Proof

Let ${\displaystyle \{x_{1},\ldots ,x_{k}\}}$ be a basis for ${\displaystyle C_{1}}$ and let ${\displaystyle \{y_{1},\ldots ,y_{l}\}}$ be a basis for ${\displaystyle C_{2}}$. Then the set

${\displaystyle \{(x_{i}\mid x_{i})\mid 1\leq i\leq k\}\cup \{(0\mid y_{j})\mid 1\leq j\leq l\}}$

is a basis for the bar product ${\displaystyle C_{1}\mid C_{2}}$.

Proof

For all ${\displaystyle c_{1}\in C_{1}}$,

${\displaystyle (c_{1}\mid c_{1}+0)\in C_{1}\mid C_{2}}$

which has weight ${\displaystyle 2w(c_{1})}$. Equally

${\displaystyle (0\mid c_{2})\in C_{1}\mid C_{2}}$

for all ${\displaystyle c_{2}\in C_{2}}$ and has weight ${\displaystyle w(c_{2})}$. So minimising over ${\displaystyle c_{1}\in C_{1},c_{2}\in C_{2}}$ we have

${\displaystyle w(C_{1}\mid C_{2})\leq \min\{2w(C_{1}),w(C_{2})\}}$

Now let ${\displaystyle c_{1}\in C_{1}}$ and ${\displaystyle c_{2}\in C_{2}}$, not both zero. If ${\displaystyle c_{2}\not =0}$ then:

${\displaystyle {\begin{aligned}w(c_{1}\mid c_{1}+c_{2})&=w(c_{1})+w(c_{1}+c_{2})\\&\geq w(c_{1}+c_{1}+c_{2})\\&=w(c_{2})\\&\geq w(C_{2})\end{aligned}}}$

If ${\displaystyle c_{2}=0}$ then

${\displaystyle {\begin{aligned}w(c_{1}\mid c_{1}+c_{2})&=2w(c_{1})\\&\geq 2w(C_{1})\end{aligned}}}$

so

${\displaystyle w(C_{1}\mid C_{2})\geq \min\{2w(C_{1}),w(C_{2})\}}$