Let $S$ be the set of triples of positive numbers $(x,y,s)$. Let $R$ be a directed relation defined between triples, such that $R(b,a)$ if at least one of the following four conditions hold:
$$a_y+a_s<b_y \bigwedge b_x<a_x \bigwedge a_x<b_x+b_s$$ $$a_x+a_s<b_x \bigwedge b_y<a_y \bigwedge a_y<b_y+b_s$$ $$a_y<b_y \bigwedge b_x<a_x \bigwedge a_x+a_s<b_x+b_s$$ $$a_x<b_x \bigwedge b_y<a_y \bigwedge a_y+a_s<b_y+b_s$$
I am trying to prove (or disprove) that $R$ is acyclic, i.e., there is no set of $k$ triples such that: $$R(a_1,a_2) \bigwedge R(a_2,a_3) \bigwedge ... \bigwedge R(a_k,a_1)$$
I managed to prove that a cycle of length 2 is impossible by way of contradiction: assume there is a cycle $R(a,b) \bigwedge R(b,a)$. Check all 16 combinations of conditions - a condition for the first relation and a condition for the second relation (some of the 16 can be easily ruled out so there are only 4 left to check). Each of these combinations leads to a contradiction.
However, when the length of the cycle grows, the number of different cases to check grows exponentially, since each leg of the cycle has 4 possible conditions...
Are there useful techniques to handle such relations?
NOTE: The triples have a geometric interpretation as squares: $x,y$ is the bottom-left corner and $s$ is the side-length, so the four conditions look like the following:
Eureka! After some playing with the formulas, I found that in each of the 4 conditions for $R(b,a)$:
$$ a_x+a_y+a_s < b_x+b_y+b_s$$
Thus, in any sequence of triples $R(a_1,a_2) \bigwedge R(a_2,a_3) ...$ the quantity $a_{ix}+a_{iy}+a_{is}$ is decreasing with $i$.
This makes it is clear that a cycle is impossible.