Showing that the Morley rank is the same.

Question

Let $A$ be an L-structure with some formula $\phi(x)$ and some $t(x)$ as an L-term, where $x=x_1,...,x_m$. I wanna show that the Morley rank of $\phi(x) \wedge y$ = $t(x)$ and $\phi(x)$ are the same. I'm wondering what this property in general would be called?(some kind of Morley rank-invariance). And, how would you go about the proof?

Answer 1

The more general property is that Morley rank is preserved under definable bijections. That is, if we have definable sets $X$ and $Y$ and a definable bijection $f: X \to Y$ (i.e. the graph of $f$ is a definable set) then the Morley rank of $X$ is the same as the Morley rank of $Y$. Definable here means definable with parameters.

In your case the following defines a bijection between $\phi(x)$ and $\phi(x') \wedge y = t(x')$, where $x'$ is a copy of the variables in $x$: $$ \phi(x) \wedge x' = x \wedge y = t(x). $$