Proof for equivalence of two regular expressions I have to show via equivalence transformation, that these two regular expressions are equivalent:
$$(ab+b^*a^*) = (a+b^*)(a^*+b)$$
Can someone give me a hint how to show this? I am only allowed to use these equivalences:
$$
R+S = S+R \\
(R+S)+T = R+(S+T) \\
(R S) T = R (S T) \\
\emptyset + R = R + \emptyset = R \\
\varepsilon R = R \varepsilon = R \\
\emptyset R = R \emptyset = \emptyset \\
R (S + T) = RS + RT \\
(S + T)R = SR + TR \\
R + R = R\\
(R^*)^* = R^* \\
(\varepsilon + R)^* = R^* \\
\emptyset^* = \varepsilon \\
\varepsilon^* = \varepsilon \\
(\varepsilon + R)R^* = R^*(\varepsilon + R) = R^*\\
RR^* = R^*R \\
R^* + R = R^* \\
\varepsilon + RR^* = R^*
$$
My best attempt so far.
 A: EDIT to clarify my answer. The core of the solution is highlighted in yellow.
I introduce the following useful concept, that permits to better reason about regular expressions.
Definition. I write $R\subset S$ if there exists $T$ such that $R+T=S$.
Lemma. If $R\subset S$, then $R+S=S$.
Proof. $R+S = R + (R+T) = (R+R)+T = R+T=S$. □

Now, back to your problem. We have
  $$
(a+b^*)(b+a^*)
= ab + aa^* + b^*b + b^*a^* ,
$$
  but $aa^*\subset b^*a^*$ and $b^*b\subset b^*a^*$, so
  $$
(a+b^*)(b+a^*)
= ab + b^*a^* .
$$

You may ask, why is $aa^*\subset b^*a^*?$ Well, that's easy 'cause
$$
b^*a^* = (\epsilon+b^*)(\epsilon+a)a^* = aa^* + \text{other terms}.
$$

For the nitpickers.
In the last formula, we used also $b^*=b^*+\epsilon$ at the beginning. Since this is not among the given equivalences, a possible derivation is as follows: $R^*+\epsilon=RR^*+\epsilon+\epsilon = RR^*+\epsilon=R^*$.
Otherwise, using our useful subset concept, notice that $\epsilon\subset R^*$, because $R^*=\epsilon+RR^*$; but then our lemma says $R^*+\epsilon=R^*$.

To conclude, I urge you to try the same exercise with a more complicate regular expression and see whether this subset thing does indeed simplify the derivations or not.
Just compare my short formulas with those of the accepted answer.
A: First, for any $R$,
$$R^*+\epsilon=\epsilon+(\epsilon+R)R^*=\epsilon+\epsilon R^*+RR^*=\epsilon+R^*+RR^*=R^*+(\epsilon+RR^*)=R^*+R^*=R^*$$
So
$$b^*a^*=(b^*+\epsilon)a^*=b^*a^*+\epsilon a^*=b^*a^*+a^*=b^*a^*+(\epsilon+a)a^*=b^*a^*+\epsilon a^*+aa^*=(b^*+\epsilon)a^*+aa^*=b^*a^*+aa^*$$
Likewise
$$b^*a^*=b^*(a^*+\epsilon)=b^*a^*+b^*\epsilon=b^*a^*+b^*=b^*a^*+(\epsilon+b)b^*=b^*a^*+\epsilon b^*+bb^*\\=b^*a^*+b^*\epsilon+b^*b=b^*(a^*+\epsilon)+b^*b=b^*a^*+b^*b$$
So that $b^*a^*=b^*a^*+aa^*+b^*b$.
Now
$$(a+b^*)(a^*+b)=(a+b^*)a^*+(a+b^*)b=aa^*+b^*a^*+ab+b^*b=ab+b^*a^*$$
