 p-deficiency at least one implies infinitude

If a group presentation has unshifted p-deficiency at least one, the p-rank of a group is at least one and gives a normal subgroup of index $p$. The index-p rewriting bound for p-deficiency gives this subgroup another presentation with p-deficiency at least one. Iterating produces subgroups of index $p^j$ for every $j$, so the group is infinite. This criterion applies even with infinitely many relators, provided the weighted sum in the definition converges.