I wanted to show that

$$ (D \setminus A) \setminus (C \setminus B) = D \setminus (A \cup (B \cup C)) $$

Since I did not know how to approach this, I let wolfram alpha do the work and got the following results: https://www.wolframalpha.com/input/?i=(D%5CA)%5C(C%5CB)%3DD%5C(A+union+B+union+C)

It says: undetermined (can be true or false).

So, it can be true or false. But on what does it depend whether the solution is true or false?

If I take for example the sets:

$$ A=\{1,2,3,4,5\} $$ $$ B=\{2,1,3,4,6\} $$ $$ C=\{3,1,2,4,7\} $$ $$ D=\{4,1,2,3,8\} $$

then $$ (D \setminus A)\setminus (C\setminus B) = \{8\} $$ $$ D \setminus (A \cup B \cup C) = \{8\} $$

But what would be cases, in which this relation does not hold?

  • $\begingroup$ In the "input interpretation" at wolfram link, the expression you wrote has changed from what you have on left of first displayed equation in the post. $\endgroup$ – coffeemath Aug 14 at 10:10
  • $\begingroup$ It still says the same thing at wolfram link-- i.e. the thing they computed was not $(D \setminus A) \setminus (C \setminus B).$ Can you re-do wolfram calculation to force the parenthesization you specified on the left side? $\endgroup$ – coffeemath Aug 14 at 10:21
  • $\begingroup$ Thanks for noticing. I tried to force brackets, but alpha just removes them every time. I don't know how to change it. $\endgroup$ – holistic Aug 14 at 10:23

Well you found one example where it is true.

And here is an example where it is false:

$$ A =\{0\}, B =\{1\}, C =\{1\}, D =\{1\} $$

Then $(D \setminus A) = \{1\} \setminus \{0\} = \{1\}$ and $(C \setminus B) = \{1\} \setminus \{1\} = \emptyset$.

Hence $(D \setminus A) \setminus (C \setminus B) = \{1\} \setminus \emptyset = \{1\}$.

However $A \cup (B \cup C) = \{0\} \cup (\{1\} \cup \{1\}) = \{0, 1\}$.

Hence $D \setminus (A \cup (B \cup C)) = \{1\} \setminus \{0, 1\} = \emptyset$.

So $(D \setminus A) \setminus (C \setminus B) = \{1\} \neq \emptyset = D \setminus (A \cup (B \cup C))$ in this particular case.

However it is true that at least one part of the equality namely $\boxed{D \setminus (A \cup (B \cup C)) \subseteq (D \setminus A) \setminus (C \setminus B)}$ does always hold.

Proof: For let $x \in D \setminus (A \cup (B \cup C)$ be an arbitrary element.

Then $x \in D$ but $x \notin A$ and $x \notin B$ and $x \notin C$.

So firstly as $x \in D$ and $x \notin A$, we have $x \in (D \setminus A)$.

And secondly as $x \notin C$ we have in particular that $x \notin (C \setminus B)$ also.

Thus combining the results we get $x \in (D \setminus A)$ and $x \notin (C \setminus B)$ are both true or to say the same thing $x \in (D \setminus A) \setminus (C \setminus B)$.

Thus $D \setminus (A \cup (B \cup C)) \subseteq (D \setminus A) \setminus (C \setminus B)$ indeed.


$A = C = \emptyset$, $B = D = \{1\}$ is a counterexample.


You're not necessarily removing the elements of B from D in the LHS, but are in the RHS.


A = {1} B = {2} C = {3} D = {1, 2, 3, 4, 5}

$C\setminus B=\{3\}$, so $(D\setminus A)\setminus(C\setminus B)=\{2,4,5\}$ but $D\setminus(A\cup B\cup C)=\{4,5\}$


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