Proof of Theorem BIS

Suppose that $C$ has more vectors than $B$. (Allowing for the possibility that $C$ is infinite, we can replace $C$ by a subset that has more vectors than $B$.) As a basis, $B$ is a spanning set for $V$ (Definition B), so Theorem SSLD says that $C$ is linearly dependent. However, this contradicts the fact that as a basis $C$ is linearly independent (Definition B). So $C$ must also be a finite set, with size less than, or equal to, that of $B$.

Suppose that $B$ has more vectors than $C$. As a basis, $C$ is a spanning set for $V$ (Definition B), so Theorem SSLD says that $B$ is linearly dependent. However, this contradicts the fact that as a basis $B$ is linearly independent (Definition B). So $C$ cannot be strictly smaller than $B$.

The only possibility left for the sizes of $B$ and $C$ is for them to be equal.