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Home/ Questions/Q 8083825
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T17:35:56+00:00 2026-06-05T17:35:56+00:00

How can I call a predicate if all calls to another predicate failed? I

  • 0

How can I call a predicate if all calls to another predicate failed?

I have:

foo(A, B, C, D) :-
    (bar1(Y,Z); bar2(L, K, M)),
    foo(A, B, C, D).

What I want :

  • at any time, If bar1/2 has succeded, bar2/3 will never be executed.
  • if all bar1/2 calls failed, then bar2/3 will eventually be executed.

Sample Backtracting tree

            root                                           root                 
                                                            |
           /   \                                            |
          /     \          all bar1 failed                  |
         /       \                                          |
        / | \     \        ===>>>======>>>>                 |
       /  |  \     \                                        |
      F   F  F     F                                        E
time  0   1  2     3                                        4 

Abbreviation:
       Bar1 failed : F
       Bar2 executed : E
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T17:35:58+00:00Added an answer on June 5, 2026 at 5:35 pm

    You’re looking for what’s known as “soft cut”,

    A *-> B ; C.
    

    This is equivalent to (A,B) ; (\+A,C): if A succeeds at least once, the above is equivalent to A,B. If not, it’s equivalent to just C. The goal A is not retried.

    Simple if construct allows the test predicate to succeed only once:

    A -> B ; C.
    

    is equivalent (almost – see you manual for details) to (once(A),B) ; (\+A,C), except that the goal A isn’t retried.

    Thus your case shall be

    foo(A, B, C, D) :-
        (bar1(Y,Z) *-> true ; bar2(L, K, M)),
        foo(A, B, C, D).
    

    addition: Some Prolog implementations might not have this construct *-> available (e.g. gprolog). In that case I see two possibilities. Either

    (A , B) ; (\+A , C)
    

    although it would retry A, or (writing goal(X) for A)

    bagof(g(X), goal(X), L) -> ( member(g(X), L), B ) ; C 
    

    Of course the order of side-effects will be changed by this. The name L of the variable should be chosen such that it does not appear free in B.

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