Classical distance gives error detection, unique decoding, and erasure recovery

The floor formula gives $2t<d$. If $He=0$, [L6] gives $e\in C$; if $e\mathrm{\ne }0$ and $\mathrm{wt}(e)<d$, [L1] is contradicted. Hence every such nonzero error is detected.

The transmitted codeword $c$ lies in the radius-$t$ ball around $r$. Any other codeword $c^{\mathrm{\prime }}$ in that ball would put $r$ in two distinct radius-$t$ balls, contradicting [L2]; therefore $c$ is the unique such word.

If two errors $e,f$ of weight at most $t$ have the same syndrome, [L5] gives $e=f$. For a received word $r=c+e$, [L3] and [L4] identify the measured syndrome as $He$, so it uniquely identifies the error within that weight range.

If $c,c^{\mathrm{\prime }}\in C$ both agree with the received word outside the known set $S$, then $c-c^{\mathrm{\prime }}$ is supported in $S$, so $\mathrm{wt}(c-c^{\mathrm{\prime }})\le \mathrm{\mid }S\mathrm{\mid }<d$. If $c\mathrm{\ne }{c}^{\mathrm{\prime }}$, this is a nonzero codeword contradicting [L1]. Thus at most one codeword is compatible. For $C=\{0\}$, the only candidate is already unique and $d=n+1$.