Secrecy and correctness are separate requirements

For ${\rho }^{(1)}$, Alice's marginal is $I\mathrm{/}2$, so it is already uniform and independent of the trivial transcript-adversary state; hence ${\epsilon }_{\mathrm{s}\mathrm{e}\mathrm{c}}=0$. Every output pair differs, so ${\epsilon }_{\mathrm{c}\mathrm{o}\mathrm{r}}=1$. The real state is supported on $\{\mathrm{\mid }0,1,t⟩,\mathrm{\mid }1,0,t⟩\}$ while the ideal state is supported on the orthogonal set $\{\mathrm{\mid }0,0,t⟩,\mathrm{\mid }1,1,t⟩\}$. Their difference therefore has positive and negative parts each of trace one, giving ${\delta }_{\mathrm{i}\mathrm{d}}=1$ by [F2]–[F3].

For ${\rho }^{(2)}$, the parties always agree, so ${\epsilon }_{\mathrm{c}\mathrm{o}\mathrm{r}}=0$. Alice's accepted state is $\mathrm{\mid }0⟩⟨0\mathrm{\mid }$, and $\mathrm{\mid }0⟩⟨0\mathrm{\mid }-I\mathrm{/}2$ has eigenvalues $1\mathrm{/}2,-1\mathrm{/}2$; hence ${\epsilon }_{\mathrm{s}\mathrm{e}\mathrm{c}}=1\mathrm{/}2$. The real-ideal output difference has the same two eigenvalues, so ${\delta }_{\mathrm{i}\mathrm{d}}=1\mathrm{/}2$. Both cases satisfy [F4] with equality, while each has exactly one of the two component errors zero.