It can be easily proved.
Let $T(x)$ denotes $x$ is a thief and $P(x)$ denotes $x$ is poor, then All thieves are poor is $\forall x(P(x) \to T(x))$
Since given statement is false, take the negation
$\neg ( \forall x(P(x) \to T(x))) = \exists x(\neg(P(x) \to T(x))$ // Rewrite $A \to B$ as $A' \vee B$
$= \exists x(\neg (P(x)' \vee T(x)))$
$=\exists x(P(x) \wedge T(x)')$
Which means " there are some thieves who are not poor"
option 2