Sub-balanced H-polyhedra inherit diameter from equality sections

Let $P\subseteq\mathbb R^d$ be an H-polyhedron defined by $n$ inequalities, with $n<2d$. Then any two vertices share a tight inequality: an extreme point is tight on at least $d$ rows (otherwise a line through it stays feasible), so two tight-row sets of size at least $d$ in a universe of size $n<2d$ cannot be disjoint. This is the complement of the spindle constraint $n\ge 2d$ for complementary tight sets.

Consequently, if every equality section $P\cap\{\langle a_i,x\rangle=b_i\}$ has padded diameter at most $B$, then so does $P$:

The walk between two vertices may be taken inside their shared equality section, whose edges remain edges of $P$. This reduces the sub-balanced regime $n<2d$ to diameter bounds on facets. It does not treat the critical balanced case $n=2d$, and it does not prove a uniform polynomial in $(n+d)$.

Formalization Note Boundedness and simplicity are not assumed. Tight-row cardinality at least $d$ holds for every extreme point of a finite H-presentation in $\mathbb R^d$.