Let $f:X\rightarrow Y$ be a function from one set $X$ to another set $Y$, and let $U,V$ be subsets of $Y$. Show that
(a) $f^{-1}(U\cup V) = f^{-1}(U)\cup f^{-1}(V)$,
(b) $f^{-1}(U\cap V) = f^{-1}(U)\cap f^{-1}(V)$,
(c) $f^{-1}(U\backslash V) = f^{-1}(U)\backslash f^{-1}(V)$.
MY ATTEMPT
(a) According to the definition of inverse image, one has \begin{align*} x\in f^{-1}(U\cup V) & \Longleftrightarrow f(x) \in U\cup V\\\\ & \Longleftrightarrow (f(x)\in U)\vee (f(x)\in V)\\\\ & \Longleftrightarrow (x\in f^{-1}(U))\vee(x\in f^{-1}(V))\\\\ & \Longleftrightarrow x\in f^{-1}(U)\cup f^{-1}(V) \end{align*}
(b) Analogously, one has \begin{align*} x\in f^{-1}(U\cap V) & \Longleftrightarrow f(x) \in U\cap V\\\\ & \Longleftrightarrow (f(x)\in U)\wedge(f(x)\in V)\\\\ & \Longleftrightarrow (x\in f^{-1}(U))\wedge(x\in f^{-1}(V))\\\\ & \Longleftrightarrow x\in f^{-1}(U)\cap f^{-1}(V). \end{align*}
(c) Finally, one has that \begin{align*} x\in f^{-1}(U\backslash V) & \Longleftrightarrow f(x)\in U\backslash V\\\\ & \Longleftrightarrow (f(x)\in U)\wedge(f(x)\not\in V)\\\\ & \Longleftrightarrow (x\in f^{-1}(U))\wedge(x\not\in f^{-1}(V))\\\\ & \Longleftrightarrow x\in f^{-1}(U)\backslash f^{-1}(V). \end{align*}
Could someone double-check my answer or provide any comments?