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.
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.
