Prove that the HartmannTzeng Bound follows from the van LintWilson Bounding Technique. Hint: See Example $4\mathrm{.}5.?$

Example $4\mathrm{.}5\mathrm{.}12$ Let $C$ be the $17,9,d$ binary cyclic code with defining set $T=\{1,2,4,8,9,13,15,16\}.$ In Example $4\mathrm{.}5\mathrm{.}7,$ we saw that the BCH Bound implies that $d3,$ and the Hartmann$-$Tzeng Bound improves this to $d5$ using $A=\{1,2\}$ and $B=\{0,7,14\}.$ The van Lint$-$Wilson Bounding Technique gives the same bound using the following argument. Let $f(x)inC$ be a nonzero codeword of weight less than $5.T$ is the $2-$cyclotomic coset $C_1.$ If $f({}^i)=0$ for some $iin{C}_3,$ then $f(x)$ is a nonzero codeword in the cyclic code with defining set $C_1{C}_3,$ which is the repetition code, and hence $wt(f(x))=17,$ a contradiction. Letting $S=\{iinN|f({}^i)=0\},$ we assume that $S$ has no elements of $C_3=\{3,5,6,7,10,11,12,14\}.$ Then the following sequence of subsets of $N$ is independent with respect to $S$ :

$156$

Cyclic codes

$I_0=\frac{O}{,}$

$I_1=I_0\{6\}=\{6\},$

$I_2=\{-7\}+I_1=\{16\}subeS,$

$I_3=I_2\{6\}=\{6,16\},$

$I_4=\{-7\}+I_3=\{9,16\}subeS,$

$I_5=I_4\{6\}=\{6,9,16\},$

$I_6=\{-7\}+I_5=\{2,9,16\}subeS,$

$I_7=I_6\{3\}=\{2,3,9,16\},$

$I_8=\{-1\}+I_7=\{1,2,8,15\}subeS,$

$I_9=I_8\{3\}=\{1,2,3,8,15\}.$

Since $wt(f(x))|I_9|=5,$ the van Lint$-$Wilson Bounding Technique shows that $d5.$ In fact, $d=5$ by $[203].$