Establishing set equivalence: Munkres exercise 1.7

Question

An exercise in Munkres asks to write this set in terms of $\cup$, $\cap$, and $-$. The set I cannot figure out is: $$F = \{x \mid x \in A \text { and } (x \in B \implies x \in C)\}.$$ Here is what I have done so far. \begin{align*} x \in A \text{ and } (x \in B \implies x \in C) & \iff x \in A \text{ and } (x \not \in B \text{ or } x \in C) \\ & \iff (x \in A \text{ and } x \not \in B) \text{ or } (x \in A \text{ and } x \in C) \\ & \iff (x \in A - B) \text{ or } (x \in A \cap C) \\ & \iff x \in (A \cap B)\cup (A - B). \end{align*} The solution to the problem, though, is $$F = A - (B - C).$$ I cannot figure out why my solution is incorrect. In the first step, I used the material implication rule , $p \implies q \equiv \neg p \lor q$. In the second, I used the fact that $\wedge$ distributes over $\lor$. All of my steps seem valid, but these resulting sets are not equivalent.

Answer 1

Your solution is correct, except that after “$(x\in A\setminus B)\vee(x\in A\cap C)$”, you should have written “$\iff x\in(A\cap C)\cup(A\setminus B)$”. And you can check that$$(A\cap C)\cup(A\setminus B)=A\setminus(B\setminus C).$$

Answer 2

$\{x: x \in B \implies x \in C\}$ is $B^\complement \cup (B \cap C)$ ( either $x \notin B$ and the implication is voidly true, or $x \in B$ is true but then also $x \in C$. You can check that is equals $(B-C)^\complement$, because the only way to refute the implication is to have $x \in B$ and $x \notin C$, or $x \in B-C$. Intersect this with $A$ (as in the question: $x \in A$ should also hold, so we have an intersection) and we get $A- (B - C)$ as expected (using $X \cap Y^\complement = X-Y$ ).

Answer 3

$$x \in A \land (x \in B \Rightarrow x \in C) \Leftrightarrow x \in A \land (x \notin B \lor x \in C) \Leftrightarrow x \in A \land \neg (x \in B \land x \notin C)$$ And now on right hand there is $A\setminus(B\setminus C)$.