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I have proved this using Venn diagram but when I am trying to prove this using the rule that "If $ A \subset B \text{ and } B \subset A $ then $ A = B $", I am having some problems with my understanding of the same,here is how I did so far:

Let $x \in A \cap (B-C) \Rightarrow x \in A \text{ and } x \in (B-C) \Rightarrow x \in A \text{ and } (x \in B \text{ and } x \notin C) $

How to proceed next? Since if I am do something like this: $ x \in A \text{ and } x \in B \text{ and } x \in A \text{ and } x \notin C $, it's not giving the correct results, what exactly I am missing here?

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4 Answers 4

up vote 8 down vote accepted

Hint: $x \notin C \Rightarrow x \notin D \cap C$ for any set $D$.

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That's sufficient for me :) –  VelvetThunder Nov 10 '10 at 15:23

Here is a different style of proof, viz. an equational and calculational one.

The idea is to start at the most complex side, translate from set theory to logic by working at the element level, then expanding the definitions, and see where there leads you.

In other words, for all $\;x\;$ we calculate: \begin{align} & x \in (A \cap B) - (A \cap C) \\ \equiv & \;\;\;\;\;\text{"expand definition of $\;-\;$, and of $\;\cap\;$ (twice)"} \\ & x \in A \land x \in B \land \lnot(x \in A \land x \in C) \\ \equiv & \;\;\;\;\;\text{"DeMorgan"} \\ & x \in A \land x \in B \land (x \not\in A \lor x \not\in C) \\ \equiv & \;\;\;\;\;\text{"use leftmost conjunct $\;x \in A\;$ in the rightmost conjunct"} \\ & x \in A \land x \in B \land (\textrm{false} \lor x \not\in C) \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & x \in A \land x \in B \land x \not\in C \\ \equiv & \;\;\;\;\;\text{"reintroduce $\;-\;$ and $\;\cap\;$"} \\ & x \in A \cap (B - C) \\ \end{align}

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Simplyfing the notation of @Marnix answer:

$$ A \cap (B - C) \\ = A \cap B \cap C' \text{ (definition of set complement)}\\ = A \cap B \cap (A' \cup C') \text{ (trick, } A \cap B \cap A' = \emptyset\text{)}\\ = A \cap B \cap (A \cap C)' \text{ (complement of intersection of sets in braces)}\\ = (A \cap B) \cap (A \cap C)' \text{ (assoc.)}\\ = (A \cap B) - (A \cap C) \text{ (definition of set complement)}\\ $$

The key is introducing a good disjoint set (with $A \cap B$) for union with C'.

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It is possible to continue your solution$$\begin{align} A\cap(B-C) & = \{x\space|\space x\in A\space\land(x\in B\space\land x\notin C)\} \\ & = \{x\space|\space (x\in A\space\land x\in B)\space\land x\notin C\} \\ & = \{x\space|\space (x\in A\space\land x\in B)\space\land (x\notin A\lor x\notin C)\} \\ & = \{x\space|\space (x\in A\space\land x\in B)\space\land \neg(x\in A\land x\in C)\} \\ & = (A\cap B)-(A\cap C) \end{align}$$

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