(a) $R_1=(0^*1 + 1^*0)^*0 + 11^*0^*$
Here, $(0^*1+1^*0)$ can generate $1,0 \implies (0^*1+1^*0)^*$ can generate all strings ie it is equivalent to $(0+1)^*$.
$R_1 \equiv (0+1)^*0+ 1^+0^*$
Here, $(0+1)^*0$ covers every string ending with $0$, so we can say every string that belongs to $1^+0^*$ and ends with $0$ is already accounted for. So, we can reduce it to $1^+$.
$R_1 \equiv (0+1)^*0 + 1^+ = R_2$
Therefore, both regular expressions are equivalent in (a).
(b) $((0+1)(0+1))^*$ can generate $1101$ but the other regular expression cannot.
Therefore, both regular expressions are not equivalent in (b).
(c) $(10+001+0^*1^*)^+$ can generate $\epsilon$ but the other regular expression cannot.
Therefore, both regular expressions are not equivalent in (c).
(d) $R_2 = (0+1)^*(\phi^+ +\epsilon) \equiv (0+1)^*$
$R_1 = (00^*1^* + 1^*10^*)^*(\phi ^* + \phi) \equiv (00^*1^*+1^*10^*)^*(\epsilon + \phi)$
Here, $(00^*1^* + 1^*10^*)$ can generate $0,1 \implies (00^*1^*+1^*10^*)^*$ can generate all strings ie it is equivalent to $(0+1)^*$.
$R_1 \equiv (0+1)^*(\epsilon + \phi) \equiv (0+1)^*$
Therefore, both regular expressions are equivalent in (d).
