# Set Proof verification

How to prove that $(A-B)\cup(A\cap B)=A$?

Here is my proof, I'm not sure if I verified it correctly.

$(A-B) \cup (A \cap B)$ simplifies into $x \in (A-B) \vee x \in (A \cap B)$.

Then we have $[x \in A$ and $x \notin B$] or $[x \in A$ and $x \in B$] $\iff x \in A$

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

• The second line is equivalent to your question, you can't assume it is true. Apr 21, 2017 at 22:40

Your proof is not correctly stated. Alternately, you can try the following proof. Denote by $U$ the universal set. We know that $A,B\subset U$. Then $A\cap U=A$ and $B\cup B^c=U$, where $B^c$ is the set complement of $B$. We also have $A-B=A\cap B^c$. Hence, \begin{align} (A-B)\cup(A\cap B)&=(A\cap B^c)\cup (A\cap B)\quad\text{then use Distributive Law to get}\\ &=A\cap (B^c\cup B)\\ &=A\cap U\\ &=A. \end{align}

I prefer the algebraic approach shown in the other answer, but if you were looking for an "element chasing" approach:

We want to show that $(A-B)\cup(A\cap B) = A$. We do this by showing $(A-B)\cup(A\cap B) \subset A$ and $A \subset (A-B)\cup(A\cap B)$

Suppose $x \in (A-B)\cup(A\cap B)$. This says that $x \in (A-B)$ or $x \in (A\cap B)$. If $x \in (A-B)$, using the definition of relative complement, $A-B = A \cap B^\complement$, we see that $x \in A$. If $x \in (A \cap B )$ then clearly $x \in A$. Either way $x \in (A-B)\cup(A\cap B) \Rightarrow x \in A$ as desired, so $(A-B)\cup(A\cap B) \subset A$.

Now for the opposite direction:

Let $x \in A$, for some other set $B$, by the principle of excluded middle, either $x \in B$ or $x \notin B$. If $x \in B$, then $x \in (A \cap B)$ so it follows that $x \in (A-B)\cup(A\cap B)$. Otherwise $x \notin B$. It then follows from the definition of relative complement that $x \in (A-B)$. This implies $x \in (A-B)\cup(A\cap B)$. Again, either way $x \in A \Rightarrow x \in (A-B)\cup(A\cap B)$. Therefore $A \subset (A-B)\cup(A\cap B)$

Since $(A-B)\cup(A\cap B) \subset A$ and $A \subset (A-B)\cup(A\cap B)$, we can conclude $(A-B)\cup(A\cap B) = A$ as desired.

Provide another proof:

Notice that $(A-B) \subset A$ and $(A\cap B) \subset A$, thus $(A-B) \cup (A\cap B) \subset A$

On the other hand, $\forall x \in A$, if $x\in B$, then $x\in A\cap B$; else if $x\notin B$, then $x\in A-B$. Thus $x\in (A-B) \cup (A\cap B)$. So we have $A\subset (A-B) \cup (A\cap B)$.

So we get $A = (A-B) \cup (A\cap B)$