# A problem from Model Theory of Chang and Keisler.

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.