Proving if a set is True or False: (A x B) $\cup$ (C x D) = (A $\cup$ C) x (B $\cup$ D) As the title suggests I'm trying to determine if the following example is true or false. AND if it is false, to provide a counter example. 
This is the example:
(A x B) $\cup$ (C x D) = (A $\cup$ C) x (B $\cup$ D)
Now here is what I have attempted:
For the L.H.S. (A x B) $\cup$ (C x D)
(x,y) $\in$ (A x B) OR (x,y) $\in$ (C x D) $\Leftarrow\Rightarrow$ x $\in$ A OR y $\in$ B OR x $\in$ C OR y $\in$ D
For the R.H.S. (A $\cup$ C) x (B $\cup$ D)
x $\in$ (A $\cup$ C) OR y $\in$ (B $\cup$ D) $\Leftarrow\Rightarrow$ (x,y) $\in$(A  $\cup$ C) x (B $\cup$ D) 
To be honest I'm not really sure if what I have done is even correct. Given my working I guess it would be a FALSE statement?
But a counter example? I just need the expertise on this forum to help me fill a few gaps. 
Thanks so much!
EDIT: It is slightly different from the linked solution as they asked
$(A \times B) \cup (C \times D) = (A \cup B) \times (C \cup D)$
And I asked
(A x B) $\cup$ (C x D) = (A $\cup$ C) x (B $\cup$ D)
I realise that's not much of a difference but just thought for peoples future reference. 

Here is your answer.
$(x,y) \in (A \times B) \cup (C \times D) \implies (x,y) \in (A \times B)\ or \ (x,y) \in (C \times D) \implies (x \in A\ and\ y \in B)\ or\ (x \in C\ and\ y \in D)$
On the other hand,
$(x,y) \in (A \cup C) \times (B \times D) \implies (x \in A\ or\ x \in C)\ and\ (y \in B\ or\ y \in D)$.
Now we use the splitting properties of the $and$ and the $or$ operators. We replace $and$ by $\cdot$ ,and $or$ by $+$. Now, let us use the following boolean variables:
$W := x \in A$
$X := y \in B$
$Y := x \in C$
$Z := y \in D$
Then the above statements are rewritten respectively as:
$$
WX+YZ=(WX+Y)(WX+Z)=(W+Y)(X+Y)(W+Z)(X+Z)
$$
and 
$$
(W+X)(Y+Z)
$$
As we see, $(W+Y)(X+Y)(W+Z)(X+Z)=(W+X)(Y+Z)$ is clearly not true all the time. So we have a good reason to assume that the statement above is false.
I decided to construct a counter example, in case you needed one: $A=\{1,2\}$,$B=\{1,4\}$,$C=\{3,7\}$,$D=\{1,3\}$. Not only is $(A \times B) \cup (C \times D) \neq (A \cup C) \times (B \times D)$, in fact their cardinalities are 8 and 12 respectively,  which are also not equal.
That brings me to the cardinalities of the  sets $(A \times B) \cup (C \times D)$ and $(A \cup C) \times (B \times D)$. If we let $a=|A|,b=|B|,c=|C|,d=|D|$, then the cardinality of the first set is less than or equal to $ab+cd$, while the cardinality of the second set is less than or equal to $(a+b)(c+d) > ab+cd$. If we find a set such that the cardinality of the second set is exactly $(a+b)(c+d)$, then we have a counterexample. My counterexample does not show this, but you can take it as an exercise and construct your own counterexample. I hope I have given you a different flavor of set theory through predicate calculus,and not merely confused you. Now you can attack any set theoretic problem using this cardinality bound method, or through direct predicate calculus.
