Associativity of product law in $R^S$ ($R$ ring, $S$ a monoid with condition)

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In Proof of associativity of polynomials product (infinite variables), I ask a question about polynomials and assume it was linked to a question of total algebra. I explicitely ask this question here because I haven't received answers for my initial question.

In detail, Bourbaki (Algebra chap. III, §10, p.455) assets that if $S$ is a monoid satisfying the condition :

(*) For all $s\in S$, there exists only a finite number of ordered pairs $(t, u)$ in $S \times S $ such that $tu = s$

then one can define a multiplicative law on $R^S$ (where $R$ is a commutative ring) by (with obvious notations) : $$ (\alpha_s)(\beta_s)=(\gamma_s) \text{ where } \gamma_s=\sum\limits_{tu=s} \alpha_t \beta_u. $$ The book says that :

"... it is associative, since, for $\alpha,\beta,\gamma$ in $A^S$, $$\sum\limits_{uvw=t} \alpha_u\beta_v\gamma_w = \sum\limits_{rw=t}\left( \left( \sum\limits_{uv=r}\alpha_u\beta_v\right) \gamma_w\right) = \sum\limits_{us=t}\left( \alpha_u\left( \sum\limits_{vw=s}\beta_v\gamma_w\right)\right)." $$

So is my question very simple : why do these equalities stand ? Explanations (more than "it's obvious" !) and/or manipulations on sums and indices to go from two sums to only one would be of great help !

EDIT (02/07/15) : 1) Lang, Algebra, II §3 "The group ring or monoid ring" writes the same statement as an evidence... ??? 2) I am looking for books about polynomials ring of infinite variables. Few books I have read deal with that (Bourbaki for instance...).

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Take $\alpha, \beta, \gamma \in R^S$, where $R$ is a ring and $S$ is a monoid. You want to show that $(1)=(2)=(3)$ for $$(1):=\sum_{uvw=t} \alpha_u \beta_v \gamma_w,\quad (2) := \sum_{rw=t} \left( \sum_{uv=r} \alpha_u \beta_v \gamma_w \right), \quad (3) := \sum_{us=t} \left( \sum_{vw=s} \alpha_u \beta_v \gamma_w \right) $$ (I've taken the liberty to apply the distributive law to $(2)$ and $(3)$ since $\alpha_u, \beta_v, \gamma_w$ are elements of the ring $R$.) I'll just show $(1)=(2)$, and the same argument proves $(1)=(3)$.

Since ring addition is commutative and associative, suffices to show that $\{ (u,v,w) \in S^3 : uvw=t \}$ is the same set as $\{ (u, v, w) \in S^3 : \exists r \in S \text{ s.t. } rw=t \text{ and } uv=r \}$, because this means that we're indexing over the same sets in $(1)$ and $(2)$. Well, let $u, v, w \in S$. You can easily see that $$uvw=t \quad \iff \quad \exists r \in S \text{ s.t. } rw=t \text{ and } uv=r $$ so the sets are indeed equal.