Not a duplicate of
Prove that $A \cup C \subseteq B \cup C$ iff $A \setminus C \subseteq B \setminus C$
Suppose $A$, $B$, and $C$ are sets. Prove that $A ∪ C ⊆ B ∪ C$ iff $A \setminus C ⊆ B \setminus C$.
This is exercise $3.5.6$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $A$, $B$, and $C$ are sets. Prove that $A\cup C\subseteq B\cup C$ iff $A\setminus C\subseteq B\setminus C$.
Here is my proof:
$(\rightarrow)$ Suppose $A\cup C\subseteq B\cup C$. Let $x$ be an arbitrary element of $A\setminus C$. This means $x\in A$ and $x\notin C$. Since $x\in A$, $x\in A\cup C$. From $A\cup C\subseteq B\cup C$ and $x\in A\cup C$, $x\in B\cup C$. Now we consider two different cases.
Case $1.$ Suppose $x\in B$. Since $x\notin C$, $x\in B\setminus C$.
Case $2.$ Suppose $x\in C$ which produces a contradiction.
From $x\in B\setminus C$ or a contradiction we obtain $x\in B\setminus C$. Thus if $x\in A\setminus C$ then $x\in B\setminus C$. Since $x$ is arbitrary, $\forall x(x\in A\setminus C\rightarrow x\in B\setminus C)$ and so $A\setminus C\subseteq B\setminus C$. Therefore if $A\cup C\subseteq B\cup C$ then $A\setminus C\subseteq B\setminus C$.
$(\leftarrow)$ Suppose $A\setminus C\subseteq B\setminus C$. Let $x$ be an arbitrary element of $A\cup C$. This means $x\in A$ or $x\in C$. Now we consider two different cases.
Case $1.$ Suppose $x\in C$. Ergo $x\in B\cup C$.
Case $2.$ Suppose $x\notin C$. From $x\in A\cup C$ and $x\notin C$, $x\in A$. Ergo $x\in A\setminus C$. Since $A\setminus C\subseteq B\setminus C$, $x\in B\setminus C$ which means $x\in B$. Thus $x\in B\cup C$.
Since the above cases are exhaustive, $x\in B\cup C$. Thus if $x\in A\cup C$ then $x\in B\cup C$. Since $x$ is arbitrary, $\forall x(x\in A\cup C\rightarrow x\in B\cup C)$ and so $A\cup C\subseteq B\cup C$. Therefore if $A\setminus C\subseteq B\setminus C$ then $A\cup C\subseteq B\cup C$.
Ergo $A\cup C\subseteq B\cup C$ iff $A\setminus C\subseteq B\setminus C$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.