[proofplan] We first prove that the proposed addition and multiplication do not depend on the chosen representatives of the cosets. Addition uses that $I$ is an additive subgroup of $R$, while multiplication uses both left and right absorption by the two-sided ideal $I$. Once the operations are well-defined, every ring axiom follows by applying the corresponding axiom in $R$ to representatives and then passing to cosets. Finally, the quotient map preserves the ring operations by definition, is surjective because every coset has a representative, and has kernel exactly $I$. [/proofplan]

[step:Show that addition on cosets is independent of representatives] Let $a,a',b,b' \in R$ satisfy $a+I=a'+I$ and $b+I=b'+I$. By equality of additive cosets, this means $a-a' \in I$ and $b-b' \in I$. Since $I$ is an additive subgroup of the additive group of $R$, it is closed under addition. Hence \begin{align*} (a+b)-(a'+b')=(a-a')+(b-b') \in I. \end{align*} Therefore $(a+b)+I=(a'+b')+I$. Thus the rule \begin{align*} (a+I)+(b+I)=(a+b)+I \end{align*} is well-defined. [/step]

[step:Show that multiplication on cosets is independent of representatives]Let $a,a',b,b' \in R$ satisfy $a+I=a'+I$ and $b+I=b'+I$. Then $a-a' \in I$ and $b-b' \in I$. Since $I$ is a two-sided ideal, $(a-a')b \in I$ and $a'(b-b') \in I$. Using distributivity in $R$, we compute \begin{align*} ab-a'b'=(a-a')b+a'(b-b'). \end{align*} The right-hand side lies in $I$ because $I$ is closed under addition. Hence $ab-a'b' \in I$, so $ab+I=a'b'+I$. Thus the rule \begin{align*} (a+I)(b+I)=ab+I \end{align*} is well-defined.[/step]

[guided]We must prove that multiplication of cosets gives the same answer no matter which representative of each coset is chosen. Suppose $a,a',b,b' \in R$ are representatives with \begin{align*} a+I=a'+I \end{align*} and \begin{align*} b+I=b'+I. \end{align*} By the definition of equality of additive cosets, these two equalities mean \begin{align*} a-a' \in I \end{align*} and \begin{align*} b-b' \in I. \end{align*} The goal is to prove that the products represent the same coset, namely that $ab+I=a'b'+I$. Again using the coset criterion, it is enough to prove \begin{align*} ab-a'b' \in I. \end{align*} We rewrite the difference in a form where the ideal property can be applied: \begin{align*} ab-a'b'=(a-a')b+a'(b-b'). \end{align*} This identity follows by expanding the right-hand side: \begin{align*} (a-a')b+a'(b-b')=ab-a'b+a'b-a'b'=ab-a'b'. \end{align*} Now the hypothesis that $I$ is two-sided is used exactly here. Since $a-a' \in I$ and $I$ absorbs multiplication on the right by arbitrary elements of $R$, we have $(a-a')b \in I$. Since $b-b' \in I$ and $I$ absorbs multiplication on the left by arbitrary elements of $R$, we have $a'(b-b') \in I$. Because $I$ is closed under addition, their sum lies in $I$: \begin{align*} (a-a')b+a'(b-b') \in I. \end{align*} Therefore $ab-a'b' \in I$, and hence $ab+I=a'b'+I$. This proves that multiplication on $R/I$ is well-defined.[/guided]

[step:Verify the ring axioms by passing the identities from $R$ to cosets] The operations are now well-defined binary operations on $R/I$. Let $a,b,c \in R$. Additive associativity follows from associativity of addition in $R$: \begin{align*} ((a+I)+(b+I))+(c+I)=((a+b)+c)+I=(a+(b+c))+I=(a+I)+((b+I)+(c+I)). \end{align*} Additive commutativity follows from commutativity of addition in $R$: \begin{align*} (a+I)+(b+I)=(a+b)+I=(b+a)+I=(b+I)+(a+I). \end{align*} The additive identity is $0_R+I$, since \begin{align*} (a+I)+(0_R+I)=(a+0_R)+I=a+I. \end{align*} The additive inverse of $a+I$ is $(-a)+I$, since \begin{align*} (a+I)+((-a)+I)=(a-a)+I=0_R+I. \end{align*} Multiplicative associativity follows from associativity of multiplication in $R$: \begin{align*} ((a+I)(b+I))(c+I)=((ab)c)+I=(a(bc))+I=(a+I)((b+I)(c+I)). \end{align*} The multiplicative identity is $1_R+I$, since \begin{align*} (1_R+I)(a+I)=(1_Ra)+I=a+I \end{align*} and \begin{align*} (a+I)(1_R+I)=(a1_R)+I=a+I. \end{align*} Left distributivity follows from left distributivity in $R$: \begin{align*} (a+I)((b+I)+(c+I))=(a(b+c))+I=(ab+ac)+I=(a+I)(b+I)+(a+I)(c+I). \end{align*} Right distributivity follows from right distributivity in $R$: \begin{align*} ((a+I)+(b+I))(c+I)=((a+b)c)+I=(ac+bc)+I=(a+I)(c+I)+(b+I)(c+I). \end{align*} Thus $R/I$ is a unital ring. [/step]

[step:Prove that the quotient map is a surjective unital ring homomorphism] Define \begin{align*} \pi:R&\to R/I \end{align*} by $\pi(a)=a+I$ for each $a \in R$. For all $a,b \in R$, the definitions of the quotient operations give \begin{align*} \pi(a+b)=(a+b)+I=(a+I)+(b+I)=\pi(a)+\pi(b) \end{align*} and \begin{align*} \pi(ab)=ab+I=(a+I)(b+I)=\pi(a)\pi(b). \end{align*} Also, \begin{align*} \pi(1_R)=1_R+I, \end{align*} which is the multiplicative identity of $R/I$. Hence $\pi$ is a unital ring homomorphism. To prove surjectivity, let $x \in R/I$. By definition of $R/I$, there exists $a \in R$ such that $x=a+I$. Therefore $x=\pi(a)$, so $\pi$ is surjective. [/step]

[step:Identify the kernel of the quotient map] The kernel of $\pi$ is \begin{align*} \ker \pi=\{a \in R:\pi(a)=0_R+I\}. \end{align*} For $a \in R$, we have \begin{align*} a \in \ker \pi \iff a+I=0_R+I \iff a-0_R \in I \iff a \in I. \end{align*} Therefore $\ker \pi=I$. This completes the construction of the quotient ring and the verification of the stated properties of the quotient map. [/step]