 As before, it follows that \(F \not\vdash G^B\). It has been assumed that \(B(x)\) strongly represents the set of theorems, so this entails \(F \vdash \neg B(\ulcorner G^B \urcorner)\), and therefore, by (D), \(F \vdash G^B\), a contradiction. Therefore, \(F\) must be undecidable. 