[proofplan] The identity is $C^\infty(U)$-linear in $\alpha$ and $\beta$ and is local, so it suffices to verify it pointwise on alternating multilinear maps. Fix $p \in U$ and evaluate both sides on an arbitrary $(k+\ell-1)$-tuple of tangent vectors $(X_1, \dots, X_{k+\ell-1})$. Using the [Shuffle Formula for the Wedge Product](/theorems/3558), the left-hand side expands as a sum over $(k,\ell)$-shuffles, with the vector $v$ inserted in the first slot. We then partition the shuffles according to whether $v$ is consumed by $\alpha$ or by $\beta$: the first family rebuilds $(\iota_v \alpha) \wedge \beta$, the second family rebuilds $\alpha \wedge (\iota_v \beta)$ with a sign $(-1)^k$ coming from moving $v$ past the $k$ slots of $\alpha$. The boundary cases $k = 0$ and $\ell = 0$ are handled separately because the shuffle decomposition collapses there. [/proofplan]

[step:Reduce the identity to a pointwise statement on alternating multilinear forms]Both sides of the asserted identity are elements of $\Omega^{k+\ell-1}(U) = \Gamma(\Lambda^{k+\ell-1} T^*U)$. By the [Equivalence of Chart-Atlas and Bundle-Section Definitions of Differential Forms](/theorems/3572), two smooth differential forms on $U$ agree if and only if for every $p \in U$ they agree as alternating multilinear maps $(T_pU)^{k+\ell-1} \to \mathbb{R}$. Fix $p \in U$ throughout the rest of the proof; the maps $\iota_v$, $\wedge$, and the operations on the right-hand side are all defined pointwise (the value at $p$ of $\iota_v \alpha$ depends only on $v(p)$ and $\alpha_p$, and the wedge of two forms at $p$ depends only on the two forms at $p$). It therefore suffices to prove, for every $p \in U$ and every $(X_1, \dots, X_{k+\ell-1}) \in (T_pU)^{k+\ell-1}$, \begin{align*} \iota_{v(p)}(\alpha_p \wedge \beta_p)(X_1, \dots, X_{k+\ell-1}) \;&=\; \bigl((\iota_{v(p)} \alpha_p) \wedge \beta_p\bigr)(X_1, \dots, X_{k+\ell-1}) \\ &\quad + (-1)^k\, \bigl(\alpha_p \wedge (\iota_{v(p)} \beta_p)\bigr)(X_1, \dots, X_{k+\ell-1}). \end{align*} For ease of notation we abbreviate $w := v(p)$ and drop the subscript $p$ on $\alpha, \beta$, writing them simply as alternating multilinear maps on $T_pU$.[/step]

[guided]Why is a pointwise reduction valid? The interior product $\iota_v$ acts fibrewise: at a point $p$, the value $(\iota_v\alpha)_p$ depends only on $v(p) \in T_pU$ and on $\alpha_p \in \Lambda^k T_p^*U$, because the defining contraction \begin{align*} (\iota_v \alpha)_p(X_1, \dots, X_{k-1}) = \alpha_p\bigl(v(p), X_1, \dots, X_{k-1}\bigr) \end{align*} involves only the values at $p$. The wedge product is likewise fibrewise: $(\alpha \wedge \beta)_p = \alpha_p \wedge \beta_p$ in $\Lambda^{k+\ell}T_p^*U$ — this is the content of the [Graded Algebra Structure on Differential Forms](/theorems/3573), where the algebra structure is defined as the pointwise wedge. Consequently, if we can show the identity holds at each fibre, that is, holds for every choice of $w \in T_pU$, $\alpha \in \Lambda^k T_p^*U$, $\beta \in \Lambda^\ell T_p^*U$ when evaluated on every $(X_1, \dots, X_{k+\ell-1}) \in (T_pU)^{k+\ell-1}$, then the global identity in $\Omega^{k+\ell-1}(U)$ follows by [Equivalence of Chart-Atlas and Bundle-Section Definitions of Differential Forms](/theorems/3572). From now on we work in the single fibre $\Lambda^* T_p^*U$ and treat $\alpha, \beta$ as constant alternating multilinear forms and $w = v(p)$ as a fixed tangent vector.[/guided]

[step:Dispose of the boundary cases where $\alpha$ or $\beta$ is a $0$-form] Assume first $k = 0$, so $\alpha \in C^\infty(U)$ is a function and $\iota_w \alpha = 0$ by definition. The wedge product specialises to scalar multiplication, $\alpha \wedge \beta = \alpha \cdot \beta$, and similarly the interior product on $\beta$ is unchanged. Thus \begin{align*} \iota_w(\alpha \wedge \beta) \;=\; \iota_w(\alpha\, \beta) \;=\; \alpha\, \iota_w \beta, \end{align*} where the last equality holds because the interior product is $C^\infty$-linear in its differential form argument (it is a fibrewise linear contraction with $w$). On the right-hand side of the asserted identity, $(\iota_w \alpha) \wedge \beta = 0$ and $(-1)^0 \alpha \wedge (\iota_w \beta) = \alpha\, \iota_w \beta$, so the identity holds. Assume next $\ell = 0$, so $\beta \in C^\infty(U)$ and $\iota_w \beta = 0$. Then $\alpha \wedge \beta = \beta\, \alpha$ and \begin{align*} \iota_w(\beta\, \alpha)(X_1, \dots, X_{k-1}) \;=\; (\beta\, \alpha)(w, X_1, \dots, X_{k-1}) \;=\; \beta\, \alpha(w, X_1, \dots, X_{k-1}) \;=\; \beta\, (\iota_w \alpha)(X_1, \dots, X_{k-1}), \end{align*} so $\iota_w(\alpha \wedge \beta) = \beta\, \iota_w \alpha = (\iota_w \alpha) \wedge \beta$. The right-hand side reads $(\iota_w \alpha) \wedge \beta + (-1)^k \alpha \wedge 0 = (\iota_w \alpha) \wedge \beta$, so the identity holds. For the remainder of the proof we assume $k \ge 1$ and $\ell \ge 1$. [/step]

[step:Expand the left-hand side using the shuffle formula]By the [Shuffle Formula for the Wedge Product](/theorems/3558), for any alternating multilinear forms $\alpha$ of degree $k$ and $\beta$ of degree $\ell$ on $T_pU$ and for any tuple $(Y_0, Y_1, \dots, Y_{k+\ell-1}) \in (T_pU)^{k+\ell}$, \begin{align*} (\alpha \wedge \beta)(Y_0, Y_1, \dots, Y_{k+\ell-1}) \;=\; \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\; \alpha\bigl(Y_{\sigma(0)}, \dots, Y_{\sigma(k-1)}\bigr)\; \beta\bigl(Y_{\sigma(k)}, \dots, Y_{\sigma(k+\ell-1)}\bigr), \end{align*} where $\operatorname{Sh}(k,\ell)$ denotes the set of $(k,\ell)$-shuffles, i.e., permutations $\sigma$ of $\{0, 1, \dots, k+\ell-1\}$ satisfying \begin{align*} \sigma(0) < \sigma(1) < \cdots < \sigma(k-1) \qquad \text{and} \qquad \sigma(k) < \sigma(k+1) < \cdots < \sigma(k+\ell-1). \end{align*} Fix $(X_1, \dots, X_{k+\ell-1}) \in (T_pU)^{k+\ell-1}$ and set $X_0 := w$ so that we have a $(k+\ell)$-tuple $(X_0, X_1, \dots, X_{k+\ell-1})$. By the definition of the interior product followed by the shuffle formula applied to this tuple, \begin{align*} \iota_w(\alpha \wedge \beta)(X_1, \dots, X_{k+\ell-1}) \;&=\; (\alpha \wedge \beta)(X_0, X_1, \dots, X_{k+\ell-1}) \\ \;&=\; \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\; \alpha\bigl(X_{\sigma(0)}, \dots, X_{\sigma(k-1)}\bigr)\; \beta\bigl(X_{\sigma(k)}, \dots, X_{\sigma(k+\ell-1)}\bigr). \tag{$\ast$} \end{align*}[/step]

[guided]The shuffle formula is the cleanest version of the wedge product because each $(k,\ell)$-shuffle corresponds to a unique way to interleave $k$ inputs (the $\alpha$-block) with $\ell$ inputs (the $\beta$-block) while preserving the internal order within each block; the antisymmetry of $\alpha$ and $\beta$ then absorbs all other permutations and produces the overall factor $\operatorname{sgn}(\sigma)$. See the [Shuffle Formula for the Wedge Product](/theorems/3558) for a derivation from the antisymmetrisation definition. The trick is to view the contraction $\iota_w(\alpha \wedge \beta)$ as the alternating multilinear form $\alpha \wedge \beta$ evaluated with $X_0 := w$ pinned in the first slot. The shuffle decomposition then asks: in each shuffle $\sigma$, which block (the $\alpha$-block or the $\beta$-block) does the index $0$ fall into? Since $\sigma(0)<\cdots<\sigma(k-1)$ and $\sigma(k)<\cdots<\sigma(k+\ell-1)$, the value $0$ — being the smallest index — must be the minimum of whichever block it lies in, so it sits at position $\sigma(0)=0$ (in the $\alpha$-block) or at position $\sigma(k)=0$ (in the $\beta$-block). These are the two cases we analyse next.[/guided]

[step:Partition the shuffles by which block consumes $X_0 = w$] The shuffle set splits as a disjoint union $\operatorname{Sh}(k,\ell) = \operatorname{Sh}_A \sqcup \operatorname{Sh}_B$, where \begin{align*} \operatorname{Sh}_A &:= \{\sigma \in \operatorname{Sh}(k,\ell) : 0 \in \{\sigma(0), \dots, \sigma(k-1)\}\}, \\ \operatorname{Sh}_B &:= \{\sigma \in \operatorname{Sh}(k,\ell) : 0 \in \{\sigma(k), \dots, \sigma(k+\ell-1)\}\}. \end{align*} Since $\sigma(0)<\cdots<\sigma(k-1)$, membership in $\operatorname{Sh}_A$ forces $\sigma(0) = 0$. Similarly, membership in $\operatorname{Sh}_B$ forces $\sigma(k) = 0$. The sum in $(\ast)$ splits as $S_A + S_B$, where $S_A$ and $S_B$ are the partial sums over $\operatorname{Sh}_A$ and $\operatorname{Sh}_B$ respectively. We compute $S_A$ and $S_B$ separately in the next two steps. [/step]

[step:Sum over $\operatorname{Sh}_A$ yields $(\iota_w \alpha) \wedge \beta$]Each $\sigma \in \operatorname{Sh}_A$ has $\sigma(0) = 0$. Define the bijection \begin{align*} \Phi_A: \operatorname{Sh}_A &\longrightarrow \operatorname{Sh}_{\{1,\dots,k+\ell-1\}}(k-1, \ell), \\ \sigma &\longmapsto \sigma|_{\{1, 2, \dots, k+\ell-1\}}, \end{align*} where the codomain is the set of $(k-1, \ell)$-shuffles of $\{1, 2, \dots, k+\ell-1\}$, i.e., bijections $\tau$ of $\{1, \dots, k+\ell-1\}$ with $\tau(1)<\cdots<\tau(k-1)$ and $\tau(k)<\cdots<\tau(k+\ell-1)$. The map $\Phi_A$ is well-defined (removing $0$ from both domain and image preserves the ordering conditions on the two blocks) and is a bijection (the inverse extends a $(k-1,\ell)$-shuffle by sending $0 \mapsto 0$). Moreover, since $\sigma$ fixes $0$, the parity of $\sigma$ equals the parity of $\Phi_A(\sigma)$, i.e., $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\Phi_A(\sigma))$. Now for each $\sigma \in \operatorname{Sh}_A$ with corresponding $\tau = \Phi_A(\sigma)$, the summand in $(\ast)$ becomes \begin{align*} \operatorname{sgn}(\sigma)\, \alpha(X_0, X_{\tau(1)}, \dots, X_{\tau(k-1)})\, \beta(X_{\tau(k)}, \dots, X_{\tau(k+\ell-1)}) \end{align*} since $\sigma(0)=0$ and $\sigma(j) = \tau(j)$ for $j \ge 1$. Recall $X_0 = w$, and use the definition $(\iota_w \alpha)(X_{\tau(1)}, \dots, X_{\tau(k-1)}) := \alpha(w, X_{\tau(1)}, \dots, X_{\tau(k-1)})$. Summing, \begin{align*} S_A \;=\; \sum_{\tau \in \operatorname{Sh}(k-1,\ell)} \operatorname{sgn}(\tau)\, (\iota_w \alpha)\bigl(X_{\tau(1)}, \dots, X_{\tau(k-1)}\bigr)\, \beta\bigl(X_{\tau(k)}, \dots, X_{\tau(k+\ell-1)}\bigr). \end{align*} Here $\operatorname{Sh}(k-1,\ell)$ denotes the $(k-1,\ell)$-shuffles of $\{1, \dots, k+\ell-1\}$, which is canonically identified (by reindexing $j \mapsto j-1$ if desired) with the standard shuffles on $\{0, \dots, k+\ell-2\}$. The expression on the right is exactly the [shuffle formula for the wedge product](/theorems/3558) $(\iota_w \alpha) \wedge \beta$ — a $(k-1)$-form wedged with an $\ell$-form, evaluated on $(X_1, \dots, X_{k+\ell-1})$. Applying the [Shuffle Formula for the Wedge Product](/theorems/3558) in reverse, \begin{align*} S_A \;=\; \bigl((\iota_w \alpha) \wedge \beta\bigr)(X_1, \dots, X_{k+\ell-1}). \end{align*}[/step]

[guided]Why is the sign preserved when we delete $0$? A permutation's sign is determined by the parity of the number of inversions. For $\sigma$ with $\sigma(0) = 0$, the inversions involving the index $0$ are: pairs $(0, j)$ with $j>0$ and $\sigma(0)>\sigma(j)$. But $\sigma(0)=0$ is the minimum value, so no such inversions exist. Hence the inversions of $\sigma$ coincide with the inversions of its restriction to $\{1, \dots, k+\ell-1\}$, and $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\tau)$. The conceptual content of Step $A$: each shuffle in $\operatorname{Sh}_A$ assigns $w$ to the leftmost slot of $\alpha$. Pulling $w$ out as the first argument of $\alpha$ converts $\alpha$ into the $(k-1)$-form $\iota_w \alpha$, and the remaining shuffle interleaves the other $k-1$ arguments of $\alpha$ with the $\ell$ arguments of $\beta$ — precisely the data of a $(k-1, \ell)$-shuffle. Re-summing gives the shuffle expansion of $(\iota_w \alpha) \wedge \beta$.[/guided]

[step:Sum over $\operatorname{Sh}_B$ yields $(-1)^k \alpha \wedge (\iota_w \beta)$]Each $\sigma \in \operatorname{Sh}_B$ has $\sigma(k) = 0$. Define \begin{align*} \Phi_B: \operatorname{Sh}_B &\longrightarrow \operatorname{Sh}_{\{1,\dots,k+\ell-1\}}(k, \ell-1), \\ \sigma &\longmapsto \tau, \end{align*} where $\tau(j) := \sigma(j)$ for $j \in \{0, \dots, k-1\}$ and $\tau(j) := \sigma(j+1)$ for $j \in \{k, \dots, k+\ell-2\}$. Equivalently, $\tau$ is the restriction of $\sigma$ to $\{0, \dots, k+\ell-1\} \setminus \{k\}$, reindexed in order. Since $\sigma$ deletes the position $k$ (which was assigned to $0$) and preserves the ordering within both blocks, $\tau$ is a $(k, \ell-1)$-shuffle of $\{1, \dots, k+\ell-1\}$. The map $\Phi_B$ is a bijection: its inverse inserts $0$ at position $k$, shifting the second block one place to the right. We now compute the sign of $\sigma$ in terms of $\operatorname{sgn}(\tau)$. The permutation $\sigma$ takes $\sigma(k) = 0$, while the values $\sigma(0), \dots, \sigma(k-1)$ are all positive integers because $\sigma$ is a bijection and only position $k$ is mapped to $0$. Therefore the index $0$ appears at position $k$ but, considering only the relative order of values, it is the smallest. The pairs $(i, k)$ with $0 \le i < k$ contribute inversions ($\sigma(i) > 0 = \sigma(k)$), and pairs $(k, j)$ with $k < j \le k+\ell-1$ contribute no inversions ($\sigma(k) = 0 < \sigma(j)$). The remaining pairs not involving $k$ have the same inversion behaviour for $\sigma$ as they do for the corresponding pairs in $\tau$. Hence \begin{align*} \operatorname{sgn}(\sigma) \;=\; (-1)^k\, \operatorname{sgn}(\tau). \end{align*} The summand in $(\ast)$ corresponding to $\sigma$ is, using $\sigma(k) = 0$ and $X_0 = w$, \begin{align*} \operatorname{sgn}(\sigma)\, \alpha\bigl(X_{\sigma(0)}, \dots, X_{\sigma(k-1)}\bigr)\, \beta\bigl(w, X_{\sigma(k+1)}, \dots, X_{\sigma(k+\ell-1)}\bigr). \end{align*} Rewriting via $\tau$: \begin{align*} \operatorname{sgn}(\sigma)\, \alpha\bigl(X_{\tau(0)}, \dots, X_{\tau(k-1)}\bigr)\, (\iota_w \beta)\bigl(X_{\tau(k)}, \dots, X_{\tau(k+\ell-2)}\bigr) \\ \;=\; (-1)^k\, \operatorname{sgn}(\tau)\, \alpha\bigl(X_{\tau(0)}, \dots, X_{\tau(k-1)}\bigr)\, (\iota_w \beta)\bigl(X_{\tau(k)}, \dots, X_{\tau(k+\ell-2)}\bigr). \end{align*} Summing over $\sigma \in \operatorname{Sh}_B$, equivalently over $\tau \in \operatorname{Sh}(k, \ell-1)$ via the bijection $\Phi_B$, \begin{align*} S_B \;=\; (-1)^k \sum_{\tau \in \operatorname{Sh}(k,\ell-1)} \operatorname{sgn}(\tau)\, \alpha\bigl(X_{\tau(0)}, \dots, X_{\tau(k-1)}\bigr)\, (\iota_w \beta)\bigl(X_{\tau(k)}, \dots, X_{\tau(k+\ell-2)}\bigr). \end{align*} Applying the [Shuffle Formula for the Wedge Product](/theorems/3558) in reverse to $\alpha$ (a $k$-form) and $\iota_w \beta$ (an $(\ell-1)$-form), \begin{align*} S_B \;=\; (-1)^k\, \bigl(\alpha \wedge (\iota_w \beta)\bigr)(X_1, \dots, X_{k+\ell-1}). \end{align*}[/step]

[guided]We now analyse the shuffles in which the input $X_0 = w$ is consumed by $\beta$. Each $\sigma \in \operatorname{Sh}_B$ satisfies $\sigma(k)=0$, because $0$ is the smallest value in the ordered second block. Define \begin{align*} \Phi_B: \operatorname{Sh}_B &\longrightarrow \operatorname{Sh}_{\{1,\dots,k+\ell-1\}}(k, \ell-1), \\ \sigma &\longmapsto \tau, \end{align*} where $\tau(j) := \sigma(j)$ for $j \in \{0, \dots, k-1\}$ and $\tau(j) := \sigma(j+1)$ for $j \in \{k, \dots, k+\ell-2\}$. This construction deletes the position $k$, the position occupied by the value $0$, and then reindexes the remaining positions in order. The ordering conditions defining a shuffle are preserved: the first block remains ordered because $\sigma(0)<\cdots<\sigma(k-1)$, and the second block remains ordered because $\sigma(k+1)<\cdots<\sigma(k+\ell-1)$. Hence $\tau$ is a $(k,\ell-1)$-shuffle of $\{1,\dots,k+\ell-1\}$. The map $\Phi_B$ is bijective. Given a $(k,\ell-1)$-shuffle $\tau$ of $\{1,\dots,k+\ell-1\}$, the inverse map inserts the value $0$ at position $k$, keeps the first $k$ positions equal to $\tau(0),\dots,\tau(k-1)$, and shifts the remaining $\ell-1$ positions one place to the right. Since $0$ is smaller than every value in $\{1,\dots,k+\ell-1\}$, the second block begins with its smallest element, and the shuffle inequalities are satisfied. We must now compute the sign introduced by inserting $0$ at position $k$. The permutation $\sigma$ takes $\sigma(k)=0$, while $\sigma(0),\dots,\sigma(k-1)$ are all positive integers because $\sigma$ is a bijection and no position other than $k$ maps to $0$. Therefore, for every $i \in \{0,\dots,k-1\}$, the pair $(i,k)$ is an inversion of $\sigma$: \begin{align*} \sigma(i) > 0 = \sigma(k). \end{align*} There are exactly $k$ such inversions. If $j \in \{k+1,\dots,k+\ell-1\}$, then $(k,j)$ is not an inversion because \begin{align*} \sigma(k)=0 < \sigma(j). \end{align*} All pairs of positions not involving $k$ have the same relative order in $\sigma$ as the corresponding pairs have in $\tau$. Hence the inversion count of $\sigma$ differs from the inversion count of $\tau$ by exactly $k$, and therefore \begin{align*} \operatorname{sgn}(\sigma) = (-1)^k\operatorname{sgn}(\tau). \end{align*} This is the source of the graded sign: $w$ must sit after the $k$ inputs assigned to $\alpha$, and that contributes one sign change for each of those $k$ inputs. For this $\sigma \in \operatorname{Sh}_B$, the corresponding summand in $(\ast)$ is \begin{align*} \operatorname{sgn}(\sigma)\, \alpha\bigl(X_{\sigma(0)}, \dots, X_{\sigma(k-1)}\bigr)\, \beta\bigl(X_{\sigma(k)}, X_{\sigma(k+1)}, \dots, X_{\sigma(k+\ell-1)}\bigr). \end{align*} Using $\sigma(k)=0$ and $X_0=w$, this becomes \begin{align*} \operatorname{sgn}(\sigma)\, \alpha\bigl(X_{\sigma(0)}, \dots, X_{\sigma(k-1)}\bigr)\, \beta\bigl(w, X_{\sigma(k+1)}, \dots, X_{\sigma(k+\ell-1)}\bigr). \end{align*} By the definition of the interior product on the $\ell$-form $\beta$, \begin{align*} (\iota_w\beta)\bigl(X_{\sigma(k+1)}, \dots, X_{\sigma(k+\ell-1)}\bigr) = \beta\bigl(w, X_{\sigma(k+1)}, \dots, X_{\sigma(k+\ell-1)}\bigr). \end{align*} Rewriting the same summand through $\tau=\Phi_B(\sigma)$ and substituting the sign relation gives \begin{align*} &\operatorname{sgn}(\sigma)\, \alpha\bigl(X_{\tau(0)}, \dots, X_{\tau(k-1)}\bigr)\, (\iota_w\beta)\bigl(X_{\tau(k)}, \dots, X_{\tau(k+\ell-2)}\bigr) \\ &\qquad = (-1)^k\operatorname{sgn}(\tau)\, \alpha\bigl(X_{\tau(0)}, \dots, X_{\tau(k-1)}\bigr)\, (\iota_w\beta)\bigl(X_{\tau(k)}, \dots, X_{\tau(k+\ell-2)}\bigr). \end{align*} Now sum this identity over all $\sigma \in \operatorname{Sh}_B$. Since $\Phi_B$ is a bijection, summing over $\sigma \in \operatorname{Sh}_B$ is the same as summing over $\tau \in \operatorname{Sh}(k,\ell-1)$, so \begin{align*} S_B = (-1)^k \sum_{\tau \in \operatorname{Sh}(k,\ell-1)} \operatorname{sgn}(\tau)\, \alpha\bigl(X_{\tau(0)}, \dots, X_{\tau(k-1)}\bigr)\, (\iota_w\beta)\bigl(X_{\tau(k)}, \dots, X_{\tau(k+\ell-2)}\bigr). \end{align*} The sum on the right is exactly the shuffle expansion of $\alpha \wedge (\iota_w\beta)$ evaluated on $(X_1,\dots,X_{k+\ell-1})$, with $\alpha$ a $k$-form and $\iota_w\beta$ an $(\ell-1)$-form. Applying the [Shuffle Formula for the Wedge Product](/theorems/3558) in reverse gives \begin{align*} S_B = (-1)^k\, \bigl(\alpha \wedge (\iota_w\beta)\bigr)(X_1, \dots, X_{k+\ell-1}). \end{align*}[/guided]

[step:Combine $S_A$ and $S_B$ to conclude] Equation $(\ast)$ together with the partition $\operatorname{Sh}(k,\ell) = \operatorname{Sh}_A \sqcup \operatorname{Sh}_B$ gives \begin{align*} \iota_w(\alpha \wedge \beta)(X_1, \dots, X_{k+\ell-1}) \;=\; S_A + S_B \;=\; \bigl((\iota_w \alpha) \wedge \beta\bigr)(X_1, \dots, X_{k+\ell-1}) \;+\; (-1)^k\, \bigl(\alpha \wedge (\iota_w \beta)\bigr)(X_1, \dots, X_{k+\ell-1}). \end{align*} Since this holds for every $(X_1, \dots, X_{k+\ell-1}) \in (T_pU)^{k+\ell-1}$ and for every $p \in U$, the pointwise identity from Step 1 is established. Combined with the boundary cases $k = 0$ and $\ell = 0$ from Step 2, we conclude that \begin{align*} \iota_v(\alpha \wedge \beta) \;=\; (\iota_v \alpha) \wedge \beta \;+\; (-1)^k\, \alpha \wedge (\iota_v \beta) \end{align*} as smooth $(k+\ell-1)$-forms on $U$, which is the asserted graded Leibniz rule. $\blacksquare$ [/step]