Answer
First Case:
Assume $\ f^-1$(C ∩ D) is true
by def of range $\ f^-1$ (y) = x , x$\in$(C ∩ D) **( x $\in$ C and x $\in$ D by def of intersection)
$\ f^-1$(y) = x , x$\in$ C implies x$\in$$\ f^-1$(C)
and
$\ f^-1$(y) = x , x$\in$ D implies x$\in$$\ f^-1$(D)
by def of intersection x $\in$ $\ f^-1$(C) ∩ $\ f^-1$(D)
Second Case :
Assume $\ f^-1$(C) ∩ x$\in$$\ f^-1$(D) is true
hence, x $\in$ $\ f^-1$(C) ∩ $\ f^-1$(D)
x$\in$ $\ f^-1$(C) $AND$ x$\in$$\ f^-1$(D) **By definition of intersection
which implies $\ f^-1$(y) = x , x $\in$ (C ∩ D)
which also means x $\in$ $\ f^-1$(C ∩ D)
In conclusion, we got from :
Case 1: $\ f^-1$(C ∩ D) $\subseteq$ $\ f^-1$(C) ∩ $\ f^-1$(D)
Case 2: $\ f^-1$(C) ∩ $\ f^-1$(D) $\subseteq$ $\ f^-1$(C ∩ D)
hence, $\ f^-1$(C ∩ D) $=$ $\ f^-1$(C) ∩ $\ f^-1$(D)
Work Step by Step
First Case:
Assume $\ f^-1$(C ∩ D) is true
by def of range $\ f^-1$ (y) = x , x$\in$(C ∩ D) **( x $\in$ C and x $\in$ D by def of intersection)
$\ f^-1$(y) = x , x$\in$ C implies x$\in$$\ f^-1$(C)
and
$\ f^-1$(y) = x , x$\in$ D implies x$\in$$\ f^-1$(D)
by def of intersection x $\in$ $\ f^-1$(C) ∩ $\ f^-1$(D)
Second Case :
Assume $\ f^-1$(C) ∩ x$\in$$\ f^-1$(D) is true
hence, x $\in$ $\ f^-1$(C) ∩ $\ f^-1$(D)
x$\in$ $\ f^-1$(C) $AND$ x$\in$$\ f^-1$(D) **By definition of intersection
which implies $\ f^-1$(y) = x , x $\in$ (C ∩ D)
which also means x $\in$ $\ f^-1$(C ∩ D)
In conclusion, we got from :
Case 1: $\ f^-1$(C ∩ D) $\subseteq$ $\ f^-1$(C) ∩ $\ f^-1$(D)
Case 2: $\ f^-1$(C) ∩ $\ f^-1$(D) $\subseteq$ $\ f^-1$(C ∩ D)
hence, $\ f^-1$(C ∩ D) $=$ $\ f^-1$(C) ∩ $\ f^-1$(D)
