Given a set $U=\{1,\ldots,c\}$, we have $t$-dimensional vector $v\_i=(x\_1,x\_2,\ldots,x\_t)$, where $x\_j$ is a subset of $U$. We can check that there are $2^{ct}$ above vectors.
Then we construct the following matrix:


$A=(a\_{ij})\_{2^{ct},2^{ct}}$, where $a\_{ij}=1$ if $\bigvee\_{1\leq l\leq t}v\_{i}(l)\wedge v\_{j}(l)=\emptyset$; $0$ otherwise.


I want to know an upper bound for the rank of the matrix $A$. I hope this upper bound is as small as possible.



Upper bound of rank of a matrix
If I understood your question correctly, it can be reformulated as follows:
The vectors $v\_i$ are in bijective correspondence with subsets of $X=\{1,\cdots, c\} \times \{1,\cdots, t\}$ by sending a vector $v\_i=(x\_1,x\_2,\cdots, x\_t)$ to the subset $S=\{(k,l)\in X:k\in x\_l\}$.
You can then see the matrix $A$ as a $2^X \times 2^X$ matrix and your condition become $A\_{S,T}=1\_{S \cap T =\emptyset}$ for all $S,T\in 2^X$.


You can factor $A=CB$ where $B\_{S,T}=1\_{S=X\setminus T}$ and $C\_{S,T}=1\_{S\subseteq T}$. Both matrices are invertible since $B$ is a permutation matrix and $C$ is upper-triangular (with respect to any linear extension of the inclusion ordering on $2^X$) with $1$'s on the diagonal. Hence your matrix $A$ always has full rank.


