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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.

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    $\begingroup$ You are correct. The sentence is true in every poset. $\endgroup$
    – drhab
    Dec 6, 2022 at 12:57
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    $\begingroup$ You are right. For whatever object one picks for $x$, one can always pick that very same $x$ for $y$, making the rest of the statement true, no matter what domain you use or how you interpret $<$ $\endgroup$
    – Bram28
    Dec 6, 2022 at 12:57
  • $\begingroup$ @drhab thanks for your response. The exercise has been amended to let the 'implication' be an 'if and only if' statement. Which is clear how to solve. $\endgroup$
    – Bessel
    Dec 6, 2022 at 14:21

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