## Appendix B Detailed Longer Proofs of Theorems, Lemmas and Corollaries
### Proofs for Section 3
Here \({\bf u},{\bf v}\) are taken as \({\bf J},{\bf J}^{\rm unif[J_{1},J_{C}]}\), which were both seen to be raised-skew-centro-symmetric and they do satisfy \(J_{1}=J_{1}^{\rm unif[J_{1},J_{C}]}\), \(J_{C}=J_{C}^{\rm unif[J_{1},J_{C}]}\); it then follows from Theorem 3 that \({\bf J}-{\bf J}^{\rm unif[J_{1},J_{C}]}\) is skew-centrosymmetric, then so is INL in (114). An alternative less abstract but more cumbersome proof follows:  
From (294), (307) \({\rm INL}_{C}=0\), and (296) we have  
\[{\rm INL}_{c}=\sum_{i=1}^{c-1}{\rm DNL}_{i},\ \ \ {\rm INL}_{C-c}=-\sum_{i=C-c}^{C-1}{\rm DNL}_{i} \tag{115}\]  
For \(c=1\) the last two equations reduce to \({\rm INL}_{1}={\rm DNL}_{1}\), \(\ \ {\rm INL}_{C-1}=-{\rm DNL}_{C-1}\) but \({\rm DNL}_{C-1}={\rm DNL}_{1}\) since the length-C-1 DNL vector is centrosymmetric; thus, \({\rm INL}_{C-1}=-{\rm DNL}_{C-1}=-{\rm DNL}_{1}=-{\rm INL}_{1}\) i.e., we have established \({\rm INL}_{C-1}=-{\rm INL}_{1}\), the first step of an inductive proof. We may continue the inductive proof of \({\rm INL}_{C+1-c}=-{\rm INL}_{c}\), assuming \({\rm INL}_{C+1-(c-1)}\)  
\[\begin{array}{l}{\rm INL}_{C+1-c}={\rm INL}_{C+1-(c-1)}-{\rm DNL}_{C+1-c}=-{ \rm INL}_{c-1}-{\rm DNL}_{C-(c-1)}\\ =-{\rm INL}_{c-1}\text{ be true. Then,}\ \  
**Proof of Lemma 6.** To show the equivalence of (314) with \(\langle\mathbf{INL}\rangle=0\), take the mean of definition (277) of the INL, which in light of (314) must come out zero:  
\[\langle\mathrm{INL}[c]\rangle\equiv\frac{\langle\delta_{\mathrm{nom}}I_{c} \rangle}{\overline{\
abla\mathbf{J}}}=\frac{\langle J_{c}-J_{c}^{\mathrm{nom} }\rangle}{\overline{\
abla J}_{c}}=\frac{\langle J_{c}\rangle-\langle J_{c}^{ \mathrm{nom}}\rangle}{\overline{\
abla J}_{c}}=\frac{\langle J_{c}\rangle- \langle J_{c}\rangle}{\overline{\
abla J}_{c}}=0\] (A56)  
Now taking the means of the two Formulas (294) yields  
\[\langle\mathrm{INL}_{c}\rangle=\sum_{i=1}^{c-1}\langle\mathrm{DNL}_{i}\rangle,\hskip 14.226378 pt\langle\mathbf{DNL}\rangle=\
abla\langle\mathbf{INL}\rangle\] (A57)  
from which it readily follows that \(\langle\mathbf{INL}\rangle=0\Leftrightarrow\langle\mathbf{DNL}\rangle=0\). \(\square\)  
**Proof of Corollary 9. **  
\[\mathbf{IEV_{J}}\equiv\delta_{\langle\mathrm{nom}\rangle}^{\widetilde{\ \ }}\equiv\frac{\delta_{\langle\mathrm{nom}\rangle}\mathbf{J}}{\langle I_{u} \rangle}=\frac{\mathbf{J}-\langle\mathbf{J}^{\mathrm{nom}}\rangle}{\langle I _{u}\rangle}=\frac{\mathbf{J}}{\langle I_{u}\rangle}-\frac{\langle\mathbf{J}^ {\mathrm{nom}}\rangle}{\langle I_{u}\rangle}=\widetilde{\ \ }-\left\langle\widetilde{\ \ }\mathbf{J}^{\mathrm{nom}}\right\rangle=\delta_{\langle\mathrm{nom}\rangle} \widetilde{\ \ }\] (A58)  
To show (330) solve the equation \(\mathbf{IEV_{J}}=\frac{1}{\langle I_{u}\rangle}[\mathbf{J}-\langle\mathbf{J}^{ \mathrm{nom}}\rangle]\) for \(\mathbf{J}\), yielding \(\mathbf{J}=\langle\mathbf{J}^{\mathrm{nom}}\rangle+\mathbf{IEV_{J}}\cdot \langle I_{u}\rangle\Leftrightarrow\langle J_{c}\rangle=\langle J_{c}^{ \mathrm{nom}}\rangle+\mathrm{IEV_{J}}[c]\cdot\langle I_{u}\rangle\). Next recall Formula (326) for the elements \(\langle J_{c}^{\mathrm{nom}}\rangle\) of the mean-nominal constellation,  
\[\langle J_{c}^{\mathrm{nom}}\rangle=\langle I_{1}\rangle+(c-1)\ \langle\overline{\
abla\mathbf{J}}\rangle=\langle J_{1}\rangle+(c-1)\ \frac{\langle\mathrm{FS_{J}}\rangle}{C-1}=\langle J_{1}\rangle+(c-1)\langle I _{u}\rangle\]  
Now, using \(\langle I_{u}\rangle\equiv\frac{\langle\mathrm{FS_{J}}\rangle}{C-1}\) the last formula is recast as  
\[\langle J_{c}^{\mathrm{nom}}\rangle=\langle J_{1}\rangle+(c-1)\langle I_{u}\rangle\] (A59)  
Using this formula proves (331):  
\[ \begin{array}{ll}\langle J_{c}\rangle=\langle J_{c}^{\mathrm{nom}}\rangle+ \mathrm{IEV_{J}}[c]\cdot\langle I_{u}\rangle&=\langle J_{1}\rangle+(c-1) \langle I_{u}\rangle+\mathrm{IEV_{J}}[c]\cdot\langle I_{u}\rangle\\ &=\langle J_{1}\rangle+\big{[}(c-1)+\mathrm{IEV_{J}}[c]\big{]}\langle I_{u} \rangle.\end{array} \] (A60)  
\(\square\)  
**Proof of Lemma 5**  
**(i)**:  
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