In my model theory lecture course, a partial type $p$ is defined to be principal when there is some $ϕ$ such that $T \models ∃vϕ(v)$ (for any n-place variable $V$) and for any $ψ ∈ p$, $T \models ∀v(ϕ(v) → ψ(v))$.
[and we define a partial type of a theory $T$ in variables $x$ to be a satisfiable set $P$ of formulas in the variables $x$, containing $T$ and closed under logical deduction.]
I'm asked to show that when $p$ is a principal (complete) type, the formula $ϕ$ which features in the definition above must be in $p$.
It's clear that since $T \models ∃vϕ(v)$, and $p$ must be closed under logical deduction, that $∃vϕ(v) \in p$ , but I don't see how it follows that $ϕ(v) \in p$, which is what I'm being asked to prove.
Any suggestions you could give me would very much be appreciated.
As pointed out in the comment this question only makes sense for complete types. So let $p$ be a principal type, with principal formula $φ \in F_n$. Since p is complete, $φ \not\in p \implies ¬φ \in p$. Add new constant symbols $c = (c_1, c_2, ..., c_n)$. The principality condition gives us that $T \cup \{φ(c)\} \models ψ(c)$ for all $ψ(x) \in p$, and so $T \cup \{φ(c)\} \models ¬φ(c)$. It follows that $T \models ¬φ(c)$, and therefore $T \models ¬\exists x φ(x)$. This contradicts the fact that $φ$ is realised in all models of T, i.e. $Τ \models \exists x φ(x)$. Thus $φ \in p$.