Repetition-code syndrome, known erasures, and unlocated errors

The codewords are $000$ and $111$, and [L2] gives $d(C)=3$. For $x=(x_0,x_1,x_2)$, multiplication gives $Hx=(x_0+x_1,x_1+x_2)$, so $Hx=0$ exactly when $x_0=x_1=x_2$ and $\ker H=C$. The first two columns of $H$ form $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, an invertible triangular matrix over $F_2$, so the rows of $H$ are independent. Therefore $H$ is a parity-check matrix by [L1]; the field operations are valid by [F1].

Direct multiplication gives the weight-at-most-one syndromes $s(000)=00$, $s(100)=10$, $s(010)=11$, and $s(001)=01$, all distinct. For $r=101$, [L3] gives $s(r)=Hr=(1+0,0+1)=11$, so the unique matching error is $e=010$ and $r-e=111$. The distinct syndromes give the one-error correction in [L4].

If coordinates $0,1$ are erased in $r=101$, the received value at coordinate $2$ is $1$, so $111$ is compatible and $000$ is not. More generally, two compatible codewords would differ only in $S$, so their distance would be at most $2<3$, contradicting [L4]. Hence $111$ is the unique completion.

If instead $000$ is sent, the two-bit error $e=110$ produces $r^{\mathrm{\prime }}=110$. Then $d_H(r^{\mathrm{\prime }},000)=2$ but $d_H(r^{\mathrm{\prime }},111)=1$, so $111$ is the unique radius-one decoding candidate. The unique-radius-one rule therefore miscorrects this word; [L4] gives no correction guarantee for the two-bit error.