= Solution
The <no-cloning theorem for two pure states> states that, for distinct nonorthogonal states $|c_0\rangle$ and $|c_1\rangle$, there is no <unitary operator> $U$ and fixed blank state $|0\rangle$ such that
$$
U\bigl(|c_j\rangle|0\rangle\bigr)
=|c_j\rangle|c_j\rangle,
\qquad j=0,1.
$$
To derive this from state discrimination, let
$$
s=|\langle c_0|c_1\rangle|,
\qquad 0<s<1,
$$
and suppose such a cloner existed. Applying it repeatedly would produce $N$ copies. The <inner product> of the two possible $N$-copy states has magnitude $s^N$, so their optimal success probability under the <Helstrom-Holevo bound> would be
$$
P_N=\frac12\left(1+\sqrt{1-s^{2N}}\right).
$$
Already for $N=2$,
$$
P_2>P_1
=\frac12\left(1+\sqrt{1-s^2}\right),
$$
because $0<s<1$ implies $s^4<s^2$. But cloning followed by the two-copy <Helstrom measurement for two pure states> would itself be a quantum procedure applied to the original single state, contradicting the optimal one-copy bound $P_1$. Hence the assumed unitary cloner cannot exist.
Equivalently, <clone-assisted asymptotic state discrimination> would give $P_N\to1$, although two nonorthogonal states are not <perfect distinguishability of pure states>[perfectly distinguishable]. Thus the Helstrom--Holevo theorem implies the no-cloning theorem.
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