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Take any consistent theory $T$ satisfying the provability conditions and the fixed point lemma. Take any sentence $q$ over $T$. $\def\imp{\rightarrow}$ $\def\eq{\leftrightarrow}$ $\def\box{\square}$ $\def\diam{\lozenge}$ Your equation By the fixed point lemma there is a sentence $p$ over $T$ such that $T \vdash p \eq \diam( p \imp q ) \land \neg \box( p \... 1 Hmm your reasoning seems weird. Take any theory$T$that satisfies the provability conditions. Take any$m,n \in \mathbb{N}$such that$m<n$. Then by Lob's theorem, if$T \vdash \square^n \bot \to \square^{n-1} \bot$then$T \vdash \square^{n-1} \bot$. Note that$T \vdash \square^m \bot \to \square^{n-1} \bot$by using (D3) for$m>0$. Thus if$T \vdash ...

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Nevermind, I just realized that in the proof of arithmetical completeness the author also needs to assume 1-consistency (To show that Bew(False) is not provable).

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Your mistake is that you mixed up the different provability operators. See the further remarks in this post. Specifically, let $\def\prov{\square}$$PA' = PA + Con(PA) = PA + \neg \prov_{PA} \bot$. Then $PA'$ is consistent and $PA' \vdash Con(PA)$ but $PA' \nvdash Con(PA') = \neg \prov_{PA'} \bot$. Lob's theorem indeed still holds for $PA'$ as expected, which ...

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Sorry for the late reply. The property $\forall x \forall y \forall z ((Rxy \land Rxz) \to (Ryx \lor Rzy))$ says that whenever there are transition from $x$ to $y$ and $z$, there is either transition from $y$ to $z$, or vice versa. So, let us assume, that in a canonical (which is reflexive and transitive) model $M$ $(w,v) \in R$ and $(w,u) \in R$, but \$(u,v) ...

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GĂ¶del/Scott themselves use both Ax. 2 and Ax. 3. The only difference is in Axiom 1 where you omit the box operator inside the scope of the allquantifier. Here is why I think this is a troublesome assumption. It might well be that it is accidentally the case that all individuals in our World that have a certain positive property A also have a Property B. For ...

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