Suppose $A$, $B$, and $C$ are sets. Prove that $A ∪ C ⊆ B ∪ C$ iff $A \setminus C ⊆ B \setminus C$. This is Velleman's exercise 3.5.6:
Suppose $A$, $B$, and $C$ are sets. Prove that $A ∪ C ⊆ B ∪ C$ iff $A \setminus C ⊆ B \setminus C$.
And here's my proof of it:
Proof. 
($\rightarrow$) Suppose $A ∪ C ⊆ B ∪ C$. Let $x$ be arbitrary and $x ∈ A$ and $x ∉ C$. From $x ∈ A$ and $A ∪ C ⊆ B ∪ C$ we have $x ∈ B ∪ C$. Since $x ∉ C$, then $x ∈ B$. Since $x$ was arbitrary, $A \setminus C ⊆ B \setminus C$.
($\leftarrow$) Suppose $A \setminus C ⊆ B \setminus C$. Let x be arbitrary and $x ∈ A ∪ C$. Now suppose $x ∉ C$. From $x ∈ A$, $x ∉ C$ and $A \setminus C ⊆ B \setminus C$, we get $x ∈ B$. Ergo $x ∈ B \lor x ∈ C$. Since $x$ was arbitrary, $A ∪ C ⊆ B ∪ C$.
Since $A ∪ C ⊆ B ∪ C$ $\Rightarrow$ $A \setminus C ⊆ B \setminus C$ and $A ∪ C ⊆ B ∪ C$ $\Leftarrow$ $A \setminus C ⊆ B \setminus C$, therefore $A ∪ C ⊆ B ∪ C$ $\iff$ $A \setminus C ⊆ B \setminus C$.
Is my proof valid?
Thanks in advance.
 A: Well, note that If $X \subset Y$ then $X\cup C \subset Y\cup C$ and $X\setminus C \subset Y \setminus C$
That's just common sense as $X\cup C$ and $X \setminus C$ are the elements of $X$ with the elements of $C$ thrown in or removed, and if $X \subset Y$ then tossing in the same to both sets or removing the same from both sets will keep them subsets.
To prove it.  If $X \subset Y$ and $x \in X \cup Y$ either $x \in X$ or $x \in C$ so either way, $x \in Y\cup C$.  And if $z \in X\setminus C$ then $z \in Y$ and $z \not \in C$ so $z \in Y \setminus C$.
Furthermore $X \setminus C = (X \cup C)\setminus C$.
That is also common sense.  $X \setminus C = $ $X$ with the elements of $C$ taken out = $X$ with the elements of $C$ put in and then immediately taken out.
To prove it.  $x \in (X\cup C)\setminus C$ then $x \in X $ or $x \in C$ but $x \not \in C$ so $x \in X$.  But $x \not \in C$ so $x \in X \setminus C$.  On the other hand if $x \in X$ then $x \in X \cup C$.  But if $x \not \in C$ then $x \not \in C$ ... so $x \in (X\cup C)\setminus C$.
So the result follows immediately...
So $A\cup C \subset B\cup C \implies  C = [A\cup C]\setminus C \subset [B\cup C]\setminus C \implies A\setminus C \subset B \setminus C$.
Of $A\setminus C \subset B \setminus C \implies (A\setminus C)\cup C \subset (B \setminus C) \cup C\implies A \cup C \subset B \cup C$.
BUT.. it is very important to notice that neither $A\cup C \subset B\cup C$ nor $A\setminus C \subset B\setminus C$ imply $A \subset B$.  (Although the implication does work the other way.)  That is because it is not true that $A \ne A \setminus C$ and $(A \setminus C) \cup C \ne A$
A: Yes this is correct. Well done.
