Proof of Theorem SLTD

Suppose to the contrary that $m=\dimension{U}<\dimension{V}=t$. Let $B$ be a basis of $U$, which will then contain $m$ vectors. Apply $T$ to each element of $B$ to form a set $C$ that is a subset of $V$. By Theorem SLTB, $C$ is a spanning set of $V$ with $m$ or fewer vectors. So we have a set of $m$ or fewer vectors that span $V$, a vector space of dimension $t$, with $m<t$. However, this contradicts Theorem G, so our assumption is false and $\dimension{U}\geq\dimension{V}$.