[proofplan] We prove the contrapositive of the desired containment statement by assuming that the set obtained from $C_1 \cup C_2$ after deleting $e$ is independent. The rank of this set then forces the rank of $C_1 \cup C_2$ to have exactly one possible value. [Submodularity of the matroid rank function](/theorems/5819) then implies that $C_1 \cap C_2$ is dependent, contradicting the fact that it is a proper subset of a circuit. Therefore the deleted union is dependent, and a minimal dependent subset of it is the required circuit. [/proofplan]

[step:Assume the deleted union is independent and compute its rank]Define the set $X \subset E$ by \begin{align*} X := (C_1 \cup C_2) \setminus \{e\}. \end{align*} Suppose, for contradiction, that $X$ is independent. Since $M$ is finite and $X$ is independent, the definition of the rank function gives \begin{align*} r(X) = |X| = |C_1 \cup C_2| - 1. \end{align*} Define $Y \subset E$ by \begin{align*} Y := C_1 \cup C_2. \end{align*} Then $Y = X \cup \{e\}$. We use the following one-element rank bound, which follows directly from the definition of rank: every independent subset of $Y$ either lies in $X$ or becomes an independent subset of $X$ after deleting $e$. Therefore \begin{align*} r(X) \le r(Y) \le r(X) + 1. \end{align*} The alternative $r(Y) = r(X) + 1 = |Y|$ is impossible, because $r(Y) = |Y|$ would mean that $Y$ is independent, and then its subset $C_1$ would be independent, contradicting that $C_1$ is a circuit. Hence \begin{align*} r(Y) = r(X) = |C_1 \cup C_2| - 1. \end{align*}[/step]

[guided]Define \begin{align*} X := (C_1 \cup C_2) \setminus \{e\} \end{align*} and assume, toward a contradiction, that $X$ is independent. The point of this assumption is that it turns the rank of $X$ into an exact cardinality: by the definition of rank in a finite matroid, \begin{align*} r(X) = |X|. \end{align*} Since $e \in C_1 \cap C_2$, deleting $e$ from $C_1 \cup C_2$ removes exactly one element, so \begin{align*} r(X) = |X| = |C_1 \cup C_2| - 1. \end{align*} Now define \begin{align*} Y := C_1 \cup C_2. \end{align*} Then $Y = X \cup \{e\}$. Adding one element to a set can increase matroid rank by at most one: an independent subset of $Y$ either avoids $e$, in which case it lies in $X$, or contains $e$, in which case deleting $e$ gives an independent subset of $X$ with one fewer element. Therefore \begin{align*} r(X) \le r(Y) \le r(X) + 1. \end{align*} There are only two possible values for $r(Y)$. We rule out the larger one. If \begin{align*} r(Y) = r(X) + 1, \end{align*} then using $r(X) = |Y| - 1$ gives \begin{align*} r(Y) = |Y|. \end{align*} For a finite matroid, $r(Y) = |Y|$ means that $Y$ itself is independent. But independence is hereditary, so every subset of $Y$ would also be independent. Since $C_1 \subset Y$, this would make $C_1$ independent, contradicting that $C_1$ is a circuit, hence dependent. Thus the larger value is impossible, and we must have \begin{align*} r(Y) = r(X) = |C_1 \cup C_2| - 1. \end{align*}[/guided]

[step:Use rank submodularity to force the intersection to be dependent] Since $C_1$ and $C_2$ are circuits, each is minimally dependent. Thus every proper subset of $C_i$ is independent, and in particular \begin{align*} r(C_i) = |C_i| - 1 \end{align*} for $i \in \{1,2\}$. We use the submodularity property of the matroid rank function: for any subsets $A, B \subset E$, one has $r(A \cup B) + r(A \cap B) \le r(A) + r(B)$. Applying this property with $A := C_1$ and $B := C_2$ gives \begin{align*} r(C_1 \cup C_2) + r(C_1 \cap C_2) \le r(C_1) + r(C_2). \end{align*} Substituting the rank values gives \begin{align*} |C_1 \cup C_2| - 1 + r(C_1 \cap C_2) \le |C_1| + |C_2| - 2. \end{align*} Using the finite-set identity \begin{align*} |C_1| + |C_2| = |C_1 \cup C_2| + |C_1 \cap C_2| \end{align*} we obtain \begin{align*} r(C_1 \cap C_2) \le |C_1 \cap C_2| - 1. \end{align*} Therefore $C_1 \cap C_2$ is dependent. [/step]

[step:Contradict minimal dependence of the two circuits] Because $C_1$ and $C_2$ are distinct circuits, neither can properly contain the other: if, for instance, $C_1 \subsetneq C_2$, then $C_1$ would be a dependent proper subset of the circuit $C_2$, contradicting minimal dependence of $C_2$. Hence $C_1 \cap C_2$ is a proper subset of both $C_1$ and $C_2$. Since every proper subset of a circuit is independent, $C_1 \cap C_2$ is independent. This contradicts the conclusion of the previous step that $C_1 \cap C_2$ is dependent. [/step]

[step:Extract a circuit contained in the deleted union] The contradiction shows that $X = (C_1 \cup C_2) \setminus \{e\}$ is dependent. Since $M$ is finite, among all dependent subsets of $X$ there exists one of minimal cardinality; denote such a subset by $C_3 \subset X$. By minimality, $C_3$ is dependent and every proper subset of $C_3$ is independent, so $C_3$ is a circuit of $M$. Since $C_3 \subset X$, we have \begin{align*} C_3 \subset (C_1 \cup C_2) \setminus \{e\}. \end{align*} This is the required circuit. [/step]