Quotient of Noetherian (Artinian) ring is also Noetherian (Artinian)

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Proposition 1: Let $0\to M' \xrightarrow{f} M \xrightarrow{g} M''\to 0$ be an exact sequence of $A$-modules. Then:

i) $M$ is Noetherian iff $M'$ and $M''$ are Noetherian.

ii) $M$ is Artinian iff $M'$ and $M''$ are Artinian.

Proposition 2: Let $A$ be a Noetherian (Artinian), $I$ an ideal of $A$. Then $A/I$ is a Noetherian (Artinian) ring.

My question is the following: I know the proof of Proposition 1. How to use it in order to prove proposition 2?

I was trying to construct some examples of exact sequences but I failed to do it.

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The exact sequence you want is $$0\to I\to A\to A/I\to 0$$ $A$ is Noetherian or Artinian, hence so is $A/I$.