Show that Goldbach’s Conjecture is true if and only if $N \not\vdash \neg Goldbach$.

Question

Let $N$ be the axioms of Robinson Arithmetic. Show that Goldbach’s Conjecture is true if and only if $N \not\vdash \neg \text{Goldbach}$.

First, I assume that $N \not\vdash \neg \text{Goldbach}$ and want to prove that Goldbach’s Conjecture is true.

Since, $N \not\vdash \neg \text{Goldbach}$ and since, Goldbach’s Conjecture is a $\Pi$-sentence.

Then, this means $N$ is not strong enough to prove that not every true $\Pi$-sentence.

Then, this means $N$ is not strong enough to prove that exists false $\Pi$-sentence.

So, this means Goldbach’s Conjecture is true.

$\mathbf{Well,~I~am~not~sure~if~the~prove~of~(\Leftarrow)~is~okay}$

$\mathbf{and~also~I~have~no~idea~on~proving~the~opposite~direction~(\Rightarrow)~which~is}$

$\mathbf{assume~that~Goldbach’s~Conjecture~is~true~and~want~to~prove~that}$ $N \not\vdash \neg Goldbach$.

Answer 1

"Only if" is just consistency: if $N$ can prove Goldbach is false then it actually is false.

For "if", suppose Goldbach is not true. Then there exists a counterexample, $n$. Since verifying that $n$ is a counterexample just requires checking finitely many numbers are composite, Robinson arithmetic can prove that $n$ is a counterexample, and hence $N\vdash\neg \text{Goldbach}$.