Proof. (2) $\Rightarrow$ (1): For each $(x_{0}, y_{0}), (x, y) \in E \times F$ and $\lambda \in \alg$,

For each $U \in \cn_{G}(0)$ circled, there exists circled neighbourhoods $V \in \cn_{E}(0)$ and $W \in \cn_{F}(0)$ such that $\lambda(V \times W) \subset U$ for all $\lambda \in \alg$. In which case, there exists $\mu > 0$ such that $y_{0} \in \mu W$ and $x_{0} \in \mu V$. Thus if $(x, y) - (x_{0}, y_{0}) \in \mu^{-1}(V \times W)$, then for every $\lambda \in \alg$,

and $\lambda(x_{0}, y - y_{0}) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_{0}, y_{0})$.$\square$