If $Q \vdash \lnot\varphi$ (Q is the Robinson arithmetic), and if I assume that $\varphi \in \Pi_1$; Can I say that $\varphi$ is a true sentence?
My thoughts are that, given that Q is $\Sigma_1$- complete, and that $Q \vdash \lnot\varphi$, I can say that $ \lnot\varphi$ is a true $\Sigma_1$ sentence($Q \vdash \lnot\varphi$ would imply that $\varphi$ is true in Q).
I also know that $\varphi \in \Delta_0 $ iff $\lnot \varphi \in \Delta_0$(Not exactly sure why?) then I could extand that to $\exists x \varphi \in \Delta_0$ iff $ \lnot\exists x \varphi \in \Delta_0$, which is then $\varphi \in \Sigma_1$ iff $\lnot \varphi \in \Pi_1$. But here is where i get stuck, what does this information tell me about whether $\phi$ is true given that $\varphi \in \Pi_1$, and $\lnot \varphi$ is true and $\in \Sigma_1$.