Let $\Delta_N := \{0 \leq \tau_1 \leq \dots \leq \tau_N \leq 1\}$ be the $N$-simplex. Let $a_j: [0, 1] \rightarrow \mathcal{A}$, $j=1, \dots, N$ be continuous functions with values in some (non-commutative) finite-dimensional Banach algebra (e.g. some matrix algebra). Consider the $\mathcal{A}$-valued integral $$ \mathrm{Int} = \sum_{\sigma \in \mathcal{S}_N} \mathrm{sgn}(\sigma) \int_{\Delta_N} a_{\sigma_N}(\tau_N) \cdots a_{\sigma_1}(\tau_1)\, \mathrm{d} \tau_1 \cdots \mathrm{d} \tau_N, $$ where $\mathcal{S}_N$ denotes the group of permutations of the numbers $\{1, \dots, N\}$. Suppose that the index set $\{1, \dots, N\}$ splits as the disjoint union of non-empty sets $I_1, \dots, I_k$ such that we elements from different sets $I_u$ anti-commute pairwise, i.e. for $i \in I_u$, $j \in I_v$, one has $$ a_i(t)a_j(s) = -a_j(s)a_i(t), ~~~~~~~~ \forall s, t \in [0, 1].$$ Clearly, if $k=N$, i.e. each set has only one element so that all the $a_j$ anti-commute, then $$\mathrm{Int} = \prod_{j=1}^N \int_{\Delta_1}a_j(\tau) \mathrm{d}\tau.$$ In the general case, we guess that one has $$ \mathrm{Int} = \prod_{j=1}^k \sum_{\sigma \in \mathcal{S}_{I_j}} \mathrm{sgn}(\sigma) \int_{\Delta_{|I_j|}} \prod_{i\in I_j} a_{\sigma_i}(\tau_i) \mathrm{d}\tau_i,$$ where $\mathcal{S}_{I_j}$ denotes the group of permutations of the set $I_j$ and the latter product is run through from the largest to smallest number $i \in I_j$.

We were unable to prove this in full generality so far, because we didn't find a way to efficiently handle the combinatorics. Does anybody have hints how to tackle this problem or probably people already looked at this so that there is a reference?