How would I go about expressing the symmetric difference if the only symbols I am allowed to use are the complement and intersect symbols? I know that $A \ominus B = (A^c\cap B) \cup (B^c\cap A)$, but how to I convert it so that the union is not needed?
[Math] Express the symmetric difference using only complement and intersect
elementary-set-theory
Related Solutions
I think there is no non-trivial way to do what the question is asking.
Let's say that a set of partial (possibly total) binary operations $S$ is complete for the symmetric difference operation $\triangle$ if
every operation in $S$ is a (possibly trivial) restriction of some composition of copies of $\triangle$, and
for every pair of sets $(A,B)$ there is a partial operation $\cdot$ that is a composition of partial operations in $S$ and has the property that $A \cdot B$ is defined and is equal to $A \mathbin{\triangle} B$. (Note that is weaker than saying that $S$ itself is a composition of partial operations in $S$, which is impossible if none of the operations in $S$ is total.)
Even with this weak version of the definition, the only sets $S$ that are complete for $\triangle$ are trivial, in a sense that we will make precise below. Note that by the first clause of the definition, and the fact that $\triangle$ is commutative, associative, and satisfies $A \mathbin{\triangle} A = \emptyset$, every partial operation $\cdot$ in $S$ is the restriction of one of the four following total operations:
The constant binary operation given by $A \cdot B = \emptyset$,
The projection given by $A \cdot B = A$,
The projection given by $A \cdot B = B$, and
The symmetric difference operation $\triangle$ itself.
The first three operations do not help us form any additional binary operations under composition, so we consider them to be trivial, and we assume that $S$ consists only of restrictions of $\triangle$.
A trivial way to get a complete set $S$ is for the union of the domains of the partial operations in $S$ to consist of all pairs $(A,B)$: for example, the proper set difference operation "$\setminus$" as defined in the question is simply the restriction of $\triangle$ to the domain $\{(A,B) : B \subseteq A\}$, so we could let $S = \{\setminus, \cdot\}$ where $\cdot$ is the restriction of $\triangle$ to the complementary domain $\{(A,B) : B \not \subseteq A\}$. More generally, we could weaken the triviality requirement to consider the domains of the partial operations in $S$ and their reverses; for example, this requirement would be satisfied by the set $S = \{\setminus, \cdot\}$ where $\cdot$ now denotes the restriction of $\triangle$ to the domain $\{(A,B) : A \not \subseteq B\}$.
However, every complete set is trivial in this way. Suppose that some pair $(A,B)$ is not in the domain of any partial operation $\cdot$ in $S$, and neither is the reversed pair $(B,A)$. Because the partial operations in $S$ are all binary, it is not hard to show that the pairs $(A,B)$ and $(B,A)$ cannot be in the domain of any composition of elements of $S$ either, so there is no way to get the symmetric difference $A \mathbin{\triangle} B$.
As you said, you can use the distributive law of union over intersection: $$ \begin{align} (A \color{red}{\cup B})\cap (\color{red}{B \cup} C\cup D)\cap(\color{red}{B \cup} C\cup D') &= \color{red}{B \cup} \left(A \cap (\color{green}{C \cup} D) \cap (\color{green}{C \cup} D')\right) \\ &= B \cup \left( A \cap \left(\color{green}{C \cup} (D \cap D') \right) \right) \\ &= B \cup (A \cap C) \end{align} $$ Note that the last equality is because $D \cap D' = \varnothing$ for any set $D$.
Best Answer
You can write a union using complements and intersections. Namely, you have to check that $A \cup B = (A^c \cap B^c)^c$. This is obvious because the complement of the union is those elements which are in neither $A$ nor $B$, ie the elements in both $A^c$ and in $B^c$.