Assume $\text{Sat }\Phi $. $\phi$ is independent of $\Phi $ iff $\text{not Sat } \Phi \cup \{\phi\}$ and $\text{not Sat } \Phi \cup \{\neg \phi\}$.
Above definitely doesn't make sense as taking any model $\mathfrak{A} $ of $\Phi$, $\text{Th}(\mathfrak{A})$ is complete thus either $\text{Sat }\Phi\cup\{\phi\}$ or $\text{Sat }\Phi\cup\{\neg\phi\}$ should hold. I got this absurd consequence while thinking about how independence from given set of formulas should be proved. Above statement is derived from the following two elemetary propositions :
- $\Phi \models \phi \text{ iff not Sat } \Phi \cup \{\neg\phi\}$
- $ \Phi \models \neg\phi \text{ iff not Sat } \Phi \cup \{\phi\} $
and thus directly follows from definition of independence and completeness, which is, if I know right, "$\phi$ is independent of $\Phi$ iff $\Phi \vdash \phi$ nor $\Phi \vdash \neg\phi$ holds".
It's a shame I can't come up with flaw myself in such short, elementary argument but would someone point out where it is that I have gone wrong.