Let $\mathfrak{A}$ is a bounded co-brouwerian lattice.
Let $x\cap a\ne 0$ for some elements of the lattice. Prove (or disprove) that the pseudodifference $a\setminus^{\ast}x\ne a$.
I recall that pseudodifference on co-brouwerian lattices is define by the formula $$a\setminus^{\ast}b=\operatorname{min}\{z\in\mathfrak{A} \,|\, a\subseteq b\cup z\}.$$
Oh, I've found a counterexample myself:
Consider the lattice of all filters (including the improper filter) on $\mathbb{R}$, with the order of filters being the reverse of set-theoretic inclusion.
Let $a$ be the principal filter corresponding to the set $\mathbb{R}$. Let $b$ be a non-trivial ultrafilter on $\mathbb{R}$.
Then $a\setminus^{*}b = a$.