proof of the dimension theorem for subspaces

Let $S$ and $T$ be subspaces of a vector space.By the rank-nullity theorem and the second isomorphism theorem (for modules)we have

$\dim (S+T)$$=\dim S+\dim ((S+T)/S)$

$=\dim S+\dim (T/(S\cap T)).$

Therefore

$\dim (S+T)+\dim (S\cap T)$$=\dim S+\dim (T/(S\cap T))+\dim (S\cap T)$

$=\dim S+\dim T,$

by the rank-nullity theorem again.