Modifying the original boolean function by de Morgan's rule in Prolog


Good day.
I need to write a program in prologue that would change parts of an expression to equivalent ones according to the de Morgan rule.
The difficulty is that the program receives as an argument a temr consisting of functors, which in turn contain more functors.
Those. an expression like ~ (A v B) & (~ (A & C) v D & B) is fed into the program as
& (~ (v (A, B)), v (~ (& (A, C)), & (D, B))).
However, I can't get the result.

transform( ~ (XX & YY), ~XX v ~YY).
transform( ~ (XX v YY), ~XX & ~YY).

exp_transform(Term, Result_Term):-
  var(Term), var(Result_Term).

exp_transform(Term, Result_Term):-
  arg(_, Term, Value_Basic),
  arg(_, Result_Term, Value_of_ResTerm),
  exp_transform(Value_Basic, Value_of_ResTerm).

exp_transform(Term, Result_Term):-
  transform(Term, Result_Term),
  exp_transform(Term, Result_Term).

I went through various options. Can you tell me what the problem is? How do I serve and process an expression?


How do I serve and process an expression?

I would probably simplify the syntax by replacing the operators with regular functors

~ -> not,

& -> and,

v -> or.

 % Упростить булеву функцию    
reduce( X, Y ) :-
     transform( X, Y ), !.
 % Не упрощается, остаётся сама собой
reduce( X, X ).

% Двойное отрицание
transform( not( not( X )), X ).
% De Morgan 
transform( not( and( X, Y )), or( not( X ), not( Y ))).
% De Morgan 
transform( not( or( X, Y )), and( not( X ), not( Y ))).
% ......

Я использую SWI-Prolog. 
Scroll to Top