Set Difference of Cartesian Products

Theorem
$$\left({S_1 \times S_2}\right) - \left({T_1 \times T_2}\right) = \left({S_1 \times \left({S_2 - T_2}\right)}\right) \cup \left({\left({S_1 - T_1}\right) \times S_2}\right)$$

Proof
Let $$\left({x, y}\right) \in \left({S_1 \times S_2}\right) - \left({T_1 \times T_2}\right)$$.

Then:

$$ $$ $$ $$ $$ $$ $$

The argument is completely reversible, so given that $$\left({x, y}\right) \in \left({S_1 \times \left({S_2 - T_2}\right)}\right) \cup \left({\left({S_1 - T_1}\right)\times S_2}\right)$$, it follows that $$\left({x, y}\right) \in \left({S_1 \times S_2}\right) - \left({T_1 \times T_2}\right)$$.

The result follows from the definition of subset and set equality.