Generative Structural Theory v1.93 — Beyond Forced and Forbidden Pairs: Feasibility at Undecided Stops

(Q26) of v1.92 asks whether there is a choice of leaf, side and deletion set D, with (N1) and (N3), where the forcing procedure of v1.92 stops undecided and no feasible set exists. This text answers it affirmatively: example C, on seven nodes, is such a choice, with a symbolic proof (Theorem ϡ3). The hypothesis of completeness, that an undecided stop always has a feasible set, is refuted, and an undecided stop cannot be read as the existence of a step. This text does not settle (Q1) of v1.76, the two-deletion chain question, or (Q22) of v1.88. At a state (A, M) of the procedure every feasible set is held between M and A, and with R = A ∖ M the feasible sets are the sets M ∪ C, C ⊆ R, where C has a pair leaving every cut of r ∪ M and holds no relative obstruction, an inclusion-minimal set O ⊆ R with which M closes a second route for a pair of H (Lemma ϡ1). At an undecided stop every cut of r ∪ M has at least two candidates in R and every relative obstruction at least two pairs, so that every constraint left is a joint one (Proposition ϡ2). In example C two cuts have two candidates each, and each of the four transversals closes a second route (Theorem ϡ3). Probing a cut W, the procedure run again from (A, M ∪ {c}) for each candidate c of W, is sound: if every run ends impossible no step exists, and a success gives a step (Proposition ϡ4); it is not shown to be complete. Every minimal strongly connected relation has a node of in-degree and out-degree one, which gives the classes of minimal relations on n nodes from those on n - 1 (Lemma ϡ5). These statements hold for every finite node set; Theorem ϡ3 is a statement about one choice. Finite checks, which try every set C ⊆ R at each undecided stop: on all minimal relations of four and five nodes with every labelled tree, the counts of v1.92 are found again and every undecided stop has a feasible set. On six and seven nodes the population is the pairs (h, S) of an isomorphism-class representative h of the minimal strongly connected relations (63 classes on six nodes, 288 on seven, made by Lemma ϡ5) and every labelled tree S: 63 · 6^4 = 81648 and 288 · 7^5 = 4840416 pairs, not classes of pairs up to isomorphism (the check meant to take the trees of one orbit of the automorphisms of h once, but that removal of repetitions did not work, and every labelled tree was taken). On six nodes, 315500 choices with (N1) and (N3) and 27762 undecided stops, every one with a feasible set; on seven, 27736992 choices and 2339392 undecided stops, of which 1178 have no feasible set, each with two or three cuts whose candidate sets have every transversal holding a relative obstruction. The 1178 are inputs counted in this population, not up to isomorphism; they are not 1178 non-isomorphic counterexamples. So, in the class of inputs taken, seven is the least number of nodes of such a choice: a computational result of the exhaustive checks on three to six nodes, kept apart from the symbolic proof of example C. The probing rule decides "impossible" at each of the 1178, and leaves undecided 936 undecided stops on six nodes and 85168 on seven, each with a feasible set: it is sound, and it is not a complete decision. These are findings of finite cases, not general statements.