[proofplan] We compare the relevant minors on their common ground set $E(M)\setminus\{e,f\}$. Two deletions commute directly from the definition of deletion. For contractions, and for one deletion together with one contraction, we compare rank functions on an arbitrary subset $A \subset E(M)\setminus\{e,f\}$ using the rank formula for contraction; the loop convention makes the same formula valid even when the contracted element is a loop. Finally, equal rank functions determine the same matroid, and the general normal form $M/C\setminus D$ follows by repeatedly swapping adjacent operations until all contractions precede all deletions. [/proofplan]

[step:Compare two successive deletions by their independent sets]Let $E := E(M)$ denote the ground set of $M$. Let $\mathcal{I}(M)$ denote the set of independent subsets of $E$. The deletion $M\setminus e$ has ground set $E\setminus\{e\}$, and its independent sets are precisely the sets $I \in \mathcal{I}(M)$ satisfying $I \subset E\setminus\{e\}$. Therefore the matroid $(M\setminus e)\setminus f$ has ground set $E\setminus\{e,f\}$ and independent sets \begin{align*} \{I \in \mathcal{I}(M) : I \subset E\setminus\{e,f\}\}. \end{align*} The same description is obtained for $(M\setminus f)\setminus e$. Hence \begin{align*} (M\setminus e)\setminus f = (M\setminus f)\setminus e \end{align*} under the identity map on $E\setminus\{e,f\}$.[/step]

[guided]Deletion simply restricts the matroid to a smaller ground set without changing which remaining subsets were independent in the original matroid. Let $E := E(M)$ and let $\mathcal{I}(M)$ be the independent-set system of $M$. After deleting $e$, the available elements are $E\setminus\{e\}$, and a subset is independent in $M\setminus e$ exactly when it was already independent in $M$ and avoids $e$. Now delete $f$ as well. A subset $I \subset E\setminus\{e,f\}$ is independent in $(M\setminus e)\setminus f$ exactly when $I$ is independent in $M\setminus e$, which is exactly when $I \in \mathcal{I}(M)$. Since $I$ already avoids both $e$ and $f$, this gives the description \begin{align*} \mathcal{I}((M\setminus e)\setminus f) = \{I \in \mathcal{I}(M) : I \subset E\setminus\{e,f\}\}. \end{align*} Performing the deletions in the opposite order imposes the same two conditions: avoid $f$, avoid $e$, and remain independent in $M$. Thus \begin{align*} \mathcal{I}((M\setminus f)\setminus e) = \{I \in \mathcal{I}(M) : I \subset E\setminus\{e,f\}\}. \end{align*} The two matroids have the same ground set and the same independent subsets, so they are equal under the identity ground-set map.[/guided]

[step:Compute the rank function after two contractions]Let $r_M$ denote the rank function of $M$. For a matroid $N$ with rank function $r_N$ and an element $g \in E(N)$, the contraction rank formula on a subset $A \subset E(N)\setminus\{g\}$ is \begin{align*} r_{N/g}(A) = r_N(A\cup\{g\}) - r_N(\{g\}). \end{align*} If $g$ is a loop in $N$, then $r_N(\{g\}) = 0$ and $r_N(A\cup\{g\}) = r_N(A)$, so this formula agrees with the stated loop convention $N/g = N\setminus g$. Let $A \subset E\setminus\{e,f\}$. Applying the contraction rank formula first in $M/e$ and then in $M$ gives \begin{align*} r_{(M/e)/f}(A) = r_{M/e}(A\cup\{f\}) - r_{M/e}(\{f\}). \end{align*} For the first term, \begin{align*} r_{M/e}(A\cup\{f\}) = r_M(A\cup\{e,f\}) - r_M(\{e\}). \end{align*} For the second term, \begin{align*} r_{M/e}(\{f\}) = r_M(\{e,f\}) - r_M(\{e\}). \end{align*} Substitution and cancellation give \begin{align*} r_{(M/e)/f}(A) = r_M(A\cup\{e,f\}) - r_M(\{e,f\}). \end{align*} Interchanging the roles of $e$ and $f$ gives the same formula: \begin{align*} r_{(M/f)/e}(A) = r_M(A\cup\{e,f\}) - r_M(\{e,f\}). \end{align*} Thus the two matroids have the same rank function on the common ground set $E\setminus\{e,f\}$.[/step]

[guided]The point of using ranks is that contraction has a clean rank formula. Let $r_M$ be the rank function of $M$. If $N$ is any matroid, $g \in E(N)$, and $A \subset E(N)\setminus\{g\}$, contraction by $g$ satisfies \begin{align*} r_{N/g}(A) = r_N(A\cup\{g\}) - r_N(\{g\}). \end{align*} This remains valid under the loop convention. Indeed, if $g$ is a loop in $N$, then $r_N(\{g\}) = 0$, and adding $g$ to a set cannot increase rank, so $r_N(A\cup\{g\}) = r_N(A)$. Hence the formula becomes $r_{N/g}(A) = r_N(A)$, which is exactly the rank function of $N\setminus g$ on $A$. Now fix an arbitrary subset $A \subset E\setminus\{e,f\}$. We compute the rank of $A$ in $(M/e)/f$. First apply the contraction formula inside the matroid $M/e$, with the element being contracted equal to $f$: \begin{align*} r_{(M/e)/f}(A) = r_{M/e}(A\cup\{f\}) - r_{M/e}(\{f\}). \end{align*} Each term on the right is now a rank in $M/e$, so we apply the contraction formula again, this time for contraction of $e$ in $M$. Since $A\cup\{f\}$ is a subset of $E\setminus\{e\}$, we have \begin{align*} r_{M/e}(A\cup\{f\}) = r_M(A\cup\{e,f\}) - r_M(\{e\}). \end{align*} Similarly, \begin{align*} r_{M/e}(\{f\}) = r_M(\{e,f\}) - r_M(\{e\}). \end{align*} Substituting these two identities into the previous display gives \begin{align*} r_{(M/e)/f}(A) = \bigl(r_M(A\cup\{e,f\}) - r_M(\{e\})\bigr) - \bigl(r_M(\{e,f\}) - r_M(\{e\})\bigr). \end{align*} The two occurrences of $r_M(\{e\})$ cancel, leaving \begin{align*} r_{(M/e)/f}(A) = r_M(A\cup\{e,f\}) - r_M(\{e,f\}). \end{align*} The same computation with $e$ and $f$ interchanged gives \begin{align*} r_{(M/f)/e}(A) = r_M(A\cup\{e,f\}) - r_M(\{e,f\}). \end{align*} Since $A$ was arbitrary, the two rank functions agree on every subset of the common ground set $E\setminus\{e,f\}$.[/guided]

[step:Compute the common rank function for one deletion and one contraction] Let $A \subset E\setminus\{e,f\}$. In $(M\setminus e)/f$, the contraction formula applied in $M\setminus e$ gives \begin{align*} r_{(M\setminus e)/f}(A) = r_{M\setminus e}(A\cup\{f\}) - r_{M\setminus e}(\{f\}). \end{align*} Since deletion of $e$ does not change ranks of subsets contained in $E\setminus\{e\}$, this becomes \begin{align*} r_{(M\setminus e)/f}(A) = r_M(A\cup\{f\}) - r_M(\{f\}). \end{align*} On the other hand, in $(M/f)\setminus e$, deletion of $e$ after contracting $f$ does not change the rank of $A$, so \begin{align*} r_{(M/f)\setminus e}(A) = r_{M/f}(A). \end{align*} Applying the contraction formula in $M$ gives \begin{align*} r_{(M/f)\setminus e}(A) = r_M(A\cup\{f\}) - r_M(\{f\}). \end{align*} Thus the rank functions of $(M\setminus e)/f$ and $(M/f)\setminus e$ agree on every subset of $E\setminus\{e,f\}$. [/step]

[step:Conclude equality from equality of rank functions] Let $N_1$ and $N_2$ be two matroids on the same finite ground set $S$, with rank functions $r_{N_1}$ and $r_{N_2}$. If $r_{N_1}(A) = r_{N_2}(A)$ for every subset $A \subset S$, then $N_1 = N_2$, because a subset $I \subset S$ is independent in $N_i$ exactly when \begin{align*} r_{N_i}(I) = |I|. \end{align*} Applying this observation to the common ground set $E\setminus\{e,f\}$ proves \begin{align*} (M/e)/f = (M/f)/e \end{align*} and \begin{align*} (M\setminus e)/f = (M/f)\setminus e. \end{align*} Together with the deletion calculation, this proves the three asserted pairwise commutation identities. [/step]

[step:Move all contractions before all deletions in an arbitrary finite minor sequence] Let a finite minor sequence from $M$ be given. Define $C \subset E$ to be the set of elements contracted in the sequence and $D \subset E$ to be the set of elements deleted in the sequence. Since an element is removed from the ground set after it is first deleted or contracted, the sets $C$ and $D$ are disjoint. If a deletion of an element $e$ occurs immediately before a contraction of a distinct element $f$, the identity \begin{align*} (M\setminus e)/f = (M/f)\setminus e \end{align*} allows these two adjacent operations to be interchanged without changing the resulting matroid. Repeating this adjacent interchange moves every contraction to the left of every deletion. The identities \begin{align*} (M/e)/f = (M/f)/e \end{align*} and \begin{align*} (M\setminus e)\setminus f = (M\setminus f)\setminus e \end{align*} show that the order among the contractions and the order among the deletions do not affect the final matroid. Hence every finite minor sequence has the form \begin{align*} M/C\setminus D \end{align*} for disjoint $C,D \subset E(M)$, independently of the order of the individual contractions and deletions. This completes the proof. [/step]