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?

> Q1. Is it true that for every $\Pi_2$ formula $\phi$, $Q\vdash_c \phi$ iff $Q\vdash_i \phi$?

Let $$Q^e=Q\cup \{x=y \lor\neg(x=y),x+z=y+z\rightarrow x=y,\neg(x=y)\leftrightarrow (x<y \lor y<x) \}$$

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] for intutionistic logic, but I'm not sure.

  Thanks.


  [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