Let $L$ be a finite lattice with a least element $0$ and a greatest element $1$, where $0\neq 1$. Fix a $t\in L$, and let $X$ be the set of non-complements of $t$, i.e., the set of all $x$ such that $x\vee t\neq 1$ or $x\wedge t\neq 0$. Note that $t\in X$.
Let $x_1,\ldots, x_n\in X$ be such that $t\vee x_1\vee\cdots\vee x_n=1$ and $t\wedge x_1\wedge\cdots\wedge x_n=0$. Is it the case that $x_1\vee\cdots\vee x_n=1$ and $x_1\wedge\cdots\wedge x_n=0$?
I'm trying to prove that, if $n_k$ is the number of $k$-element subsets of $X$ whose join is $1$ and whose meet is $0$, then $\sum_k (-1)^k n_k=0$. I'm trying to do this by using the parity-reversing bijection that either removes $t$ from a subset if it's there or adds it in if it's not. Obviously, if $\bigvee S=1$ and $\bigwedge S=0$, then adding $t$ to $S$ won't change that, but I need to show that removing $t$ won't change it either.
Counterexample for your question: Let $L$ be the lattice of all subsets of $\{1,2,3,4,5,6\}$ and let $n=2,\ t=\{1,3,5\},\ x_1=\{2,3,6\},\ x_2=\{4,5,6\}.$
Counterexample to the identity you're trying to prove: Consider the $7$-element lattice $L=\{0,1,a,b,c,d,t\}$ where
$$a\wedge b=0,\ a\vee b=t,\ c\wedge t=a,\ d\wedge t=b,\ c\vee d=c\vee t=d\vee t=1.$$ Then $X=L\setminus\{t\}=\{0,1,a,b,c,d\}$ and $$-n_1+n_2-n_3+n_4-n_5+n_6=-0+4-15+15-6+1=-1$$ if I counted right.