Q: I want to replace some regular-language A to some other regular language B as long as there is no regular-language C after A.

I was expecting to be able to do something like this:

A -> B || _ ?* ~C;

However, this does not do what I want.

A: No, it won't work like that. Two key points to remember:

1. A rule of the form

   A -> B || L _ R ;

   is compiled as

   A -> || ?* L _ R ?* ;

   with the left and right context being extended out infinitely to the left and right, respectively. (You can block the infinite extension by overtly indicating the word boundaries with .#.)

2. Remember that ~C is the complement of C, and if C does not include the empty language, then ~C will include the empty language. Thus ~C can match the empty string and so will match the empty language. Thus

   A -> B || _ ?* ~C

   will match with any right context. I.e. you basically have

   A -> B || ?* _ ?* ~C ?*

   where ~C includes the empty language, so the rule reduces to

   A -> B || ?* _ ?*

   and does not constrain the replacement at all.

   The following should perform the constraint that you want:

   A -> B || _ ~$[C] .#.

   I.e. the right context must not include C. Be sure to indicate the .#. or the automatic extension of the contexts with ?* will get you in trouble again.