Kaplansky's zero-divisor conjecture asserts that the group ring of a torsion-free group over a field has no zero divisors. It holds for every group with the unique-product property, so a counterexample can only come from a torsion-free group without unique products. In lectures in 2021, Gardam announced that the torsion-free $\tilde{A}_2$ lattice $Γ= \langle a, b \mid a b a^2 b^{-1} a^2 b^{-2}, a b^3 a b^4 a^{-1} b \rangle$ does not have unique products and presented it as a new candidate group: it has property (T), and the known methods for proving the conjecture do not apply to it. To our knowledge, no proof of the announcement has been published. We give a proof checked by the Lean 4 kernel and stated against Mathlib's UniqueProds class. The witness is an explicit pair of finite subsets with $|A| = 32$ and $|B| = 28$ in which each of the 896 products coincides with another product. For 658 products the certificate is an identity in the free group; the other 238 certificates are explicit products of conjugated relators, 970 conjugates in all, checked by free reduction. A homomorphism onto $\mathbb{Z}/42$ shows that each pair $(u,v)$ differs from its partner $(u',v')$ as a pair of group elements, which is all the theorem requires. Together with a homomorphism onto the alternating group $A_4$ it also shows that the listed words are pairwise distinct, so the sets have exactly 32 and 28 elements. The witness and certificates come from an untrusted search program and are re-checked by Lean. The development uses only the axioms propext, Classical.choice and Quot.sound, with no sorry and no native_decide. The mathematical statement is Gardam's. To our knowledge this is the first verification in a proof assistant of a unique-product failure in a torsion-free group; torsion-freeness of $Γ$ is taken from Gardam and is not formalized here.