Given a class $C$ of arithmetical sentences, 
an arithmetical theory $T$ is said to be $C$-sound if 
all the theorems of $T$ which are in $C$ are true. 
For instance, $T$ is $\Sigma\_1$-sound if all the $\Sigma\_1$ theorems of $T$ are true. 


Now, for some classes $C$, like the class of $\Sigma\_1$ sentences, 
the statement "$T$ is $C$-sound" is expressible in the language of arithmetic.
For other classes, like the class of all arithmetical sentences, $C$-soundness 
(which is just soundness) can't be definable in the language of arithmetic.
But there are some classes, like the class of $\Pi\_1$ sentences, 
which have a rarer property: 
not only is $C$-soundness definable in the language of arithmetic, it can be defined within $C$ itself. 


However, I'm looking for an even rarer property:



> 
> does there exist a class $C$ for which 
>  $C$-soundness is definable within $C$, *and* 
>  which is also closed under negation?
> 
> 
> 


Or, failing that:



> 
> does there exist a class $C$ for which 
>  "$T$ is **not** $C$-sound" is expressible within $C$?
> 
> 
> 



Can a class of arithmetical statements containing its own soundness condition be closed under negation?
The answer to both questions is negative. 


**Theorem.** There is no class $C$ of formulas in the language of
arithmetic, such that
the assertion "$T$ is not $C$-sound" is uniformly expressible in $C$
for c.e. theories $T$ (regarded as an index of $T$ as a c.e. set).


Proof. Suppose $C$ is like that. By the Gödel fixed-point lemma, there is a sentence $\psi$
such that $$\text{PA}\vdash\psi\longleftrightarrow
(\ \{\text{PA}+\psi\}\text{ is not }C\text{-sound }).$$ The assumption on $C$ ensures that the
assertion "$T$ is not $C$-sound" is expressible in $C$, and so it follows that
$\psi$ is PA-provably equivalent to a sentence in $C$, and we may
assume without loss that $\psi$ is actually in $C$.


On the one hand, if $\psi$ is true, then $\text{PA}+\psi$ is true
and therefore also sound, since every true theory is sound. But
since $\psi$ asserts that $\text{PA}+\psi$ is not $C$-sound, and
since this assertion is in $C$, then it must be true that
$\text{PA}+\psi$ is not sound, a contradiction.


On the other hand, if $\psi$ is false, then because $\psi$ is in
$C$, it is true that $\text{PA}+\psi$ is not $C$-sound. But in
this case, $\psi$ would be true, contrary to our assumption. QED


