Is the quotient of a complete ring, complete?

If $(R,\mathfrak m)$ is a complete local ring (with respect to the $\mathfrak m$-adic topology) and $I$ a prime ideal in $R$, is $R/I$ complete (with respect to the $\mathfrak m/I$-adic topology)?

It seems too strong, but I am unable to give a counterexample.

Proposition. Let $(R, \mathfrak{m})$ be a complete Noetherian local ring. Let $I$ be a proper ideal of $R$. Then $(R/I, \mathfrak{m}/I)$ is complete.

Proof. The following sequence of $R$-modules is exact

$$0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0.$$

Since $R$ is Noetherian, the following sequence is exact

$$0 \rightarrow \widehat{I} \rightarrow \widehat R \rightarrow \widehat{R/I} \rightarrow 0,$$

where $\widehat I$ etc. are the completions of $I$ etc. with respect to $\mathfrak{m}$-adic topology.

Since $R$ is complete, $\widehat R = R$. It is well known that $I$ is a closed submodule of $R$ with respect to $\mathfrak{m}$-adic toplogy. Since $R$ is complete, so is $I$, hence $\widehat I = I$. Therefore we get the following exact sequence

$$0 \rightarrow I \rightarrow R \rightarrow \widehat{R/I} \rightarrow 0,$$

hence $R/I$ is complete.