Question: Suppose $F$ and $G$ are families of sets. Prove that $\bigcup F$ and $\bigcup G$ and are disjoint iff for all $A∈F$ and $B∈G$ , $A$ and $B$ are disjoint.

It has been suggested to use contrapositive idea
$$P→Q \\ ¬Q→¬P$$

But I'm not sure how to apply it to this problem

Answer: 
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Formally, you're asked to prove that $\bigcup F \cap \bigcup G = \emptyset$ is equivalent to $\langle \forall A,B :: A \cap B = \emptyset \rangle$, where I've left $\;A \in F\;$ and $\;B \in G\;$ implicit, which will reduce the visual noise in the rest of this answer.

Here is how I would prove this: expand the definitions, simplify, and then try to reach the goal from there. Let's start from the former expression, and see where this approach leads us.

$$\calc
\bigcup F \cap \bigcup G = \emptyset
\calcop{\equiv}{definitions of $\;\emptyset\;$ and $\;\cap\;$}
\lnot \langle \exists x :: x \in \bigcup F \land x \in \bigcup G \rangle
\calcop{\equiv}{definition of $\;\bigcup\;$, twice, leaving $\;A \in F\;$ and $\;B \in G\;$ implicit}
\lnot \langle \exists x :: \langle \exists A :: x \in A \rangle \;\land\; \langle \exists B :: x \in B \rangle \rangle
\calcop{\equiv}{logic: simplify by merging independent quantifications}
\tag{\*} \lnot \langle \exists x,A,B :: x \in A \land x \in B \rangle
\calcop{\equiv}{logic: split quantifications -- to get $\;A,B\;$ at top level}
\lnot \langle \exists A,B :: \langle \exists x :: x \in A \land x \in B \rangle \rangle
\calcop{\equiv}{logic: DeMorgan -- to prepare for the next step}
\langle \forall A,B :: \lnot \langle \exists x :: x \in A \land x \in B \rangle \rangle
\calcop{\equiv}{definitions of $\;\cap\;$ and $\;\emptyset\;$}
\langle \forall A,B :: A \cap B = \emptyset \rangle
\endcalc$$

That completes the proof that was requested.

Note that $\text{(\*)}$ is the point where we have done all simplifications, and start working towards our goal.
**Hint:**

The contrapositive of that statement is:

> 
> Suppose $F, G$ are families of sets. If $\exists \; A \in F, B \in G$ so that $A$ and $B$ are not disjoint iff $\bigcup F$ and $\bigcup G$ are not disjoint.
> 
> 
> 

A *somewhat* accurate way to create the contrapositive that has helped me in the past is by changing all the $\forall$ to $\exists$, flipping the order of the statement and changing *are* (*is*) to *are not* (*is not*).

In this case that above blanket statement doesn't quite work since we don't say that $\exists A \not\in \ldots$ we say $\exists A \in \ldots$. So let me try to explain how I found the contrapositive more accurately:

First notice that the statement

> 
> Let $F,G$ be families of sets.
> 
> 
> 

needs to stay the same. For this entire problem we are interested, intuitively, given two families if sets whether or not the union of the families being disjoint has anything to do with the elements of the families being disjoint, so no matter the way which we prove this statement we will need to work with families of sets.

Next we can put our *hypothesis* (I put this in italics because I'm unsure if this is the correct terminology) $P$ to be the statement:

> 
> $\bigcup F, \bigcup G$ are disjoint
> 
> 
> 

The logical opposite of this statement can be seen to be that these two sets $\bigcup F, \bigcup G$ are not disjoint so we form $\lnot P$ to be

> 
> $\bigcup F, \bigcup G$ are not disjoint
> 
> 
> 

Now our second statement $Q$ can be considered to be

> 
> $\forall [ A \in F, B \in G ]$ are disjoint
> 
> 
> 

Negating this intuitively we want to show that this statement is false. This statement will be false if $\exists$ an $A \in F, B \in G$ so that $A,B$ are not disjoint. So now we find $\lnot Q$ to be

> 
> $\exists [ A \in F, B \in G ]$ where $A,B$ are not disjoint
> 
> 
> 

Finally we flip the order of the negated statements to arrive at

> 
> Let $F,G$ be families of sets. $\exists [ A \in F, B \in G ]$ where $A,B$ are not disjoint iff $\bigcup F, \bigcup G$ are not disjoint.
> 
> 
> 

Now pay attention to the fact that this is not simply an implication but an *iff* statement. That means we need to show both
$$
P \to Q \land Q \to P
$$
Since we have that
$$
\lnot Q \to \lnot P \iff P \to Q
$$
we can equivalently prove
$$
\lnot Q \to \lnot P \land \lnot P \to \lnot Q
$$