(a_i {\ensuremath{\otimes}}a_i) 
    & = \sum_{i=1}^n \alpha_i \delta(a_i) \\
    & = \delta(a_0) \\
    & = (\sum_{i=1}^n \alpha_i a_i) {\ensuremath{\otimes}}(\sum_{j=1}^n \alpha_j a_j) \\
    & = \sum_{i,j=1}^n \alpha_i \alpha_j (a_i {\ensuremath{\otimes}}a_j).
  \end{aligned}$$ By minimality, $\{a_1,\ldots,a_n\}$ is linearly independent. Hence $\alpha_i^2=\alpha_i$ for all $i$, and $\alpha_i \alpha_j=0$ for $i \neq j$. So $\alpha_i=0$ or $\alpha_i=1$ for all $i$. If $\alpha_j=1$, then $\alpha_i=0$ for all $i \neq j$, so $a_0=a_j$. By minimality, then $j=1$ and $\{a_0,a_j\} = \{a_0\}$, which is impossible. So we must have $\alpha_i=0$ for all $i$. But then $a_0=0$, which is likewise a contradiction.

\[lem:normal\] Assuming only (M), nonzero copyable elements have unit norm.

Let $a$ be a copyable element. Then $\delta(a) = a {\ensuremath{\otimes}}a$. Hence $$\|a\| = \|\delta(a)\| = \|a
  {\ensuremath{\otimes}}a\| = \|a\|^2.$$ It follows that $\|a\|$ is either 0 or 1. Therefore, if $a$ is a nonzero, then $\|a\|=1$.

\[prop:orthogonal\] Assuming only (F), copyable elements are pairwise orthogonal.

[@coeckepavlovicvicary:bases Corollary 4.7] Let $a,b$ be copyables. Then: $$\begin{aligned}
        {\ensuremath{\langle a\,|\,a \rangle}} \cdot {\ensuremath{\langle a\,|\,a \rangle}} \cdot {\ensuremath{\langle b\,|\,a \rangle}}
    & = {\ensuremath{\langle a {\ensuremath{\otimes}}a {\ensuremath{\otimes}}b\,|\,a {\ensuremath{\otimes}}a {\ensuremath{\otimes}}a \rangle}} \\
    & = {\ensuremath{\langle (\delta {\ensuremath{\otimes}}{\ensuremath{\mathrm{id}_{}}})(a {\ensuremath{\otimes}}b)\,|\,({\ensuremath{\mathrm{id}_{}}}{\ensuremath{\otimes}}\delta)(a {\ensuremath{\otimes}}a) \rangle}} \\
    & = {\ensuremath{\langle a {\ensuremath{\otimes}}b\,|\,(\delta^\dag {\ensuremath{\otimes}}{\ensuremath{\mathrm{id}_{}}}) {\circ}({\ensuremath{\mathrm{id}_{}}}{\ensuremath{\otimes}}\delta)(a {\ensuremath{\otimes}}a) \rangle}} \\
    & = {\ensuremath{\langle 