I've been scratching my head at this for quite a while, the question is as follows:
(pardon my weird wording, I'm not studying math in English so I might get some proof wording conventions wrong)
let $X$ be a set, and $\mathcal{R}$ be the set of all non-strict partial orders over set $X$.
let $\trianglelefteq,\leqslant \in \mathcal{R} $
We'll define Partial order $\sqsubseteq$ over $\mathcal{R} $ as follows:
$\leqslant \sqsubseteq \trianglelefteq \ \Leftrightarrow \forall x,y\in X\ \left( x\leqslant y\ \rightarrow x\trianglelefteq y\right)$
is $(\mathcal{R},\sqsubseteq)$ a poset?
The first thing that baffled me is that if I want to disprove this, I have to give a counterexample, but to do that, I need to give a partial order over $X$, only I don't know what $X$ is.
On the other hand, to prove that this is a poset, I've tried showing that it's antisymmetric to no avail, and got stuck on that front. for example:
$let\,\, R_1,R_2\in\mathcal{R}\, such\ that\, R_{1} \sqsubseteq R_{2}$ By the assumption, for all $x,y$ the condition is satisfied, in particular for $\hat{x} ,\hat{y} \in X\ \left(\hat{x} R_{1}\hat{y} \ \rightarrow \hat{x} R_{2}\hat{y}\right)$
then in order for $R_{2} \sqsubseteq R_{1}\,$ to also occur, the condition must also apply for all $x,y\,$, in particular $\hat{x} R_{2}\hat{y} \ \rightarrow \hat{x} R_{1}\hat{y}$
But this is where I got stuck, I can't seem to get to a conclusion from this point.
Much appreciated
There's another way of looking at this that makes it very easy. If we think of a relation as a set of ordered pairs, then the relation $\sqsubseteq$ is the same as $\subseteq$. $$xR_1y\rightarrow xR_2y$$ means the same thing as $$(x,y)\in R_1\rightarrow (x,y)\in R_2$$ which, of course, is equivalent to $$R_1\subseteq R_2$$
So, the answer is "yes".