Answer
True.
Work Step by Step
By definition, if $\{\vec{v}_{1},\vec{v}_{2},\vec{v}_{3},\vec{v}_{4}\}$ is a linearly independent set of vectors, then the only solution to the vector equation $c_{1}\vec{v}_{1}+c_{2}\vec{v}_{2}+c_{3}\vec{v}_{3}+c_{4}\vec{v}_{4}=\vec{0}$ is $c_{1}=c_{2}=c_{3}=c_{4}=0$. Hence, $c_{1}\vec{v}_{1}+c_{2}\vec{v}_{2}+c_{3}\vec{v}_{3}=\vec{0}$ cannot have a nontrivial solution: otherwise, we could add $0\vec{v}_{4}$ to the equation and have a nontrivial solution in the original four vectors, meaning those four vectors would not be linearly independent, contrary to the given information.
