Define $\triangle$ as $A\triangle B = (A\cup B - A\cap B )$, show associativity of $\triangle$

Question

We define $\triangle$ as $A\triangle B = (A\cup B - A\cap B )$ and want to prove axiom G1 for the group (G, $\triangle)$.

We want to show that $(A\triangle B)\triangle C = A\triangle (B\triangle C)$

$(A\triangle B)\triangle C = ((A\cup B - A\cap B)\cup C) - ((A\cup B - A\cap B)\cap C) = $ $(A\cup B \cup C - A \cap B) - ((A\cup B)\cap C) + A\cap B = $

$A\cup B \cup C - (A\cup B)\cap C$

$A\triangle (B\triangle C) = (A\cup (B\cup C - B\cap C)) - (A\cap (B\cup C - B\cap C)) = $ $(A\cup B \cup C - B \cap C) - (A\cap (B\cup C)) + B\cap C = $

$A\cup B \cup C - (B\cup C)\cap A$

The problem I'm running into now is that I believe that $A\cap (B\cup C) \neq (A\cup B)\cap C$ in general. I'm not sure if I'm taking wrong steps in my derivation (probably). Does anyone know what steps are wrong, or if there's a better way to prove this?

Answer 1

You're making a mistake here:

$((A\cup B - A\cap B)\cup C) - ((A\cup B - A\cap B)\cap C) = $ $(A\cup B \cup C - A \cap B) - ((A\cup B)\cap C) + A\cap B = ...$

I am not even exactly sure what you tried to do here, but I assume you tried to go from $((A\cup B - A\cap B)\cup C)$ to $(A\cup B \cup C - A \cap B)$, and from $((A\cup B - A\cap B)\cap C)$ to $((A\cup B)\cap C) + A\cap B$

Now, first of all, there is no set-operator $+$.

Second, I would urge you use use parentheses to avoid any kind of confusion.

So, just focusing on the first term, what you have is:

$$((A\cup B) - (A\cap B))\cup C$$

Now, it looks like you tried some kind of Distribution on this, but you executed that rather poorly. Here is what you would get if you do Distribution:

$$((A\cup B \cup C) - ((A\cap B)\cup C)$$

However, I wouldn't use Distribution in combination with the setminus operator at all: yes, Distribution of $\cup$ over $-$ happens to work out ok, but Distribution of $\cap$ over $-$ is actually a little weird, since the setminus operator corresponds to a $\cap$ operator itself:

$$A - B =_{df} A \cap B^C$$

Anyway, I recommend using that very identity to rewrite your expressions in terms of $\cap$, $\cup$, and complement, and now you can use all the 'typical' equivalences.

To give you a start:

$$(((A\cup B) - (A\cap B))\cup C) - (((A\cup B) - (A\cap B))\cap C) = $$

$$(((A\cup B) \cap (A\cap B)^C)\cup C) \cap (((A\cup B) \cap (A\cap B)^C)\cap C)^C = $$

$$(((A\cup B) \cap (A^C \cup B^C))\cup C) \cap (((A\cup B) \cap (A\cap B)^C)^C\cup C^C) = ...$$

Yes, it's a slog, but you'll get there eventually!