= Solution
Taking <partial inner products> with $\langle0|$ and $\langle1|$ on the first wire gives $(|0\rangle+|1\rangle)/2$ and $(-|0\rangle+|1\rangle)/2$. They are nonzero and <orthogonal>, hence <linearly independent>. This contradicts the necessary product-state property proved in part (a), so \b[the second state is <entangled>].
Equivalently, its <tensor-product coefficient matrix> is $C=\tfrac12\bigl(\begin{smallmatrix}1&1\\-1&1\end{smallmatrix}\bigr)$, with $\det C=1/2\ne0$. The <two-qubit product-state determinant criterion> gives the same conclusion. In fact $CC^\dagger=I/2$, so its <Schmidt coefficients> are both $1/\sqrt2$: it is maximally <entangled>.
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