Let $\mathcal{N}=(\mathbb{N}, ...)$ be the standard model of Q (Robinson Arithmetic), and let $\mathcal{N}^{\ast}=(N, ...)$ be an arbitrary model. Let $\varphi$ be a $\Sigma_1$-theorem, and let $\psi$ be a $\Pi_1$-theorem.
Show, that if $\mathcal{N}\models\varphi$, then $\mathcal{N}^{\ast}\models\varphi$. If $\mathcal{N}^{\ast}\models\psi$, then $\mathcal{N}\models\psi$
Hello,
I want to show this statement, and need some help. My problem is, that our definition of $\Sigma_1$ and $\Pi_1$-theorems is a little bit odd. I hope these are common notations and I would be thankfull if you could give me maybe a link to a definition, which is probably more clear, than our.
Otherwise we might use induction on the complexity of the formula to show this results. $\mathcal{N}^{\ast}$ is an arbitrary model. Does that mean that just the carrier-set $N$ can be different of $\mathbb{N}$, or does it also mean $\mathcal{N}^{\ast}$ can contain other functions and relations?
Thanks in advance.
Note that $\lnot\psi$ is a $\Sigma_1$ statement. If $\cal N\not\models\psi$, then it satisfies $\lnot\psi$. Now use the fact about $Q$ and true $\Sigma_1$ sentences to conclude that $\cal N^*$ cannot satisfy $\psi$ either.