Formula and its Negation Cannot Both Cause Forking

From ProofWiki
Jump to navigation Jump to search


Let $T$ be a complete $\mathcal{L}$-theory.

Let $\mathfrak{C}$ be a monster model for $T$.

Let $A\subseteq B$ be subsets of the universe of $\mathfrak{C}$.

Let $\pi(\bar x)$ be an $n$-type over $B$.

If $\pi$ does not fork over $A$, then for any formula $\phi(\bar x, \bar b)$, either $\pi \cup \{\phi\}$ or $\pi \cup \{\neg\phi\}$ does not fork over $A$.


We prove the contrapositive.

Suppose both $\pi \cup \{\phi\}$ and $\pi \cup \{\neg\phi\}$ fork over $A$.

By definition of forking,

$\pi\cup\{\phi\}$ implies $\phi_1 (\bar x, \bar c_1) \vee \cdots \vee \phi_k (\bar x, \bar c_k)$ and
$\pi\cup\{\neg\phi\}$ implies $\psi_1 (\bar x, \bar d_1) \vee \cdots \vee \psi_k (\bar x, \bar d_h)$,

where each $\phi_i$ and $\psi_j$ divide over $A$.

Then $\pi$ implies the disjunction

$\bigvee_i \phi(\bar x, \bar c_i) \vee \bigvee_j \psi(\bar x, \bar d_i)$

with each component formula dividing over $A$.

By definition, this means that $\pi$ forks over $A$.