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\begin{align} c. & \quad (A-B)-C \subseteq A-C\\ & \quad x \in [(A-B)-C]\text{ is equivalent to }x\in A\wedge x\not\in B\wedge x\not\in C,\text{ by definition of difference. }\\ &\quad x\in A-C\text{ is equivalent to }x\in A\wedge x\not\in C,\text{ by definition of difference.} \end{align}

Basically, I'm not sure how to proceed. I can see this two ways. One with $(A - B) - C$ having at least one element, and on in which it has no elements. Should I show how the left side is a subset of the right in both cases? Or is there something more simple I could do here? Also, is there a named law for going from $x$ is in $A$ and $x$ is in $B$ therefore $x$ is in $A$? Thanks!

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Is this homework? –  Jeel Shah Feb 1 '13 at 23:38
    
yes this is homework –  papercuts Feb 1 '13 at 23:39
    
Show if $A \subset A'$, then $A \setminus C \subset A' \setminus C$. Start by choosing $x \in A \setminus C$. This means $x \in A$, $x \notin C$. How does that relate to $A' \setminus C$? –  copper.hat Feb 1 '13 at 23:43
    
A` = the complement of the set A? –  papercuts Feb 1 '13 at 23:46
    
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. –  Julian Kuelshammer Feb 1 '13 at 23:52

2 Answers 2

up vote 3 down vote accepted

So every $x \in (A - B) - C$ is in $A-C$, yes? (You've done most of the "element chasing" in your preliminary work, save for the conclusion), i.e.,

$x \in (A - B) - C \implies x \in (A - B) \land x\notin C \implies x\in A \land x \notin B \land x \notin C.$ From the fact that $x \in A \land x \notin B \land x \notin C$ it is sufficient to know that $x \in A \land x \notin C$ from which it follows that $ x \in (A - C)$.

Therefore, by definition of set inclusion $$(A-B)-C \subseteq (A - C)$$

We've shown that any $x$ in $(A - B) - C$, if there is such an $x$, must be in $A - C$. That's it. We are not trying to prove set EQUIVALENCE.

If there is no $x \in (A - B) - C$, it still holds that $(A - B) - C) \subseteq (A - C)$, as the empty set is the subset of every set.

The fact that $x \notin B$ (from the first set) doesn't matter, since logically, from the fact that $p \land \lnot q \land \lnot r,$ we can conclude $p \land \lnot r$ (and leave $\lnot q$ alone, since when each of $p$, $\lnot q$, $\lnot r$ are true, so is any one or any two of them true.

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papercuts: Is this any clearer? Most of the work was already done! –  amWhy Feb 2 '13 at 0:21
    
Yeah this helped big time. Thanks again! –  papercuts Feb 2 '13 at 0:26

First, begin with $(A-B)-C$ and proceed by a chain of equalities:

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

So we can work with the expression $(A - C) \cap B^c$ rather than the original one. Now in general (meaning: for all sets $X$ and $Y$), it holds that $$X \cap Y \subseteq X.$$ Thus $$(A - C) \cap B^c \subseteq A-C.$$

But recall that $(A-B)-C = (A - C) \cap B^c$. Thus $$(A - B) - C \subseteq A-C.$$

EDIT: The moral of the story is to use intersections, unions and complements. These operations have nice properties, while set-subtraction does not.

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