Extend non-principal n-type Is the following true? Let T be a complete theory in some elementary language. Let $n$ be a natural number and suppose $\Gamma$ is a non-principal $n$-type of T. Let $\Delta$ ne an $n+1$-type of T containing $\Gamma$. Then $\Delta$ too is non-principal. Give a proof or a counterexample. During the lecture we got the following theorem: Given an elementary language and T a complete theory in this language with at least one model. Let $n$ be a natural number: T has infinitely $n$-types iff T has a non-prinical $n$-type. So i thought we can prove by contradiction: Suppose $\Delta$ is wel principal, then T has only finitely many $n+1$-types, but $\Gamma\subset\Delta$ and $\Gamma$ non-principal, thus it contains infinitely many n-types, thus contradiction. Is this argumentation good? And so not why not and how can i solve it? In general: How can i decide whether something is prinicpal or not? Thank you for help.

With the given definition of principal $n$-type, your idea of proof by contraposition seems to be correct. However, your idea of applying the theorem fails in that, while $\Delta$ may be principal, there could be a different $n+1$-type $\Delta'$ that is non-principal. Thus it isn't guaranteed the theorem applies.

To solve the problem, you can try to use contraposition, using explicitly what it means for $\Delta$ to be principal. For example, you can then apply the standard "trick" that consistency is preserved under replacing variables by fresh constants, and Craig's Interpolation Theorem to deduce formally that $\Gamma$ is principal.

In general, it's easier to show some $\Gamma$ principal (just find an appropriate $\psi$) than non-principal, like it is easier to prove that a number is algebraic (find a polynomial of which it is a root) than it is prove a number is transcendental.