Let $\sigma = \exists x\forall y (x\leq y\lor y\leq x)$. We wish to find posets $A$ and $B$ such that $A \vdash \sigma$ and $B \vdash\neg\sigma.$ I have found the poset $A$. The natural numbers with $\leq$ does this. There exists an $x$ (in fact, for all $x$, but we don't require this), such that every other natural number is either bigger than or equal to $x$ or less than or equal to $x$. This is obvious.
I am struggling with the second portion. Here is my attempt at how I did it. Firstly, I formed the negation of $\sigma$. $\neg\left[ \exists x\forall y (x\leq y\lor y\leq x)\right] \equiv \forall x\exists y(x>y\land x<y).$ In other words, we want a poset such that for all elements in the set, there's another element that's BOTH bigger than it AND less than it. This feels..... impossible. Maybe this is because I'm restricting my view to very common posets like $\mathbb{N},\mathbb{Q},\mathbb{R},...$ but is this possible? This seems to violate the notion of the $\leq$ sign. One element cannot be bigger than AND less than another element. Can someone help with this?
Or.... perhaps this answer is that it doesn't exist. Van Dalen does not word the question in a way such that this is possible.
Poset:
Thus, for $\mathfrak A \vDash \sigma$ we may assume $A = \{ 0,1 \}$ with $\le^{A} = \{ (0,0), (0,1), (1,1) \}$, while for $\mathfrak B \vDash \lnot \sigma$ we may assume $B = \{ 0,1 \}$ with $\le^{B} = \{ (0,0), (1,1) \}$.
The first case is obvious.
For the second case: $\forall x \exists y \ [\lnot (x \le y) \land \lnot (y \le x)]$, we have that $\lnot (0 \le 1)$ as well as $\lnot (1 \le 0)$.