[proofplan] We unpack both congruence hypotheses using the definition of congruence modulo $n$ as divisibility by $n$. The additive compatibility follows by rewriting the difference of the sums as the sum of two integer multiples of $n$. The multiplicative compatibility follows by rewriting $ab-a'b'$ as a sum of two terms, each visibly divisible by $n$. [/proofplan] [step:Translate the congruence hypotheses into divisibility statements] By the definition of congruence modulo $n$, the assumptions $a \equiv a' \pmod{n}$ and $b \equiv b' \pmod{n}$ mean that \begin{align*} n \mid (a-a') \end{align*} and \begin{align*} n \mid (b-b'). \end{align*} Thus there exist integers $r,s \in \mathbb{Z}$ such that \begin{align*} a-a' = nr \end{align*} and \begin{align*} b-b' = ns. \end{align*} [/step] [step:Show that the sums differ by a multiple of $n$] Using the integers $r,s \in \mathbb{Z}$ defined above, compute \begin{align*} (a+b)-(a'+b') = (a-a')+(b-b'). \end{align*} Substituting $a-a'=nr$ and $b-b'=ns$ gives \begin{align*} (a+b)-(a'+b') = nr+ns = n(r+s). \end{align*} Since $r+s \in \mathbb{Z}$, this proves $n \mid ((a+b)-(a'+b'))$. Therefore, by the definition of congruence modulo $n$, \begin{align*} a+b \equiv a'+b' \pmod{n}. \end{align*} [guided] We want to prove a congruence between the two sums. By definition, this means we must prove that their difference is divisible by $n$. The relevant difference is \begin{align*} (a+b)-(a'+b'). \end{align*} Rearranging terms in $\mathbb{Z}$ gives \begin{align*} (a+b)-(a'+b') = (a-a')+(b-b'). \end{align*} From the hypotheses, we already introduced integers $r,s \in \mathbb{Z}$ satisfying \begin{align*} a-a' = nr \end{align*} and \begin{align*} b-b' = ns. \end{align*} Substituting these two equalities into the difference of sums yields \begin{align*} (a+b)-(a'+b') = nr+ns. \end{align*} Factoring out $n$ gives \begin{align*} (a+b)-(a'+b') = n(r+s). \end{align*} Because $\mathbb{Z}$ is closed under addition, $r+s \in \mathbb{Z}$. Hence $(a+b)-(a'+b')$ is an integer multiple of $n$, so \begin{align*} n \mid ((a+b)-(a'+b')). \end{align*} Applying the definition of congruence modulo $n$ again, we conclude \begin{align*} a+b \equiv a'+b' \pmod{n}. \end{align*} [/guided] [/step] [step:Show that the products differ by a multiple of $n$] Again using $a-a'=nr$ and $b-b'=ns$, decompose the product difference as \begin{align*} ab-a'b' = ab-a'b+a'b-a'b'. \end{align*} Factoring the two pairs of terms gives \begin{align*} ab-a'b' = b(a-a')+a'(b-b'). \end{align*} Substituting the divisibility representations gives \begin{align*} ab-a'b' = bnr+a'ns. \end{align*} Factoring out $n$ gives \begin{align*} ab-a'b' = n(br+a's). \end{align*} Since $b,r,a',s \in \mathbb{Z}$, closure of $\mathbb{Z}$ under multiplication and addition gives $br+a's \in \mathbb{Z}$. Hence $n \mid (ab-a'b')$, and therefore \begin{align*} ab \equiv a'b' \pmod{n}. \end{align*} [/step] [step:Combine the two compatibility conclusions] The previous two steps prove both required congruences: \begin{align*} a+b \equiv a'+b' \pmod{n} \end{align*} and \begin{align*} ab \equiv a'b' \pmod{n}. \end{align*} This completes the proof. [/step]