Using double inclusion to show $(A △ B) ∪ C = (A ∪ C) △ (B \setminus C)$ Could someone please tell me if I have used double inclusion correctly to prove $(A △ B) ∪ C = (A ∪ C) △ (B \setminus C)$?
(NOTE: "$△$" denotes symmetric difference or exclusive or, i.e. $A △ B = (A ∪ B) \setminus (A ∩ B)  = (A \setminus B) \cup (B \setminus A)$.)
Showing $L ⊆ R$:
Let $x ∈ (A △ B) ∪ C$.
By the definition of union, this means that either x is in $C$ (case 3) or in $A \setminus B$ (case 1) or in $B \setminus A$ (case 2).

*

*If $x ∈ C$, then $x ∈ A ∪ C$, therefore, $x ∈ (A ∪ C) △ (B \setminus  C)$.

*If $x ∈ A \setminus  B$, then x ∈ A ∪ C, therefore, $x ∈ (A ∪ C) △ (B \setminus C)$.

*If $x ∈ B \setminus  A$, then x ∈ B, but $x \notin A$, which implies $x \notin A ∪ C$, which implies $x \notin C$, which implies $x ∈ B \setminus C$. Therefore, $x ∈ (A ∪ C) △ (B \setminus  C)$
Showing $L ⊇ R$:
Let $x ∈ (A ∪ C) △ (B \setminus C)$.
By the definition of symmetric difference, this means that x is in $(A ∪ C) \setminus (B \setminus C)$ (case 1) or x is in $(B \setminus C) \setminus (A ∪ C)$ (case 2).

*

*By De Morgan's law, $x ∈ (A ∪ C) \setminus (B \setminus C) \iff x ∈ (A ∪ C) \land (\lnot(x \in B) \lor x \in C)$. So, if $x ∈ (A ∪ C) \setminus (B \setminus C)$, then either $x ∈ A \setminus B$ (case 1.1) or $x \in C$ (case 1.2).


*

*1.1. If $x ∈ A \setminus B$, then $x ∈ (A △ B)$, which implies $x ∈ (A △ B) ∪ C$.

*1.2. If $x ∈ C$, then $x ∈ (A △ B) ∪ C$.



*By De Morgan's law, $x ∈ (B \setminus C) \setminus (A ∪ C) \iff x ∈ (B \setminus C) \land \lnot(x \in A ∪ C)$. $x ∈ (B \setminus C)$ implies $x ∈ B$ and $\lnot(x \in A ∪ C)$ implies $x \notin A$. So we have $x \in B$ and $x \notin A$, which implies $x \in (A △ B)$ and therefore, $x \in (A △ B) ∪ C$.


I'm especially unsure about step 2 of $L ⊇ R$. Any guidance would be greatly appreciated, as this is my first time using double inclusion.

EDIT: Here's my second attempt. I’ve only included the first half for now, since it ended up being way longer and I’m not sure if it’s correct. But I’ve tried to be more systematic this time.
Some questions that came up:

*

*Is it ok to use logical equivalences to simplify expressions as I have done in cases 1 and 2? None of the examples of double inclusion I've come across use them, so I don't know if it's cheating.

*I’m not sure about cases 1.3, 2.3 and 3.3. They each simplify to either $x ∈ C$ or $x ∉ C$, but since they contain contradictions, shouldn't they evaluate to the empty set? If so, then we still have $x ∈ (A ∪ C) △ (B \setminus C)$ in each of these cases, since the empty set is a subset of every set.

Proof: (first-half only)
We want to show $(A △ B) ∪ C = (A ∪ C) △ (B \setminus C)$
We start by showing $L ⊆ R$.
Let $x ∈ (A △ B) ∪ C$. Then either $x ∈ (A △ B)$ (case 1) or $x ∈ C$ (case 2) or both (case 3).

*

*Case 1: x ∈ (A △ B) ∧ x ∉ C.
\begin{align*} 
x ∈ (A △ B) ∧ x ∉ C 
& \iff (x ∈ A \setminus B ∧ x ∉ C) \lor (x ∈ B \setminus A ∧ x ∉ C)                  &&\text{(distributivity)}
\\& \iff (x ∈ A ∧ x ∉ B ∧ x ∉ C) \lor (x ∈ B ∧ x ∉ A ∧ x ∉ C)      &&\text{(def of $\setminus$)}
\end{align*}

*

*Case 1.1: $x ∈ A ∧ x ∉ B ∧ x ∉ C$
Since $x ∈ A$, $x ∈ A ∪ C$.
Since $x ∉ B$, $x ∉ B \setminus C$.
So, we have $x ∈ (A ∪ C) △ (B \setminus C)$.


*Case 1.2: $x ∈ B ∧ x ∉ A ∧ x ∉ C$
Since $x ∈ B$ and $x ∉ C$, $x ∈ B \setminus C$.
Since $x ∉ A$ and $x ∉ C$, $x ∉ (A ∪ C)$.
So, we have $x ∈ (A ∪ C) △ (B \setminus C)$.


*Case 1.3: $(x ∈ A ∧ x ∉ B ∧ x ∉ C) ∧ (x ∈ B ∧ x ∉ A ∧ x ∉ C)$.
\begin{align*} 
    (x ∈ A ∧ x ∉ B ∧ x ∉ C) ∧ (x ∈ B ∧ x ∉ A ∧ x ∉ C) \iff x ∉ C         &&\text{(from associativity, idempotent and contradiction laws)}
    \end{align*}
Not sure about this. I feel like the contradictions in the unsimplified expression mean that we would end up with $x ∈ Ø$, and so again we have $x ∈ (A ∪ C) △ (B \setminus C)$, since $Ø \subseteq (A ∪ C) △ (B \setminus C)$.




*Case 2: $x ∉ (A △ B) ∧ x ∈ C$.
\begin{align*} 
x ∉ (A △ B) ∧ x ∈ C 
& \iff  ¬(x ∈ A \setminus B \lor x ∈ B \setminus A) ∧ x ∈ C  
\\& \iff ¬[(x ∈ A ∧ ¬(x ∈ B)) \lor (x ∈ B ∧ ¬(x ∈ A))] ∧ x ∈ C &&\text{(def of $\setminus$)}
\\& \iff ¬(x ∈ A ∧ ¬(x ∈ B)) ∧ ¬(x ∈ B ∧ ¬(x ∈ A)) ∧ x ∈ C &&\text{(De Morgan)}
\\& \iff (¬(x ∈ A) \lor x ∈ B) ∧ (¬(x ∈ B) \lor x ∈ A) ∧ x ∈ C      &&\text{(De Morgan)}
\\& \iff [¬(x ∈ A) ∧ (¬(x ∈ B) \lor x ∈ A)] \lor [x ∈ B ∧ (¬(x ∈ B)             \lor x ∈ A)] ∧ x ∈ C &&\text{(distributivity)}
\\& \iff [(¬(x ∈ A) ∧ ¬(x ∈ B)) \lor (¬(x ∈ A) ∧ x ∈ A)] \lor [(x ∈ B ∧     ¬(x ∈ B)) \lor (x ∈ B ∧ x ∈ A)] ∧ x ∈ C &&\text{(distributivity)}
\\& \iff [(¬(x ∈ A) ∧ ¬(x ∈ B)) \lor (x ∈ B ∧ x ∈ A)] ∧ x ∈ C      &&\text{(P ∨ ⊥ ⟺ P)}
\\& \iff [(¬(x ∈ A) ∧ ¬(x ∈ B)) ∧ x ∈ C] \lor [(x ∈ B ∧ x ∈ A) ∧ x ∈ C] &&\text{(distributivity)}
\\& \iff [¬(x ∈ A) ∧ ¬(x ∈ B) ∧ x ∈ C] \lor [x ∈ B ∧ x ∈ A ∧ x ∈ C]     &&\text{(associativity)}
\end{align*}

*

*Case 2.1: $¬(x ∈ A) ∧ ¬(x ∈ B) ∧ x ∈ C$
Since $x ∈ C$, $x ∈ A ∪ C$.
Since $x ∉ B$, $x ∉ B \setminus C$.
So, we have $x ∈ (A ∪ C) △ (B \setminus C)$.


*Case 2.2: $x ∈ B ∧ x ∈ A ∧ x ∈ C$
Since $x ∈ A$ and $x ∈ C$, $x ∈ A ∪ C$.
Since $x ∈ B$ and $x ∈ C$, $x ∉ B \setminus C$.
So, we have $x ∈ (A ∪ C) △ (B \setminus C)$.


*Case 2.3: $[¬(x ∈ A) ∧ ¬(x ∈ B) ∧ x ∈ C] ∧ [x ∈ B ∧ x ∈ A ∧ x ∈ C]$
Like case 1.3, this simplifies to $x ∈ C$, but shouldn't it be the empty set?




*Case 3: $x ∈ (A △ B) ∧ x ∈ C$.
\begin{align*} 
x ∈ (A △ B) ∧ x ∈ C 
& \iff (x ∈ A \setminus B ∧ x ∈ C) \lor (x ∈ B \setminus A ∧ x ∈ C)      &&\text{(distributivity)}
\\& \iff (x ∈ A ∧ x ∉ B ∧ x ∈ C) \lor (x ∈ B ∧ x ∉ A ∧ x ∈ C)  &&\text{(def of $\setminus$)}
\end{align*}

*

*Case 3.1: $x ∈ A ∧ x ∉ B ∧ x ∈ C$
Since $x ∈ A$ and $x ∈ C$, $x ∈ A ∪ C$.
Since $x ∉ B$, $x ∉ B \setminus C$.
So, we have $x ∈ (A ∪ C) △ (B \setminus C)$.


*Case 3.2: $x ∈ B ∧ x ∉ A ∧ x ∈ C$
Since $x ∈ B$ and $x ∈ C$, $x ∉ B \setminus C$.
Since $x ∈ C$, $x ∈ A ∪ C$.
So we have $x ∈ (A ∪ C) △ (B \setminus C)$.


*Case 3.3: $(x ∈ A ∧ x ∉ B ∧ x ∈ C) ∧ (x ∈ B ∧ x ∉ A ∧ x ∈ C)$
Same as case 2.3.
 A: There are a number of errors in the first part of your proof, so I'm going to just deal with those in this answer.


*

*If $x ∈ C$, then $x ∈ A ∪ C$, therefore, $x ∈ (A ∪ C) △ (B \setminus  C)$.


This is sloppy.  Symmetric differences remove part of the sets, so $x\in A$ does not imply that $x\in A\Delta B$. $x\in A\cup C$ is OK but you also need to note that $x\not\in B\setminus C$ before you can plonk it into that symmetric difference.



*If $x ∈ A \setminus  B$, then x ∈ A ∪ C, therefore, $x ∈ (A ∪ C) △ (B \setminus C)$.


Similar, you need to note that it's not in $B$, which is obvious, but the symmetric difference does not contain all of $A\cup C$, so you do need to note it.



*If $x ∈ B \setminus  A$, then x ∈ B, but $x \notin A$, which implies $x \notin A ∪ C$, which implies $x \notin C$

No it does not imply $x\notin A\cup C$, you've completely derailed here. It does imply that $x\notin A\cap C$ though.

, which implies $x ∈ B \setminus C$. Therefore, $x ∈ (A ∪ C) △ (B \setminus  C)$

