I'm a new one to the theory of C$^*$-algebras, and I'm really missing something significant.
According to Blackadar,"Operator algebras", page 153 Brown-Green-Rieffel theorem - For $\sigma$-unital C$^*$-algebras Morita equivalence is the same as stable isomorphism.
According to Walter Beer, "On Morita equivalence of nuclear C$^*$-algebras", page 262, theorem 3.7 - All non-type one, separable, nuclear C$^*$-algebras are Morita equivalent.
So that, since separabillity implies $\sigma$-unitallity, we have that all separable, nuclear, non-type one C$^*$-algebras are stably isomorphic, that is false.
For example take two Cuntz algebras $\mathcal{O}_n$, $\mathcal{O}_m$. They are: separable (because they are finitely generated), nuclear, non-type one. Then they are stably isomorphic according to the result.
$$ \mathcal{O}_n \otimes \mathbb{K} \cong \mathcal{O}_m \otimes \mathbb{K} $$
Then take $K_0$. Since $K_0$ is stable, we have:
$$ \mathbb{Z}_{n-1} \cong K_0(\mathcal{O}_n) \cong K_0(\mathcal{O}_n \otimes \mathbb{K}) \cong K_0(\mathcal{O}_m \otimes \mathbb{K}) \cong K_0(\mathcal{O}_m) \cong \mathbb{Z}_{m-1} $$
That holds for all $n,m$.
Where am I wrong?