shiftDown_top_unitStep
plain-language theorem explainer
If an integer rung field is unit-step on an edge set and M upper-bounds every rung on those edges, descending every cell at rung M by one keeps the field unit-step. Cosmology and cost-meter arguments cite this as the unconditional top-rung case of the cut criterion. The proof feeds the cut lemma and closes the cut by forcing each top neighbour to sit exactly at M−1.
Claim. Let $k:V\to\mathbb{Z}$ be a rung field and $E$ a finite set of ordered edges. Suppose every edge changes $k$ by at most one (the unit-step property), and every endpoint of every edge in $E$ has rung $\le M$. Then the field obtained by subtracting $1$ from every vertex with $k(v)=M$ (and leaving all other vertices fixed) is again unit-step on $E$.
background
Phase 59 studies integer-rung descent as the positive complement to Phase 58. The live engine does not move rungs by real means; it descends a region by exactly one rung when a distinction is forced (T-3). The invariant that makes the Phase-56 graded cost law well-posed is UnitStep: every adjacency changes the integer rung by at most one, i.e. $k(p_1)-k(p_2)\in{0,+1,-1}$.
The elementary move is shiftDown S k, which sends every cell in a decidable set $S$ to $k(v)-1$ and leaves the complement fixed. The general preservation theorem is the cut criterion: descending $S$ preserves unit-step on all of $E$ exactly when every cut edge (one endpoint in $S$, one outside) still has gap in ${0,\pm 1}$ after the move. Interior and exterior edges are free, since both endpoints shift by the same amount or by none.
This declaration specialises that criterion to the top rung. Taking $S={v:k(v)=M}$ with $M$ an upper bound on every rung appearing in $E$ makes the cut condition automatic: a top cell cannot neighbour anything above it, and unit-step forbids a neighbour two or more below, so the neighbour sits at $M-1$.
proof idea
One application of the cut criterion shiftDown_unitStep_of_cut with $S={v:k(v)=M}$, followed by a case split that discharges hcut.
For a cut edge $p\in E$, the two cut orientations are handled symmetrically. If the first endpoint is in $S$ and the second is not, the upper bound plus $k(p_2)\ne M$ gives $k(p_2)<M$; unit-step then forces $k(p_2)=M-1$ by omega. Evaluating the shifted field via shiftDown_pos on the top endpoint and shiftDown_neg on the exterior endpoint yields values $M-1$ and $M-1$, so the new gap is $0$. The opposite cut orientation is identical with the roles reversed. Non-cut edges are already covered by the cut lemma.
why it matters
This is the unconditional top-rung discharge that Phase 59 needs. The module doc states the point directly: descending the coarsest (top) rung, the natural parameter-free relaxation move, provably keeps the field unit-step, so the Phase-56 cost law applies to the next cycle.
It is the first conjunct of the Phase-59 headline t59_rung_descent_preservation, which pairs this positive fact with the necessity counterexample on the chain $0,1,2$ (descending a non-top cell breaks unit-step). It is also the engine of exists_top_descent_unitStep: once $M$ is taken to be the realised maximum rung on the edge vertices, this theorem supplies a nontrivial preserving descent with no external parameter.
In the broader Recognition picture this closes the honest answer to invariant preservation under the active dynamics: the generic real mean move does not preserve unit-step globally (Phase 58), but forced top-rung integer descent does. That is exactly the update the cost meter charges when T-3 descends one rung at a time.
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