reduced density matrix for the given composite system Given the composite system of two qubits
$$
|\psi^{AB}\rangle=\frac{1}{\sqrt{2}}(|0^{A}\rangle \otimes|0^{B}\rangle+|1^{A}\rangle\otimes|1^{B}\rangle)
$$
with the density matrix of the composite system 
\begin{equation}
\begin{split}
\rho^{AB}=|\psi^{AB}\rangle\langle\psi^{AB}|=\frac{1}{2}[{|0^{A}\rangle \otimes|0^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|+|1^{A}\rangle \otimes|1^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|+|0^{A}\rangle \otimes|0^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|+|1^{A}\rangle \otimes|1^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|}]
\end{split}
\end{equation}
The reduced density operator of $A$, i.e,  $\rho^{A}$ is
\begin{equation}
\begin{split}
\rho^{A}=\text{tr}_{B}(\rho^{AB})=\langle0^{B}|\rho^{AB}|0^{B}\rangle+\langle1^{B}|\rho^{AB}|1^{B}\rangle=\frac{1}{2}[{\langle0^{B}|0^{A}\rangle \otimes|0^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|0^{B}\rangle+\langle0^{B}|1^{A}\rangle \otimes|1^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|0^{B}\rangle+\langle0^{B}|0^{A}\rangle \otimes|0^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|0^{B}\rangle+\langle0^{B}|1^{A}\rangle \otimes|1^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|0^{B}\rangle}+{\langle1^{B}|0^{A}\rangle \otimes|0^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|1^{B}\rangle+\langle1^{B}|1^{A}\rangle \otimes|1^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|1^{B}\rangle+\langle1^{B}|0^{A}\rangle \otimes|0^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|1^{B}\rangle+\langle1^{B}|1^{A}\rangle \otimes|1^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|1^{B}\rangle}]\\ =\frac{1}{2}[\langle0^{B}|0^{A}\rangle \otimes|0^{B}\rangle\langle0^{A}|\otimes I_{B}+\langle0^{B}|1^{A}\rangle \otimes|1^{B}\rangle\langle0^{A}|\otimes I_{B}+\langle1^{B}|0^{A}\rangle \otimes|0^{B}\rangle\langle1^{A}|\otimes I_{B}+\langle1^{B}|1^{A}\rangle \otimes|1^{B}\rangle\langle1^{A}|\otimes I_{B}]
\end{split}
\end{equation}
I think the expansion is correct so far.
How do I proceed further and prove that $\rho^{A}=\frac{|0^{A}\rangle\langle0^{A}|+|1^{A}\rangle\langle1^{A}|}{2}$ and thus the reduced state is mixed.
 A: \begin{split}
\langle0^{B}|\rho^{AB}|0^{B}\rangle=\frac{1}{2}\bigg(\langle0^{B}|\big[{|0^{A}\rangle \otimes|0^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|+|1^{A}\rangle \otimes|1^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|+|0^{A}\rangle \otimes|0^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|+|1^{A}\rangle \otimes|1^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|}\big]|0^{B}\rangle\Big)\\=\frac{1}{2}\Big({|0^{A}\rangle \otimes\langle0^{B}||0^{B}\rangle\langle0^{A}|\otimes\langle0^{B}||0^{B}\rangle+|1^{A}\rangle \otimes\langle0^{B}||1^{B}\rangle\langle0^{A}|\otimes\langle0^{B}||0^{B}\rangle+|0^{A}\rangle \otimes\langle0^{B}||0^{B}\rangle\langle1^{A}|\otimes\langle1^{B}||0^{B}\rangle+|1^{A}\rangle \otimes\langle0^{B}||1^{B}\rangle\langle1^{A}|\otimes\langle1^{B}||0^{B}\rangle}\Big)
\end{split}
Since $\langle1^{B}||0^{B}\rangle=0$ and $\langle0^{B}||0^{B}\rangle=\langle1^{B}||1^{B}\rangle=\mathcal{I}$
\begin{split}
\langle0^{B}|\rho^{AB}|0^{B}\rangle=\frac{1}{2}\Big(|0^{A}\rangle \otimes\mathcal{I}\langle0^{A}|\otimes\mathcal{I}\Big)=\frac{1}{2}\Big(|0^{A}\rangle \otimes\langle0^{A}|\Big)
\end{split}
Similarly,
\begin{split}
\langle1^{B}|\rho^{AB}|1^{B}\rangle=\frac{1}{2}\Big(|1^{A}\rangle \otimes\langle1^{A}|\Big)
\end{split}
The reduced density operator, $\rho^{A}=Tr_{B}(\rho^{AB})=\frac{|0^{A}\rangle \otimes\langle0^{A}|+|1^{A}\rangle \otimes\langle1^{A}|}{2}$
