[Math] Prove $A+B= A \cup B$ if and only if $A \cap B = \emptyset$ using the definition of $A+B$

elementary-set-theorylogicproof-verification

Let $A$ and $B$ be sets. Define the symmetric difference of $A$ and $B$, written $A+B$, by $A+B=(A \cup B) \backslash (A \cap B)$.

Prove $A+B= A \cup B$ if and only if $A \cap B = \emptyset$.

My attempt:

Part 1b. If $A+B = A \cup B$, then $ A \cap B = \emptyset$

$A+B=(A \cup B) \backslash (A \cap B)$.

There shouldn't be any elements inside $A \cap B$.

$A+B=(A \cup B) \backslash (\emptyset)$

Therefore, $A \cap B = \emptyset$.

Edit: What if I put…

Definition: Let $(A \cup B)$ and $(A \cap B)$ be sets. The complement of $(A \cup B)$ relative to $(A \cap B)$, written $ (A \cup B)\backslash (A \cap B)$ is the set $(A \cup B) \backslash (A \cap B) = [x: x \in (A \cup B) \land x \notin (A \cap B)]$

While the elements are inside $(A \cup B)$, they don't belong in $(A \cap B)$

hmmmm they isn't any element in common…so it's an emptyset… wait a sec… emptyset means that there aren't any elements.

Part 2b. If $A \cap B = \emptyset$, then $A+B = A \cup B$

$A+B=(A \cup B) \backslash (\emptyset)$

Definition: Let $(A \cup B)$ and $\emptyset$ be sets. The complement of $(A \cup B)$ relative to $\emptyset$, written $ (A \cup B)\backslash \emptyset$ is the set $(A \cup B) \backslash \emptyset = [x: x \in (A \cup B) \land x \notin \emptyset]$

There are elements inside $A \cup B$, but not in $A \cap B$.

As a result, $A+B = (A \cup B)$

Did I do this correctly?

Best Answer

I will try to show you the result exploiting the definitions.

Define the symmetric difference of $A$ and $B$ (written $A+B$), by

$A+B = (A \cup B) \backslash (A \cap B)$.

We have that

$x \in A+B \quad$ iff $\quad ( x \in A \lor x \in B) \quad$ and $\quad \lnot (x \in A \land x \in B)$

But we have also that

$x \in A \cup B \quad$ iff $\quad ( x \in A \lor x \in B)$

Now we want that $A + B$ must be equal to $A \cup B$.

If we "compare" the two above condition, we can write them as

$P \land Q$ and $P$, and we want that $P \land Q \equiv P$.

By truth-table, we can check that $P \land Q \equiv P$ iff $Q$ is identically true.

So, we want that our condition $\lnot (x \in A \land x \in B)$ must be identically true.

This is simply :

$\forall x \lnot (x \in A \land x \in B)$.

By definition, $x \in A \cap B$ iff $x \in A \land x \in B$.

So, our condition is : $\forall x (x \notin A \cap B)$ and this, in turn, means that $A \cap B$ is $\emptyset$ (the empty set).

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