Changes

|}

|}

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 u = v = 1, namely, du = dv = 0.

To put it another way, the Table characterizes Orbit 1

by means of the data:  <u, v, du, dv>  = <1, 1, 0, 0>.

Last but not least, yet another way to convey the same

information is by means of the (first order) extended

proposition:  u v (du)(dv).

</pre>

{| align="center" cellpadding="8" style="text-align:center"

{| align="center" cellpadding="8" style="text-align:center"