Set Theory Proof of Set Difference I am just learning set theory proofs and I am struggling with the following:
Prove that $(S_1 \setminus S_2) \cup (S_2 \setminus S_3) \subseteq (S_1 \cup S_2) \setminus(S_1 \cap S_2 \cap S_3)$
Here is my attempt so far:

*

*Let $x \in (S_1 \setminus S_2) \cup (S_2 \setminus S_3)$ so we know by definition that $x\in S_1 $and $ x \notin S_2$ or $x\in S_2 $and $ x \notin S_3$.

*So $x \in S_1$ or $x\in S_2$ and $x \notin S_2$ and $x \notin S_3$.

*$(S_1 \cup S_2) \setminus(S_2\cap S_3)$
I am unsure on how to continue this proof or if my steps above are even correct. Any help or insights are greatly appreciated. Thank you!
 A: Fix $x\in (S_1\setminus S_2)\cup (S_2\setminus S_3)$.
Suppose first that $x\in S_1\setminus S_2$. Then $x\in S_1$, and so $x\in S_1\cup S_2$. Also, $x\not\in S_2$, so $x\not \in S_1\cap S_2\cap S_3$. Therefore, $x\in (S_1\cup S_2)\setminus (S_1\cap S_2 \cap S_3)$.
The case $x\in S_2\setminus S_3$ is analogous.
A: You can also use set algebra to simplify the inclusion you have to prove. Recall that by definition $A\backslash B=A\cap B^c$.
So we have
\begin{equation}
\begin{split}
(S_1\cup S_2)\backslash(S_1\cap S_2\cap S_3)&=(S_1\cup S_2)\cap(S_1\cap S_2\cap S_3)^c\\&=(S_1\cup S_2)\cap({S_1}^c\cup {S_2}^c\cup {S_3}^c)
\end{split}\tag{1}
\end{equation}
And for the other set,
\begin{equation}
\begin{split}
(S_1\backslash S_2)\cup(S_2\backslash S_3)&=(S_1\cap {S_2}^c)\cup(S_2\cap {S_3}^c)\\
&=(S_1\cup S_2)\cap(S_1\cup {S_3}^c)\cap({S_2}^c\cup S_2)\cap ({S_2}^c\cup {S_3}^c)\\
&=(S_1\cup S_2)\cap\big((S_1\cap {S_2}^c)\cup{S_3}^c\big)
\end{split}\tag{2}
\end{equation}
So from (1) and (2), it suffices to prove that $$\big((S_1\cap {S_2}^c)\cup{S_3}^c\big)\subseteq({S_1}^c\cup {S_2}^c\cup {S_3}^c)$$
hint: $(S_1\cap {S_2}^c)\subseteq S_2^c$
