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I am trying to prove $ A - \left (B \cap C \right)= \left (A - B \right) \cup \left (A - C \right)$ I came up with this proof:

  • Let $ S \{ x | x \in A - \left(B \cap C \right)\}$ Let $ Q \{y | y \in \left (A - B \right) \cup \left (A - C \right)\} $
  • All $x \in A $ and $ x \notin B,C$.
  • All $y \in A $ and $ y \notin B,C$.
  • Because all $x$ fit the definition of $Q$ then we say $S \subseteq Q$
  • Because all $y$ fit the definition of $S$ then we say $Q \subseteq S$
  • Since $ S \subseteq Q$ and $Q \subseteq S$ then $ S = Q \implies A - \left (B \cap C \right)= \left (A - B \right) \cup \left (A - C \right)$ $ \blacksquare $

But then quickly realized it was wrong because the $x$ from set $S$ must meet the following criteria:

  • $x \in A $ and $ x \notin B \wedge C $

whereas the $y$ from set $Q$ must meet different criteria:

  • $y \in A $ and $ y \notin B \vee C$

If you make a Venn diagram for the three sets and let it represent $A$, $B$, and $C$ then you see that when you take away $\left (B \cap C \right)$ from $A$ what you are taking away is the center where all three sets meet(i.e $A \cap B \cap C$) If you evaluate $\left (A - B \right) \cup \left (A - C \right)$ you take away from $A$ the center ($A \cap B \cap C$) and both $A \cap B$ and $A \cap C$, which would render the theorem I am trying to prove wrong.

I don't know if I am thinking about it incorrectly, but the theorem is out of Calculus Vol. 1 By Tom M. Apostol.

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up vote 2 down vote accepted

One can produce a watertight proof by showing a double inclusion.

First $A\setminus (B\cup C)\subseteq (A\setminus C)\cup (A\setminus B)$

P Pick $x\in A\setminus (B\cup C)$. Then $x\in A$ and $x\notin B\cap C$. This means that $x\in A$ and $x\notin B$ or $x\notin C$. In any case, $x\in A\setminus B$ or $x\in A\setminus C$ (why?), so $x\in (A\setminus B)\cup (A\setminus C)$.

Second, $A\setminus (B\cup C)\supseteq (A\setminus C)\cup (A\setminus B)$

P Suppose $x\in (A\setminus C)\cup (A\setminus B)$. Then $x\in A\setminus C$ or $x\in A\setminus B$. Thus $x\in A$ and $x\notin C$; or $x\in A$ and $x\notin B$. But this is the same as saying $x\in A$ and, $x\notin C$ or $x\notin B$. And $x\notin C$ or $x\notin B$ is the same as $x\notin B\cap C$. Thus $x\in A,x\notin B\cap C$, i.e. $x\in A\setminus (B\cap C)$

Note we use the De Morgan Laws $$\mathscr C\left(\bigcup A\right)=\bigcap \mathscr CA$$ $$\mathscr C\left(\bigcap A\right)=\bigcup \mathscr CA$$

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(+1) What is that word "watertight"? I never heard it. – S. Snape Jul 22 '13 at 17:33
@BabakS. See here (figurative, of course) – Pedro Tamaroff Jul 22 '13 at 17:41

I think we need to recall DeMorgan's law:

$x\in (B\cap C) \iff x\in A \;\text{AND} \;x \in B$

$x\notin (B\cap C) \iff \;\text{it is NOT the case that}\; (x\in A\; \text{AND}\; x \in B) \iff x \notin B \;\text{OR} \;x\notin C$

Your work initial work seems to show this understanding, but I think your "second guessing" is slightly confused. Let's expand your initial work:

I'll include your work below, and "highlight" in blue the expansions:

Let $ S \{ x | x \in A - \left(B \cap C \right)\}$ Let $ Q \{y | y \in \left (A - B \right) \cup \left (A - C \right)\} $

  • $\color{blue} S = $ All $x \in A $ and $ x \notin B,C$:
    $\text{All x such that}\;\color{blue}{x \in A\land \lnot(x \in B\land x \in C)}$
  • $\color{blue} Q = $ All $y \in A $ and $ y \notin B,C$: $\text{ALL y such that}\;\;\color{blue}{(y \in A \land y \notin B) \lor (y \in A \land y\notin C) \iff y \in A \land (y \notin B \lor y \notin C) \\ \iff y \in A \land \lnot(y \in B \land y \in C)} $

Now we see that $x\in S \implies x\in Q$ and $y \in Q \implies y \in S$.

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Can we use $&$ instead of writing $\text{AND}$ in maths texts, Amy? Is it legal in Maths notations? – S. Snape Jul 27 '13 at 16:05
Hi, @Babak! & is often used in computer science, e.g., in programming, to denote boolean AND. In math texts, best to stick with the word "and" or $\land$ ;-) – amWhy Jul 27 '13 at 16:10
I hope I answered your question, dear friend, @Babak! – amWhy Jul 27 '13 at 17:43
Yes you did kindly. Are every things stable? – S. Snape Jul 27 '13 at 17:48
Glad of hearing that. Indeed, being at the hospital or clinic is a bit boring. Just doing some routine boring job... I pray for you. :) – S. Snape Jul 27 '13 at 17:54

Recall that $$A-B=A\cap B^c$$ and $$(A\cap B)^c=A^c\cup B^c$$

so $$A-(B\cap C)=A\cap(B\cap C)^c=A\cap(B^c\cup C^c)\\=(A\cap B^c)\cup (A\cap C^c)=(A-B)\cup(A-C)$$

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When you look at $(A - B) \cup (A - C)$ you are not taking away $A \cap B$ and $A \cap C$. For instance, an element which is in $A \cap C$ but not in $A \cap B \cap C$ is in $A - B$, so by definition of union, it is in $(A-B) \cup (A - C)$.

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