore \begin{align*} r_{M/A}(S \setminus X) = r_M(E \setminus X) - r_M(A). \end{align*} Substituting into the dual rank formula gives \begin{align*} r_{(M/A)^*}(X) = |X| - r_M(E) + r_M(E \setminus X). \end{align*} For the matroid deletion $M^* \setminus A$, ranks of subsets of $S$ are unchanged from $M^*$. Thus \begin{align*} r_{M^* \setminus A}(X) = r_{M^*}(X). \end{align*} Applying the dual rank formula for $M^*$ gives \begin{align*} r_{M^* \setminus A}(X) = |X| - r_M(E) + r_M(E \setminus X). \end{align*} Hence \begin{align*} r_{(M/A)^*}(X) = r_{M^* \setminus A}(X) \end{align*} for every $X \subset S$. [/step]

[step:Conclude the second matroid identity from equality of ranks] Both $(M/A)^*$ and $M^* \setminus A$ are matroids on the ground set $S = E \setminus A$. The preceding step proves that their rank functions agree on every subset of $S$. Since a matroid is determined by its rank function, we conclude that \begin{align*} (M/A)^* = M^* \setminus A. \end{align*} This proves both deletion-contraction duality identities. [/step]