I'm trying to solve the exercise 2.3.1 of Chang-Keisler book "Model Theory".
If $\phi(x_1, \cdots, x_n)$ is a complete formula in a theory $T$ with respect to $x_1,\cdots, x_n$, then $\exists x_n \phi (x_1, \cdots, x_{n-1}, x_n)$ is a complete formula in $T$ with respect to $x_1, \cdots, x_{n-1}$.
By a complete formula, we mean that for all formula $\psi(x_1,\cdots, x_n)$, we have exactly one between $T \vdash \phi \rightarrow \psi$ and $T \vdash \phi \rightarrow \neg \psi$.
I have no idea how to approach this problem. Help?
Let $\psi(x_1,\ldots,x_{n-1})$ be a formula. $\psi$ is trivially also a formula in $x_1,\ldots,x_n$ with no dependence on $x_n$, which I will also denote by $\psi$. Let $\tilde{\phi}(x_1,\ldots,x_{n-1})=\exists x_n \phi(x_1,\ldots,x_n)$.
Then we have either $T\vdash\phi \to \psi$ or $T\vdash \phi \to \lnot \psi$ since $\phi$ is complete. Suppose we have $T\vdash \phi \to\psi$, then assuming $\exists x_n \phi(x_1,\ldots,x_n)$, we have $\psi(x_1,\ldots,x_{n-1},x_n)$, which is equivalent to $\psi(x_1,\ldots,x_{n-1})$, since $\psi$ doesn't depend on $x_n$. Thus given $T\vdash \phi \to \psi$, we have $T\vdash \tilde{\phi}\to \psi$.
Symmetrically, if we have $T\vdash \phi\to\lnot\psi$, we know $T\vdash \tilde{\phi}\to \lnot\psi$. Thus either $T\vdash \tilde{\phi}\to \psi$ or $T\vdash \tilde{\phi}\to \lnot \psi$. Therefore $\tilde{\phi}$ is complete.