[proofplan] We express $\mathbb{P}(A \cap B^c)$ by writing $A = (A \cap B) \cup (A \cap B^c)$ as a disjoint union, then use the independence of $A$ and $B$ to verify the product formula $\mathbb{P}(A \cap B^c) = \mathbb{P}(A)\,\mathbb{P}(B^c)$. [/proofplan] [step:Decompose $A$ into $(A \cap B) \cup (A \cap B^c)$ and solve for $\mathbb{P}(A \cap B^c)$] The sets $A \cap B$ and $A \cap B^c$ are disjoint (an element of $A$ either belongs to $B$ or does not), and their union is $A$. By finite additivity, \begin{align*} \mathbb{P}(A) = \mathbb{P}(A \cap B) + \mathbb{P}(A \cap B^c), \end{align*} so $\mathbb{P}(A \cap B^c) = \mathbb{P}(A) - \mathbb{P}(A \cap B)$. [/step] [step:Substitute independence of $A$ and $B$ and factor] Since $A$ and $B$ are independent, $\mathbb{P}(A \cap B) = \mathbb{P}(A)\,\mathbb{P}(B)$. Substituting: \begin{align*} \mathbb{P}(A \cap B^c) &= \mathbb{P}(A) - \mathbb{P}(A)\,\mathbb{P}(B) \\ &= \mathbb{P}(A)\bigl(1 - \mathbb{P}(B)\bigr) \\ &= \mathbb{P}(A)\,\mathbb{P}(B^c), \end{align*} where the last equality uses $\mathbb{P}(B^c) = 1 - \mathbb{P}(B)$. This is the product formula for $A$ and $B^c$, so $A$ and $B^c$ are independent. [/step]