Let $A,B,X$ be finite sets. Prove that $|A \triangle X| + |X \triangle B| = |A \triangle B| \iff A \cap B \subseteq X \subseteq A \cup B$ I'm trying to do the following exercise:

Let $A,B,X$ be finite subsets of a set $U$. Prove that $|A \triangle X| + |X \triangle
 B| = |A \triangle B| \iff A \cap B \subseteq X \subseteq A \cup B$

I am allowed to assume that the cardinality of the symmetric difference of finite sets is a metric.
So I can use the following properties without proving them:

Let $A,B,C$ be finite subsets of a set $U$, then:
P.1) $|A \triangle B| \ge 0$
P.2) $|A \triangle B| = 0 \iff A=B$
P.3) $|A \triangle B| = |B \triangle A| $
P.4) $|A \triangle C| \le |A \triangle B|  + |B \triangle C|$

This is how I tried to prove the implication $A \cap B \subseteq X \subseteq A \cup B \implies |A \triangle X| + |X \triangle
 B| = |A \triangle B|$
I'm assisting myself with the following two lemmas:
Lemma 1: $A \cap B \subseteq X \subseteq A \cup B \implies (A \triangle X) \cap (X \triangle B) = \emptyset$
Proof:
Let $A,B,X$ be finite sets such that $A \cap B \subseteq X \subseteq A \cup B$
Let's assume there exists an element $t \in (A \triangle X) \cap (X \triangle B)$
There are four possible cases:
i) $t \in (A-X) \ \land \ t \in (X-B)$
ii) $t \in (A-X) \ \land \ t \in (B - X)$
iii) $t \in (X-A) \ \land \ t \in (X-B)$
iv) $t \in (X-A) \ \land \ t \in (B-X)$
Assuming i):
$t \in (A-X) \ \land \ t \in (X-B) \implies (t \in A \ \land \ t \notin X) \land (t \in X \ \land t \notin B) \implies t \notin X \land t \in X$ , which is a contradiction.
Assuming ii):
$t \in (A-X) \ \land \ t \in (B - X) \implies (t \in A \ \land t \notin X) \land (t \in B \ \land \ t \notin X)$
$\implies (t \in A \ \land \ t \in B) \land t \notin X \implies t \in A \cap B \ \land \ t \notin X$, which contradicts the hypothesis $A \cap B \subseteq X$.
Assuming iii):
$t \in (X-A) \ \land \ t \in (X-B) \implies (t \in X \ \land \ t \notin A) \land (t \in X \ \land \ \ t \notin B)$
$\implies t \in X \ \land \ t \notin A \ \land t \notin B \implies t \in X \ \land \ t \notin A \cup B$, which contradicts the hypothesis $X \subseteq A \cup B$.
Assuming iv):
$t \in (X-A) \ \land \ t \in (B-X) \implies (t \in X \ \land \ t \notin A) \land (t \in B \ \land \ t \notin X) \implies t \in X \land t \notin X$ , which is a contradiction.
We get a contradiction in all four cases, so there does not exist an element $t$ such that $t \in (A \triangle X) \cap (X \triangle B)$.
Therefore $(A \triangle X) \cap (X \triangle B) = \emptyset$
$ \blacksquare$
Lemma 2: If $A,B,X$ are finite subsets of $U$ then $(A \triangle X) \cup (X \triangle B) = (A \cup X \cup B) - (A \cap X \cap B)$
Proof:
By definition of symmetric difference we have:
$(A \triangle X) \cup (X \triangle B) = [(A \cup X) \cap \overline{(A \cap X)}] \cup [(X \cup B) \cap \overline{(X \cap B)}]$ 
Applying the distributive laws of union and intersection:
$(A \triangle X) \cup (X \triangle B) = [((A \cup X) \cap \overline{(A \cap X)} \ ) \cup (X \cup B)] \cap [((A \cup X) \cap \overline{(A \cap X)} \ ) \cup \overline{(X \cap B)}]$ 
$= [((A \cup X) \cup (X \cup B)) \cap ( \ \overline{(A \cap X)} \  \cup (X \cup B))] \cap [((A \cup X) \cup \overline{(X \cap B)} \ ) \cap ( \ \overline{(A \cap X)} \ \cup \ \overline{(X \cap B)} \ )]$
By de Morgan laws and idempotence and associativity of union and intersection:
$(A \triangle X) \cup (X \triangle B) = [(A \cup X \cup B) \cap ( ( \ \overline{A} \cup \overline{X} \ ) \cup (X \cup B))] \cap [((A \cup X) \cup ( \ \overline{X} \cup \overline{B} \ )) \cap ( \overline{A} \cup \overline{X} \cup \overline{B})]$
$ = [(A \cup X \cup B) \cap (  \ \overline{A} \cup ( \overline{X} \  \cup X ) \cup B)] \cap [(A \cup (X \cup  \ \overline{X}) \cup \overline{B} \ ) \cap ( \overline{A} \cup \overline{X} \cup \overline{B})]$
By complementation and absorption laws:
$(A \triangle X) \cup (X \triangle B) = [(A \cup X \cup B) \cap (  \ \overline{A} \cup U \cup B)] \cap [(A \cup U \cup \overline{B} \ ) \cap ( \overline{A} \cup \overline{X} \cup \overline{B})]$
$= [(A \cup X \cup B) \cap U] \cap [U \cap ( \overline{A} \cup \overline{X} \cup \overline{B})]$
$= (A \cup X \cup B) \cap ( \overline{A} \cup \overline{X} \cup \overline{B})$
$= (A \cup X \cup B) \cap ( \overline{A \cup X \cup B} )$
$= (A \cup X \cup B) - ( A \cup X \cup B )$
$\blacksquare$
Now, going back to the original implication we have:
$A \cap B \subseteq X \subseteq A \cup B \implies A \cup B \cup X = A \cup B \ \land \ A \cap B \cap X = A \cap B$
$\implies (A \cup B \cup X ) - (A \cap B \cap X) = (A \cup B) -(A \cap B)$
$\implies (A \cup B \cup X ) - (A \cap B \cap X) = A \triangle B$
So, applying lemma 2: 
$(A \triangle X) \cup (X \triangle B) = A \triangle B \implies |(A \triangle X) \cup (X \triangle B) | = |A \triangle B|$
$\implies |(A \triangle X)| + |(X \triangle B) | - |(A \triangle X) \cap (X \triangle B)|= |A \triangle B|$
And finally, by lemma 1:
$|(A \triangle X)| + |(X \triangle B) | - |\emptyset|= |A \triangle B|$
$|(A \triangle X)| + |(X \triangle B) | - 0= |A \triangle B|$
$|(A \triangle X)| + |(X \triangle B) |= |A \triangle B|$
$\blacksquare$
Is this part of the proof correct? And how do I prove that $|A \triangle X| + |X \triangle
 B| = |A \triangle B| \implies A \cap B \subseteq X \subseteq A \cup B$ ?? I don't know what to do on that part.
 A: I have not checked your proof/effort, but reach unto you an alternative route.

In the first place draw a Venn-diagram.
Discern the following $7$ sets:


*

*$S_{1}=A\cap B\cap X$

*$S_{2}=A\cap B\cap X^{\complement}$

*$S_{3}=A\cap B^{\complement}\cap X$

*$S_{4}=A\cap B^{\complement}\cap X^{\complement}$

*$S_{5}=A^{\complement}\cap B\cap X$

*$S_{6}=A^{\complement}\cap B\cap X^{\complement}$

*$S_{7}=A^{\complement}\cap B^{\complement}\cap X$
The sets are disjoint and:


*

*$A\Delta X=S_{2}\cup S_{4}\cup S_{5}\cup S_{7}$

*$X\Delta B=S_{2}\cup S_{3}\cup S_{6}\cup S_{7}$

*$A\Delta B=S_{3}\cup S_{4}\cup S_{5}\cup S_{6}$
So the equality $\left|A\Delta X\right|+\left|X\Delta B\right|=\left|A\Delta B\right|$
can be written as:
$$\left|S_{2}\right|+\left|S_{4}\right|+\left|S_{5}\right|+\left|S_{7}\right|+\left|S_{2}\right|+\left|S_{3}\right|+\left|S_{6}\right|+\left|S_{7}\right|=\left|S_{3}\right|+\left|S_{4}\right|+\left|S_{5}\right|+\left|S_{6}\right|\tag1$$
Equivalent with $(1)$ are the following statements:


*

*$\left|S_{2}\right|+\left|S_{7}\right|=0$

*$S_{2}=\varnothing=S_{7}$

*$A\cap B\subseteq X\subseteq A\cup B$
