Proof verification for $(A-B)-C = (A-C)-(B-C)$

Question

Let $A$, $B$, and $C$ be sets. Prove that $(A-B)-C=(A-C)-(B-C)$.

I attempted to prove this with the following. I am very new to proof writing, and I feel I may have glossed over something or have done something just completely invalid. Any assistance with revisions or a superior method would be greatly appreciated.

Suppose $x \in (A-B)-C$. This would mean $x \in A$, $x \notin B$, $x \notin C$. The statements $x \in A$ and $x \notin C$ could be written as $(A-C)$. The statements $x \notin B$ and $x \notin C$ could be written as $-(B-C)$. Together these statements could be written as $(A-C)-(B-C)$. This shows that $(A-B)-C \subseteq (A-C)-(B-C)$.

Suppose $x \in (A-C)-(B-C)$. This would mean that $x \in A$, $x \notin B$, $x \notin C$. The statements $x \in A$ and $x \notin B$ could be written as $(A-B)$. In addition $x \notin C$, this combined with the earlier statement that $(A-B)$ would result in $(A-B)-C$. This shows that $(A-C)-(B-C) \subseteq (A-B)-C$.

Concluding the proof that $(A-B)-C = (A-C)-(B-C)$.

Answer 1

The main idea is good, but the way you write it can be improved.

In the first part, after taking an element $x$, your goal is to show that $x \in (A - C) - (B - C)$.

$x \in A$ and $x \notin C$ could be written as $(A - C)$

I see what you are trying to say, but this is not mathematically correct. "$x \in A$ and $x \notin C$" is a statement and $A - C$ is a set. They are not equivalent. However, what you tried to say is probably "$x \in A$ and $x \notin C$, so $x \in (A - C)$".

The statements $x$ $\notin$ $B$ and $x$ $\notin$ $C$ could be written as $-(B-C)$.

This is similar. The first problem with this statement is that you are equating a sentence and a set. The second problem is that $-(B - C)$ is not a correct expression. What you meant is probably $X - (B - C)$ where $X$ is the universe.

Instead of this, you can just say $x \notin B$, so $x \notin (B - C)$.

Putting these together will result in something like...

Since $x \in (A - B) - C$, we know that $x \in A, x \notin B, x \notin C$. Since $x \in A$ and $x \notin C$, $x \in (A - C)$. Since $x \notin B$, $x \notin (B - C)$. Since $x \in (A - C)$ and $x \notin (B - C)$, $x \in (A - C) - (B - C)$.

Thus $(A - B) - C \subset (A - C) - (B - C)$.

The opposite direction has two issues:

• The first one is the same as above. You can't say $x \in A$ and $x \notin B$ could be written as $A - B$ because one is a statement and the other one is a set. You can say $x \in A$ and $x \notin B$, so $x \in A - B$.
• The second issue is that it is not obvious that $x \in (A - C) - (B - C)$ implies that $x \notin B$. (It is true, but it's not obvious.)

Answer 2

Below, a proof using only the laws of the algebra of sets.

Remark - If $X$ is a set, the symbol $X'$ ("$X$ prime") denotes here the complement of the set $X$.