Let $C$ be an axis-aligned orthogonal polygon. Define an *anti-rectangle* in $C$ as a set of small squares in $C$, such that no two of them are covered by a single large rectangle in $C$. Define a *maximum anti-rectangle* as an anti-rectangle of maximal size. For example, the following L-shape has a maximum anti-rectangle of size 2:

![enter image description here][1]

When I try to build an anti-rectangle, I usually start with putting squares in the corners of $C$, because intuitively, the corners have the least chance of being covered by a rectangle. This raises the following conjecture:

**For every $C$, there is a maximum anti-rectangle, which contains a corner square.**

(- a *corner square* is a square with at least two adjacent sides that are in the boundary of $C$).

I found in [Chaikan et al (1981)](http://dx.doi.org/10.1137/0602042) a proof that the conjecture is true when $C$ is *linearly-convex* (- contains every vertical or horizontal line that connects two of its points; like the L-shape above). 

Their proof is not valid when $C$ is not linearly-convex, but I also cannot find a counter-example. What do you say?

  [1]: https://i.sstatic.net/FjvTp.png