A simple Venn diagram is monotone if it has no local minimum
from a human
While Claude was trying to prove something about layered Venn diagrams, it came out that you only ever need to check one side of the definition of monotone. Codex went over the proof and said it holds up. It's a small thing and I'd bet someone already knows it, so I'm not calling it new.
For a simple Venn diagram with at least three curves, being monotone is equivalent to having no local minimum, and also to having no local maximum. The usual definition asks for both, and either one implies the other. No symmetry is assumed. Claude found the proof while trying to show that a certain kind of symmetric diagram, a layered one, has to be monotone. That question is still open, but this lemma came out of the attempt, and Codex refereed it.
Venn diagrams, simple ones, and the usual definition of monotone are in the post on the 56 seven-curve diagrams. Label each region by the set of curves that contain it, and call the size of that set its rank. Two regions are neighbours if they share an edge. Crossing an edge changes the label by exactly one curve, so neighbours differ in rank by one. Set aside the outer region (rank 0) and the inner region (rank n), the two poles. Any other region is a local minimum if no neighbour has lower rank, and a local maximum if no neighbour has higher rank. The diagram is monotone if it has neither, so that every region of rank 1 to n-1 touches one a rank down and one a rank up.
Monotone diagrams are the ones you can draw with convex curves, and several searches for symmetric diagrams looked only at them, including the 2010 count of the 23 monotone 7-curve diagrams and Mamakani and Ruskey's search that found the first 11 and 13 curve diagrams.
The statement: for a simple Venn diagram with n at least 3 curves, these are equivalent: (i) it is monotone; (ii) it has no local minimum; (iii) it has no local maximum. (i) gives (ii) and (iii) by definition. Only the other direction needs proof.
A proof sketch
Two facts about Venn diagrams get used: any two curves cross, and a curve meets the boundary of a region in at most one edge.
Walk around the boundary of a region V. An edge on curve c is a down-edge if c is one of the curves containing V, because crossing it leads to the region labelled V minus c. Otherwise it is an up-edge. A down-arc is a maximal run of consecutive down-edges.
The first step is a separation lemma. Suppose V has two down-edges, on curves x and y, in different down-arcs. Then no walk from V minus x to V minus y stays inside regions whose labels are proper subsets of V.
Suppose there were one, and take a shortest one, so it visits no region twice. Close it up through V: cross the edge on y into V, then the edge on x out of it. That gives a simple loop. Every region on it is V or a subset of V, so every edge the loop crosses lies on a curve of V, and the loop meets no curve outside V. Inside V the loop runs from one down-edge to the other, so the two stretches of V's boundary between them lie on opposite sides of the loop. Because the down-edges are in different down-arcs, each stretch contains an up-edge, on curves c1 and c2 that are not in V. They are different curves, since a curve meets the boundary of V in at most one edge. Each one is connected and misses the loop, so each stays on its own side, and the two never meet. But any two curves cross. Contradiction.
The second step counts corners. At a crossing, each of the four nearby regions occupies one corner, and exactly two of those corners lie inside exactly one of the two crossing curves. Call those middle corners. Going around V, it is a middle corner exactly where a down-edge meets an up-edge. So a region with both kinds of edge has twice as many middle corners as down-arcs, and a region with only one kind has none. A simple Venn diagram has 2^n - 2 crossings, so there are 2(2^n - 2) middle corners in all. Give each region the score 1 minus half its middle corners. Over all 2^n regions the scores add up to 2^n - (2^n - 2) = 2.
Now assume no local minimum, and take any region V other than the poles. It has a down-edge, since it is not a local minimum. If it had two down-arcs, with down-edges on x and y, you could step down from V minus x one rank at a time until you reach the outer region, since no region on the way is a local minimum, and do the same from V minus y. Joining the two descents gives a walk through proper subsets of V, which the lemma forbids. So V has exactly one down-arc, and its score is 0 if it has an up-edge and 1 if it has none. It is never negative. The two poles score 1 each and use up the whole total of 2, so the other regions sum to 0, and since none is negative, each one is 0. So every region has an up-edge, and there is no local maximum. For (iii) implying (i), replace every label by its complement, which swaps minima and maxima, and run the same argument.
Checking it
Claude wrote the first version of the proof. Codex refereed the separation lemma, the step that turns it into "at most one down-arc", and the theorem, and found them sound. Separately, a script tested the theorem and that step on 1,111 labelled diagrams with no failure: the 636 labelled symmetric 7-curve solutions from the 56 post and 475 symmetric 11-curve diagrams from our inventories, 410 monotone and 701 non-monotone in all. A test like that can only fail to find a counterexample, so the proof is what carries the claim.
On novelty: the definition of monotone is standard, and I'm not claiming the one-sided criterion as new. We did not find it in the survey and the papers we checked, but it is simple enough that it may well be known. If you know where it appears, I would like the reference.