We prove that every fair-division instance with four agents, additive valuations over the non-negative reals, and at most nine indivisible goods admits a \emph{complete} allocation that is envy-free up to any good in the strong, zero-tolerant sense ($\EFXo$). The case $m=9=n+5$ lies beyond the previously known frontier for complete EFX with four agents ($m\le n+3$). The proof combines a small set of hand-proven reduction lemmas with a machine-verified certificate corpus. The valuation polytope is covered by a collection of smaller polytopes. For each smaller polytope $P$, a family $F$ of allocations is found that contains an $\EFXo$ allocation for every valuation in $P$. The check that $F$ suffices for $P$ is a quantifier-free linear-arithmetic unsatisfiability verdict, re-derived and solved from scratch by an independent certifier, corroborated per clause, and re-verifiable by a independent small third implementation. The $m=8$ case is established twice: by an earlier independent project at that size and as a one-paragraph padding corollary of the $m=9$ theorem. We additionally give a possible explanation why the problem is hard: difficulty concentrates on near-identical valuations, where only ${\approx}0.14\%$ of all $4^9$ allocations are $\EFXo$, and explicit valuation pairs inside a single region force opposite mandatory allocation structure, evidence relevant to the general conjecture independently of any solver stack.

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