Prove that for any sets $A$, $B$, and $C$, $A×(B\setminus C) = (A×B)\setminus (A×C)$. This is Velleman's exercise 4.1.8 ("$×$" means the Cartesian product):
Prove that for any sets $A$, $B$, and $C$, $A×(B\setminus C) = (A×B)\setminus (A×C)$.
Cartesian product of $A$ and $B$, denoted $A × B$ is equal to $\{(a, b) | a ∈ A and b ∈ B\}$ ("$(a, b)$" is an ordered pair).
And here's my proof of it:
Proof. 
($\rightarrow$) Let $(x, y)$ be an arbitrary element of $A×(B\setminus C)$, then $x ∈ A$ and $y ∈ (B\setminus C)$ which means that $y ∈ B$ and $y ∉ C$. From $x ∈ A$ and $y ∈ B$, we get $(x, y) ∈ (A×B)$. From $x ∈ A$ and $y ∉ C$, we get $(x, y) ∉ (A×C)$. Thus $(x, y) ∈ (A×B)\setminus (A×C)$. Therefore $A×(B\setminus C) ⊆ (A×B)\setminus (A×C)$.
($\leftarrow$) Let $(x, y)$ be an arbitrary element of $(A×B)\setminus (A×C)$, then $(x, y) ∈ (A×B)$ which means that $x ∈ A$ and $y ∈ B$ and $(x, y) ∉ (A×C)$ which means that either $x$ is not an element of $A$ or $y$ is not an element of $C$ or both. But since we saw that $x ∈ A$, therefore $y ∉ C$. From $y ∈ B$ and $y ∉ C$, we get $y ∈ (B\setminus C)$. From $x ∈ A$ and $y ∈ (B\setminus C)$, we get $(x, y) ∈ A×(B\setminus C)$. Therefore $(A×B)\setminus (A×C) ⊆ A×(B\setminus C)$.
From ($\rightarrow$) and ($\leftarrow$), we get $A×(B\setminus C) = (A×B)\setminus (A×C)$.
Is my proof correct (particularly "From $x ∈ A$ and $y ∉ C$, we get $(x, y) ∉ (A×C)$" in the forward direction)? 
 A: 
($\rightarrow$) Let $(x, y)$ be an arbitrary element of $A×(B\setminus C)$, then $x ∈ A$ and $y ∈ (B\setminus C)$ which means that $y ∈ B$ and $y ∉ C$. From $x ∈ A$ and $y ∈ B$, we get $(x, y) ∈ (A×B)$ $(x,y)\in A\times B$. From $x ∈ A$ and $y ∉ C$, we get $(x, y) ∉ (A×C)$. (Enough to say "From y ∉ C, ...") Thus $(x, y) ∈ (A×B)\setminus (A×C)$. Therefore $A×(B\setminus C) ⊆ (A×B)\setminus (A×C)$.
($\leftarrow$) Let $(x, y)$ be an arbitrary element of $(A×B)\setminus (A×C)$, then $(x, y) ∈ (A×B)$ (Redundant parentheses) which means that $x ∈ A$ and $y ∈ B$ and $(x, y) ∉ (A×C)$ (Redundant parentheses) which means that either $x$ is not an element of $A$ or $y$ is not an element of $C$ or both. But since we saw that $x ∈ A$, therefore $y ∉ C$. From $y ∈ B$ and $y ∉ C$, we get $y ∈ (B\setminus C)$ (Redundant parentheses). From $x ∈ A$ and $y ∈ (B\setminus C)$, we get $(x, y) ∈ A×(B\setminus C)$. Therefore $(A×B)\setminus (A×C) ⊆ A×(B\setminus C)$.
From ($\rightarrow$) and ($\leftarrow$), we get $A×(B\setminus C) = (A×B)\setminus (A×C)$.

(Just several redundant parentheses for sets. Your proof is correct.)
