# 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.

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What is the definition of a principal $n$-type in this context? – Lord_Farin Oct 29 '12 at 14:07
By definition: a n-type is principal if there exists a $\psi=\psi(x_0,...,x_{n-1})$ such that for all $\phi=\phi(x_0,...,x_{n-1})$ holds: the sentence $\forall x_0\cdots\forall x_{n-1}[\psi\rightarrow\phi]$ is a logical consequence of the theory T. – Model of the Theory Oct 29 '12 at 14:15

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.
Okay i tried it, so suppose $\Delta$ is $n+1$-type, thus there exists $\psi$ (see above) such that for all $\phi$ in $\Delta$ holds: $\forall x_0\cdots\forall x_n[\psi\rightarrow\phi]$. I don't know preciesly what you mean with replacing variables by fresh constants. I know that you can do this but why can we apply this here?! Moreover: The craig-interpolation theorem holds only for sentences and not for sets with sentences or not, and why does this implies that $\Gamma$ is principal? Sorry that i ask so much questions but i want to understand this complete. – Model of the Theory Oct 29 '12 at 17:12
Well, WLOG say that $\Delta$ contains variables $x_0\ldots x_{n-1}$ (so not variable $x_n$). I merely tried to sketch a motivation why the quantification over $x_n$ can be moved into $\psi$ (whether existentially or universally) or plainly omitted (via a Craig IT(-like) argument). Apparently I didn't succeed in doing this in a clear way. Plainly, (a small modification of) $\psi$ will also entail all $\phi$ from $\Delta$, because the last quantification doesn't matter ($x_n$ does not occur in any formula of $\Delta$). Apologies for excessive parentheses. – Lord_Farin Oct 29 '12 at 17:27
I thought over this exercise again and i got a conclusion, but i don't know whether the deduction is good: We want a contradiction thus suppose $\Delta$ is principal then there exists $\psi=\psi(x_0,\cdots,x_n)$ such that for every $\phi=\phi(x_0,\cdots,x_n)$ holds: $forall x_0\cdots\forall x_n[\psi\rightarrow\phi]$ logical consequence T. Thus also: $forall x_0\cdots\exists x_n[\psi\rightarrow\phi]$, thus also $forall x_0\cdots\forall x_{n-1}[\exists\psi\rightarrow\exists\phi]$, thus also: $forall x_0\cdots\forall x_{n-1}[\psi\rightarrow\phi]$ thus $\Gamma$ principal. Contradiction. – Model of the Theory Oct 31 '12 at 15:52