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Basically, if $\varphi$ is irreducibly self-referential via $\chi$, then (modulo a special consideration for the role of $\mathsf{PA}$) reading $\varphi$ as "I have property $\chi$" is both appropriate and optimal.

Basically, if $\varphi$ is irreducibly self-referential via $\chi$, then (modulo a special consideration for the role of $\mathsf{PA}$) reading $\varphi$ as "I have property $\chi$" is completely justified: not only is there no information lost (corresponding to inequivalent fixed points), but there is also no unnecessary self-reference added.

• Non-example: Let $\chi=\neg\mathsf{Prov}_{\mathsf{PA}}$ be the usual unprovability predicate for $\mathsf{PA}$, and let $\varphi$ be a fixed point for that predicate. It is known that any two fixed points of $\neg\mathsf{Prov}_\mathsf{PA}$ are $\mathsf{PA}$-provably equivalent, so $\varphi$ satisfies the 1/2-equivalence. However, taking $\psi=\perp$ we see that $\varphi$ fails to satisfy the 1/3-equivalence: $\mathsf{PA}\vdash\neg\mathsf{Prov_{PA}}(\perp)\leftrightarrow\varphi$ is true (indeed this is how we establish the 1/2 equivalence for $\varphi$) but $\mathsf{PA}\vdash\perp\leftrightarrow\varphi$ is (hopefully!) not. So the usual Godel sentence is not irreducibly self-referential via $\neg\mathsf{Prov}_{\mathsf{PA}}$ ... although in principle it could be irreducibly self-referential via some other formula.

• Silly examples: On the other hand, the unmodified provability formula $\mathsf{Prov_{PA}}(-)$ does show that every $\mathsf{PA}$-theorem is irreducibly self-referential. Suppose $\mathsf{PA}\vdash\varphi$ and $\psi$ is some other sentence. By Lob's Theorem, condition (2) is equivalent to $\mathsf{PA}\vdash\psi$, which in turn is equivalent to condition $(1)$ since $\mathsf{PA}\vdash\varphi$, and we clearly have condition $(1)$ implying condition $(3)$. Meanwhile, soundness of $\mathsf{PA}$ + $\mathsf{PA}\vdash\varphi$ also shows that $(3)$ implies $\mathsf{PA}\vdash\psi$, which gives us $(1)$. A related argument via non-disprovability gives the i.s.r.-ness of every $\mathsf{PA}$-disprovable sentence. (An earlier version of this claim was bonkers; thanks to Fedor Pakhomov for pointing this out!)

• Non-example: Let $\chi=\neg\mathsf{Prov}_{\mathsf{PA}}$ be the usual unprovability predicate for $\mathsf{PA}$, and let $\varphi$ be a fixed point for that predicate. It is known that any two fixed points of $\neg\mathsf{Prov}_\mathsf{PA}$ are $\mathsf{PA}$-provably equivalent, so $\varphi$ satisfies the 1/2-equivalence. However, taking $\psi=\perp$ we see that $\varphi$ fails to satisfy the 1/3-equivalence: $\mathsf{PA}\vdash\neg\mathsf{Prov_{PA}}(\perp)\leftrightarrow\varphi$ is true (indeed this is how we establish the 1/2 equivalence for $\varphi$) but $\mathsf{PA}\vdash\perp\leftrightarrow\varphi$ is (hopefully!) not. So the usual Godel sentence is not irreducibly self-referential via $\neg\mathsf{Prov}_{\mathsf{PA}}$ ... although in principle it could be irreducibly self-referential via some other formula.

• Silly examples: On the other hand, the unmodified provability formula $\mathsf{Prov_{PA}}(-)$ does show that every $\mathsf{PA}$-theorem is irreducibly self-referential. Suppose $\mathsf{PA}\vdash\varphi$ and $\psi$ is some other sentence. By Lob's Theorem, condition (2) is equivalent to $\mathsf{PA}\vdash\psi$, which in turn is equivalent to condition $(1)$ since $\mathsf{PA}\vdash\varphi$, and we clearly have condition $(1)$ implying condition $(3)$. Meanwhile, soundness of $\mathsf{PA}$ + $\mathsf{PA}\vdash\varphi$ also shows that $(3)$ implies $\mathsf{PA}\vdash\psi$, which gives us $(1)$. A related argument via non-disprovability gives the i.s.r.-ness of every $\mathsf{PA}$-disprovable sentence. (Thanks to Fedor Pakhomov and Sridhar Ramesh for fixing very silly errors in this part!)

• Non-example: Let $\chi=\neg\mathsf{Prov}_{\mathsf{PA}}$ be the usual unprovability predicate for $\mathsf{PA}$, and let $\varphi$ be a fixed point for that predicate. It is known that any two fixed points of $\neg\mathsf{Prov}_\mathsf{PA}$ are $\mathsf{PA}$-provably equivalent, so $\varphi$ satisfies the 1/2-equivalence. However, taking $\psi=\perp$ we see that $\varphi$ fails to satisfy the 1/3-equivalence: $\mathsf{PA}\vdash\neg\mathsf{Prov_{PA}}(\perp)\leftrightarrow\varphi$ is true (indeed this is how we establish the 1/2 equivalence for $\varphi$) but $\mathsf{PA}\vdash\perp\leftrightarrow\varphi$ is (hopefully!) not. So the usual Godel sentence is not irreducibly self-referential via $\neg\mathsf{Prov}_{\mathsf{PA}}$ ... although in principle it could be irreducibly self-referential via some other formula.

• Silly examples: On the other hand, the unmodified provability formula $\mathsf{Prov_{PA}}(-)$ does show that every $\mathsf{PA}$-theorem is irreducibly self-referential. Suppose $\mathsf{PA}\vdash\varphi$ and $\psi$ is some other sentence. By Lob's Theorem, condition (2) is equivalent to $\mathsf{PA}\vdash\psi$, which in turn is equivalent to condition $(1)$ since $\mathsf{PA}\vdash\varphi$, and we clearly have condition $(1)$ implying condition $(3)$. Meanwhile, Lob's Theorem + $\mathsf{PA}\vdash\varphi$ also shows that $(3)$ implies $\mathsf{PA}\vdash\psi$, which gives us $(1)$. A related argument via non-disprovability gives the i.s.r.-ness of every $\mathsf{PA}$-disprovable sentence. (An earlier version of this claim was bonkers; thanks to Fedor Pakhomov for pointing this out!)

• Non-example: Let $\chi=\neg\mathsf{Prov}_{\mathsf{PA}}$ be the usual unprovability predicate for $\mathsf{PA}$, and let $\varphi$ be a fixed point for that predicate. It is known that any two fixed points of $\neg\mathsf{Prov}_\mathsf{PA}$ are $\mathsf{PA}$-provably equivalent, so $\varphi$ satisfies the 1/2-equivalence. However, taking $\psi=\perp$ we see that $\varphi$ fails to satisfy the 1/3-equivalence: $\mathsf{PA}\vdash\neg\mathsf{Prov_{PA}}(\perp)\leftrightarrow\varphi$ is true (indeed this is how we establish the 1/2 equivalence for $\varphi$) but $\mathsf{PA}\vdash\perp\leftrightarrow\varphi$ is (hopefully!) not. So the usual Godel sentence is not irreducibly self-referential via $\neg\mathsf{Prov}_{\mathsf{PA}}$ ... although in principle it could be irreducibly self-referential via some other formula.

• Silly examples: On the other hand, the unmodified provability formula $\mathsf{Prov_{PA}}(-)$ does show that every $\mathsf{PA}$-theorem is irreducibly self-referential. Suppose $\mathsf{PA}\vdash\varphi$ and $\psi$ is some other sentence. By Lob's Theorem, condition (2) is equivalent to $\mathsf{PA}\vdash\psi$, which in turn is equivalent to condition $(1)$ since $\mathsf{PA}\vdash\varphi$, and we clearly have condition $(1)$ implying condition $(3)$. Meanwhile, soundness of $\mathsf{PA}$ + $\mathsf{PA}\vdash\varphi$ also shows that $(3)$ implies $\mathsf{PA}\vdash\psi$, which gives us $(1)$. A related argument via non-disprovability gives the i.s.r.-ness of every $\mathsf{PA}$-disprovable sentence. (An earlier version of this claim was bonkers; thanks to Fedor Pakhomov for pointing this out!)