shiftUp_neg

Phase 59 studies integer-rung updates that the live cost meter actually charges. The engine does not move rungs by real means; it raises or lowers a region by exactly one rung when a distinction is forced. The dual of descent is the upward shift: on a decidable set $S$, every $v \in S$ is sent to $k(v)+1$ and every $v \notin S$ is left alone.

The invariant being protected is the forced minimal-distinction property UnitStep: across every edge, the rung gap lies in ${-1,0,+1}$. A unit recognition step resolves at most one $\varphi$-rung, and T-3 descends (or raises) one rung at a time, so the cost law of Phase 56 applies only while this holds.

This lemma is the outside-set evaluation rule for the upward shift. Its sibling for the inside set adds one; together they feed the cut-edge case analysis that proves bottom-rung raise preserves unit-step.

One-line term proof. The upward shift is defined by if S v then k v + 1 else k v. Under the hypothesis $\neg S(v)$, if_neg selects the else branch and yields equality with $k(v)$. No further lemmas are needed.

Feeds shiftUp_bot_unitStep, the dual of top-rung descent: if $m$ lower-bounds every rung on the edge set and $S$ is the bottom level ${v : k(v)=m}$, raising $S$ by one preserves UnitStep unconditionally. A bottom cell's neighbour is forced one rung up, so after the raise every cut gap collapses to $0$.

That theorem is the positive half of Phase 59 on the raise side. Together with top-rung descent and the necessity counterexample (descending a non-top cell can open a gap of $2$), it closes the item that the active dynamics preserves the graded-rung invariant cycle to cycle along the forced extremal moves, so the Phase-56 cost law remains theorem-backed. The present lemma is the trivial outside-set arm of that case split.