Show that $A \setminus (A \cap B) = (A \cup B) \setminus B$

Question

Help with missing step in the following proof:

In order to show that the equality above holds for all A,B , we need to show that (1) $$A \setminus (A \cap B) \subseteq (A \cup B) \setminus B$$ and (2) $$ (A \cup B) \setminus B \subseteq A \setminus (A \cap B)$$

For a given $x$ , $A \setminus (A \cap B) = \{ x \in A \ | \ x \notin B \} $ . Given that $ x \notin (A \cap B) \iff x \notin A \text{ or } x \notin B$ , it follows that: $x$ is in $A$ and $x$ is not in $A$ , or $x$ is not in $B$. Therefore, since $A \cap A^{c} = \emptyset$ , $x$ is in A and $x$ is not in $B$.

Conversely, for (2) we have, $$x \in (A \cup B) \ | \ x \notin B$$

Then $x$ is in $A$ or $x$ is in $B$, bt $x$ is not in $B$. Therefore, given that $B \cap B^{c} = \emptyset$ , we have that $x$ is in $A$ but not in $B$.

Two big problems I see:

1. I haven't actually shown that they are subsets of each other, though I can clearly see it. A similar proof presented in lecture, the lecturer translates the "natural language" sentence back into formal set logic showing that they are equal.

2. I was told that starting from the L.H.S, and then starting from R.H.S and arriving at some intermediate conclusion which is the same for both is not a valid proof method. And this seems to be exactly what happened here.

For completeness, here is the kind of proof the lecturer is expecting:

Answer 1

Intended answer can be found here. In particular , the intermediate step for (1) is the fact that

Since x is in $A$ it is in $∪$

Answer 2

In this case, a direct approach is probably easier and more instructive.

$A \setminus (A \cap B) = A \cap (A \cap B)^c = A \cap (A^c \cup B^c) = A \cap B^c $.

$(A \cap B)\setminus B = (A \cap B) \cap B^c = A \cap B^c$.

Answer 3

We first select an item $x\in A-(A\cap B)$ (I'm using - as difference because I can't seem to get a \ in this editor.

Notice that $x\in A$, however $x\not \in A\cap B$. As such we can say that $x \in A, x \not \in B$. So $x \in A - B$ and $x\in (A\cup B) - B$, since adding in $B$ will not change that $x$ is in this set. Therefore

$$ A-(A\cap B) \subseteq (A\cup B)-B $$

Now for the reverse. Assume $y \in (A\cup B)-B$. That means that $y\in A\cup B$ but $y \not \in B$. As such $y \in A$. Since $y\in A, y\not\in B \implies y\not\in A\cap B$ so $y \in A-(A\cap B)$. Therefore

$$ (A\cup B)-B \subseteq A-(A\cap B) $$

and since both are subsets of each other, then

$$ (A\cup B)-B = A-(A\cap B) $$

Q.E.D.