The preimage of the complement is the complement of the preimage

up:: Preimage of Function

Let $f:X\to Y$ a function, and $B\subseteq X,D\subseteq Y$.

Let $y\in f(X)\setminus f(B)$. Then, by definition of a Set Complement,

$$(y\in f(X))\wedge (y\notin f(B))⟹(\exists x\in X\mid y=f(x))\wedge (\nexists x\in B\mid f(x)=y)⟹\exists x\in X\setminus B\mid f(x)=y⟹y\in f(X\setminus B)$$

Thus, $f(X)\setminus f(B)\subseteq f(X\setminus B)$.