Given sets $A$, $B$, and $C$, express each of the following sets in terms of $A$, $B$ and $C$, using the symbols $\cup$, $\cap$, and $-$.
$$F=\{x\mid x\in A\text{ and }(x\in B\implies x\in C)\}$$
My solution:
$x\in B\implies x\in C$
can be rewritten as
$\neg(x\in B)$ or $x\in C$
i.e. $x\notin B$ or $x\in C$
i.e. $x\in B^c$ or $x\in C$
$\begin{aligned} F&=A\cap(B^c\cup C) \\ &=(A\cap B^c)\cup(A\cap C) \\ &=(A-B)\cup(A\cap C) \end{aligned}$
I don't like this solution as I have to introduce $B^c$ and the concept of a universal set, which we haven't covered in the textbook.
Can someone think of a better solution?
This is essentially what you've already written, but slightly rephrased:
First, we rewrite the condition for being in $F$ as $$(x\in A\mbox{ and }x\not\in B)\quad\mbox{or}\quad(x\in A\mbox{ and }x\in C).$$ Now we just note that the first disjunct is exactly the definition of "$x\in A-B$."