By Friedman translation $HA$ and $PA$ prove the same $\Pi_2$ formulas. Is it true for Intutionistic Robinson arithmetic(Robinson axioms with intutionistic logic) and classic Robinson arithmetic?
Axioms of $Q$ are:
- $\neg(Sx=0)$
- $Sx=Sy\rightarrow x=y$
- $y=0 \lor \exists x(Sx=y)$
- $x+0=x$
- $x+Sy=S(x+y)$
- $x\cdot 0=0$
- $x\cdot Sy=(x\cdot y)+x$
- $\neg(x<0)$
- $0=x\lor 0<x$
- $x<y \leftrightarrow (Sx<y \lor Sx=y)$
- $x<Sy \leftrightarrow (x<y\lor x=y)$
Q1. Is it true that for every $\Pi_2$ formula $\phi$, $Q\vdash_c \phi$ iff $Q\vdash_i \phi$?
Let $$Q^e=Q\cup \{x=y \lor\neg(x=y),\neg(x=y)\leftrightarrow (x<y \lor y<x) \}$$
What happens to Q1 if we replace $Q$ by $Q^e$?
Q2. Is it true that for every $\Pi_2$ formula $\phi$, $Q^e\vdash_c \phi$ iff $Q^e\vdash_i \phi$?
I think the second question can be proved by strong completeness of [Beth model][1]Beth model for intutionistic logic, but I'm not sure.
Thanks.
Edit:
Definition. The set $\Delta^+_0$ formula is the smallest set such that:
- $s=t\in \Delta^+_0$ for every term $s$ and $t$,
- $s<t\in \Delta^+_0$ for every term $s$ and $t$,
- if $\phi,\psi\in \Delta^+_0$, then $\phi\circ \psi\in\Delta^+_0$ where $\circ\in \{\lor,\land \}$,
- if $\phi\in \Delta^+_0$, then $\exists x(x<s \land \phi(x))\in \Delta^+_0$ where $s$ is a term.
- if $\phi\in \Delta^+_0$, then $\forall x(x<s \rightarrow \phi(x))\in \Delta^+_0$ where $s$ is a term.
By $\Pi_2$ formula $\phi$ I mean $\phi=\forall{\bf x}\exists{\bf y}\psi({\bf x},{\bf y})$ where $\psi\in \Delta^+_0$. [1]: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwixw6LikZvKAhXJuhoKHTK-CGYQFggiMAA&url=http%3A%2F%2Faleteya.cs.buap.mx%2F~jlavalle%2Fpapers%2Fvan%2520Dalen%2FIntuitionistic%2520Logic.pdf&usg=AFQjCNGTB77-ykRxaWnWNCgjp8K43TEmcA&sig2=d4SvNkos7aH1KtwXhnxwMA
