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 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 $\psi\in \Delta^+_0$.