Let $F/K$ be a field extension of transcendence degree $r$ and let $\{x_1,\ldots,x_n\}$ be a set of elements which are algebraically independent over $K$. Then is it true that $tr.deg. F(x_1,\ldots,x_n)/K(x_1,\ldots,x_n)=r$?
If $\{t_1,\ldots,t_r\}$ is a transcendence base of the extension $F/K$ then we know $F/K(t_1,\ldots,t_r)$ is algebraic and $K(t_1,\ldots,t_r)/K$ is purely transcendental. Thus $F(x_1,\ldots,x_n)/K(x_1,\ldots,x_n)(t_1,\ldots,t_r)$ is also algebraic. Now it will be enough to show that $\{t_1,\ldots,t_r\}$ is algebraically independent over $K(x_1,\ldots,x_n)$. How should I approach?
My above queries are from the book of Lang's Algebra book in the context of being free from one field to another. I have added a photo from Lang's Algebra book: I want to understand why $tr. deg. F(y)/K(y)=r$ as shown in the diagram.