Proving $(A - B) \times (C - D) = (A \times C - B \times C) - A \times D$

Question

I am trying to prove $(A - B) \times (C - D) = (A \times C - B \times C) - A \times D$ using biconditionals, but cannot quite get there. For any ordered tuple $(\alpha, \beta)$, we have: \begin{align*} (\alpha, \beta) \in (A - B) \times (C - D) & \iff \alpha \in (A - B) \land \beta \in (C - D) \\ & \iff (\alpha \in A \land \alpha \not \in \beta) \land (\beta \in C \land \beta \not \in D) \\ & \iff ((\alpha \in A \land \alpha \not \in \beta) \land \beta \in C) \land (\beta \not \in D) \\ & \iff ((\alpha \in A \land \alpha \not \in \beta) \land \beta \in C) \land (\alpha \in A \land \beta \not \in D) \\ & \iff ((\alpha \in A \land \beta \in C) \land (\alpha \not \in B \land \beta \in C)) \land (\alpha \in A \land \beta \not \in D) \\ & \iff (((\alpha, \beta) \in A \times B) \land (\alpha \not \in B \land \beta \in C)) \land (\alpha, \beta) \not \in A - D. \end{align*} At this point, I do not know how to finish. $\alpha \not \in B$ and $\beta \in C$ does imply that $\alpha \not \in B \times C$, but the converse isn't true since we may have $\beta \not \in C$.

Answer 1

Very ugly, but here is: \begin{align} (\alpha,\beta) \in [(A \times C) &\setminus (B \times C)] \setminus (A \times D) \\ &\Leftrightarrow \quad (\alpha,\beta) \in (A \times C) \setminus (B \times C) \ \ \& \ \ (\alpha,\beta) \notin A \times D \\ &\Leftrightarrow \quad (\alpha,\beta) \in A \times C \ \ \& \ \ (\alpha,\beta) \notin B \times C \ \ \& \ \ (\alpha,\beta) \notin A \times D \\ &\Leftrightarrow \quad [\alpha \in A \ \ \& \ \ \beta \in C] \ \ \& \ \ [\alpha \notin B \ \ \vee \ \ \beta \notin C] \ \ \& \ \ [\alpha \notin A \ \ \vee \ \ \beta \notin D] \\ &\Leftrightarrow \quad [\alpha \in A \ \ \& \ \ \beta \in C] \ \ \& \ \ [\alpha \notin B] \ \ \& \ \ [\beta \notin D] \\ &\Leftrightarrow \quad (\alpha,\beta) \in (A \setminus B) \times (C \setminus D). \end{align}

Note : $\&$ stands for $``\textrm{ and}"$ and $\vee$ for $``\textrm{or}"$.

Answer 2

Hint

For convenience, I will use $a,b,c,d$ as appreviations for the logical statements $\alpha\in A$, $\alpha\in B$, $\beta \in C$ and $\beta\in D$.

It turns out it is much easier to start with $(\alpha,\beta)\in (A\times C-B\times C)-A\times D$ and show it is equivalent to the other side.

\begin{align} (\alpha,\beta)\in (A\times C-B \times C)-A\times D &\iff a\wedge c\wedge \neg(b\wedge c)\wedge \neg (a\wedge d)\\ &\iff a\wedge c\wedge (\neg b\vee \neg c)\wedge (\neg a\vee \neg d) \end{align} Next, use the the fact that $\wedge$ distributes over $\vee$ to expand out the above statement, in the same way you would expand $x\cdot z\cdot (-y-z)\cdot (-x-w)$ to a sum of four terms each with three variables. You will find many "cancellations."