Answer
See the solution.
Work Step by Step
We know for propositions $p$ and $q$, the statement "if $p$, then $q$, is equivalent to $q$ unless $\lnot p$. Hence to prove the statement in the problem, we will prove for nonempty sets $A$ and $B$, if $A \neq B$, then $A \times B \neq B \times A$.
$Proof.$
Let $A$ and $B$ be nonempty sets, and suppose $A\neq B$. Then either $A$ is not a subset of $B$ or $B$ is not a subset of $A$. First, suppose $A$ is not a subset of $B$. Then there exists an element $a$ in $A$ such that $a \notin B$. Also, since $B$ is nonempty, there exists an element $b$ in $B$. This means the ordered pair $(a,b)$ is in $A\times B$, but, since $a \notin B$, $(a,b)$ cannot be in $B \times A$. Hence $A \times B \neq B \times A$. Next, suppose $B$ is not a subset of $A$. Then there exists an element $b$ in $B$ such that $b \notin A$. Also, since $A$ is nonempty, there exists an element $a$ in $A$. This means the ordered pair $(b,a)$ is in $B\times A$, but, since $b \notin A$, $(b,a)$ cannot be in $A \times B$. Hence $A \times B \neq B \times A$. Thus we have shown if $A \neq B$, then $A \times B \neq B \times A._\Box$
