First part :

I want to prove the following De Morgan's law : ref.(dfeuer)

$A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) $

Second part:

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


Let $y\in (A\setminus B) \cup (A\setminus C)$

$(A\setminus B) \cup (A\setminus C) = (y \in A\; \land y \not\in B\;) \vee (y \in A\; \land y \not\in C\;)$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A\ \land ( y \not\in B\; \vee \; y \not\in C ) $

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A \land (\lnot ( y\in B) \lor \lnot( y\in C) )$

According to De-Morgan's theorem :

$\lnot( B \land C) \Longleftrightarrow (\lnot B \lor \lnot C)$ thus

$y \in A \land (\lnot ( y\in B) \lor \lnot( y\in C) ) = y \in A \land y \not\in (B \land C)$

We can conclude that $(A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C)$

  • $\begingroup$ That is not very clear English. Perhaps you could parenthesize it or something? $\endgroup$ – dfeuer Dec 7 '13 at 23:49
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    $\begingroup$ But really, what you generally want to do is to show that for all $x$, if $x\in A\setminus (B\cap C)$ then $x\in $(A\setminus B)\cup (A \setminus C)$ and the other way around. $\endgroup$ – dfeuer Dec 7 '13 at 23:50
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    $\begingroup$ $y\in A\land (y\notin B\lor y\notin C)$ is not equivalent to $y\in A\land y\notin (B\cup C)$. $\endgroup$ – dfeuer Dec 8 '13 at 2:52
  • $\begingroup$ possible duplicate of proof of $ A - \left (B \cap C \right)= \left (A - B \right) \cup \left (A - C \right)$ $\endgroup$ – user91500 Dec 8 '13 at 5:25

I'll guide you through one direction. You will need to figure the other out yourself.

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

Then by the definition of $\setminus$, $x\in A$ and $x\notin B\cap C$. The latter statement can be translated into $\lnot (x \in B \cap C)$.

By the definition of $B\cap C$, $x\in B\cap C \iff (x \in B) \land (x\in C)$.

Thus we conclude that $\lnot ((x \in B) \land (x\in C))$.

By De Morgan's laws for logic, we can rewrite this as $\lnot(x\in B)\lor\lnot(x\in C)$.

Suppose that $\lnot (x\in B)$. Then since $x\in A$, $x\in A\setminus B$.

Suppose instead that $\lnot (x \in C)$. Then since $x\in A$, $x\in A \setminus C$.

As one of these must be true, $(x\in A\setminus B)\lor (x\in A \setminus C)$.

By the definition of union, $x\in (A\setminus B)\cup(A \setminus C)$.

  • $\begingroup$ excuse me if it's a silly a question. when we say that $x\notin B\cap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets $\endgroup$ – Hani Gotc Dec 8 '13 at 0:06
  • $\begingroup$ Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great $\endgroup$ – Hani Gotc Dec 8 '13 at 0:11
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    $\begingroup$ @HaniGotc, let $B = \{1,2\}$ and let $C = \{2,3\}$. Is $1\in B$? Is $1\in C$? Is $1\in B\cap C$? Is it true that $1\notin B\cap C$? $\endgroup$ – dfeuer Dec 8 '13 at 0:13
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    $\begingroup$ Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $x\in (a\setminus B)\cup(A \setminus C)$ then we can conclude also that $x\in A \setminus (B\cap C)$? $\endgroup$ – dfeuer Dec 8 '13 at 0:15
  • $\begingroup$ $ 1 \in B\; $ yes and $1 \not\in C\;$ yes. So $1\;$ doesn't belong to the intersection yes true. OK let me edit it $\endgroup$ – Hani Gotc Dec 8 '13 at 0:17

You write,

Can we say that:

$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$

In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.

To show that sets $S$ and $T$ are equal, a typical strategy is to show $S \subseteq T$ and $T\subseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.

In your example, $A\setminus (B\cap C)$ is your $S$. So begin by explicitly writing:

"Let $x\in A\setminus (B\cap C)$."

Now what do we know about $x$? Well, if $x\in A\setminus (B\cap C)$ then $x\in A$ but $x\notin (B\cap C)$. Unpacking the second part of that last sentence, if $x\notin (B\cap C)$ then either $x\notin B$ or $x\notin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $x\notin B$ or $x\notin B$. Suppose that $x\notin B$. We know that $x\in A$ so this means that $x\in A\setminus B$, and so $x\in (A\setminus B)\cup (A\setminus C)$. In the second case, we know that $x\in A$ and $x\notin C$, which means $x\in A\setminus C$, which further implies that $x\in (A\setminus B)\cup (A\setminus C)$. Since these are the only two possible cases, we conclude that $x\in (A\setminus B)\cup (A\setminus C)$.

To go to the other direction, again, begin by explicitly writing

"Let $y\in (A\setminus B) \cup (A\setminus C)$"

And from there, determine what you can say about $y$. I'll let you finish the other direction.

  • $\begingroup$ I finished the other direction. Can you please just downgrade or tell me if it's right. $\endgroup$ – Hani Gotc Dec 8 '13 at 2:48

If $x \in A \setminus (B \cap C ) $. then this occur

$$ \iff x \notin B\cup C \iff x \notin B \; \; \text{or} \; \; x \notin C \iff$$

$$ \iff x \in A \setminus B \; \; \text{or} \; \; x \in A \setminus C \iff x \in(A \setminus B ) \cup (A \setminus C )$$

$$ \therefore A \setminus (B \cap C ) = (A \setminus B ) \cup (A \setminus C ) $$

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    $\begingroup$ I think that $x \notin B\cup C \iff x\notin B \,\wedge \, x\notin C $, because $x \notin B\cup C \iff (B\cup C)^C\iff B^C \cap C^C $ $\endgroup$ – Voyager Dec 8 '13 at 9:13

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