For finite subextensions in a common field, $[E{E}^{\mathrm{\prime }}:F]\le [E:F][E^{\mathrm{\prime }}:F]$

Choose $F$-bases $(u_1,\dots ,u_m)$ of $E$ and $(v_1,\dots ,v_n)$ of $E^{\mathrm{\prime }}$. The $F$-span $A$ of the $mn$ products $u_i{v}_j$ contains $E$ and $E^{\mathrm{\prime }}$ and is closed under addition and multiplication, because products are reduced separately in the two bases.

By [L4], every $u_i$ and $v_j$ is algebraic over $F$, so [L5] makes $E{E}^{\mathrm{\prime }}=F(u_1,\dots ,u_m,v_1,\dots ,v_n)$ finite over $F$. Every $r\in A\subseteq E{E}^{\mathrm{\prime }}$ is therefore algebraic over $F$.

If $0\mathrm{\ne }r\in A$, take a nonzero annihilating polynomial, factor out its largest power of $t$, and cancel the corresponding nonzero power of $r$ in the ambient field. This gives $c_0+c_{1}r+\cdots +c_d{r}^d=0$ with $c_0\mathrm{\ne }0$. Then $r^{-1}=-c_0^{-1}(c_1+c_{2}r+\cdots +c_d{r}^{d-1})\in A$, so $A$ is a field.

Since $A$ is a field containing $E\cup E^{\mathrm{\prime }}$, [L1] gives $E{E}^{\mathrm{\prime }}\subseteq A$; the reverse inclusion is clear from the product span, so $A=E{E}^{\mathrm{\prime }}$.

The $mn$ products span $E{E}^{\mathrm{\prime }}$. By [L3] they contain a basis of at most $mn$ elements, so [L2] yields $[E{E}^{\mathrm{\prime }}:F]\le mn=[E:F][E^{\mathrm{\prime }}:F]$.