We need to construct a poset where the sentence $\phi$ is false. Let $\phi$ be given by $$\phi=\forall x\exists y\forall z(z < x \to z < y ∨ z = y).$$
However, I think that this sentence is true in all posets. Since we can choose $x=y$.
The negation is given by $$\exists x \forall y \exists z(z<x\wedge\neg(z<y ∨ z=y) ).$$
But this doesn't seem to help.