I just read that in modal logic the axiomatic system KB4 is equal to KB5, but there was no proof. I tried to show that from. $$A \rightarrow \Box \Diamond A$$ and from $$\Box A \rightarrow \Box \Box A$$ We can get $$\Diamond A \rightarrow \Box \Diamond A$$
In order to show that from KB4 we can derive KB5 and then show that from $$A \rightarrow \Box \Diamond A$$ and $$\Diamond A \rightarrow \Box \Diamond A$$ we can derive $$\Box A \rightarrow \Box \Box A$$ in order to show that KB4 is derivable from KB5. But I can't really figure out how to do this . Any help?
You cannot prove these by using only formulas of modal degree 2. You need to go through intermediate formulas of modal degree 3, using the trick I show in this answer:
First note that $\Diamond\Diamond A \rightarrow \Diamond A$ is equivalent to axiom 4 and $\Diamond \Box A \rightarrow A$ is equivalent to axiom B (by substituting $A$ with $\neg A$ and applying operator duality). Then you also need to apply the operator monotony property (from $p \rightarrow q$ infer $\Box p \rightarrow \Box q$ etc.) as shown below.
1) From $B$ and $4$ infer $5$ (left hand = substitution, right hand = monotony):
$$ \begin{align*} \frac{A\rightarrow \Box\Diamond A}{\Diamond A\rightarrow \Box\Diamond\Diamond A} & \qquad & \frac{\Diamond\Diamond A \rightarrow \Diamond A}{\Box\Diamond\Diamond A \rightarrow \Box\Diamond A} \\ \\ \hline \end{align*} $$ $$\Diamond A \rightarrow \Box\Diamond A$$
2) From $5$ and $B$ infer $4$ (left hand = monotony, right hand = substitution) :
$$ \begin{align*} \frac{\Diamond A \rightarrow \Box\Diamond A}{\Diamond\Diamond A \rightarrow \Diamond\Box\Diamond A} & \qquad & \frac{\Diamond\Box A\rightarrow A}{\Diamond\Box\Diamond A \rightarrow \Diamond A} \\ \\ \hline \end{align*} $$ $$\Diamond\Diamond A \rightarrow \Diamond A$$