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A quick inspection of the first Table suggests a rule to cover the case when <math>\texttt{u~=~v~=~1},</math> namely, <math>\texttt{du~=~dv~=~0}.</math>  To put it another way, the Table characterizes Orbit 1 by means of the data:  <math>(u, v, du, dv) = (1, 1, 0, 0).\!</math>  Another way to convey the same information is by means of the extended proposition:  <math>\texttt{u~v~(du)(dv)}.</math>

A quick inspection of the first Table suggests a rule to cover the case when <math>\texttt{u~=~v~=~1},</math> namely, <math>\texttt{du~=~dv~=~0}.</math>  To put it another way, the Table characterizes Orbit&nbsp;1 by means of the data:  <math>(u, v, du, dv) = (1, 1, 0, 0).\!</math>  Another way to convey the same information is by means of the extended proposition:  <math>\texttt{u~v~(du)(dv)}.</math>

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A more fine combing of the second Table brings to mind a rule that partly covers the remaining cases, that is, <math>\texttt{du~=~v}, ~\texttt{dv~=~(u)}.</math> To vary the formulation, this Table characterizes Orbit&nbsp;2 by means of the following vector equation:  <math>(\texttt{du}, \texttt{dv}) = (\texttt{v}, \texttt{(u)}).</math>  This much information about Orbit&nbsp;2 is also encapsulated by the extended proposition, <math>\texttt{(uv)((du, v))(dv, u)},</math> which says that <math>u\!</math> and <math>v\!</math> are not both true at the same time, while <math>du\!</math> is equal in value to <math>v,\!</math> and <math>dv\!</math> is opposite in value to <math>u.\!</math>

A more fine combing of the second Table brings to mind

a rule that partly covers the remaining cases, that is,

du = v, dv = (u).  To vary the formulation, this Table

characterizes Orbit 2 by means of the following vector

equation:  <du, dv> = <v, (u)>. This much information

about Orbit 2 is also encapsulated by the (first order)

extended proposition, (uv)((du, v))(dv, u), which says

that u and v are not both true at the same time, while

du is equal in value to v, and dv is the opposite of u.

</pre>

==Note 21==

==Note 21==