If $G$ and $H$ are arbitrary equivalence relations in $A$, prove that
$$A/(G\circ H) ≈ (A/G)/(G\circ H/G).$$
How to think $(G\circ H)/G$ intuitively, especially if $(G\circ H)$ is empty set, and in that case how $G$ will cancel out.
Please, help me prove it, too.
Let $A = \{2, 3, 5, 15\}, G$ is the equivalence relation divisible by 3 and $H$ is the equivalence relation divisible by 5.
Now the partition $A/G = \{\{3, 15\}, \{5\}, \{2\}\}$
partition $A/H = \{\{5, 15\}, \{3\}, \{2\}\}$.
$G = \{\{2, 2\}, \{3, 3\}, \{5, 5\}, \{15, 15\}, \{3, 15\}, \{15, 3\} \}$
$H = \{\{2, 2\}, \{3, 3\}, \{5, 5\}, \{15, 15\}, \{5, 15\}, \{15, 5\} \}$
$G\circ H = \{\{2, 2\}, \{3, 3\}, \{5, 5\}, \{15, 15\}, \{15, 3\}, \{5, 15\}, \{5, 3\}, \{15, 5\} \}$
In the above example will there be any partions formed by $A/(G\circ H)$ ? if there is a partition or not, how to calculate it ? and also how to get $(G\circ H)/G$ ? please explain me the intuition behind it.