Prove $\Diamond p \rightarrow \lnot\Diamond\lnot\Diamond p$ in modal logic

Question

I need to prove $\Diamond p \rightarrow \lnot\Diamond\lnot\Diamond p$ in B axiomatic, which contains next conversion rules:
1.$(p\land q)\rightarrow(q\land p)$
2.$(q\land p)\rightarrow p$
3.$p\rightarrow(p\land p)$
4.$p\land(q\land r)\rightarrow(p\land q)\land r$
5.$p\rightarrow \lnot(\lnot p)$
6.$(p\land q)\land(q\land r)\rightarrow(p\land r)$
7.$(p\land(p\rightarrow q))\rightarrow q$
8.$\Diamond(p\land q)\rightarrow\Diamond p$
9.$p\rightarrow p$
10.$\Box p\rightarrow\Box\Box p$
11.$p\rightarrow\lnot\Diamond\lnot\Diamond p$
12.$p\rightarrow q\equiv \not\Diamond(p\land \lnot q) $