I'm trying to prove the converse of what was shown here.
Namely, I'm trying to prove B-axioms of modal logic ($\vdash \phi\to \square \diamond \phi$ or $\vdash\diamond\square\phi\to\phi$, whatever is easier) in K if $\vdash\phi \land \diamond\psi \to \diamond(\psi\land\diamond \phi)$ is given, and one is allowed to use any propositional tautology and standard inference rules.
I suppose I need to apply the distribution axiom in a clever way. Originally I was been trying to prove what I need without it (the distribution axiom); it took quite a few pages, but I don't think I made any progress. But after seeing the answer in the question cited, I believe I have to use that axiom (however, again, I'm not sure which sentences to apply it to). Any help is appreciated.
Best Answer
Suppose for contradiction that $\varphi\land \lozenge\square\lnot\varphi$. Then by our assumption, $\lozenge(\square\lnot\varphi\land \lozenge\varphi)$, equivalently $\lozenge\lnot(\square\lnot\varphi\rightarrow\square\lnot\varphi)$, equivalently $\lnot\square(\square\lnot\varphi\rightarrow\square\lnot\varphi)$.
But we have $\square\lnot\varphi\rightarrow\square\lnot\varphi$, so by necessitation $\square(\square\lnot\varphi\rightarrow\square\lnot\varphi)$, contradiction.
It's interesting that I had to use distribution to prove your axiom from B, and conversely I had to use necessitation to prove B from your axiom. Probably there would be a good explanation for this if I understood modal logic better...