Morley Rank of Conjunction Let $M$ be an $L$-structure. Let $\varphi ( x )$ and $\psi (x)$ be $L_{ M }$-formulas, where $x$ is some finite tuple of variables. With $\mbox{RM}$ we mean the Morley rank with respect to $M$ and $x$. (To recall the definition, see for example page 86 in http://people.math.sc.edu/mcnulty/modeltheory/Henson.pdf)
Then, is it true that
$$ \mbox{RM} ( \varphi (x) \wedge \psi (x) ) = \min \{ \mbox{RM} ( \varphi (x) ) , \mbox{RM} ( \psi (x) ) \} \; ? $$
I suspect this is true. I tried to prove this, and came up with something which might be correct, though I'm not sure. I will sketch the proof now.
First, it is sufficient to prove that for all ordinals $\alpha$ it holds that
$$  \mbox{RM} ( \varphi (x) \wedge \psi (x) ) \geq \alpha \quad \Leftrightarrow \quad [ \mbox{RM} ( \varphi (x) ) \geq \alpha \; \mbox{ and } \; \mbox{RM} ( \psi (x) ) \geq \alpha ]. $$
(Is this true?)
Second, we show this equivalence. I'm not sure if it's correct, and I'd be glad if somebody could confirm that my proof is true/false.


*

*Left-to-right: Since $M \models (\varphi (x) \wedge \psi (x)) \to \varphi (x) $ and $ M \models (\varphi (x) \wedge \psi (x)) \to \psi (x) \, $,  this is trivial.

*Right-to-left: This is the harder case. Again, we need only do the successor case $\alpha = \beta +1$. By definition, there is an elementary extension $B_1$ of $M$ and a sequence $(\varphi_{1,k} \, | \, k \in \mathbb N )$ of $L_{B_1}$-formulas, and similarly an elementary extension $B_2$ of $M$ and a sequence $(\varphi_{2,k} \, | \, k \in \mathbb N )$ of $L_{B_2}$-formulas, such that 
(a) $B_1 \models \varphi_{1,k} (x) \to \varphi (x)$ for all $k \in \mathbb N  $, and $B_2 \models \varphi_{2,k} (x) \to \psi (x)$ for all $k \in \mathbb N$;
(b) $B_1 \models \forall x \, \neg ( \varphi_{1,k} (x) \wedge \varphi_{1, \, l} (x) )$ for all distinct $k,l \in \mathbb N$, and $B_2 \models \forall x \, \neg ( \varphi_{2,k} (x) \wedge \varphi_{2, \, l} (x) )$ for all distinct $k,l \in \mathbb N$;
(c) $\mbox{RM} (B_1,  \varphi_{1,k} (x) ) \geq \beta$ for all $k \in \mathbb N$, and $\mbox{RM} (B_2,  \varphi_{2,k} (x) ) \geq \beta$ for all $k \in \mathbb N$.
Now, applying Robinson's Consistency Theorem (do I really need this?) we can find an $L \cup B_1 \cup B_2$-structure $N$ which is an elementary extension of both $B_1$ and $B_2$.
For each $k$, define $\varphi_k (x)$ to be $\varphi_{1,k} (x) \wedge \varphi_{2,k} (x)$. From (c) and the induction hypothesis we know $\mbox{RM} (N,  \varphi_{k} (x) ) \geq \beta$ for all $k \in \mathbb N$. Also, it is clear that $N \models \varphi_{k} (x) \to ( \varphi (x) \wedge \psi (x) )$ for all $k \in \mathbb N$ (so (a) holds), and $N \models \forall x \, \neg ( \varphi_{k} (x) \wedge \varphi_{ l} (x) )$ for all distinct $k,l \in \mathbb N$ (so (b) holds). Therefore, $\mbox{RM} ( \varphi (x) \wedge \psi (x) ) \geq \beta +1$.
Is the proof correct?
 A: As others have noted, the result is false. I'll give a conceptual reason why we shouldn't expect it to be true. Rather than think about Morley rank as applying to formulas, you can think of Morley rank as applying to complete types. Thinking of formulas as (clopen) sets of types, the Morley rank of a formula is just the supremum (equivalently, max) of the Morley rank of the types in it. The conjunction of two formulas corresponds to taking an intersection of the sets of types, and similarly a disjunction corresponds to a union.
I suspect you made this conjecture because the Morley rank of a disjunction is the max of the rank of the disjuncts. That fact follows easily from the above characterization: in a union, the type achieving maximum Morley rank must lie in one of the components. There is no dual line of reasoning for an intersection, however: if you intersect two sets of types, the types realizing the maximum Morley rank in each component may not lie in the intersection.
A: As Pece notes in the comments, what you're trying to prove is not true. Any two disjoint definable sets provide a counterexample: if $\varphi(x)$ and $\psi(x)$ are both consistent, but $\varphi(x)\land\psi(x)$ is inconsistent, then $\text{RM}(\varphi(x)) \geq 0$ and $\text{RM}(\psi(x))\geq 0$ but $\text{RM}(\varphi(x)\land \psi(x)) = -\infty$.
In fact, this is the only error in your proof: you've forgotten to do the base case $\alpha = 0$!
