Let $K\subset E\subset\bar{K}$ be field extensions, $\bar{K}$ an algebraic closure of $K$. Denote $E\_s$ the field of separable elements of $E$ over $K$, denote $\tilde{E}\subset\bar{K}$ the normal closure of $E/K$ (the minimal subfield of $\bar{K}$ which contains $E$ and is normal over $K$). Is it true that
$$
(E\_s)^{\sim}=(\tilde{E})\_s?
$$
(The meaning of the notation is obvious.) It's easy to see that the RHS is larger and both are Galois over $K$. It's also not hard to see this is true in the following cases: $E/K$ is separable, or normal, or purely inseparable. I want to deduce the general case from these but I can't. Any ideas or counterexample?



Normal closure and separable elements
Yes, it is true. 


We will denote by $p(a,K)\in K[x]$ the minimal polynomial of $a\in \overline{K}$ over $K$: it is monic, irreducible and has $a$ as a root. An irreducible polynomial is called separable if it has no multiple roots on $\overline{K}$


Then $$E\_s:=\{a\in E\ | \ p(a,K) \mbox{ separable }\}$$ and $$\widetilde{E}:=\{ b\in \overline{K}\ | \ p(b,K) \mbox{ has a root in } E\}.$$


Then $$(\widetilde{E})\_s=\{b \in \overline{K}\ |\ \ p(b,K) \mbox{ is separable and it has a root in } E\}$$
while 
$$\widetilde{E\_s}=\{b \in \overline{K}\ |\ \ p(b,K) \mbox{ has a root in } E\_s\}.$$


But an irreducible polynomial that is separable and it has a root $a$ in $E$, then $a\in E\_s$ by definition; the converse is also easily seen to be true. Hence they are equal. 


