Two binary operations $\oplus$ and $^ \ast $ are defined over the set $\{a, e, f, g, h\}$ as per the following tables
|
$\oplus$
|
a
|
e
|
f
|
g
|
h
|
|
a
|
a
|
e
|
f
|
g
|
h
|
|
e
|
e
|
f
|
g
|
h
|
a
|
|
f
|
f
|
g
|
h
|
a
|
e
|
|
g
|
g
|
h
|
a
|
e
|
f
|
|
h
|
h
|
a
|
e
|
f
|
g
|
|
$\ast$
|
a
|
e
|
f
|
g
|
h
|
|
a
|
a
|
a
|
a
|
a
|
a
|
|
e
|
a
|
e
|
f
|
g
|
h
|
|
f
|
a
|
f
|
h
|
e
|
g
|
|
g
|
a
|
g
|
e
|
h
|
f
|
|
h
|
a
|
h
|
g
|
f
|
e
|
Thus, according to the first table $f \oplus g = a$, while according to the second table $g \ast h=f$, and so on. Also, let $f^2 = f \ast f, \: g^3 = g \ast g \ast g$, and so on.
Upon simplification, $ f \oplus \left[f \ast \{ f \oplus (f \ast f )\} \right]$ equal
- $e$
- $f$
- $g$
- $h$