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Prove that $A - (B \cap C) = (A - B) \cup (A - C)$

How do I go about solving this proof? I've tried a few things but I'm stuck. I don't need the exact proof step by step, just a few hints on how to solve it would really help me.

edit: I tried to solve considering $3$ cases.

If $x$ is an element of $A−(B\cap C)$, then $x$ belongs to $A$ and $x$ doesn't belong to $B\cap C$.

Then, there I considered $3$ possibilites:

  • $x\in A$ and ($x\in B$ or $x\in C$)

  • $x\in A$ and ($x\notin B$ or $x\in C$)

  • $x\in A$ and ($x\in B$ or $x\notin C$)

After that I found that all of these cases satisfied the definition of set difference which states that $A-B$ is equal to the elements of $A$ that are not in $B$. In that case my $B$ is "($B\cap C$)". After that I tried to prove the other side "$(A−B)\cup (A−C)$" but I got lost and didn't know what to do.

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  • $\begingroup$ Have you drawn a Venn diagram? $\endgroup$ – Patrick Stevens Mar 24 '18 at 20:03
  • $\begingroup$ @PatrickStevens Many instructors, I included, don't accept Venn diagrams as a "proof". They can give us good intuition, but they can also be wrong in some situations. $\endgroup$ – N. S. Mar 24 '18 at 20:07
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    $\begingroup$ Can you show us the steps you did try? $\endgroup$ – Air Conditioner Mar 24 '18 at 20:17
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    $\begingroup$ @N.S. I quote: "I don't need the exact proof step by step, just a few hints on how to solve it would really help me." The result falls out of an appropriate Venn diagram. $\endgroup$ – Patrick Stevens Mar 24 '18 at 20:30
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    $\begingroup$ @N.S. "Venn diagrams ... can also be wrong in some situations" Which situations do you have in mind? $\endgroup$ – Did Mar 24 '18 at 20:44
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Let $A, B, C$ be sets and $x \in A \setminus (B \cap C)$.

$x \in A \setminus (B \cap C)$

$\Leftrightarrow x \in A \wedge x \notin (B \cap C)$

$\Leftrightarrow x \in A \wedge (x \notin B \vee x \notin C)$

$\Leftrightarrow (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)$

$\Leftrightarrow x \in (A \setminus B) \cup (A \setminus C)$.

The equivalencies give us that the LHS and the RHS are subsets of each other, so they are equal.

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    $\begingroup$ This site is not a homework completion site. Please don't attract users who only want us to do their homework, by rewarding them, so they come back and again, ask homework questions. $\endgroup$ – amWhy Mar 24 '18 at 20:20
  • $\begingroup$ I started the proof like you but then I considered 3 cases: x∈A∧(x∉B∨x∉C) or x∈A∧(x∉B∨x∈C) or x∈A∧(x∈B∨x∉C) then I concluded that for these cases indeed x∈A∖(B∩C). But then after that I didn't know what to do. Is it necessary to consider these 3 cases or I just created more work for myself? $\endgroup$ – Sara Mar 24 '18 at 20:26
  • $\begingroup$ You created more work because $\vee$ includes your two additional cases. $\endgroup$ – Jan Mar 24 '18 at 20:28
  • $\begingroup$ I see now, thank you! $\endgroup$ – Sara Mar 24 '18 at 21:00
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Hint To prove an equality of sets $X =Y$ you need to show that $X \subseteq Y$ and $Y \subseteq X$, where $X,Y$ are the expressions in that particular problem.

Hint 2 To show that $X \subseteq Y$ you start with "Let $x \in X$ be arbitrary." You explain what this means and try to get that $x \in Y$.

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  • $\begingroup$ Now I see I was all around the place with my way of solving. This will also help me solve other similar proofs. Just a question though, when you say "explain what this means" should I follow a particular strategy? If so, which one? Because I tried to explain what x meant but I got really lost and ended up getting nowhere close to proofing that x also belonged to Y. $\endgroup$ – Sara Mar 24 '18 at 21:06
  • $\begingroup$ @Sara In this case when you start with $x \in A-(B \cup C)$ for example, by explain I mean you write the definition of the operation(s) you have: $x \in A - (B \cap C)$ means $x \in A$ AND $x \notin (B \cap C)$. For the last, again you explain the definition of $B \cap C$ to get that $x \notin B \cap C$ means $x \notin B$ Or $x \notin C$. This way, you explained $x \in A -(B \cap C)$ in terms of $x$ belonging/not belonging to each set. At this point, look at your goal and try to get there. This is almost every time the strategy. $\endgroup$ – N. S. Mar 24 '18 at 21:11

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