Is true arithmetic + $\lnot Con (TA)$ consistent?

Question

Is the theory $TA+\lnot Con(TA)$ consistent?

In particular, for every $TA \vdash \phi$, we take as an axiom $\phi$, and $TA \vdash \phi$. We also assert $TA \vdash 0 = 1$. We call this theory $TA + \lnot Con(TA)$.

Note that the theory we are talking about does not include the statement that $TA$ is true (i.e. $TA \vdash \phi \iff \phi$). The theory can only see what $TA$ implies, and that $TA$ is a theory of first order arithmetic.

Is this theory consistent?

Note: TA + $\lnot Con(TA)$ is not a theory in the language of arithmetic. We are adjoining a symbol for $TA \vdash$ to the language (as an unary predicate on godel codes). You can also interpret it as a weak set theory.

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EDIT: This is an attempt by someone other than the OP to describe the theory in question in a clearer way; this should not be taken as fully accurate until the OP weighs in.

Let $T_0$ be the theory, in the language of arithmetic together with a new unary predicate symbol $P$, consisting of the following axioms:

• Each $\varphi\in $ TA.

• Each sentence of the form "$P(n)$" for $n$ the Godel number of a sentence in TA.

• $P(\ulcorner 0=1\urcorner)$.

This is (I believe) the theory the OP describes.

Meanwhile, here are some sentences which are not in our theory (I'm explicitly mentioning them to make it clear that the above really is a complete description of the OP's theory in question):

• Internal modus ponens (IMP): The sentence $$\forall x, y[P(x)\wedge P(implies(x, y)\implies P(y)].$$ Here "$implies$" is an abbreviation for the usual primitive recursive binary function given by $(\ulcorner p\urcorner, \ulcorner q\urcorner)\mapsto \ulcorner p\implies q\urcorner$. The point is that we could write this as an axiom scheme, ranging over all actual Godel numbers of appropriate sentences, but we can actually just get it with a single sentence. This isn't a big deal, and if one prefers one can use the scheme version without any impact here.

• Reflection (R): the scheme "$P(\ulcorner \varphi\urcorner)\implies\varphi$" for arithmetic $\varphi$.

• Completeness (C): the scheme "$\varphi\implies P(\ulcorner\varphi\urcorner)$" for arithmetic $\varphi$.

Answer 1

Usually we use the notation $\newcommand\TA{\text{TA}}\TA$ to denote the theory of true arithmetic, meaning the theory of the standard model of arithmetic $\langle\mathbb{N},+,\cdot,0,1,<\rangle$ in the usual language of arithmetic, or in other words, the collection of all sentences that are true in that structure.

Tarski's theorem on the non-definability of truth, however, tells us that the set of Gödel codes of assertions in $\TA$ is not arithmetically definable. It follows that we cannot express $\text{Con}(\TA)$ directly in the language of arithmetic, in the way that we are able to express $\text{Con}(\text{PA})$ or $\text{Con}(T)$ for other arithmetically definable theories $T$.

For this reason, with the ordinary usage of $\text{Con}$, there is no such assertion as $\text{Con}(\TA)$, and it is not sensible to write such things as $\TA+\neg\text{Con}(\TA)$.

If one wants truly to refer to the consistency of $\TA$, one must therefore take far more care in the details of the formalization, to make it clear exactly how it is that one will try to refer to the consistency of this theory and what it is that is meant. I don't see that you have really done that extra work sufficiently in your question, and I think that this is the reason your question has not had a positive reception.

Meanwhile, let me point you toward the Mostowski reflection theorem, which I believe can be taken as a kind of negative answer to a version of the question that one might have asked.

Namely, it is not difficult to see that for any standard natural number $n$, one can write down a definition of a $\Sigma_n$ truth predicate $\text{Tr}_n$, and $\newcommand\PA{\text{PA}}\PA$ proves that indeed this definition fulfills the Tarskian recursion for all $\Sigma_n$ formulas. Thus, $\text{Tr}_n$ is an arithmetically definable predicate referring to the $\Sigma_n$ fragment of true arithmetic. This is an approximation to $\TA$.

Theorem.(Mostowski reflection theorem) For any standard natural number $n$, the theory $\PA$ proves $\text{Con}(\text{Tr}_n)$.

See Mostowski, A., On models of axiomatic systems, Fundam. Math. 39, 133-158 (1953). ZBL0053.20102.

This theorem can be seen as a stronger version of the often mentioned fact that $\PA$ proves the consistency of all of its particular finite subtheories. In particular, this implies that $\PA$ is not finitely axiomatizable.

Answer 2

I think you haven't written what you want to.

At present your theory has a simple model: it's the expansion of standard arithmetic by interpreting $P$ as exactly the set of Godel numbers of sentences of true arithmetic and of "$0=1$."

One thing missing from this is an appropriate "deduction theorem" axiom - you don't have a rule saying "$P(p), P(p\implies q)\implies P(q)$" or similar (conflating sentences and their Godel numbers). Adding this lets us prove "$\forall x(P(x))$" since "$0=1\implies p$" is in TA for every sentence $p$.

However, we still have a consistent theory! Namely, if $\mathcal{M}$ is any model of TA then the expansion of $\mathcal{M}$ by interpreting $P$ as the set of all natural numbers gives a model of this stronger theory. This reflects the really important problem with the theory you've described: you don't have any kind of reflection. That is, just knowing "$P(n)$" doesn't let you conclude anything about arithmetic, and so there's no interaction between the two parts of the theory.

In order to get your "internal version" of TA to work properly, the most natural thing to add is the reflection scheme $$P(\ulcorner p\urcorner)\implies p$$ as $p$ ranges through the sentences of true arithmetic. However, at this point we get an immediate contradiction since $P(\ulcorner 0=1\urcorner)$ and the reflection scheme gives $0=1$. On the other hand, without a reflection scheme there's no connection between $P$ and truth in the model, so no reason to believe that anything weird is happening.

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EDIT: In the language of my edit to the OP, here is a summary of the situation:

• $T_0$ is consistent: for example, the natural numbers together with $P$ interpreted as the whole universe satisfies $T_0$ (note that no axiom in $T_0$ implies $\exists x(\neg P(x))$).

• $T_0$ + C + IMP is also consistent: the model above is also a model of this theory (again, because neither C nor IMP provide any negative facts about $P$).

• $T_0$ + R is inconsistent. Indeed, the tiny subtheory R$_{0=1}$ + $P(\ulcorner 0=1 \urcorner)$ + $\neg 0=1$ is inconsistent (where "R$_\varphi$" denotes the instance of reflection for $\varphi$).

The first bulletpoint means that the answer to your question as asked (assuming I've interpreted it correctly) is "yes," but the example model indicates that $T_0$ probably isn't what you actually want; meanwhile, the third bulletpoint suggests that the answer to the question I think you want to ask is "no." The second bulletpoint has no direct relevance, I just think it's worth mentioning in the context of considering theories like this.

Answer 3

To elaborate on what everyone else is already saying: if T is an effective (i.e. recursively enumerable) arithmetic theory, then Con(T) is basically an arithmetic sentence asserting that the Turing machine enumerating the theorems of T never generates the sentence "0=1". T's recursive enumerability in turn means (by definition) that a finite sized Turing machine exists that generates those theorems. And Gödel brilliantly showed that this statement (about the Turing machine not generating 0=1, translated from computability jargon) can be encoded as an arithmetic sentence.

Gödel's incompleteness theorem can be seen as proving that True Arithmetic (TA) is not recursively enumerable (since if it was, it would be incomplete, and thus wouldn't be TA). That means there can be no such sentence asserting TA's consistency (since it would have to encode a Turing machine that generates all of TA's sentences, but by incompleteness, there is no such machine).

So the problem is not just that "TA + ¬Con(TA) is not a theory in the language of arithmetic". It's that Con(TA) and ¬Con(TA) aren't even sentences in the language of arithmetic.

That's why your question doesn't seem to make any sense.