Proof: By Induction

In the following proof, $(R, + ,\cdot)$ is an integral domain.

Let $(a)\lhd R$ and $(b)\lhd R$ be two principal ideals with $(a)=(b)$.
By the equality of sets, whe have $(a)\subseteq (b)$ and $(b)\subseteq (a).$
By the definition of divisibility of ideals, it follows $(b)\mid (a)$ and $(a)\mid (b).$
By the lemma of divisibility of principal ideals, it follows $a\mid b$ and $b\mid a.$
Following the definition of divisors there are elements $c,d\in R$ with $ac=b$ and $bd=a.$
It follows with the definition of associates that $a\sim b.$

Let $a\sim b.$
By the criterion for associates there is a unit $c\in R^\ast$ with $a=bc.$
Since $(c)=R$ (see principal ideal generated by a unit) it follows $(a)=aR=bcR=bR=(b).$

If $(a)=(0)$ then $a\sim 0.$ Therefore, there exist $b,c\in R$ with $ab=0$ and $c0=a.$ In particular, $a=0.$

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