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I need to show that $n(A-B) = n(A)-n(A\cap B)$.

Having the truth table, I can find these sets, but don't know how to show, I know that is it, but show is a little bit hard.

$$\begin{array}{|l|l|l|} \hline A&B\\\hline V & V & n(A\cap B) \\ V & F & n(a) \\ F & V & \text{~} \\ F & F & \text{~} \\\hline \end{array}$$

How can I show that?

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@Daniel-Rust yes, is it.. – Matheus Silva Aug 18 '13 at 23:24
Thank you but Brian did most of the work :). – Dan Rust Aug 18 '13 at 23:25
You might want a more informative title... – Thomas Andrews Aug 18 '13 at 23:27
up vote 2 down vote accepted

Here's a proof based on the assumption that you know: $n(S\cup T)=n(S)+n(T)$ if $T\cap S = \emptyset$.

We try to show $(A-B)\cup (A\cap B) = A$ and $(A-B)\cap(A\cap B)=\emptyset$. Then with $S=A-B$ and $T=A\cap B$ in the above, we have $n(A) = n(S\cup T) = n(S) + n(T) = n(A-B) + n(A\cap B)$, so $n(A - B) = n(A) - n(A\cap B)$.

$\underline{\mathbf{A=(A-B) \cup (A\cap B)}}$:

$x\in A \Leftrightarrow x\in A \wedge (x\in B\vee x\notin B) \Leftrightarrow (x\in A \wedge x\in B) \vee (x\in A \wedge x\notin B) \Leftrightarrow (x\in A\cap B)\vee (x\in A-B) \Leftrightarrow x\in (A\cap B)\cup (A-B)$

$\underline{\mathbf{\emptyset = (A-B) \cap (A\cap B)}}$:

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

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