Associativity on Indexing Set

Theorem
Let $$\left({S, \circ}\right)$$ be a semigroup.

Let $$\left \langle {x_\alpha} \right \rangle_{\alpha \in A}$$ be a family of terms of $S$ indexed by a finite nonempty set $$A$$.

Let $$\left \langle {B_k} \right \rangle_{1 \le k \le n}$$ be a family of distinct subsets of $$A$$ forming a partition of $$A$$. Then:


 * $$\prod_{k=1}^n \left({\prod_{a \in B_k} x_\alpha}\right) = \prod_{\alpha \in A} x_\alpha$$

Proof
For each $$k \in \N^*$$, let $$\left|{B_k}\right| = p_k$$.

Let $$r_0 = 0$$, $$\forall k \in \N^*: r_k = \sum_{j=1}^k {p_j}$$, and let $$p = r_n$$.

Then $$r_k - r_{k-1} = p_k$$, so by Isomorphism to Closed Interval, both $$\left[{1 \,. \, . \, p_k}\right]$$ and $$\left[{r_{k-1}+1 \,. \, . \, r_k}\right]$$ have $$p_k$$ elements.

By Unique Isomorphism between Finite Totally Ordered Sets, there is a unique isomorphism $$\tau_k: \left[{1 \,. \, . \, p_k}\right] \to \left[{r_{k-1}+1 \,. \, . \, r_k}\right]$$ as both are totally ordered.

The orderings on both of these are those induced by the ordering on $$\N$$.

It is clear that $$\tau_k$$ is defined as:


 * $$\forall j \in \left[{1 \, . \, . \, p_k}\right]: \tau_k \left({j}\right) = r_k + j$$

For each $$k \in \left[{1 \,. \, . \, n}\right]$$, let $$\rho_k: \left[{1 \,. \, . \, p_k}\right] \to B_k$$ be a bijection.

By Strictly Increasing Sequence induces Partition, the mapping $$\sigma: \left[{1 \,. \, . \, p}\right] \to A$$ defined by:


 * $$\forall j \in \left[{r_{k-1}+1 \, . \, . \, r_k}\right]: \forall k \in \left[{1 \, . \, . \, n}\right]: \sigma \left({j}\right) = \rho_k \left({\tau_k^{-1} \left({j}\right)}\right)$$

is a bijection.

Let $$\forall j \in \left[{1 \,. \, . \, p}\right]: y_j = x_{\sigma \left({j}\right)}$$.

By definition:


 * $$\prod_{\alpha \in A} {x_\alpha} = \prod_{j=1}^p {x_{\sigma \left({j}\right)}} = \prod_{j=1}^p {y_j}$$

Also:


 * $$\forall k \in \left[{1 \, . \, . \, n}\right]: \prod_{\alpha \in B_k} {x_\alpha} = \prod_{i=1}^{p^k} {x_{\rho_k \left({i}\right)}}$$

Also by definition:


 * $$\prod_{j = r_{k-1}+1}^{r_k} y_j = \prod_{i=1}^{p_k} y_{\tau_k \left({i}\right)} = \prod_{i=1}^{p_k} x_{\sigma \left({\tau_k \left({i}\right)}\right)} = \prod_{i=1}^{p_k} x_{\rho_k \left({i}\right)}$$

So by the General Associativity Theorem:

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