Corollary 2

— Let L be an extension of K and E, F subextensions of L. If F is algebraic over K, the subring $K[E, F]$ of L generated by $E \cup F$ is a field coinciding with $E(F)$ and algebraic over E.

For each element of $F$, being algebraic over $K$, is also algebraic over $E$ ($V$, p. 17, Cor. 2) hence $E(F)$ is an algebraic extension of $E$, and we have $E(F) = E[F]$ by Cor. 1.