Reification of term equality/inequality

Question

Pure Prolog programs that distinguish between the equality and inequality of terms in a clean manner suffer from execution inefficiencies ; even when all terms of relevance

Answer 1

Here's an even shorter logically-pure implementation of occurrences/3.

We build it upon the meta-predicate tfilter/3, the reified term equality predicate (=)/3, and the coroutine allEqual_to__lazy/2 (defined in my previous answer to this question):

occurrences(E,Xs,Es) :-
   allEqual_to__lazy(Es,E),
   tfilter(=(E),Xs,Es).

Done! To ease the comparison, we re-run exactly the same queries I used in my previous answer:

?- Fs = [1,2], occurrences(E,Es,Fs).
false.

?- occurrences(E,[1,2,3,1,2,3,1],Fs).
Fs = [1,1,1],     E=1                     ;
Fs = [2,2],                 E=2           ;
Fs = [3,3],                           E=3 ;
Fs = [],      dif(E,1), dif(E,2), dif(E,3).

?- occurrences(1,[1,2,3,1,2,3,1],Fs).
Fs = [1,1,1].

?- Es = [E1,E2], occurrences(E,Es,Fs).
Es = [E, E ], Fs = [E,E],     E1=E ,     E2=E  ;
Es = [E, E2], Fs = [E],       E1=E , dif(E2,E) ;
Es = [E1,E ], Fs = [E],   dif(E1,E),     E2=E  ;
Es = [E1,E2], Fs = [],    dif(E1,E), dif(E2,E).

?- occurrences(1,[E1,1,2,1,E2],Fs).
    E1=1 ,     E2=1 , Fs = [1,1,1,1] ;
    E1=1 , dif(E2,1), Fs = [1,1,1]   ;
dif(E1,1),     E2=1 , Fs = [1,1,1]   ;
dif(E1,1), dif(E2,1), Fs = [1,1].

?- occurrences(1,L,[1,2]).
false.

?- L = [_|_],occurrences(1,L,[1,2]).
false.

?- L = [X|X],occurrences(1,L,[1,2]).
false.

?- L = [L|L],occurrences(1,L,[1,2]).
false.

At last, the most general query:

?- occurrences(E,Es,Fs).
Es = Fs, Fs = []      ;
Es = Fs, Fs = [E]     ;
Es = Fs, Fs = [E,E]   ;
Es = Fs, Fs = [E,E,E] % ... and so on ad infinitum ...

We get the same answers.