Finite Direct Product of Modules is Module/Proof 2

Proof
We need to show that:

$\forall x, y, \in G, \forall \lambda, \mu \in R$:


 * $(1): \quad \lambda \circ \paren {x + y} = \paren {\lambda \circ x} + \paren {\lambda \circ y}$


 * $(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} + \paren {\mu \circ x}$


 * $(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$

Checking the criteria in order:

Criterion 1

 * $(1): \quad \lambda \circ \paren {x + y} = \paren {\lambda \circ x} + \paren {\lambda \circ y}$

Let $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in G$.

So $(1)$ holds.

Criterion 2

 * $(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} + \paren {\mu \circ x}$

Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.

So $(2)$ holds.

Criterion 3

 * $(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$

Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.

So $(3)$ holds.

Hence the result.