I don't know much about this topic so I am sorry if my question is silly. As far as I understand if neutrinos are Majorana particles, one consequence is that neutrinos are their own antiparticles. This can be observed, for example, in neutrinoless double beta decay. However, if we take the following reaction: $$\nu+p\to e^++n$$ we know from experiment that when $\nu$ is what we identify as an antineutrino the reaction is observed, but when $\nu$ is what we call a neutrino, the reaction doesn't take place. If the neutrino and antineutrino were the same particles, shouldn't both reaction take place equally often? Isn't this a clear evidence that neutrino is not its own antiparticle and hence not a Majorana particle? Of course I am missing something but I am not sure what. Can someone enlighten me please?
In $\overline \nu + p \to e^+ + n$ the (anti) neutrino is right-handed and interacts via a virtual $W$ to give a right-handed positron, and this reaction happens.
In $\nu + p \to e^+ + n$ the neutrino is left handed and would have to interact via a virtual $W$ to give a left-handed positron. But the weak interaction only occurs for right handed particles and left handed antiparticles, so this doesn't happen.
So the selection rules for lepton number remain, but are due to the selection rules for handedness and there is no need for an additional quantum number to keep track of the distinction between $\nu$ and $\overline \nu$. That's all contained in whether the particle is right-handed or left-handed.