# Uniqueness of super godel numbers of $\varphi$ and $\neg \varphi$

Let $e_{0},e_{1},...,e_{n}$ be a sequence of wffs or other expressions. Code each $e_{i}$ by a regular godel number $g_{i}$, to yield a sequence of numbers $g_{0},g_{1},...,g_{n}$. Then encode this sequence of regular godel numbers using a super godel number, to get $$2^{g_{0}} \cdot 3^{g_{1}} \cdot 5^{g_{2}} \cdot ... \cdot \pi_{n}^{g_{n}}$$ where $\pi_{n}$ is the $n+1$-th prime number. Then, define $Prf(m,n)$ to hold just if $m$ is the super godel number of a sequence of wffs that is a $\mathsf{PA}$ proof (Peano Arithmetic) of the closed wff with regular godel number $n$.

I am working on a much wider question to do with Rosser provability, but I am stuck inside of a fifth subproof, where I simply need to show that from $Prf(k,\ulcorner \neg \urcorner \star \ulcorner 0 =1 \urcorner) \wedge Prf(c,\ulcorner 0=1 \urcorner)$, that I can prove $c \neq k$, where $m \star n$ is the standard concatenation function. Here is the start of my attempt, although I am certain there is a very simple way!

Assume $c=k$. Further assume that $(\ulcorner \neg \urcorner \star \ulcorner 0=1 \urcorner)=(\ulcorner 0=1 \urcorner)$. Then, $$2^{1} \cdot 3^{21} \cdot 5^{15} \cdot 7^{23} \cdot 11^{21} = 2^{21} \cdot 3^{15} \cdot 5^{23} \cdot 7^{21}$$ by the standard godel coding of, $\neg: 1, 0: 21, =:15,S:23$, where $S$ is the successor function. This contradicts the fundamental theorem of arithmetic, so $(\ulcorner \neg \urcorner \star \ulcorner 0=1 \urcorner) \neq (\ulcorner 0=1 \urcorner)$.

Here is where I am stuck. I know previously in the proof that $Prf(k,\ulcorner \neg \urcorner \star \ulcorner 0 =1 \urcorner)$ and $Prf(c,\ulcorner 0=1 \urcorner)$, so how can I derive a contradiction to conclude that $c \neq k$?

Any help is greatly appreciated and if necessary I can explain more of the background problem, but I am certain that this part of the proof can be solved independently without relying on anything else other than simple logic and the definitions I have provided. I suspect that it has something to do with the uniqueness of the super godel number; namely, that there does not exist a number $n$ which is the super godel number of both $\ulcorner \neg \urcorner \star \ulcorner \varphi \urcorner$ and $\ulcorner \varphi \urcorner$. Yet I cannot represent my intuition formally!

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Cross-posted on math.SE: math.stackexchange.com/questions/122419/… – Asaf Karagila Mar 20 '12 at 8:52

More specifically, one can prove in PA right from the definition you have given that if $\text{Prf}(c,n)$, then $n$ is the exponent of the largest prime dividing $c$. From this it follows immediately that $\text{Prf}(c,n)\wedge\text{Prf}(k,m)\wedge n\neq m\implies c\neq k$, since if $c=k$, then both $n$ and $m$ would be that exponent, contradiction. This answers the question, since it is easy to prove that $\neg 0=1$ and $0=1$ have distinct Gödel codes.