Skip to main content
MathOverflow

Return to Answer

added 2 characters in body
Source Link
Zuhair Al-Johar
Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47

Noah's answer may be the best here. But I'll add this as an additional answer. In particular, the theory in question is not fully formalized in the language of arithmetic.

Let's take theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$, this is an arithmetically unsound $\omega$-consistent arithmetic theory. Now, to break $\omega$-consistency we can add a primitive constant $c$ to it's language, then add the schema: $$ [\mathcal S_n(0) < c]_{n=0,1,2,..}$$ Where $ \mathcal S_n(0)$ denotes the $n$-iterated successor of $0$, where of course $n$ being concrete natural. This would break $\omega$-consistency at $c$ for the property $(x < c)$.

So, this theory is both arithmetically unsound and $\omega$-inconsistent.

Now, the base theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$, being $\omega$-consistent, cannot prove its own inconsistency, and since the above theory is obtained through compactness from the base theory then it cannot prove a sentence in the language of the base theory that the base theory itself cannot prove. Hence, $T \not \vdash \neg \operatorname{Con}(T)$.

Noah's answer may be the best here. But I'll add this as an additional answer. In particular, the theory in question is not fully formalized in the language of arithmetic.

Let's take theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$, this is an arithmetically unsound $\omega$-consistent arithmetic theory. Now, to break $\omega$-consistency we can add a primitive constant $c$ to it's language, then add the schema: $$ [\mathcal S_n(0) < c]_{n=0,1,2,..}$$ Where $ \mathcal S_n(0)$ denotes the $n$-iterated successor of $0$, where of course $n$ being concrete natural. This would break $\omega$-consistency at $c$ for the property $(x < c)$.

So, this theory is both arithmetically unsound and $\omega$-inconsistent.

Now, the base theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$ being $\omega$-consistent cannot prove its own inconsistency, and since the above theory is obtained through compactness from the base theory then it cannot prove a sentence in the language of the base theory that the base theory itself cannot prove. Hence, $T \not \vdash \neg \operatorname{Con}(T)$.

Noah's answer may be the best here. But I'll add this as an additional answer. In particular, the theory in question is not fully formalized in the language of arithmetic.

Let's take theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$, this is an arithmetically unsound $\omega$-consistent arithmetic theory. Now, to break $\omega$-consistency we can add a primitive constant $c$ to it's language, then add the schema: $$ [\mathcal S_n(0) < c]_{n=0,1,2,..}$$ Where $ \mathcal S_n(0)$ denotes the $n$-iterated successor of $0$, where of course $n$ being concrete natural. This would break $\omega$-consistency at $c$ for the property $(x < c)$.

So, this theory is both arithmetically unsound and $\omega$-inconsistent.

Now, the base theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$, being $\omega$-consistent, cannot prove its own inconsistency, and since the above theory is obtained through compactness from the base theory then it cannot prove a sentence in the language of the base theory that the base theory itself cannot prove. Hence, $T \not \vdash \neg \operatorname{Con}(T)$.

Source Link
Zuhair Al-Johar
Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47

Noah's answer may be the best here. But I'll add this as an additional answer. In particular, the theory in question is not fully formalized in the language of arithmetic.

Let's take theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$, this is an arithmetically unsound $\omega$-consistent arithmetic theory. Now, to break $\omega$-consistency we can add a primitive constant $c$ to it's language, then add the schema: $$ [\mathcal S_n(0) < c]_{n=0,1,2,..}$$ Where $ \mathcal S_n(0)$ denotes the $n$-iterated successor of $0$, where of course $n$ being concrete natural. This would break $\omega$-consistency at $c$ for the property $(x < c)$.

So, this theory is both arithmetically unsound and $\omega$-inconsistent.

Now, the base theory $\sf PA + \neg \, \omega$-$\sf \operatorname{Con}(PA)$ being $\omega$-consistent cannot prove its own inconsistency, and since the above theory is obtained through compactness from the base theory then it cannot prove a sentence in the language of the base theory that the base theory itself cannot prove. Hence, $T \not \vdash \neg \operatorname{Con}(T)$.