= Solution
The point needing proof is that a corrected error's residual state cannot retain dependence on the logical input. Let $W_j$ denote the complete linear map from the logical input to all five output wires for error $E_j$. Protection gives $W_j|0\rangle=|0\rangle|\eta_{j,0}\rangle$ and $W_j|1\rangle=|1\rangle|\eta_{j,1}\rangle$. Applying protection to $|+\rangle$ and comparing its two top-qubit components yields
$$
\frac{|0\rangle|\eta_{j,0}\rangle+|1\rangle|\eta_{j,1}\rangle}{\sqrt2}
=|+\rangle|\eta_{j,+}\rangle,
$$
so $|\eta_{j,0}\rangle=|\eta_{j,1}\rangle=|\eta_{j,+}\rangle$. Denote the common vector by $|\eta_j\rangle$. Linearity then gives $W_j|\psi\rangle=|\psi\rangle|\eta_j\rangle$ for every input.
The encoding and all operations surrounding the error are fixed. Replacing the error by $E=\alpha E_1+\beta E_2$ consequently replaces the overall map by $\alpha W_1+\beta W_2$, giving
$$
\boxed{W_E|\psi\rangle=|\psi\rangle\otimes\left(\alpha|\eta_1\rangle+\beta|\eta_2\rangle\right).}
$$
Because $E$ and the surrounding circuit are unitary, the output is normalized for every normalized input; the residual vector in parentheses thus has norm one. This proves <linearity of coherent quantum error correction> and hence protection against $E$. Orthogonality of $|\eta_1\rangle$ and $|\eta_2\rangle$ is not required for this proof.
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