 The proof of (2) runs like this. First we prove the implication from left to right. If \(\vdash \exists x\)Bew\((x, \langle \phi \rangle)\) then there is some \(n\) such that \(\not\vdash \neg\)Bew\((n, \langle \phi \rangle)\), by \(\omega\)-consistency. By completeness we get \(\vdash\)Bew\((n, \langle \phi \rangle)\) for this \(n\). By (1) above we get that \(n\) denotes a proof of \(\phi\). That is, \(\phi\) is provable, so we have \(\vdash \phi\). To prove the implication from right to left, note that if \(\vdash \phi\) then there must be an \(n\) such that \(\vdash\) Bew\((n, \langle \phi \rangle)\), by (1). From this we get \(\vdash \exists x\)Bew\((x, \langle \phi \rangle)\), as required. This concludes the proof of (2). Now, when in a complete theory we have (2), we must also have: