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

Question

I think I have figured out this proof, but was hoping someone could verify it.

\begin{align*} x \in (A \cap B) \cup (A - B) & \iff x \in A \cap B \land x \in A - B \\ & \iff (x \in A \land x \in B) \lor(x \in A \land x \not \in B) \\ & \iff x \in A \land (x \in B \lor x \not \in B) \\ & \iff x \in A. \end{align*} The first line is the definition of union. The second is the definition of interection and set difference. The third uses the rule from propositional logic that $p \wedge (q \lor r) \equiv (p \wedge q) \lor (p \wedge r)$. Finally, $p \lor \neg p$ is a tautology that is always true, and $p \wedge T \equiv p$.

Answer 1

The retranscript from set operations to propositions is immediate:

$$(a\land b)\lor(a\land\lnot b)$$

and this can be rewritten

$$a\land(b\lor\lnot b)=a.$$

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You can also use a "membership" table,

$$\begin{array}{|c|c|c|c|c|} A&B&A\cap B&A\setminus B&(A\cap B)\cup(A\setminus B) \\\hline \in&\in&\in&\notin&\in \\\in&\notin&\notin&\in&\in \\\notin&\in&\notin&\notin&\notin \\\notin&\notin&\notin&\notin&\notin \\\hline \end{array}$$

Answer 2

You did a fine job, and proved the biconditional. You also provided good justification: the "third" property can be justified by the "distributive law/rule" of "and" over "or".

In general, distributivity also holds among set union/intersection, so $$A\cap (B\cup C) = (A\cap B) \cup (A\cap C).$$

Also note that one can write $(A\cap B)\cup (A-B)$ as $(A\cap B)\cup (A\cap B^C)$, where $B^C$ is the complement of $B$.

So considere your proof verified enthusiastically.

Answer 3

Yes, your solution is correct.

Now, let me do it in a less formal way.

To prove a set identity like $X=Y$, we can prove it in two steps: $X\subset Y$ and $Y \subset X$.

Now, by definition of "$\subset$", you can easily check:

• $A\cap B\subset A$,
• $A-B\subset A$
• $(A\cap B)\cup (A-B)\subset A$

Secondly, for any element $x\in A$, $x$ either in $B$ or (exclusively) not in $B$. This tells you $$ A\subset (A\cap B)\cup (A-B)\;. $$