Exercise 12

Show that if $\{\hskip 1.00006pt\vec{x},\vec{y},\vec{z}\hskip 1.00006pt\}$ is linearly independent then so are all of its proper subsets: $\{\hskip 1.00006pt\vec{x},\vec{y}\hskip 1.00006pt\}$, $\{\hskip 1.00006pt\vec{x},\vec{z}\hskip 1.00006pt\}$, $\{\hskip 1.00006pt\vec{y},\vec{z}\hskip 1.00006pt\}$, $\{\hskip 1.00006pt\vec{x}\hskip 1.00006pt\}$,$\{\hskip 1.00006pt\vec{y}\hskip 1.00006pt\}$, $\{\hskip 1.00006pt\vec{z}\hskip 1.00006pt\}$, and $\{\hskip 1.00006pt\hskip 1.00006pt\}$. Is that ‘only if’ also?