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Let $p(x)$ be a complete type over some set of parameters $B$, and let $A$ be a subset of $B$. One says that $p(x)$ splits over $A$ if $\phi(x;b_1) \in p(x), \quad \phi(x;b_2) \notin p(x)$ for some formula $\phi(x;y)$, and $b_1, b_2 \in B$ having the same type over $A$. Splitting is a weaker condition than dividing, so not splitting is a stronger condition than not dividing. If $M$ is a sufficiently saturated model containing $A$, (for example, the monster model), then $p \in S(M)$ doesn't split over $A$ if and only if $p$ is $\operatorname{Aut}(M/A)$-invariant.