Rigorously prove that $(A \setminus B) \cup A = A$

Question

Rigorously prove that (A/B) U A = A

So I started off with

(x ∈ A ^ x ∉ B) v (x ∈ A)

Applying Distributivity

(x ∈ A v x ∈ A) ^ (x ∈ A v x ∉ B)

Applying Idempotency

(x ∈ A) ^ (x ∈ A v x ∉ B)

That's the point I reached, i.e. I'm stuck. I don't know how to break down further (x ∈ A v x ∉ B). Should I be using dominance or complement?

Thanks in Advance

Answer 1

Your last line is equivalent to just $x \in A$ because if the first part ($x \in A$) is satisfied, the second part ($x \in A \text{ or } x \notin B$) is always satisfied as well, hence the second part can be removed.

Answer 2

Firstly, note that for any $x \in A$, we have $x \in C \cup A$ for any set $C$, so in particular, we have $x \in (A \setminus B) \cup A$. Thus, $A \subseteq (A \setminus B) \cup A$.

For the converse, for any $x \in (A \setminus B) \cup A$, either $x \in A$, or $x \in A \setminus B \subseteq A$, so in either case, $x \in A$. Thus, $(A \setminus B) \cup A \subseteq A$.

Combining the two, we have $(A \setminus B) \cup A = A$, as required.

Answer 3

$A\setminus B$ and $A$ are both subsets of $A$ and consequently the union of these sets is also a subset of $A$.

So:$$(A\setminus B)\cup A\subseteq A\tag1$$

$A$ is a subset of $A$ hence also a subset of the union $(A\setminus B)\cup A$.

So:$$A\subseteq(A\setminus B)\cup A\tag2$$Combining $(1)$ and $(2)$ we are allowed to conclude that:$$(A\setminus B)\cup A= A$$