[proofplan] Both $(\alpha \wedge \beta) \wedge \gamma$ and $\alpha \wedge (\beta \wedge \gamma)$ are alternating $(p+q+r)$-multilinear forms, so it suffices to evaluate them on an arbitrary tuple $(v_1, \dots, v_{p+q+r})$. We expand each side via two nested applications of the shuffle formula. The left side first sums over $(p+q, r)$-shuffles and then splits the first block via $(p, q)$-shuffles; the right side first sums over $(p, q+r)$-shuffles and then splits the second block via $(q, r)$-shuffles. We show that both nested sums are reindexed by the same set $\mathrm{Sh}(p, q, r)$ of $(p, q, r)$-shuffles — permutations of $\{1, \dots, p+q+r\}$ that are increasing on each of the three blocks $\{1, \dots, p\}$, $\{p+1, \dots, p+q\}$, $\{p+q+1, \dots, p+q+r\}$ — and that the sign of the composed permutation matches in each case. Both expansions therefore equal the same triple sum, proving the identity. [/proofplan] [step:Reduce associativity to an identity of values on vector tuples] By definition, $\Lambda^p(V^*)$, $\Lambda^{q+r}(V^*)$, etc. are subspaces of the spaces of multilinear forms on the appropriate Cartesian powers of $V$. Two multilinear forms on $V^{p+q+r}$ are equal if and only if they take the same value on every tuple $(v_1, \dots, v_{p+q+r}) \in V^{p+q+r}$. Hence to prove \begin{align*} (\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma) \quad \text{in } \Lambda^{p+q+r}(V^*), \end{align*} it suffices to fix an arbitrary $(v_1, \dots, v_{p+q+r}) \in V^{p+q+r}$ and prove \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(v_1, \dots, v_{p+q+r}) = \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(v_1, \dots, v_{p+q+r}). \end{align*} For the rest of the proof, fix such a tuple and write $n := p + q + r$ for the total arity. [guided] The wedge product $\wedge: \Lambda^p(V^*) \times \Lambda^q(V^*) \to \Lambda^{p+q}(V^*)$ outputs an alternating multilinear form. Multilinear forms are determined by their values on arbitrary tuples of vectors: if two multilinear forms $\omega_1, \omega_2: V^n \to k$ satisfy $\omega_1(v_1, \dots, v_n) = \omega_2(v_1, \dots, v_n)$ for every $(v_1, \dots, v_n) \in V^n$, then $\omega_1 = \omega_2$ as elements of $\Lambda^n(V^*)$ (since equality of functions on a set $A$ is, by definition, pointwise equality). So the entire proof reduces to checking, for an arbitrary but fixed tuple $(v_1, \dots, v_n)$ with $n = p + q + r$, that \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(v_1, \dots, v_n) = \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(v_1, \dots, v_n). \end{align*} This converts an identity in $\Lambda^{p+q+r}(V^*)$ into an identity of scalars in $k$, which we will establish by expanding both sides via the shuffle formula. [/guided] [/step] [step:Expand the left-hand side as a double sum over $(p{+}q, r)$- and $(p, q)$-shuffles] Write $\omega := \alpha \wedge \beta \in \Lambda^{p+q}(V^*)$. Applying the shuffle formula to $\omega \wedge \gamma \in \Lambda^{n}(V^*)$ at $(v_1, \dots, v_n)$, \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(v_1, \dots, v_n) &= (\omega \wedge \gamma)(v_1, \dots, v_n) \\ &= \sum_{\tau \in \mathrm{Sh}(p+q, r)} \mathrm{sgn}(\tau)\, \omega(v_{\tau(1)}, \dots, v_{\tau(p+q)})\, \gamma(v_{\tau(p+q+1)}, \dots, v_{\tau(n)}). \end{align*} For each fixed $\tau \in \mathrm{Sh}(p+q, r)$, apply the shuffle formula a second time to expand \begin{align*} \omega(v_{\tau(1)}, \dots, v_{\tau(p+q)}) = (\alpha \wedge \beta)(w_1, \dots, w_{p+q}), \end{align*} where we set $w_j := v_{\tau(j)}$ for $j = 1, \dots, p+q$. The shuffle formula gives \begin{align*} (\alpha \wedge \beta)(w_1, \dots, w_{p+q}) = \sum_{\rho \in \mathrm{Sh}(p, q)} \mathrm{sgn}(\rho)\, \alpha(w_{\rho(1)}, \dots, w_{\rho(p)})\, \beta(w_{\rho(p+1)}, \dots, w_{\rho(p+q)}). \end{align*} Substituting $w_j = v_{\tau(j)}$ and combining, we obtain the double sum \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(v_1, \dots, v_n) &= \sum_{\tau \in \mathrm{Sh}(p+q, r)} \sum_{\rho \in \mathrm{Sh}(p, q)} \mathrm{sgn}(\tau)\, \mathrm{sgn}(\rho) \\ &\qquad \times \alpha\bigl(v_{\tau(\rho(1))}, \dots, v_{\tau(\rho(p))}\bigr)\, \beta\bigl(v_{\tau(\rho(p+1))}, \dots, v_{\tau(\rho(p+q))}\bigr)\, \gamma\bigl(v_{\tau(p+q+1)}, \dots, v_{\tau(n)}\bigr). \tag{$L$} \end{align*} [guided] We unpack the left-hand side by applying the shuffle formula twice. First, treat $\alpha \wedge \beta$ as a single $(p+q)$-form $\omega$ and apply the formula to $\omega \wedge \gamma$: \begin{align*} (\omega \wedge \gamma)(v_1, \dots, v_n) = \sum_{\tau \in \mathrm{Sh}(p+q, r)} \mathrm{sgn}(\tau)\, \omega(v_{\tau(1)}, \dots, v_{\tau(p+q)})\, \gamma(v_{\tau(p+q+1)}, \dots, v_{\tau(n)}). \end{align*} Here $\tau$ ranges over permutations of $\{1, \dots, n\}$ that are strictly increasing on the first $p+q$ positions and on the last $r$ positions: $\tau(1) < \dots < \tau(p+q)$ and $\tau(p+q+1) < \dots < \tau(n)$. Now we have to expand $\omega = \alpha \wedge \beta$ itself. For each fixed $\tau$, the vectors fed into $\omega$ are $(v_{\tau(1)}, \dots, v_{\tau(p+q)})$. To apply the shuffle formula to $\alpha \wedge \beta$, it is helpful to rename these vectors $w_j := v_{\tau(j)}$ — purely a notational convenience, not a mathematical step. Then \begin{align*} \omega(w_1, \dots, w_{p+q}) = (\alpha \wedge \beta)(w_1, \dots, w_{p+q}) = \sum_{\rho \in \mathrm{Sh}(p, q)} \mathrm{sgn}(\rho)\, \alpha(w_{\rho(1)}, \dots, w_{\rho(p)})\, \beta(w_{\rho(p+1)}, \dots, w_{\rho(p+q))}. \end{align*} Here $\rho$ is a permutation of $\{1, \dots, p+q\}$ that is increasing on $\{1, \dots, p\}$ and on $\{p+1, \dots, p+q\}$. Re-substituting $w_j = v_{\tau(j)}$ inside $\alpha$ and $\beta$, the indices of the vectors become $\tau(\rho(j)) = (\tau \circ \rho)(j)$ — function composition is what naturally arises. Putting both expansions together: \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(v_1, \dots, v_n) &= \sum_{\tau \in \mathrm{Sh}(p+q, r)} \sum_{\rho \in \mathrm{Sh}(p, q)} \mathrm{sgn}(\tau)\, \mathrm{sgn}(\rho) \\ &\qquad \times \alpha\bigl(v_{(\tau\rho)(1)}, \dots, v_{(\tau\rho)(p)}\bigr)\, \beta\bigl(v_{(\tau\rho)(p+1)}, \dots, v_{(\tau\rho)(p+q)}\bigr)\, \gamma\bigl(v_{\tau(p+q+1)}, \dots, v_{\tau(n)}\bigr). \tag{$L$} \end{align*} Note that we extend $\rho \in S_{p+q}$ to a permutation $\tilde\rho \in S_n$ by setting $\tilde\rho(j) = j$ for $j > p + q$; we will continue to write $\rho$ for this extension, so that $\tau \circ \rho$ is a permutation of $\{1, \dots, n\}$ acting as $\rho$ on the first $p + q$ indices and as the identity on the last $r$ indices, followed by $\tau$. This makes the indexing $\gamma(v_{\tau(p+q+1)}, \dots, v_{\tau(n)}) = \gamma(v_{(\tau\rho)(p+q+1)}, \dots, v_{(\tau\rho)(n)})$ correct. [/guided] [/step] [step:Identify the composition $\tau \circ \rho$ as a $(p, q, r)$-shuffle and check the sign matches] Define \begin{align*} \mathrm{Sh}(p, q, r) := \{ \sigma \in S_n : \sigma(1) < \dots < \sigma(p),\; \sigma(p+1) < \dots < \sigma(p+q),\; \sigma(p+q+1) < \dots < \sigma(n)\}, \end{align*} the set of $(p, q, r)$-shuffles — permutations of $\{1, \dots, n\}$ that are strictly increasing on each of the three blocks $B_1 = \{1, \dots, p\}$, $B_2 = \{p+1, \dots, p+q\}$, $B_3 = \{p+q+1, \dots, n\}$. For each $\rho \in \mathrm{Sh}(p, q)$, extend it to $\tilde\rho \in S_n$ by $\tilde\rho(j) = j$ for $j \in B_3$ (so $\tilde\rho$ fixes the last $r$ positions). With this extension, $\mathrm{sgn}(\tilde\rho) = \mathrm{sgn}(\rho)$, since adjoining $r$ fixed points contributes no inversions. We claim the map \begin{align*} \Phi: \mathrm{Sh}(p+q, r) \times \mathrm{Sh}(p, q) &\to \mathrm{Sh}(p, q, r) \\ (\tau, \rho) &\mapsto \tau \circ \tilde\rho \end{align*} is a bijection, and $\mathrm{sgn}(\tau \circ \tilde\rho) = \mathrm{sgn}(\tau)\, \mathrm{sgn}(\rho)$. We prove these in two sub-claims. [claim:$\tau \circ \tilde\rho \in \mathrm{Sh}(p, q, r)$ for every $(\tau, \rho) \in \mathrm{Sh}(p+q, r) \times \mathrm{Sh}(p, q)$] [proof] Let $\sigma := \tau \circ \tilde\rho$. We verify the three monotonicity conditions. On $B_1 = \{1, \dots, p\}$: by definition of $\rho \in \mathrm{Sh}(p, q)$, $\rho(1) < \rho(2) < \dots < \rho(p)$, and these values all lie in $\{1, \dots, p+q\}$. Since $\tau$ restricted to $\{1, \dots, p+q\}$ is strictly increasing (the first half of being a $(p+q, r)$-shuffle), the composition $\tau \circ \tilde\rho$ is strictly increasing on $B_1$: $\sigma(1) < \dots < \sigma(p)$. On $B_2 = \{p+1, \dots, p+q\}$: by definition of $\rho$, $\rho(p+1) < \dots < \rho(p+q)$, with values in $\{1, \dots, p+q\}$. By the same monotonicity of $\tau$ on $\{1, \dots, p+q\}$, $\sigma(p+1) < \dots < \sigma(p+q)$. On $B_3 = \{p+q+1, \dots, n\}$: $\tilde\rho$ fixes $B_3$ by construction, so $\sigma(j) = \tau(j)$ for $j \in B_3$. Since $\tau \in \mathrm{Sh}(p+q, r)$ is strictly increasing on $B_3$ by definition, so is $\sigma$. Hence $\sigma \in \mathrm{Sh}(p, q, r)$. [/proof] [/claim] [claim:Every $\sigma \in \mathrm{Sh}(p, q, r)$ has a unique preimage under $\Phi$] [proof] Fix $\sigma \in \mathrm{Sh}(p, q, r)$. We construct the unique $(\tau, \rho)$ with $\tau \circ \tilde\rho = \sigma$. Let $A := \sigma(B_1 \cup B_2) = \{\sigma(1), \dots, \sigma(p+q)\} \subseteq \{1, \dots, n\}$, a subset of cardinality $p+q$, and let $A^c := \{1, \dots, n\} \setminus A = \sigma(B_3)$. **Construction of $\tau$.** Define $\tau \in S_n$ as follows: $\tau$ maps $\{1, \dots, p+q\}$ bijectively and increasingly onto $A$, and maps $\{p+q+1, \dots, n\}$ bijectively and increasingly onto $A^c$. Explicitly, if we list $A = \{a_1 < a_2 < \dots < a_{p+q}\}$ and $A^c = \{b_1 < b_2 < \dots < b_r\}$, set $\tau(j) = a_j$ for $1 \le j \le p+q$ and $\tau(p+q+k) = b_k$ for $1 \le k \le r$. By construction $\tau \in \mathrm{Sh}(p+q, r)$, and the choice of $\tau$ is forced once $A$ is known (the increasing enumeration of $A$ is unique). Also $\sigma(B_3) = A^c$ is the increasing enumeration of $A^c$ (because $\sigma$ is increasing on $B_3$), which coincides with $\tau(B_3)$, so $\sigma|_{B_3} = \tau|_{B_3}$. **Construction of $\rho$.** Define $\tilde\rho := \tau^{-1} \circ \sigma$. Since $\sigma(B_3) = \tau(B_3) = A^c$ and both $\sigma|_{B_3}$ and $\tau|_{B_3}$ enumerate $A^c$ increasingly, $\sigma|_{B_3} = \tau|_{B_3}$, hence $\tilde\rho|_{B_3} = \mathrm{id}|_{B_3}$. So $\tilde\rho$ is the extension by identity of its restriction $\rho := \tilde\rho|_{\{1, \dots, p+q\}}$. It remains to check $\rho \in \mathrm{Sh}(p, q)$. By definition of $\tau$, $\tau^{-1}$ maps $A$ bijectively and increasingly onto $\{1, \dots, p+q\}$. Since $\sigma$ is increasing on $B_1$ with image inside $A$, the composition $\rho = \tau^{-1} \circ \sigma$ is increasing on $B_1$. Similarly increasing on $B_2$. So $\rho \in \mathrm{Sh}(p, q)$. **Uniqueness.** Conversely, if $\tau \circ \tilde\rho = \sigma$, then $\sigma(B_1 \cup B_2) = \tau(\tilde\rho(B_1 \cup B_2)) = \tau(\{1, \dots, p+q\}) = A$ (since $\tilde\rho$ permutes $\{1, \dots, p+q\}$ and fixes $B_3$), which determines $A$ from $\sigma$, hence determines $\tau$ uniquely (as the unique $(p+q, r)$-shuffle with $\tau(\{1, \dots, p+q\}) = A$), and then $\tilde\rho = \tau^{-1} \circ \sigma$ is determined. Hence $\Phi$ is a bijection. [/proof] [/claim] The sign identity $\mathrm{sgn}(\tau \circ \tilde\rho) = \mathrm{sgn}(\tau)\, \mathrm{sgn}(\tilde\rho) = \mathrm{sgn}(\tau)\, \mathrm{sgn}(\rho)$ follows from the [Sign Homomorphism](/theorems/778) being a group homomorphism $\mathrm{sgn}: S_n \to \{\pm 1\}$ together with $\mathrm{sgn}(\tilde\rho) = \mathrm{sgn}(\rho)$. [guided] We now justify rigorously why the double indexing set on the left-hand side is exactly the same as the set of three-block shuffles. Define \begin{align*} \mathrm{Sh}(p, q, r) := \{ \sigma \in S_n : \sigma(1) < \dots < \sigma(p),\; \sigma(p+1) < \dots < \sigma(p+q),\; \sigma(p+q+1) < \dots < \sigma(n)\}. \end{align*} Thus a permutation $\sigma \in \mathrm{Sh}(p, q, r)$ is increasing on each of the three blocks \begin{align*} B_1 := \{1, \dots, p\}, \qquad B_2 := \{p+1, \dots, p+q\}, \qquad B_3 := \{p+q+1, \dots, n\}. \end{align*} For each $\rho \in \mathrm{Sh}(p, q)$, define its extension $\tilde\rho \in S_n$ by \begin{align*} \tilde\rho(j) := \rho(j) \quad \text{for } 1 \leq j \leq p+q, \qquad \tilde\rho(j) := j \quad \text{for } p+q+1 \leq j \leq n. \end{align*} This extension is needed because $\rho$ originally permutes only the first $p+q$ positions, whereas $\tau \in \mathrm{Sh}(p+q,r)$ is a permutation of all $n$ positions. We claim that the map \begin{align*} \Phi: \mathrm{Sh}(p+q, r) \times \mathrm{Sh}(p, q) &\to \mathrm{Sh}(p, q, r) \\ (\tau, \rho) &\mapsto \tau \circ \tilde\rho \end{align*} is a bijection, and that \begin{align*} \mathrm{sgn}(\tau \circ \tilde\rho) = \mathrm{sgn}(\tau)\, \mathrm{sgn}(\rho). \end{align*} First prove that $\Phi$ is well-defined. Fix $(\tau, \rho) \in \mathrm{Sh}(p+q, r) \times \mathrm{Sh}(p, q)$ and set \begin{align*} \sigma := \tau \circ \tilde\rho. \end{align*} We check the three defining monotonicity conditions for membership in $\mathrm{Sh}(p, q, r)$. On $B_1$, the condition $\rho \in \mathrm{Sh}(p,q)$ gives \begin{align*} \rho(1) < \rho(2) < \dots < \rho(p), \end{align*} and every value $\rho(j)$ lies in $\{1, \dots, p+q\}$. Since $\tau \in \mathrm{Sh}(p+q,r)$, the restriction of $\tau$ to $\{1, \dots, p+q\}$ is strictly increasing. Therefore \begin{align*} \sigma(1) = \tau(\rho(1)) < \tau(\rho(2)) < \dots < \tau(\rho(p)) = \sigma(p). \end{align*} On $B_2$, the condition $\rho \in \mathrm{Sh}(p,q)$ gives \begin{align*} \rho(p+1) < \rho(p+2) < \dots < \rho(p+q), \end{align*} again with all values in $\{1, \dots, p+q\}$. Applying the same monotonicity of $\tau$ on the first $p+q$ positions gives \begin{align*} \sigma(p+1) < \sigma(p+2) < \dots < \sigma(p+q). \end{align*} On $B_3$, the extension $\tilde\rho$ fixes every element, so $\sigma(j) = \tau(j)$ for $j \in B_3$. Since $\tau$ is increasing on $B_3$, we get \begin{align*} \sigma(p+q+1) < \sigma(p+q+2) < \dots < \sigma(n). \end{align*} Hence $\sigma \in \mathrm{Sh}(p,q,r)$, so $\Phi$ is well-defined. Next prove surjectivity and uniqueness of the preimage. Fix $\sigma \in \mathrm{Sh}(p,q,r)$. Define \begin{align*} A := \sigma(B_1 \cup B_2) = \{\sigma(1), \dots, \sigma(p+q)\} \subseteq \{1, \dots, n\}, \end{align*} and define its complement in $\{1, \dots, n\}$ by \begin{align*} A^c := \{1, \dots, n\} \setminus A. \end{align*} Because $\sigma$ is a permutation and $B_3$ is the complement of $B_1 \cup B_2$ in $\{1, \dots, n\}$, we have \begin{align*} A^c = \sigma(B_3). \end{align*} List the elements of these two finite sets in increasing order: \begin{align*} A = \{a_1 < a_2 < \dots < a_{p+q}\}, \qquad A^c = \{b_1 < b_2 < \dots < b_r\}. \end{align*} Define $\tau \in S_n$ by \begin{align*} \tau(j) := a_j \quad \text{for } 1 \leq j \leq p+q, \qquad \tau(p+q+k) := b_k \quad \text{for } 1 \leq k \leq r. \end{align*} By construction, $\tau$ is increasing on $\{1, \dots, p+q\}$ and increasing on $B_3$, so \begin{align*} \tau \in \mathrm{Sh}(p+q,r). \end{align*} Moreover, because $\sigma \in \mathrm{Sh}(p,q,r)$ is increasing on $B_3$ and $\sigma(B_3) = A^c = \{b_1 < \dots < b_r\}$, the restrictions agree: \begin{align*} \sigma|_{B_3} = \tau|_{B_3}. \end{align*} Now define \begin{align*} \tilde\rho := \tau^{-1} \circ \sigma. \end{align*} The equality $\sigma|_{B_3} = \tau|_{B_3}$ implies \begin{align*} \tilde\rho(j) = j \quad \text{for every } j \in B_3. \end{align*} Thus $\tilde\rho$ is the extension by the identity on $B_3$ of its restriction \begin{align*} \rho := \tilde\rho|_{\{1, \dots, p+q\}}. \end{align*} It remains to verify that $\rho \in \mathrm{Sh}(p,q)$. The map $\tau^{-1}$ sends $A$ increasingly onto $\{1, \dots, p+q\}$, because $\tau$ was defined as the increasing enumeration of $A$. Since $\sigma$ is increasing on $B_1$ with image contained in $A$, the composition \begin{align*} \rho = \tau^{-1} \circ \sigma \end{align*} is increasing on $B_1$. The same argument applied to $B_2$ shows that $\rho$ is increasing on $B_2$. Hence \begin{align*} \rho \in \mathrm{Sh}(p,q). \end{align*} The construction also gives \begin{align*} \tau \circ \tilde\rho = \tau \circ (\tau^{-1} \circ \sigma) = \sigma, \end{align*} so every $\sigma \in \mathrm{Sh}(p,q,r)$ lies in the image of $\Phi$. The same construction proves uniqueness. If $(\tau,\rho)$ satisfies \begin{align*} \tau \circ \tilde\rho = \sigma, \end{align*} then \begin{align*} \sigma(B_1 \cup B_2) = \tau(\tilde\rho(B_1 \cup B_2)) = \tau(\{1, \dots, p+q\}). \end{align*} Thus $A = \sigma(B_1 \cup B_2)$ determines the image of the first block of $\tau$. Since $\tau$ must be a $(p+q,r)$-shuffle, it is forced to enumerate $A$ increasingly on $\{1, \dots, p+q\}$ and $A^c$ increasingly on $B_3$. Hence $\tau$ is unique. Once $\tau$ is known, \begin{align*} \tilde\rho = \tau^{-1} \circ \sigma \end{align*} is unique, and therefore its restriction $\rho$ to $\{1, \dots, p+q\}$ is unique. This proves that $\Phi$ is a bijection. Finally check the sign. The [Sign Homomorphism](/theorems/778) states that \begin{align*} \mathrm{sgn}: S_n \to \{\pm 1\} \end{align*} is a group homomorphism. Applying this to the composition $\tau \circ \tilde\rho$ gives \begin{align*} \mathrm{sgn}(\tau \circ \tilde\rho) = \mathrm{sgn}(\tau)\,\mathrm{sgn}(\tilde\rho). \end{align*} Since $\tilde\rho$ acts as $\rho$ on $\{1, \dots, p+q\}$ and fixes $B_3 = \{p+q+1, \dots, n\}$ pointwise, adjoining those fixed points adds no inversions. Therefore \begin{align*} \mathrm{sgn}(\tilde\rho) = \mathrm{sgn}(\rho), \end{align*} and hence \begin{align*} \mathrm{sgn}(\tau \circ \tilde\rho) = \mathrm{sgn}(\tau)\,\mathrm{sgn}(\rho). \end{align*} This is the precise reason the double sum indexed by $(\tau,\rho)$ can be reindexed by the single variable $\sigma \in \mathrm{Sh}(p,q,r)$ without changing either the vector arguments or the sign factor. [/guided] [/step] [step:Reindex the left-hand side as a single sum over $(p, q, r)$-shuffles] Combining Step 2 and Step 3, set $\sigma := \tau \circ \tilde\rho$ and reindex the double sum $(L)$: \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(v_1, \dots, v_n) &= \sum_{\sigma \in \mathrm{Sh}(p, q, r)} \mathrm{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)})\, \gamma(v_{\sigma(p+q+1)}, \dots, v_{\sigma(n)}). \tag{$*$} \end{align*} Here we have used: - the bijection $\Phi: \mathrm{Sh}(p+q, r) \times \mathrm{Sh}(p, q) \to \mathrm{Sh}(p, q, r)$ from Step 3 to re-index the sum; - the sign identity $\mathrm{sgn}(\tau)\, \mathrm{sgn}(\rho) = \mathrm{sgn}(\sigma)$; - the relations $(\tau \circ \tilde\rho)(j) = \tau(\rho(j))$ for $j \in B_1 \cup B_2$ and $(\tau \circ \tilde\rho)(j) = \tau(j)$ for $j \in B_3$, which let us replace the indices inside $\alpha$, $\beta$, $\gamma$ by $\sigma(j)$ uniformly. [/step] [step:Expand the right-hand side and reindex it as the same sum] We now apply the symmetric argument to the right-hand side. Set $\omega' := \beta \wedge \gamma \in \Lambda^{q+r}(V^*)$ and apply the shuffle formula to $\alpha \wedge \omega'$: \begin{align*} \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(v_1, \dots, v_n) = \sum_{\tau' \in \mathrm{Sh}(p, q+r)} \mathrm{sgn}(\tau')\, \alpha(v_{\tau'(1)}, \dots, v_{\tau'(p)})\, \omega'(v_{\tau'(p+1)}, \dots, v_{\tau'(n)}). \end{align*} For each $\tau' \in \mathrm{Sh}(p, q+r)$, expand $\omega' = \beta \wedge \gamma$ using a $(q, r)$-shuffle. Setting $w'_k := v_{\tau'(p+k)}$ for $k = 1, \dots, q+r$, \begin{align*} \omega'(w'_1, \dots, w'_{q+r}) = \sum_{\rho' \in \mathrm{Sh}(q, r)} \mathrm{sgn}(\rho')\, \beta(w'_{\rho'(1)}, \dots, w'_{\rho'(q)})\, \gamma(w'_{\rho'(q+1)}, \dots, w'_{\rho'(q+r)}). \end{align*} Extend $\rho' \in S_{q+r}$ to $\tilde\rho' \in S_n$ by setting $\tilde\rho'(j) = j$ for $j \in B_1 = \{1, \dots, p\}$ and $\tilde\rho'(p + k) = p + \rho'(k)$ for $k = 1, \dots, q+r$. Then $\mathrm{sgn}(\tilde\rho') = \mathrm{sgn}(\rho')$, and the double sum becomes \begin{align*} \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(v_1, \dots, v_n) &= \sum_{\tau' \in \mathrm{Sh}(p, q+r)} \sum_{\rho' \in \mathrm{Sh}(q, r)} \mathrm{sgn}(\tau')\, \mathrm{sgn}(\rho') \\ &\qquad \times \alpha\bigl(v_{(\tau'\tilde\rho')(1)}, \dots, v_{(\tau'\tilde\rho')(p)}\bigr)\, \beta\bigl(v_{(\tau'\tilde\rho')(p+1)}, \dots, v_{(\tau'\tilde\rho')(p+q)}\bigr)\, \gamma\bigl(v_{(\tau'\tilde\rho')(p+q+1)}, \dots, v_{(\tau'\tilde\rho')(n)}\bigr). \tag{$R$} \end{align*} The same argument as in Step 3, with the roles of the first and last blocks interchanged, shows that \begin{align*} \Psi: \mathrm{Sh}(p, q+r) \times \mathrm{Sh}(q, r) &\to \mathrm{Sh}(p, q, r) \\ (\tau', \rho') &\mapsto \tau' \circ \tilde\rho' \end{align*} is a bijection, with $\mathrm{sgn}(\tau' \circ \tilde\rho') = \mathrm{sgn}(\tau')\, \mathrm{sgn}(\rho')$. Concretely: a $(p, q, r)$-shuffle $\sigma$ is determined by the set $A' := \sigma(B_2 \cup B_3) \subseteq \{1, \dots, n\}$ (which recovers $\tau'$ as the $(p, q+r)$-shuffle with $\tau'(\{p+1, \dots, n\}) = A'$), together with the $(q, r)$-shuffle $\rho'$ on the last $q + r$ positions (which records how $A'$ is split between the second and third blocks). The verification is identical in structure to the two sub-claims in Step 3 — the only change is that the roles of "first block of size $p+q$" and "last block of size $q+r$" are swapped. Reindexing $(R)$ with $\sigma := \tau' \circ \tilde\rho'$, \begin{align*} \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(v_1, \dots, v_n) &= \sum_{\sigma \in \mathrm{Sh}(p, q, r)} \mathrm{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)})\, \gamma(v_{\sigma(p+q+1)}, \dots, v_{\sigma(n)}). \tag{$**$} \end{align*} [guided] The argument for the right-hand side is the mirror image of the argument for the left. Where the left side bracketed off the first $p + q$ inputs and then split them into $p + q$, the right side brackets off the last $q + r$ inputs and splits them into $q + r$. The two bijections \begin{align*} \Phi&: \mathrm{Sh}(p+q, r) \times \mathrm{Sh}(p, q) \to \mathrm{Sh}(p, q, r) \quad \text{(used on the LHS)} \\ \Psi&: \mathrm{Sh}(p, q+r) \times \mathrm{Sh}(q, r) \to \mathrm{Sh}(p, q, r) \quad \text{(used on the RHS)} \end{align*} parametrise $(p, q, r)$-shuffles in two different ways — by which $p + q$ values appear in the first two blocks, and by which $q + r$ values appear in the last two blocks, respectively — but they both produce the same set on the right. **Why the sign identity transfers.** Just as on the LHS, the [Sign Homomorphism](/theorems/778) gives $\mathrm{sgn}(\tau' \tilde\rho') = \mathrm{sgn}(\tau')\, \mathrm{sgn}(\tilde\rho')$, and $\mathrm{sgn}(\tilde\rho') = \mathrm{sgn}(\rho')$ since extending by the identity on $B_1$ contributes no inversions. **Verifying $\Psi$ is a bijection.** The same construction as in Step 3 works *mutatis mutandis*: given $\sigma \in \mathrm{Sh}(p, q, r)$, set $A' = \sigma(B_2 \cup B_3)$ and reconstruct $\tau'$ as the unique $(p, q+r)$-shuffle with $\tau'(B_1) = \{1, \dots, n\} \setminus A'$ enumerated increasingly and $\tau'(\{p+1, \dots, n\}) = A'$ enumerated increasingly; then $\tilde\rho' = (\tau')^{-1} \circ \sigma$ is the identity on $B_1$ (because both $\sigma$ and $\tau'$ are increasing on $B_1$ with the same image), and its restriction $\rho'$ to $\{p+1, \dots, n\}$ — re-indexed to $\{1, \dots, q+r\}$ via the obvious shift — is a $(q, r)$-shuffle by the same monotonicity argument. The conclusion is that $(R)$ reindexes to $(**)$, which is **literally the same expression as $(*)$**. [/guided] [/step] [step:Conclude equality of the two expansions] By $(*)$ from Step 4 and $(**)$ from Step 5, \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(v_1, \dots, v_n) &= \sum_{\sigma \in \mathrm{Sh}(p, q, r)} \mathrm{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)})\, \gamma(v_{\sigma(p+q+1)}, \dots, v_{\sigma(n)}) \\ &= \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(v_1, \dots, v_n). \end{align*} Since $(v_1, \dots, v_n) \in V^n$ was arbitrary, the reduction in Step 1 gives \begin{align*} (\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma) \quad \text{in } \Lambda^{p+q+r}(V^*), \end{align*} as claimed. [/step]