[proofplan] We prove the equivalence by unpacking the quotient-ring notation: $a+I$ and $b+I$ are additive cosets of the ideal $I$ in the additive group of $R$. Equality of cosets first implies that $a$ lies in $b+I$, which gives $a-b \in I$. Conversely, if $a-b \in I$, then the additive subgroup properties of the ideal show both inclusions $a+I \subset b+I$ and $b+I \subset a+I$. [/proofplan]

[step:Unpack equal cosets to obtain the difference condition]Assume $a+I=b+I$. Since $0_R \in I$, we have $a=a+0_R \in a+I$. Therefore $a \in b+I$, so there exists an element $i \in I$ such that \begin{align*} a=b+i. \end{align*} Adding $-b$ to both sides in the additive group of $R$ gives \begin{align*} a-b=i. \end{align*} Since $i \in I$, it follows that $a-b \in I$.[/step]

[guided]Assume $a+I=b+I$. The notation $a+I$ denotes the additive coset \begin{align*} a+I=\{a+i : i \in I\}. \end{align*} Because $I$ is an ideal, it is in particular an additive subgroup of $R$. Hence $0_R \in I$, and therefore \begin{align*} a=a+0_R \in a+I. \end{align*} The hypothesis $a+I=b+I$ says these two subsets of $R$ are equal, so every element of $a+I$ is an element of $b+I$. Thus $a \in b+I$. By the definition of $b+I$, there exists an element $i \in I$ such that \begin{align*} a=b+i. \end{align*} Now subtract $b$ from both sides using the additive group structure of the ring: \begin{align*} a-b=i. \end{align*} Since $i \in I$, this proves $a-b \in I$.[/guided]

[step:Use the difference condition to prove both coset inclusions] Assume $a-b \in I$. To prove $a+I \subset b+I$, let $x \in a+I$. Then there exists $i \in I$ such that \begin{align*} x=a+i. \end{align*} Rewrite $x$ as \begin{align*} x=b+((a-b)+i). \end{align*} Since $a-b \in I$ and $i \in I$, additive closure of $I$ gives $(a-b)+i \in I$. Hence $x \in b+I$, and therefore $a+I \subset b+I$. For the reverse inclusion, let $y \in b+I$. Then there exists $j \in I$ such that \begin{align*} y=b+j. \end{align*} Since $a-b \in I$ and $I$ is closed under additive inverses, $b-a=-(a-b) \in I$. Hence \begin{align*} y=a+((b-a)+j), \end{align*} and additive closure of $I$ gives $(b-a)+j \in I$. Thus $y \in a+I$, so $b+I \subset a+I$. The two inclusions imply $a+I=b+I$. Therefore \begin{align*} a+I=b+I \quad \iff \quad a-b \in I. \end{align*} [/step]