Do filters on a Boolean algebra also make a Boolean algebra?

Let $\mathfrak{B}=(B,\bot,\top,\lnot,\wedge,\vee)$ be a boolean algebra. $B_F$ be the set of all filters on $\mathfrak B$. And for all filter $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ in which $\mathbf C$ denotes the filter closure operator; $F \vee_{B_F} G \colon= F \cap G$; $0 \colon= B$, $1 \colon= \{\top\}$. Then $(B_F,0,1,\wedge_{B_F},\vee_{B_F})$ makes a complete lattice.

My question is can we add a negation in order to make it be a Boolean algebra?

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The ideals of a Boolean algebra form a complete, distributive lattice, but they do not, in general, form a Boolean algebra. To give an example, it is helpful to introduce some terminology. An ideal is maximal if it is a proper ideal that is not properly included in any other proper ideal. We shall see in the next chapter that an infinite Boolean algebra $B$ always has at least one maximal ideal that is not principal. Assume this result for the moment. A "complement" of such an ideal $M$ in the lattice of ideals of $B$ would be an ideal $N$ with the property that $$M\wedge N=\{0\} \qquad\text{and}\qquad M\vee N=B.$$ Suppose the first equality holds. If $q$ is any element in $N$, then $p \wedge q = 0$, and therefore $p \le q'$, for every element $p$ in $M$, by Lemma 1. In other words, the ideal $M$ is included in the principal ideal $(q')$. The two ideals must be distinct, since $M$ is not principal. This forces $(q')$ to equal $B$, by the maximality of $M$. In other words, $q' = 1$, and therefore $q = 0$. What has been shown is that the meet $M\wedge N$ can be the trivial ideal only if $N$ itself is trivial. In this case, of course, $M \vee N$ is $M$, not $B$. Conclusion: a maximal, non-principal ideal does not have a complement in the lattice of ideals.