Proof for the symmetric difference? I want to prove that:
a)
$A \Delta B = \emptyset \Leftrightarrow A = B $
Prove for $\Leftarrow$
Then:
$A  \Delta B = (A$ \ $A) \cup (A $\ $A) = \emptyset$
Prove for $\Rightarrow$
Then(proof by contrapositive):
$A  = B \rightarrow (A$ \ $B) \cup (A $\ $B) = (A$ \ $B) \cup (A $\ $B)= \emptyset$
b)
$A \Delta B = A \cup B \Leftrightarrow A \cap B = \emptyset$
Prove for $\Leftarrow$
$A \Delta B \rightarrow (A \cup B)$ \ $(A \cap B) = ( A \cup B) \rightarrow ( A \cap B) = \emptyset$
Prove for $\Rightarrow$
Then(proof by contrapositive):
$A \Delta B \rightarrow (A \cup B)$ \ $\emptyset = ( A \cup B) \rightarrow ( A \cap B) = ( A \cap B) $
I am not sure if these two proves are correct. I really appreciate your answer!
 A: For a) $\Rightarrow)$ and since you want use the contrapositive argument you must prove that if $A\neq B$ then $A\Delta B\neq \emptyset$, so suppose that $A\neq B$ then there's $x\in A$ and $x\not\in B$ or  there's $x\in B$ and $x\not\in A$ so $A\setminus B\ne\emptyset$ or $B\setminus A\ne\emptyset$ and then $A\Delta B=(A\setminus B)\cup (B\setminus A)\ne\emptyset$.
For b) if $A\Delta B=A\cup B$ then $(A\cup B)\setminus (A\cap B)=(A\cup B)$ so $(A\cap B)\subset(A\cup B)^c $ and since clearly $(A\cap B)\subset(A\cup B) $ then
$$(A\cap B)\subset(A\cup B)\cap(A\cup B)^c=\emptyset $$
A: Let $X$ be a set.
Let $F_2 = \mathbb{Z}/2\mathbb{Z} = \{0, 1\}$ be the field consisting of $2$ elements.
Let $F_2^X$ be the set of maps $X \rightarrow F_2$.
$F_2^X$ is a ring with pointwise addition and multiplication.
Let $A$ be a subset of $X$.
Let $\chi_A$ be the characteristic function of $A$.
We regard $\chi_A$ as an element of $F_2^X$.
Now let $A, B$ be subsets of $X$.
Clearly $\chi_{A\Delta B} = \chi_A + \chi_B$.
Suppose $A\Delta B = \emptyset$.
Then $0 = \chi_A + \chi_B$.
Hence $\chi_A = -\chi_B = \chi_B$.
Hence $A = B$.
Conversely suppose $A = B$.
Then $\chi_A = \chi_B$.
Hence $\chi_{A\Delta B} = \chi_A + \chi_B = 0$.
Hence $A\Delta B = \emptyset$.
A: Here is an alternative, calculational proof which does not have two separate directions, using the fact (an alternative definition, actually) that for all $\;x\;$,
$$x \in A \Delta B \equiv x \in A \not\equiv x \in B$$
Now we try to simplify $\;A \Delta B = \emptyset\;$:
\begin{align}
& A \Delta B = \emptyset \\
\equiv & \;\;\;\;\;\text{"basic property of $\;\emptyset\;$: it contains no elements"} \\
& \langle \forall x :: \lnot(x \in A \Delta B) \rangle \\
\equiv & \;\;\;\;\;\text{"the above definition of $\;\Delta\;$"} \\
& \langle \forall x :: \lnot(x \in A \not\equiv x \in B) \rangle \\
\equiv & \;\;\;\;\;\text{"logic"} \\
& \langle \forall x :: x \in A \equiv x \in B \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;=\;$ on sets, i.e., set extensionality"} \\
& A = B \\
\end{align}
If you want to use the definition $\;A \Delta B = (A \setminus B) \cup (B \setminus A)\;$, then you can use a similar approach, using
\begin{array}\\
x \in A \setminus B & \equiv & x \in A \land \lnot(x \in B) \\
x \in A \cup B & \equiv & x \in A \lor x \in B \\
\end{array}
This makes the calculation slightly longer, of course.  Hint: use DeMorgan, and the fact that $\;\lnot P \lor Q\;$ is the same as $\;P \Rightarrow Q \;$.
A: a) ($\rightarrow$) Suppose $A \Delta B = \emptyset$. Let x be an arbitrary element of A. Suppose $x \notin B$. Then since $x \in A$ and $x \notin B$, $x \in B\setminus A$, so $x \in A \Delta B$, and therefore $x \in \emptyset$. But this last statement is clearly a contradiction, so we can conclude that $x \in B$. Since x was an arbitrary element of $A$, $A \subseteq B$. A similar argument shows that $B \subseteq A$. Then since $A \subseteq B$ and $B \subseteq A$, $A = B$. 
($\leftarrow$) Suppose $A = B$. Suppose $A \Delta B \neq \emptyset$. Then we can choose some $x$ such that either $x \in A\setminus B$ or $x \in B\setminus A$. We consider these cases separately.
Case 1. $x \in A \setminus B$. Then since $A = B$, $x \in A \setminus A$, so $x \in A$ and $x \notin A$, which is a contradiction.
Case 2. $x \in B \setminus A$. Similarly, this leads to a contradiction.
Thus we can conclude that $A \Delta B = \emptyset$.
b) ($\rightarrow$) Suppose $A \Delta B = A \cup B$. Suppose $A \cap B \neq \emptyset$. Then we can choose some $x$ such that $x \in A \cap B$. This means that $x \in A$ and $x \in B$. Since $x \in A$, it follows that $x \in A \cup B$. But then since $A \Delta B = A \cup B$, $x \in A \Delta B$, so $x \in A \cup B$ and $x \notin A \cap B$. This contradicts the fact that $x \in A \cap B$. Therefore $A \cap B = \emptyset$.
($\leftarrow$) Suppose $A \cap B = \emptyset$. Let $x$ be an arbitrary element of $A \Delta B$. Then $x \in (A \cup B) \setminus (A \cap B)$. This means that $x \in A \cup B$ and $x \notin A \cap B$, so in particular $x \in A \cup B$. Since $x$ was an arbitrary element of $A \Delta B$, $A \Delta B \subseteq A \cup B$. 
Now let $x$ be an arbitrary element of $A \cup B$. Then either $x \in A$ or $x \in B$. 
Case 1. $x \in A$. Since $A \cap B = \emptyset$, $x \notin B$. Since $x \in A$ and $x \notin B$, $x \in A \setminus B$, so $x \in A \Delta B$. 
Case 2. $x \in B$. A similar argument shows that $x \in A \Delta B$. 
Since x was an arbitrary element of $A \cup B$, $A \cup B \subseteq A \Delta B$. But then since $A \Delta B \subseteq A \cup B$ and $A \cup B \subseteq A \Delta B$, we can conclude that $A \Delta B = A \cup B$.
