 One elegant and simple way of demonstrating the undecidability of extensions of Q goes, roughly, as follows: Let \(F\) be any consistent theory that contains Q.  Assume then that the set of its theorems is decidable, that is (by the Church-Turing thesis), recursive. It would then follow that the set (of the Gödel numbers) of the theorems of \(F\) is strongly representable in \(F\) itself. Recall that this means that there is some formula \(B(x)\) of the language of \(F\) such that not only \(F \vdash B(\ulcorner A\urcorner)\) whenever \(F \vdash A\) (which even weak representability guarantees), but also that \(F \vdash \neg B(\ulcorner A\urcorner)\) whenever \(F \not\vdash A\). However, the technique used in the proof of the first incompleteness theorem also shows that there are always sentences for which the latter does not hold: it is possible to construct a Gödel sentence \(G^B\) relative to \(B(x)\) for \(F\) such that: