In classical logic, $\ c\Rightarrow a\ \&\ c\Rightarrow b\ \ .\mbox{always}\Longrightarrow.\ \ c\Rightarrow a\&b\ $, although in linear logic that's typically not true (e.g., the vending machine model, https://en.wikipedia.org/wiki/Linear_logic#The_resource_interpretation).
So, correspondingly for posets, does or doesn't $\ a\sqsubseteq c\ \&\ b\sqsubseteq c\ \ .\mbox{always}\Longrightarrow.\ \ a\sqcup b\sqsubseteq c\ $? (that's the main question)
I'm hoping "doesn't always", and if so can you give me an example where it doesn't?
Note that it "always does" for a set-theoretic poset order, where $a\sqsubseteq c$ means $c\subseteq a$, and where $a\sqcup b$ means $a\cap b$.
So we'd necessarily need some other kind of interpretation of poset order for an example where it "doesn't always".