Extend definable types I am self-studying Pillay's lecture notes on stability theory (https://www3.nd.edu/~apillay/pdf/lecturenotes.stability.pdf) and having trouble solving Exercise 2.13.
Let $T$ be a complete $L$-theory with models $M \subset N$. Let $p(x) ∈ S(M)$ be a definable type with defining schema $d: \delta(x,y) \mapsto \psi_\delta(y)$. How to prove that $d(N) := \{\delta(x,b):$ for $\delta(x, y) ∈ L$ and $b ∈ N$ such that $N \models d(\delta)(b)\}$ is a complete type over $N$ extending $p(x)$?
This result is used to show that any complete type over a model of a totally transcendental theory has Morley degree 1 in the notes.
My attempt: $p(x)$ is realized by $a$ in a monster model $\overline M$. Consider an elementary embedding $f: N \rightarrow \overline M$ such that $(i: M \rightarrow \overline M) = f \circ (i: M \rightarrow N)$ (does such $f$ exist?). Maybe $d(N) = \{\delta(x,b): \delta(x,f(b)) \in \text{tp}_\overline M(a/f(N))\}$, but I don't think I am right.
Thanks in advance for any help.
 A: First a remark about the monster model convention. The point of working with a monster model $\bar{M}$ is that we assume all elements and sets that we consider to live in $\bar{M}$. Furthermore, any models that we consider will be elementary submodels. This can indeed be done exactly because an $f$ like in your question does exist (exercise, use saturation). In this answer I will follow this convention and leave $\bar{M}$ implicit.
We have to prove two things about $d(N)$: namely that it is a consistent set of formulas and that it is complete.
Consistent. Let $\delta_1(x, b_1), \ldots, \delta_n(x, b_n) \in d(N)$. For any $c_1, \ldots, c_n \in M$ such that $M \models d(\delta_i)(c_i)$ for all $1 \leq i \leq n$ we have that $\delta_i(x, c_i) \in p(x)$ for all $1 \leq i \leq n$, which implies $\delta_1(x, c_1) \wedge \ldots \wedge \delta_n(x, c_n) \in p(x)$ and thus $\models \exists x(\delta_1(x, c_1) \wedge \ldots \wedge \delta_n(x, c_n))$ and hence $M \models \exists x(\delta_1(x, c_1) \wedge \ldots \wedge \delta_n(x, c_n))$ as $M$ is an elementary submodel of the monster. We thus have that
$$
M \models \forall y_1 \ldots y_n ( d(\delta_1)(y_1) \wedge \ldots \wedge d(\delta_n)(y_n) \to \exists x(\delta_1(x, y_1) \wedge \ldots \wedge \delta_n(x, y_n)) ).
$$
As $M$ is an elementary submodel of the monster this sentence then also holds in the monster model (or as there are no parameters: is even a consequence of the theory). By assumption we have, for all $1 \leq i \leq n$, that $\delta_i(x, b_i) \in d(N)$ and so $\models d(\delta_i)(b_i)$. From the above it then follows that $\models \exists x(\delta_1(x, b_1) \wedge \ldots \wedge \delta_n(x, b_n))$. We conclude that $d(N)$ is finitely satisfiable and is thus consistent by compactness.
Complete. Let $\delta(x, b)$ be any formula with parameters in $N$ (i.e. $b \in N$). We will show that either $\delta(x, b) \in d(N)$n or $\neg \delta(x, b) \in d(N)$. For this we claim that $d(\delta)(y)$ is equivalent to $\neg d(\neg \delta)(y)$. This would indeed be enough, because then there are two cases:

*

*$\models d(\delta)(b)$, so $\delta(x, b) \in d(N)$;

*$\models \neg d(\delta)(b)$, so $\models d(\neg \delta)(b)$ and thus $\neg \delta(b) \in d(N)$.

To verify the claim we let $c \in M$ be arbitrary. Then we have that $M \models d(\delta)(c)$ iff $\delta \in p(x)$ iff $\neg \delta \not \in p(x)$ iff $M \models \neg d(\neg \delta)(c)$. So we see that $M \models \forall y(d(\delta)(y) \leftrightarrow \neg d(\neg \delta)(y))$, and we conclude by using the fact that $M$ is an elementary submodel of the monster.
