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Let $A,B$ and $C$ be any sets.

To prove $ \space B-(A-C) \subseteq (B-C) \cup A \Leftrightarrow B \cap C \subseteq A \space$ I began proving the implication $ \space B-(A-C) \subseteq (B-C) \cup A \Rightarrow B \cap C \subseteq A \space$. I started with the direct method, but after a few trys I changed the method.

So, let $B \cap C \subseteq A$ be false, then $B \cap C \nsubseteq A$. There exists a $x \in B \cap C$ such that $x \notin A$. My goal is to prove that $B-(A-C) \subseteq (B-C) \cup A$ is also false.

Let be $\space x \in B \cap C$, so $x \in B \space $ and $ \space x \in C$. Then, by the hypothesis $x \in B-(A-C)$. But if $x \in B \space $,$ \space x \in C$ and $ \space x \notin A \space$, then $ \space x \notin (B-C) \cup A$.

So $B-(A-C) \subseteq (B-C) \cup A$ is false and then the implication is true.

My problem is to prove the reverse implication. $B \cap C \subseteq A \Rightarrow B-(A-C) \subseteq (B-C) \cup A$. I'm not figure it out how to start.

Thanks again.

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up vote 1 down vote accepted

Start with an arbitrary $x\in B\setminus(A\setminus C)$; you want to show that $x\in(B\setminus C)\cup A$, assuming that $B\cap C\subseteq A$.

You know that $x\in B$ and $x\notin A\setminus C$, so either $x\notin A$, or $x\in C$. If $x\in A$, then $x\in(B\setminus C)\cup A$; what if $x\notin A$? Then $x\in C$. We also know that $x\in B$, so $x\in B\cap C\subseteq A$. That’s a contradiction; it shows that $x$ must be in $A$, and therefore $x\in(B\setminus C)\cup A$. Thus, the assumption that $B\cap C\subseteq A$ does imply that $B\setminus(A\setminus C)\subseteq(B\setminus C)\cup A$, as desired.

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$$ B−(A−C)⊆(B−C)∪A \tag{1}$$

First, rewrite it using $X - Y = X \cap Y'$:

$$B \cap (A' \cup C) \subseteq (B \cap C') \cup A$$

Then use distributive law:

$$(B \cap A') \cup (B \cap C) \subseteq (B \cap C') \cup A$$

Observe that $B \cap A'$ cannot be contained in $A$ and $B \cap C$ cannot be contained in $C'$ and so even more in $B \cap C'$. So this expression implies that

\begin{align} B \cap A' &\subseteq B \cap C', \tag{2}\\ B \cap C &\subseteq A\tag{3}. \end{align}

However, by monotocity of $\cup$ (that is we can sum the formulas side-by-side) the $\Leftarrow$ implication is also true. Hence, we have proved, that $(1) \iff (2) \land (3)$.

We have shown that $(1) \Rightarrow (2)$. To prove the other implication, consider $B \cap C \subseteq A$. This implies that $(B \cap C)' \supseteq A'$ that is $A' \subseteq B' \cup C'$, however $\cap$ is monotonic, and therefore $B \cap A' \subseteq B \cap (B' \cup C')$ which is equivalent to $B\cap A' \subseteq B \cap C'$. So, we have that $(2) \Rightarrow (3)$ and that ends the proof of the whole thing.


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For problems like these, I find that resolving this on the logic level makes them straightforward. Basically, we are asked to simplify $ B-(A-C) \subseteq (B-C) \cup A$. By the definition of $\subseteq$, that means that we want to simplify, for all $x$, $$ \begin{align} & x \in B-(A-C) \Rightarrow x \in (B-C) \cup A \\ \equiv & \;\;\;\;\;\text{"expand definitions of $-$ and $\cup$"} \\ & x \in B \land \lnot(x \in A \land \lnot (x \in C)) \Rightarrow (x \in B \land \lnot(x \in C)) \lor x \in A \\ \equiv & \;\;\;\;\;\text{"expand $\Rightarrow$, using DeMorgan on its left hand side"} \\ & \lnot(x \in B) \lor (x \in A \land \lnot (x \in C)) \lor (x \in B \land \lnot(x \in C)) \lor x \in A \\ \equiv & \;\;\;\;\;\text{"logic: assume negation of $\lnot(x \in B)$ and $x \in A$ in other branches of $\lor$"} \\ & \lnot(x \in B) \lor (\textrm{false} \land \lnot (x \in C)) \lor (\textrm{true} \land \lnot(x \in C)) \lor x \in A \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & \lnot(x \in B) \lor \textrm{false} \lor \lnot(x \in C) \lor x \in A \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & \lnot(x \in B) \lor \lnot(x \in C) \lor x \in A \\ \equiv & \;\;\;\;\;\text{"logic: introduce $\Rightarrow$, using DeMorgan -- suggested by the shape of $B \cap C \subseteq A$"} \\ & x \in B \land x \in C \Rightarrow x \in A \\ \equiv & \;\;\;\;\;\text{"introduce $\cap$ using its definition"} \\ & x \in B \cap C \Rightarrow x \in A \\ \end{align} $$ Therefore, again by the definition of $\subseteq$, we have simplified the original expression to $B \cap C \subseteq A$, as we were asked to prove.

Note how the middle three steps perform the essence of the simplification, and the other steps are just for unpacking and packing of the definitions of set theory. Also note that this proves both directions together, instead of having separate proofs for $\Rightarrow$ and $\Leftarrow$.

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