[proofplan] We prove the left-annihilator criterion by packaging the annihilators of $I$ into a closed two-sided ideal. If $I$ is essential and $bI=\{0\}$, then the ideal $K=\{x\in B:xI=\{0\}\}$ is nonzero and closed, so it must meet $I$ nontrivially; that intersection is forced to be zero, a contradiction. Conversely, if $I$ is not essential, a nonzero ideal disjoint from $I$ has zero product with $I$, so any nonzero element of it annihilates $I$ on the left. The right-annihilator statement follows by applying the left-annihilator statement to adjoints. [/proofplan] [step:Package the left annihilator into a closed ideal] Assume that $I$ is essential, so every nonzero closed two-sided ideal of $B$ has nonzero intersection with $I$. [guided] The goal is to convert the hypothesis $bI=\{0\}$ into the statement that a nonzero closed two-sided ideal annihilates $I$ on the left. Once that ideal is built, essentiality forces it to meet $I$ nontrivially, and that intersection will collapse to $\{0\}$ because anything in it both lies in $I$ and kills $I$ on the left. Let $b\in B$ satisfy $bI=\{0\}$. If $b=0$, there is nothing to prove. Suppose instead that $b\neq 0$, and define $c:=b^*b\in B$. Since $\|c\|_B=\|b^*b\|_B=\|b\|_B^2>0$ by the $C^*$-identity, we have $c\neq 0$. Now define the left annihilator of $I$ by \begin{align*} K:=\{x\in B:xI=\{0\}\}. \end{align*} This set is a closed two-sided ideal: it is linear by construction, it is closed because multiplication in a $C^*$-algebra is continuous, and it is a two-sided ideal because if $x\in K$, $a,d\in B$, and $i\in I$, then $(axd)i=ax(di)=0$ since $di\in I$. Moreover $c\in K$, because for every $i\in I$ we have $ci=b^*(bi)=0$. Thus $K$ is a nonzero closed two-sided ideal. Since $I$ is essential, $K\cap I\neq \varnothing$. Choose $x\in K\cap I$. Because closed two-sided ideals in a $C^*$-algebra are closed under adjoints, we have $x^*\in I$. Since $x\in K$, we have $xI=\{0\}$, so in particular $xx^*=0$ by taking $i=x^*\in I$. The $C^*$-identity now gives $\|x\|_B^2=\|xx^*\|_B=0$, hence $x=0$. This contradicts the fact that $K\cap I$ was chosen as a nonzero intersection. Therefore the assumption $b\neq 0$ is impossible, and so $b=0$. [/guided] Let $b\in B$ satisfy $bI=\{0\}$. We show that $b=0$. If $b=0$, there is nothing to prove. Otherwise define $c:=b^*b\in B$. Then $c\neq 0$ because the $C^*$-identity gives $\|c\|_B=\|b^*b\|_B=\|b\|_B^2>0$. Define \begin{align*} K:=\{x\in B:xI=\{0\}\}. \end{align*} The set $K$ is a closed two-sided ideal: it is linear by construction, it is closed because $x_n\to x$ and $x_nI=\{0\}$ imply $xI=\{0\}$ by continuity of multiplication, and it is two-sided because if $x\in K$, $a,d\in B$, and $i\in I$, then $(axd)i=ax(di)=0$ since $di\in I$. Moreover $c\in K$, because for every $i\in I$ we have $ci=b^*(bi)=0$. Thus $K$ is a nonzero closed two-sided ideal. Since $I$ is essential, $K\cap I\neq \varnothing$. Choose $x\in K\cap I$. Because closed two-sided ideals in a $C^*$-algebra are $*$-closed, $x^*\in I$. Since $x\in K$, we have $xI=\{0\}$, so in particular $xx^*=0$. The $C^*$-identity now gives $\|x\|_B^2=\|xx^*\|_B=0$, hence $x=0$, contradicting $x\in K\cap I$ being nonzero. Therefore the assumption $b\neq 0$ is impossible, and $b=0$. [/step] [step:Use a disjoint ideal to produce a nonzero left annihilator] Assume that for every $b\in B$, the condition $bI=\{0\}$ implies $b=0$. Suppose that $I$ is not essential. Then there exists a nonzero closed two-sided ideal $J\trianglelefteq B$ such that $J\cap I=\{0\}$. For any $j\in J$ and $i\in I$, the product $ji$ lies in $J$ because $J$ is a two-sided ideal, and it lies in $I$ because $I$ is a two-sided ideal. Hence $ji\in J\cap I=\{0\}$, so $ji=0$ for all $j\in J$ and $i\in I$. Therefore $JI=\{0\}$. Choose $b\in J$ with $b\neq 0$. Then $bI=\{0\}$, contrary to the hypothesis. Hence $I$ must be essential. [/step] [step:Pass to adjoints for the right annihilator formulation] Assume the left-annihilator criterion has been established. Suppose that $Ib=\{0\}$ for some $b\in B$. For every $i\in I$, we have $ib=0$, so taking adjoints gives $(ib)^*=b^*i^*=0$. Since $I$ is a closed two-sided ideal in a $C^*$-algebra, it is $*$-closed, so $i^*$ ranges over all of $I$. Thus $b^*I=\{0\}$, and the left-annihilator criterion applied to $b^*$ yields $b^*=0$. Hence $b=0$. Conversely, if the right-annihilator criterion holds and $bI=\{0\}$, then the same argument applied to $b^*$ shows that $Ib^*=\{0\}$, so $b^*=0$ and therefore $b=0$. Thus the left- and right-annihilator formulations are equivalent. [/step] Oops final included typo and unfinished. Need final complete. Can't edit? We are in final already. But can still provide final? I need replace entire final. Use final content. Ensure no code fences. Need no raw ; OK align* no line breaks? Some align multiple lines? Avoid & maybe okay single. Need no row. Step tags balanced. Use guided in first step. Check "No code fences". Need mention J explicit: closure of span {x c y}. In nonunital if B lacks unit, c may not be of form xcy with x,y? Actually without unit, c may not be in B c B. The closed two-sided ideal generated by c is closure of span {α c + x c + c y + x c y?}. In nonunital two-sided ideal generated by c includes finite sums a c d plus maybe c. Since no unit. Simplest define as intersection of all closed two-sided ideals containing c. For JI=0 need show all elements in this intersection? Hard. Instead define K = {x in B: x I =0}? Is K closed left? If b in K. Need ideal generated by c contained K? K is closed left ideal? Also right multiplication: if xI=0, (a x)I=0; (x a)I = x(aI)=0 since aI subset I. so two-sided ideal. Closed. cI=0, so closed ideal generated by c subset K. Nice. Need J nonzero since contains c. Then for x∈J∩I, x∈K so xI=0 and x∈I, then x*x? Since x in I, and x∈K gives xI=0, choose i? x^*? Not in I? I star closed. x^*∈I, so x x^*=0? xI=0 with i=x^* gives x x^*=0, so x=0. Or if K left annihilator. Good. Need but bI=0, cI= b*(bI)=0, so c∈K. Great. Simpler. For converse JI. Right equivalently: if Ib=0 then (b* I=0?) For all i∈I, b* i = ? From Ib=0: for i, i b=0; adjoint gives b* i* =0. Since i* ranges over I, b* I=0. Left criterion gives b*=0. Conversely right implies left similarly. Since I *-closed. Need state "closed two-sided ideals are self-adjoint" maybe use as standard property? We can prove inline? Probably enough in C*-algebra. But rigorous: "Since closed two-sided ideals in C*-algebras are closed under adjoint" no citation. acceptable? Could prove? Might be long. Could avoid? The theorem statement itself calls ideal in C*-algebra, many define as self-adjoint. Formal statement can say closed two-sided *-ideal? But original says closed two-sided ideal; don't change? In C*-algebras automatic. We'll say "closed two-sided ideal (so $I^*=I$)". Need "product of JI subset" in converse. Need guided self-contained for first step. Ensure exact outside enough. Final answer only. Need ensure no "clearly". No citations. Since no external. Need no "all elements $xcb$" typo. Let's output clean.