Prove or disprove $(A + B) \cap C = (A \cap C) +(B \cap C)$
I want to disprove this statement.
$(A+B)$ is the symmetric difference and has the form of $(A \cup B) \backslash (A \cap B)$
I am starting on the left which is $(A + B) \cap C $
If I take the complement definition of $(A+B)$, I would have
$[x: x \in A \cup B \land x \notin A \cap B]$
So I am left with $A \cup B$
now I'm going to use the distributive law $\cap C$ on $A \cup B$
The result would be $(A \cup C) \cap (B \cup C)$
$(A \cup C) \cap (B \cup C) \neq (A \cap C) +(B \cap C)$
because in the middle we have $\cap$ on the left and $+$ on the right... but that's wrong.
What if I let $ C = \emptyset$ ?
Then I would have $(A +B) \cap \emptyset = (A \cap \emptyset) +(B \cap \emptyset )$
I'm going to start at the right this time .. because I've seen some properties already and it's similar to what I did weeks before
$= (A \cap \emptyset) +(B \cap \emptyset )$
Universal Bound Law $A \cap \emptyset = \emptyset$
$= (\emptyset) +(\emptyset )$
If I take the symmetric difference of $= (\emptyset) +(\emptyset )$ it's an empty set.
hmmm now that I think about it...if I approach it this way it looks like I'm proving that they are indeed equal.
$(A + B) \cap \emptyset$ [ distributive law]
$(A \cap \emptyset ) + (B \cap \emptyset)$ [universe bound laws]
$( \emptyset ) + ( \emptyset)$ [symmetric difference]
that's an empty set...
I want to disprove this statement... but how?
edit: another attempt at this problem using set intersection definition
Prove or disprove $(A + B) \cap C = (A \cap C) +(B \cap C)$
Suppose $x \in (A +B) \cap C$, then $[x \in (A +B)] \cap C$ and we have $(x \in A + x \in B) \cap C$
For $x \in A$, by set intersection definition, we have $x \in A \land C$ and $x \in B \land C$
[maybe for $x \in C$, by set intersection definition, we have $x \in A \land C$ and $x \in B \land C$ since C is being distributed, not A. ]
By symmetric difference definition, we have $x \in A \land C + x \in B \land C$
Therefore, $(A \cap C) +(B \cap C)$
Is this correct?!