(Q25) of v1.91 asks, for the one-sided re-hangings of a transit leaf under the deletion bound two, for conditions or procedures that decide the absence of a step without the full search over the sets B of tree pairs of v1.91 Theorem ϝ2. This text does not settle (Q1) of v1.76, the two-deletion chain question, or (Q22) of v1.88; it gives a procedure that updates in turn the pairs every feasible set must hold and the pairs no feasible set may hold, proves what its decisions mean, and leaves its completeness open. For a fixed side and deletion set D, with (N1) and (N3) of v1.89 holding, a set B ⊆ U of tree pairs is feasible when r ∪ B is strongly connected on V' (r = H ∪ {v}) and B holds no obstruction (v1.91); by v1.91 Theorem ϝ2 a step exists exactly when a feasible set exists. The forcing procedure keeps a state (A, M), M ⊆ A ⊆ U, from (U, ∅), and applies in rounds four rules: (a) r ∪ A not strongly connected: impossible; (b) a pair c of A whose removal leaves r ∪ (A minus {c}) not strongly connected is forced into M; (c) a second route for a pair e of H in H_e ∪ M: impossible; (d) a pair c with which M closes such a route is forbidden and taken out of A; it stops when a round changes nothing, with success if r ∪ M is strongly connected and undecided otherwise. Lemma ϟ1: every feasible set B keeps M ⊆ B ⊆ A, and the two decisions of impossibility are sound. Theorem ϟ2: the procedure stops after at most |U| + 1 rounds, with at most (|U| + 1)^2 + 1 tests of strong connectivity and at most (|U| + 1)^2 |H| searches of reachability; at a success M is feasible, is held by every feasible set and has no proper subset keeping strong connectivity, so that H ∪ M itself gives a step, with nothing taken away from M; an undecided stop decides nothing. Proposition ϟ3: the outcome does not depend on the order of the rules, and when it is not "impossible" neither does the final state. Corollary ϟ4: the first round decides "impossible" exactly where (N4) or (N5) fails, the pairs first forced being the set Q of v1.91. These statements hold for every finite node set. The procedure itself tries no set B; it is not shown to be complete. Finite checks, which try every set B to compare: on all minimal relations of four nodes (decided directly), on all of five nodes (the step taken as the existence of a feasible set, v1.91 Theorem ϝ2, with no direct test), on the records of v1.88 at six nodes (364 instances, 534 leaves and 7492 choices of leaf, side and deletion set; decided directly; not every relation on six nodes) and in the sample on six nodes of v1.91 at its 19509 choices with (N1)-(N4) (decided directly; its other 21897 choices are not decided), every choice without a step stops as impossible, every success has a step, and no undecided stop without a step is found. The 73 choices of that sample where (N5) holds and no step exists stop as impossible by rule (a) in the second round, the pairs forbidden in the first round leaving a cut with no pair, as in example B of v1.91, whose certificate has five lines; the 16 choices of v1.90 stop in the first round by rule (c), (N5) failing there. These are findings of finite cases, not general statements; the 21192 residual instances are not enumerated or run again here.
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Koji Okino (2026) studied this question.
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