Proof.
Assume (1). If A is zero, then (2) holds. If A is not zero, then there exists a smallest nonzero object A_1 \subset A by the Artinian property. Of course A_1 is simple. If A_1 = A, then we are done. If not, then we can find A_1 \subset A_2 \subset A minimal with A_2 \not= A_1. Then A_2/A_1 is simple. Continuing in this way, we can find a sequence 0 \subset A_1 \subset A_2 \subset \ldots of subobjects of A such that A_ i/A_{i - 1} is simple. Since A is Noetherian, we conclude that the process stops. Hence (2) follows.
Assume (2). We will prove (1) by induction on n. If n = 1, then A is simple and clearly Noetherian and Artinian. If the result holds for n - 1, then we use the short exact sequence 0 \to A_{n - 1} \to A_ n \to A_ n/A_{n - 1} \to 0 and Lemmas 12.9.4 and 12.9.5 to conclude for n.
\square
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