Proof Involving De Morgan's Law and Cartesian Product of Sets Please check/critique the following proof. I think it is correct, but a bit verbose/overexplained.
Let $A$ and $B$ be sets. Show, in general, that $\overline{(A \times B)} \neq \overline{A} \times \overline{B}$.
Let $(x,y)\in \overline{A \times B}$
$\implies (x,y)\not\in A \times B$
$\implies \lnot ((x,y) \in A \times B)$
$\implies \lnot(x\in A \land y\in B)$
$\implies x\not\in A \lor y\not\in B$ 
$\implies (x \in A \land y\not\in B) \lor (x\not\in A \land y\in B) \lor (x\not\in A \land y\not\in B)$
Let $(x \in A \land y\not\in B) \neq (x\not\in A \land y\not\in B) = (x\in \overline{A} \land y\in \overline{B}) = (x,y)\in \overline{A} \times \overline{B}$.
Thus, $\overline{A \times B} \not\subseteq \overline{A} \times \overline{B}$ and $\overline{A \times B} \neq \overline{A} \times \overline{B}$.
Thanks
 A: Pick a point $a\in A$ and a point $c\notin B$ 
The pair $(a,c)$ is in $\overline{(A \times B)}$ but it is not in $\bar{A} \times \bar{B}$
Thus the two sets are not necessarily equal. 
A: Give a concrete example, that's enough to refute a general statement.
So take $X=\{1,2\}$, $Y=\{3,4\}$, $A=\{1\}$, $B=\{3\}$ (keep it small and simple if you can.) Then
$A \times B=\{(1,3)\}$ so $\overline{A \times B} = \{(1,4),(2,3),(2,4)\}$.
$\overline{A}=\{2\}, \overline{B}=\{4\}$ so $\overline{A} \times \overline{B} = \{(2,4)\}$. It's clear that these sets are unequal: $(1,4) \in \overline{A \times B}$ while $(1,4) \notin \overline{A} \times \overline{B}$.
What is true in general: $\overline{A} \times \overline{B} \subseteq \overline{A \times B} $, which is easy to see: $(x,y) \in \overline{A} \times \overline{B}$ means that $x \notin A$ and $y \notin B$ so certainly $(x,y) \notin A \times B$ or equivalently $(x,y) \in \overline{ A \times B}$, showing the inclusion.
Now try to show in general that for $A \subseteq X, B \subseteq Y$ we have
$$\overline{A \times B}=(\overline{A} \times Y) \cup (X \times \overline{B})$$
and maybe try to generalise to larger products.
A: Note : By  S' is meant here : the complement of set S. 
We'll try to show that the two sets are not identical ingeneral, since they are only identical in the special case where A and B are the same set. 



*

*The complement of the cartesian product A cross B is 


{ (x,y) | x does not belong to A OR y does not belong to B } 
Note : this can be explained using DeMorgan's law. 


*

*The cartesian product A' cross B' is the set 


{ (x,y) | x does not belong to A AND y does not belong to B} 


*

*One can see that in the defining formulas of the sets in question, the same propositions are involved, but linked by " OR" in the first case, and by "AND" in the second case. 

*Let : X = x does not belong to A , and Y = y does not belong to B. 

*The question is : in which case "X OR Y" is identical ( or equivalent) "X AND Y"; in other words, in which cases do these  2 expressions have the same truth value. 

*A truth table will show that " X OR Y" and " X AND Y" have the same truth value just in case X and Y are both true , or both false, that is, just in case X and Y are equivalent. 

*But the equivalence of X and of Y is tentamount to the equivalence of their negations, namely, of " x belongs to A" and " y belongs to Y". 

*Conclusion : the 2 sets we are talking about ( "complement of A cross B" and " complement of A cross complement of B") are equal ( identical) if and only if set A and set B are identical. 

