Proof:

We will prove the equivalence of the two statements.

Definitions:

1. Commutative ring: A ring $R$ where multiplication is commutative: $ab = ba$ for all $a, b \in R$.
2. Integral domain: A commutative ring $R$ with no zero divisors, meaning if $ab = 0$, then $a = 0$ or $b = 0$.

Part 1: Prove that $R$ is an integral domain $\implies$ $ab = ac$ and $a \neq 0$ implies $b = c$.

1. Assume $R$ is an integral domain.
2. Let $a, b, c \in R$ such that $a \neq 0$ and $ab = ac$.
3. Subtract $ac$ from $ab$: $ab - ac = 0.$
4. By distributivity: $a(b - c) = 0.$
5. Since $R$ is an integral domain and $a \neq 0$, it follows that $b - c = 0$.
6. Therefore: $b = c.$
7. Thus, if $R$ is an integral domain, $ab = ac$ and $a \neq 0$ imply $b = c$.

Part 2: Prove that $ab = ac$ and $a \neq 0 \implies b = c$ $\implies$ $R$ is an integral domain.

1. Assume $ab = ac$ and $a \neq 0$ imply $b = c$ for all $a, b, c \in R$.
2. Suppose $R$ is not an integral domain. Then there exist $x, y \in R$ with $x \neq 0$, $y \neq 0$, and $xy = 0$.
3. Set $a = x$, $b = y$, and $c = 0$. Clearly, $ab = ac$ because $xy = x \cdot 0 = 0$.
4. However, $b \neq c$ since $y \neq 0$. This contradicts the assumption that $ab = ac$ and $a \neq 0$ imply $b = c$.
5. Therefore, $R$ must be an integral domain.

Conclusion:

A commutative ring $R$ is an integral domain if and only if for all $a, b, c \in R$, $a \neq 0$ and $ab = ac$ imply $b = c$.

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Do you want further clarification or additional examples? Here are some related questions for exploration:

1. What are some examples of commutative rings that are not integral domains?
2. How does the property of "no zero divisors" relate to the cancellation law in $R$?
3. Can you generalize this proof for non-commutative rings?
4. What is the significance of integral domains in algebraic number theory?
5. How does this property fail in $\mathbb{Z}/n\mathbb{Z}$ for composite $n$?

Tip: In proofs involving cancellation laws, always leverage the absence of zero divisors effectively.