'if' in prolog?

Question

Is there a way to do an if in prolog, e.g. if a variable is 0, then to do some actions (write text to the terminal). An else isn\'t even needed, but I can\'t find any docume

Answer 1

First, let's recall some classical first order logic:

"If P then Q else R" is equivalent to "(P and Q) or (non_P and R)".

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How can we express "if-then-else" like that in Prolog?

Let's take the following concrete example:

If X is a member of list [1,2] then X equals 2 else X equals 4.

We can match above pattern ("If P then Q else R") if ...

• condition P is list_member([1,2],X),
• negated condition non_P is non_member([1,2],X),
• consequence Q is X=2, and
• alternative R is X=4.

To express list (non-)membership in a pure way, we define:

list_memberd([E|Es],X) :-
   (  E = X
   ;  dif(E,X),
      list_memberd(Es,X)
   ).

non_member(Es,X) :-
   maplist(dif(X),Es).

---

Let's check out different ways of expressing "if-then-else" in Prolog!

1. (P,Q ; non_P,R)

   ?-      (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4).
   X = 2 ; X = 4.
   ?- X=2, (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=2.
   X = 2 ; false.
   ?-      (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=2.
   X = 2 ; false.
   ?- X=4, (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=4.
   X = 4.
   ?-      (list_memberd([1,2],X), X=2 ; non_member([1,2],X), X=4), X=4.
   X = 4.
   

   Correctness score 5/5. Efficiency score 3/5.

2. (P -> Q ; R)

   ?-      (list_memberd([1,2],X) -> X=2 ; X=4).
   false.                                                % WRONG
   ?- X=2, (list_memberd([1,2],X) -> X=2 ; X=4), X=2.
   X = 2.
   ?-      (list_memberd([1,2],X) -> X=2 ; X=4), X=2.
   false.                                                % WRONG
   ?- X=4, (list_memberd([1,2],X) -> X=2 ; X=4), X=4.
   X = 4.
   ?-      (list_memberd([1,2],X) -> X=2 ; X=4), X=4.
   false.                                                % WRONG
   

   Correctness score 2/5. Efficiency score 2/5.

3. (P *-> Q ; R)

   ?-      (list_memberd([1,2],X) *-> X=2 ; X=4).
   X = 2 ; false.                                        % WRONG
   ?- X=2, (list_memberd([1,2],X) *-> X=2 ; X=4), X=2.
   X = 2 ; false.
   ?-      (list_memberd([1,2],X) *-> X=2 ; X=4), X=2.
   X = 2 ; false.
   ?- X=4, (list_memberd([1,2],X) *-> X=2 ; X=4), X=4.
   X = 4.
   ?-      (list_memberd([1,2],X) *-> X=2 ; X=4), X=4.
   false.                                                % WRONG
   

   Correctness score 3/5. Efficiency score 1/5.

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(Preliminary) summary:

1. (P,Q ; non_P,R) is correct, but needs a discrete implementation of non_P.

2. (P -> Q ; R) loses declarative semantics when instantiation is insufficient.

3. (P *-> Q ; R) is "less" incomplete than (P -> Q ; R), but still has similar woes.

---

Luckily for us, there are alternatives: Enter the logically monotone control construct if_/3!

We can use if_/3 together with the reified list-membership predicate memberd_t/3 like so:

?-      if_(memberd_t(X,[1,2]), X=2, X=4).
X = 2 ; X = 4.
?- X=2, if_(memberd_t(X,[1,2]), X=2, X=4), X=2.
X = 2.
?-      if_(memberd_t(X,[1,2]), X=2, X=4), X=2.
X = 2 ; false.
?- X=4, if_(memberd_t(X,[1,2]), X=2, X=4), X=4.
X = 4.
?-      if_(memberd_t(X,[1,2]), X=2, X=4), X=4.
X = 4.

Correctness score 5/5. Efficiency score 4/5.