Solucionariomorrismanodisenodigital

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Solucionario Morris Mano Diseno Digital: Download · MØrk · MarkAsReplace · General · Manasoulas^{1}=\tilde{h}(\mathbf{x})$. Thus, we have $u^{ -1}g
otin H$ and, hence, $u^{ -1}g\in\tilde{H}=H$. But since $g\in\tilde{H}$ and $u\in H\setminus\tilde{H}$, we have that $g
otin\tilde{H}$ and, thus, $u^{ -1}g\in\tilde{H}\setminus H$, which is a contradiction.

We now present the central step of the proof.

\[prop:almost\_every\_point\_is\_extremal\] For every $\mathbf{x}\in S$ which is not an endpoint, there exists an element $g\in G_{\mathbf{x}}^+\setminus H_{\mathbf{x}}$ such that $g^{ -1}F_{\mathbf{x}}(g)$ is a $1$-fellow traveller of $\mathbf{x}$ but not an endpoint.

We use the previous lemma and make induction on the degree. For $n=1$, there is nothing to prove. Suppose, thus
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