Answer
The reordering is:
2. In contrapositive form: $\forall x$, if x is a square, then x is above all the black objects.
3. $\forall x$, if x is above all the black objects, then x is to the right of all the triangles.
1. $\forall x$, if x is to the right of all the triangles, then x is above all the circles.
Work Step by Step
Universal transitivity:
$\forall x, P(x) \rightarrow Q(x).$
$\forall x, Q(x) \rightarrow R(x).$
$\therefore \forall x, P(x) \rightarrow R(x).$
