Partitioning a definable set of Morley degree $d$ into $d$ definable disjoint sets of Morley degree $1$. The following is taken from Marker's Introduction to Model theory (Corollary 6.3.12) Here $\mathbb M$ denotes a monster model.
Let $\mathcal M \models T$ (complete, $\omega$-stable theory), and let $\phi(v)$ be an $L_M$-formula of Morley rank $\alpha$ and degree $d$.
i) There is an $L_M$-formula $\theta(v)$ such that $\theta(\mathbb M) \subseteq \phi(\mathbb M)$, with Morley rank $\alpha$ and Morley degree $1$.
ii) There are $L_M$-formulas $\theta_1 ,\dots, \theta_d$ each of Morley rank $\alpha$ and Morley degree $1$ such that $\phi(\mathbb M)$ is the disjoint union of $\theta_1(\mathbb M) ,\dots, \theta_d(\mathbb M)$.
The proof of part i) comes from a lengthy discussion about non-forking extensions in $\omega$-stable theories. I am however, a bit lost on the proof of part ii).
Intuitively, this should not be too hard, we are just lowering the degree one by one until the last set remaining must have degree $1$.
"We inductively define $\theta_1 ,\dots, \theta_d$. Given $\theta_1, \dots , \theta_{m-1}$ for $m < d$, we use i) to obtain a rank $\alpha$, degree 1, subset of
$$\phi(\mathbb M) \setminus \bigcup_{i=1}^{m-1} \theta_i(\mathbb M)$$
Then take
$$\theta_d = \phi \land \bigwedge_{i=1}^{d-1}\theta_i$$
My first question: Why does $\phi \land \neg \theta_1\land\dots \land \neg \theta_{m-1}$ have rank $\alpha$? Is it in general true that removing a degree 1 subset of a larger degree one keeps the rank unchanged?
My second question is about $\theta_d$. This must surely be a typo since it isn't even disjoint from the rest. Perhaps it should contain $\neg \theta_i$ instead?
Any help would be appreciated, I found this chapter a bit lacking on properties of the degree (They only prove it exists), is there maybe another, more explicit reference to study this from?
 A: I think the key fact that you're missing is Marker's Lemma 6.2.16(iii):

For any $L_A$-formula $\varphi(x)$ of Morley rank $\alpha$, the Morley degree of $\varphi(x)$ is equal to the number of complete types in $S_x(\mathbb{M})$ of Morley rank $\alpha$ which contain $\varphi(x)$.

[Marker doesn't give a proof, leaving it as an exercise for the reader. You should try to prove it!]
First question: Yes, if $X\subseteq Y$ are definable sets, $Y$ has Morley rank $\alpha$ and Morley degree $d>1$, and $X$ has Morley rank $\alpha$ and Morley degree $1$, then $Y\setminus X$ has Morley rank $\alpha$ and Morley degree $d-1$.
Proof: By the Lemma, $Y$ contains $d$ types in $S_x(\mathbb{M})$ of Morley rank $\alpha$, and $X$ contains $1$ such type. So $Y\setminus X$ contains $d-1$ such types. Since $(d-1)\geq 1$, the rank of $Y\setminus X$ is at least $\alpha$. But also $(Y\setminus X)\subseteq Y$, so the rank of $Y\setminus X$ is equal to $\alpha$. By the Lemma, the degree of $Y\setminus X$ is $d-1$.
Second question: Yes, it's a typo. $\theta_d$ should be $\varphi\land \bigwedge_{i=1}^{d-1} \lnot \theta_i$.
Third question: The book A Course in Model Theory by Tent and Ziegler has a thorough treatment of Morley rank and degree in Chapter 6.
