My problem is following: prove that $\Pi_{i \in I} \mathbf{A}_i \cong \Pi_{j \in J} (\Pi_{i \in I_j} \mathbf{A}_i)$, where $\langle \mathbf{A}_i \ | \ i \in I \rangle$ is an indexed set of similar algebras and $\{I_j \ | \ j \in J\}$ is a partition of I.
It is so called generalized associative law for the product of algebras.
I was informed that the isomorphism is given by: $ a \mapsto \langle \langle a(i) \ | \ i \in I_j \rangle \ | \ j \in J\rangle$
How to prove that this map (say $h$) is a homomorphism, that is $$h(F^{\Pi_{i \in I} \mathbf{A}_i} (a_1, \dots, a_n)) = F^{\Pi_{j \in J} (\Pi_{i \in I_j} \mathbf{A}_i)} (h(a_1), \dots, h(a_n)),$$ where $F$ is a functional symbol from the signature.
My calculations are so massive and elaborative that I hardly believe they are correct and sincerely speaking I'm not quite sure I can see the idea behind them.
In the fist step I apply $h$ to the argument $F^{\Pi_{i \in I} \mathbf{A}_i} (a_1, \dots, a_n)$ and obtain from the definition of $h$: $$\langle \langle F^{\Pi_{i \in I} \mathbf{A}_i} (a_1, \dots, a_n)(i) \ | \ i \in I_j \rangle \ | \ j \in J \rangle$$ which by the definition of product is $$\langle \langle F^{\mathbf{A}_i} (a_1(i), \dots, a_n(i)) \ | \ i \in I_j \rangle \ | \ j \in J \rangle.$$ Because $i \in I_j$ and $I_j \subseteq I$ ($I_j$ is a block of our partition), $a(i) = \langle a(i) \ | \ i \in I_j \rangle (i)$ so what we've got equals $$\langle \langle F^{\mathbf{A}_i} (\langle a_1(i) \ | \ i \in I \rangle (i), \dots, \langle a_n(i) \ | \ i \in I \rangle (i)) \ | \ i \in I_j \rangle \ | \ j \in J \rangle,$$ which in turn equals $$ \langle F^{\Pi_{ i \in I_J} \mathbf{A}_i} (\langle a_1(i) \ | \ i \in I \rangle, \dots, \langle a_n(i) \ | \ i \in I) \rangle) \ | \ j \in J \rangle.$$
From now my reasoning is vague. Is the presented argument correct? If it is, how to carry on? Could you give me a hint? Thanks!