Trivial K-theory implies trivial K-theory of hereditary corners?

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Let $A$ be a $C^*$-algebra with trivial $K$-theory, that is $K_0(A)=K_1(A)=0$. Let $p$ be a projection in $A$. Does it follow that the hereditary corner $pAp$ has trivial $K$-theory?

I think in the direction of counter example. Notice that a counter example must be a non- simple $C^*$-algebra, otherwise we get that $pAp$ is full-hereditary and therefore $pAp$ and $A$ are stably isomorphic. Using stability of the $K$-groups, it follows that $K_*(pAp)=0$.

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See Example 9.9.4. in "An introduction to K-theory for C*-algebras" by Rordam, Larsen and Lausten.

Take $\mathcal T_0 \subset \scr L(\ell^2(\mathbb N))$ the reduced Toeplitz-algebra. It contains the compacts as an ideal and has trivial K-theory. However, the corner induced by a rank-one projection is isomorphic to $\mathbb C$.