# A poset in which the sentence $\phi$ is false

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

But this doesn't seem to help.

• You are correct. The sentence is true in every poset. Dec 6, 2022 at 12:57
• 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 $<$ Dec 6, 2022 at 12:57
• @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. Dec 6, 2022 at 14:21