@user21820 Another question that occurred to me: Robinson arithmetic with a doubling function rather than multiplication: D0 = 0 and D(Sx) = S(S(Dx)). How strong is it? Is it essentially undecidable?
If that's too weak, what about a doubling and a tripling function T0 = 0 and T(Sx) = S(S(S(Tx)))?
@user41805 That's what I was thinking. You can replace any instance of Dx with x+x and Tx with x+x+x. But what about some other unary function? Say, modulo 2? M0 = 0, M(S0) = S0, M(S(Sx)) = Mx.
