I am trying to prove in a Hilbert system $(\lozenge \square p \land \lozenge \square q) \to \square \lozenge (p \land q)$ in $B$. I am stuck with $(\square p \land \square q) \to \square \lozenge (p \land q)$ and I don't know how to get $\lozenge \square$ for each conjunct in the antecedent. Any help is appreciated.
I've tried substitutions of the $B$ axiom using $\lozenge \square p \land \lozenge \square q / p$ to no avail. I have derived $\lnot(\lozenge \square p \land \lozenge \square q) \to \square \lozenge \lnot (p \land q)$ so far starting from $\square(p \land q) \equiv \square p \land \square q$. I am unsure where to go from here.
Here's a proof I found:
$\qquad\begin{array}{rl} 1. &(♢◻p∧♢◻q)→♢◻p\\ 2. &♢◻p→p \\ 3. &(♢◻p∧♢◻q) →p\\ 4. &(♢◻p∧♢◻q) →q\\ 5. & (♢◻p∧♢◻q) →(p∧q)\\ 6. &♢(♢◻p∧♢◻q) →♢(p∧q)\\ 7. &(♢◻p∧♢◻q) →◻♢(p∧q)\end{array}$
(7 follows from 6 from the theorem derivable in B: $⊢♢p→q \implies ⊢p→◻q$)