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Re: Friendly amendment to Motion 25

Nate Hayes wrote:

Evaluating the expression above, e.g.:

   sqrt( X ) + X
       = sqrt( (X,dd0) ) + (X,dd0)   // promote X to decorated interval
       = (sqrt(X),inf(S(sqrt,X),dd0)) + (X,dd0)
       = (sqrt(X),inf(dd1,dd0)) + (X,dd0)
       = (sqrt(X),dd2) + (X,dd0)
       = (sqrt(X)+X,inf(S(+,sqrt(X),X),dd2,dd0))
       = (sqrt(X)+X,inf(dd3,dd2,dd0))
       = (sqrt(X)+X,dd4)

where dd0 is the "native" decoration for X, and dd1, ... dd4 are computed
decorations of intermediate steps of computation; then X is the independent 
variable of the entire expression, because the decoration dd0 of X is not
given by T. The decoration dd1 of sqrt(X) is given by T, as is the
decoration dd3 of the addition operation. So neither of those results are
independent variables of the expression. The decoration dd4 of the entire
expression is also given by T, and is an dependent variable.


I'm sorry, type-o:

   -- The decoration dd2 of sqrt(X) is given by T
   -- The decoration dd4 of the addition operation is given by T
The decorations dd1 and dd3 are given by S, and are the decorations of the "current" operations. 

Nate