# Prove or disprove: $A-(B-C) = (A-B)-C$

This is what I have so far:

First, we will attempt to show $A-(B-C) \subseteq (A-C) - C.$ Let $x \in A -(B-C)$. Then $x\in A$ and $x \notin (B-C)$. By DeMorgan's we have that $$x \notin (B-C) = x \notin B \vee x \in C.$$ We have that $x \in A \wedge (x \notin B \vee x \in C)$. Then we have that $(x \in A \wedge x \notin B) \vee (x \in A \wedge x \in C)$.

So by definition of $\cup$ and $\cap$, we have $x \in (A-B) \cup (x\in (A \cap C))$. I'm kind of stuck here.

I can see a contradiction here in that LHS says $x \in C$ but the RHS says $x \notin C$.

Can anyone tell me what to do from here? I'm not quite sure how to formally state that the statement is false. I feel like I'm not being complete here.

• In order to disprove this statement, it suffices to find a counterexample, i.e. sets $A,B,C$ that don't satisfy the claimed identity. See below for one such example. Nov 4, 2016 at 4:29
• @testpilot No, if $A,B,C$ are disjoint (in fact, if $A$ is disjoint from $B$ and $C$), then this identity always holds. Both sides just evaluate to $A$. Nov 4, 2016 at 4:32
• just need one element. If you have x in C then x won't be in the LHS but might be in the right hand side under what conditions? Well, it has to be in A of course but if it isn't in B it won't be omitted. So just need x in A and x in C but X not in B. Let A = {1} B = emptyset C = {1}. RHS = {1}, LHS = empty set. Nov 4, 2016 at 4:41
• If $B=\emptyset$ then the left side is $A$ and the right side is $A-C.$ Do you think $A=A-C$ is a valid identity?
– bof
Nov 4, 2016 at 4:42
• The natural numbers are an instructive special case. Nov 4, 2016 at 5:12

Let $A=B=C = \{1\}$. Then $$(A - B) - C = \emptyset,$$ but $$A-(B-C) = \{1\} - \emptyset = \{1\} \neq \emptyset.$$

You can use your observation to create a counterexample. I claim that if $x \in A \cap C$, then $x \in A-(B-C)$ (the left-hand side), but $x \notin (A-B)-C$ (the right hand side). Do you see why?

To create a counterexample, come up with (very simple) sets $A, B,$ and $C$ and an element $x \in A \cap C$. Then explicitly compute what $A-(B-C)$ and $(A-B)-C$ are, and show they are not the same.

Why would you start trying to prove some random identity without first testing it in some easily computed cases to see if it's right or wrong??? Because I like to look for easy solutions before hard ones, I always try plugging in zero for one of the variables, or in this case because we're dealing with sets, the empty set.

If $A=\emptyset$ then $A-(B-C)=(A-B)-C$ turns into $\emptyset=\emptyset,$ check.

If $C=\emptyset$ it turns into $A-B=A-B,$ check.

If $B=\emptyset$ it turns into $A=A-C.$ Now that looks fishy, doesn't it? What if $C=A,$ then we get $A=\emptyset.$ That sure isn't an identity . . .

Counterexample: Let $B=\emptyset$ and $A=C=\mathbb R$ (or any nonempty set). Then $$A-(B-C)=\mathbb R-(\emptyset-\mathbb R)=\mathbb R-\emptyset=\mathbb R$$ while $$(A-B)-C=(\mathbb R-\emptyset)-\mathbb R=\mathbb R-\mathbb R=\emptyset.$$

Note subtraction is not associative in the natural numbers. If A,B,C are finite with natural cardinalities and C is a subset of B is a subset of A then the cardinality on either side is determined precisely by this subtraction on the cardinalities. But the cardinalities not being equal,the corresponding sets are not equal.

Let X <=Y if x is a subset of Y. Note if X<=Y, Z-X>=Z-Y. B-C <=B . Thus A-(B-C)>=A-B . A-(B-C)-nullset>=A-B-C similarly. It is easily shown when we have one strict inequality between B-C and B or the nullset and C our final inequality is strict and the theorem fails. When we have no such strict inequality by antisymmetry of <= the equation holds. That is, when C is disjoint from B and C is not the nullset.