= Solution
Let $D=\mathcal C_{\mathrm{ev}}$, and write $\mathcal C_{\mathrm{odd}}=t+D$ for any odd-weight word $t\in\mathcal C_H$. The parity functional on the four-dimensional binary <linear code> $\mathcal C_H$ is nonzero, so its kernel $D$ has dimension three and its other fibre is the coset $t+D$. For example $t=(1,\ldots,1)$ belongs to the standard length-seven <Hamming code> and has odd <Hamming weight>. Denote the two logical basis states by $|\psi_0\rangle$ and $|\psi_1\rangle$, corresponding to these two cosets. The eight computational <orthonormal basis> vectors in each sum are distinct, so both logical states have norm one; their disjoint supports make them <orthogonal>.
For a binary vector $w\in\mathbb F_2^7$, define the product of <Pauli Z gates>
$$
Z(w)=\bigotimes_{j=1}^7 Z_j^{w_j},\qquad
Z(w)|x\rangle=(-1)^{w\cdot x}|x\rangle,
$$
where the dot product is computed modulo two. The key diagonal overlap, with $t_0=0$ and $t_1=t$, is
$$
\langle\psi_a|Z(w)|\psi_a\rangle
=\frac{(-1)^{w\cdot t_a}}8\sum_{x\in D}(-1)^{w\cdot x}.
$$
If $w\in D^\perp$, every summand is one. If $w\notin D^\perp$, choose $x_0\in D$ with $w\cdot x_0=1$. Translation $x\mapsto x+x_0$ permutes $D$ and negates every summand. The sum must then be zero. Thus
$$
\sum_{x\in D}(-1)^{w\cdot x}
=\begin{cases}8,&w\in D^\perp,\\0,&w\notin D^\perp.\end{cases}
$$
Off-diagonal logical overlaps vanish for every $w$:
$$
\langle\psi_0|Z(w)|\psi_1\rangle=0,
$$
because a diagonal <Pauli operator> cannot move a computational basis word from one coset to the disjoint other coset.
The possible errors are $E_0=I$ and $E_j=Z(e_j)$ for $j=1,\ldots,7$. Set $e_0=0$. Each <phase flip> is its own adjoint and inverse, so
$$
E_i^\dagger E_j=Z(e_i+e_j).
$$
If $i\ne j$, the vector $e_i+e_j$ is nonzero and has <Hamming weight> one or two. By the given <dual code> identity $D^\perp=\mathcal C_H$ and the minimum <Hamming distance> three of the <Hamming code>, it cannot belong to $D^\perp$. Hence all its diagonal logical overlaps vanish. If $i=j$, the product is the identity and its logical overlaps are $\delta_{ab}$. Combining both cases gives
$$
\boxed{\langle\psi_a|E_i^\dagger E_j|\psi_b\rangle
=\delta_{ij}\delta_{ab}\quad(a,b\in\{0,1\},\ i,j\in\{0,\ldots,7\}).}
$$
This proves the <Knill--Laflamme condition> with an identity error-overlap <matrix>, not merely a possibly degenerate scalar <matrix>. In particular the eight two-dimensional error subspaces $E_i\mathcal X_{\mathrm{Steane}}$ are mutually <orthogonal>. This is precisely <nondegenerate phase-flip correction in the Steane code>.
To exhibit recovery rather than only the criterion, let $P$ be the <orthogonal projector> onto the code and $P_i=E_iPE_i^\dagger$. The <projective measurement> consisting of these eight mutually <orthogonal projectors>, completed by $I-\sum_iP_i$, determines which error occurred. For an arbitrary encoded <pure state> $|\chi\rangle=\alpha|\psi_0\rangle+\beta|\psi_1\rangle$ affected by $E_j$, outcome $j$ occurs with certainty and leaves $E_j|\chi\rangle$ unchanged. Applying $E_j$ again recovers $|\chi\rangle$. The outcome gives the <error syndrome> and does not reveal $\alpha$ or $\beta$, so recovery preserves logical <quantum superpositions>.
An equivalent explicit <error syndrome> uses three commuting X-type <stabilizer generators>. Choose a <parity-check matrix> for $\mathcal C_H$ with the seven different nonzero binary columns:
$$
H=\begin{pmatrix}
1&0&1&0&1&0&1\\
0&1&1&0&0&1&1\\
0&0&0&1&1&1&1
\end{pmatrix}.
$$
Its rows span $\mathcal C_H^\perp=D$. For a row $h_l$, the operator $X(h_l)=\bigotimes_jX_j^{(h_l)_j}$ permutes each of $D$ and $t+D$, and therefore has eigenvalue $+1$ on both logical states. A <phase flip> at position $j$ changes its eigenvalue to $(-1)^{H_{lj}}$, since $X(h_l)Z_j=(-1)^{H_{lj}}Z_jX(h_l)$. Measuring the three commuting <stabilizer generators> gives the $j$th column of $H$ as a binary <syndrome>; no error gives the zero column. All eight outcomes are different, and applying the identified $Z_j$ restores the code state. Measuring these <stabilizer generators> never distinguishes the two logical states.
Finally, if a noise <Kraus operator> is a coherent linear combination $F=\sum_jc_jE_j$, the same syndrome measurement separates its mutually <orthogonal> components $c_jE_j|\chi\rangle$. Conditional recovery returns $|\chi\rangle$ in every nonzero branch. Linearity then also corrects any <quantum channel> whose <Kraus operators> lie in this span. The argument therefore corrects arbitrary unknown single-site <phase flips> nondegenerately, including coherent phase-error amplitudes, rather than only a known error on a basis codeword.
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