I'm working through a proof in Dixmier's book on $C^*$-algebras and I'm stuck on part of a proof. I'm given a Banach algebra $\mathcal{A}$ which has norm $\lVert\cdot\rVert$ and a semi-norm $\lVert\cdot\rVert'$ was constructed satisfying $\lVert ab\rVert' \le \lVert a\rVert'\lVert b\rVert'$, $\lVert a^*\rVert' = \lVert a\rVert'$ and $\lVert a^*a\rVert' = \lVert a\rVert'^2$.
Then we define $\mathcal{J}=\{a\in\mathcal{A}:\lVert a\rVert'=0\}$. This is clearly a closed, self-adjoint, two-sided ideal in $\mathcal{A}$. What I want to show is that the $C^*$ identity holds on $\mathcal{A}/\mathcal{J}$, i.e. for $a+\mathcal{J}$ then $\lVert(a+\mathcal{J})^*(a+\mathcal{J})\rVert'=\lVert a+\mathcal{J}\rVert'^2$. I have no idea how to show this in this case. I was able to show that $\lVert(a+\mathcal{J})^*(a+\mathcal{J})\rVert'\le\lVert a+\mathcal{J}\rVert'^2$ but can't show the other inequality.
Any help would be greatly appreciated.
For any $a\in A$, $b\in \mathcal J$, we have $$ \|a\|'=\|a\|'-\|b\|'\leq\|a+b\|'\leq\|a\|'+\|b\|'=\|a\|'. $$ So $\|a+\mathcal J\|'=\|a\|'$. Then $$ \|a^*a+\mathcal J\|'=\|a^*a\|'=\|a\|'^2=\|a+\mathcal J\|'^2. $$