= 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.
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