@Punit Sharma
$+1$ Thanks man.
One more thing is it a procedure in $mealy/moore$ that initially we have to separate them based on $output$ only and not on any other thing $?$
$Set\ 1:\ [ABEF]$$[CDH]$
$A\overset{0}{\rightarrow}$$B$
$B\overset{0}{\rightarrow}$$F$
$A\overset{1}{\rightarrow}$$H$
$B\overset{1}{\rightarrow}$$D$
$B\ \&\ F$ same set on input 0 and $H\ \&\ D$ same set on input 1. Hence $AB$ will be in same set.
We can observe $E$ either with $A\ or\ B$ as both are in same set, and we found out that E will not be in the set with $AB$ because:
$A\overset{0}{\rightarrow}$$B$
$E\overset{0}{\rightarrow}$$D$
$B\ \&\ D$ are on different sets. Hence $A\ \&\ E$ will be in different sets and $So\ on.....$