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We prove that if $I(D)$ is vertex-splittable, then $HS_1(I(D))$ has linear quotients. Furthermore, we show that $\\sqrt{HS_k(I(D))} = HS_k(I(G))$, for all $k\\geq 1$, where $G$ is the underlying simple graph of $D$. We show that if $I(D)$ has homological linear quotients, then $I(G)$ also has homological linear quotients. 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