Forking implies that the Morley Rank became smaller I am trying to prove the following statement: 
Let $\phi$ be a formula over $A$ with Morley Rank, and $\psi$ be a subclass of $\phi(U)$ (where $U$ is a monster model). If $\psi$ forks over $A$ then $MR(\psi)< MR(\phi)$, where $MR$ denotes the Morley Rank of a formula. 
What I have tried is more or less the following: Let's say that $\psi$ divides over $A$, then one can find and indiscernible sequence $b_{i}$ over $A$ such that $\{ \psi(x, b_{i}) \ | \ i \in \omega\}$ is $k$-inconsistent. Without loss of generality, we can take $k \in \omega$ the smallest integer such that the inconsistency is given. Now, let $\theta_{1}(x)= \bigwedge_{j=1}^{k-1} \psi_{j}(x,b_{j})$, then $\theta_{2}(x):= \bigwedge_{j=k}^{2k-1}\psi(x,b_{j})$, and so on. So each of these formulas has the same Morley rank of $\psi$ and they must induce a partition inside $\phi(x)$ of par wise disjoint formulas. But, how could this be generalize for the forking case?
 A: To say that $\psi$ forks over $A$ is to say that $\psi\rightarrow \bigvee_{i=1}^k \psi_i$, where each $\psi_i$ divides over $A$. Then we have $\psi(U)\subseteq \bigcup_{i=1}^k \psi_i(U)$, so $\mathrm{MR}(\psi)\leq \max(\mathrm{MR}(\psi_1),\dots,\mathrm{MR}(\psi_k))$. So if you've shown that $\mathrm{MR}(\psi_i)<\mathrm{MR}(\phi)$ for all $1\leq i\leq k$, it follows immediately that $\mathrm{MR}(\psi)<\mathrm{MR}(\phi)$.
At this point, it looks like you're done, since you've argued that if $\psi\rightarrow \phi$ and $\psi$ divides over $A$, then $\mathrm{MR}(\psi)< \mathrm{MR}(\phi)$. But I'm not convinced by your argument: how do you know that $\bigwedge_{j=1}^{k-1} \psi(x,b_j)$ has the same Morley rank as $\psi$?
Instead, try this: you know that each formula $\psi(x,b_j)$ has Morley rank $\alpha$, so you can extend each one to a complete type $p_j$ of Morley rank $\alpha$ over an $|A|^+$-saturated model containing $(b_j)_{j\in \omega}$. Some of the $p_j$ might be equal (i.e. it's possible that $p_0 = p_1$). But you can show that the set $\{p_j\mid j\in \omega\}$ must be infinite, and then use these infinitely many types of Morley rank $\alpha$ to show that $\mathrm{MR}(\phi)\geq \alpha+1$.
