'if' in prolog?

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 documentation of if.

Yes, there is such a control construct in ISO Prolog, called ->. You use it like this:

( condition -> then_clause ; else_clause )

Here is an example that uses a chain of else-if-clauses:

(   X < 0 ->
    writeln('X is negative.  That's weird!  Failing now.'),
    fail
;   X =:= 0 ->
    writeln('X is zero.')
;   writeln('X is positive.')
)

Note that if you omit the else-clause, the condition failing will mean that the whole if-statement will fail. Therefore, I recommend always including the else-clause (even if it is just true).

A standard prolog predicate will do this.

   isfive(5). 

will evaluate to true if you call it with 5 and fail(return false) if you run it with anything else. For not equal you use \=

isNotEqual(A,B):- A\=B.

Technically it is does not unify, but it is similar to not equal.

Learn Prolog Now is a good website for learning prolog.

Edit: To add another example.

isEqual(A,A). 

Prolog predicates 'unify' -

So, in an imperative langauge I'd write

function bazoo(integer foo)
{
   if(foo == 5)
       doSomething();
   else
       doSomeOtherThing();
}

In Prolog I'd write

bazoo(5) :-  doSomething.
bazoo(Foo) :- Foo =/= 5, doSomeOtherThing.

which, when you understand both styles, is actually a lot clearer.
"I'm bazoo for the special case when foo is 5"
"I'm bazoo for the normal case when foo isn't 5"

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)".

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.

(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.

I found this helpful for using an if statement in a rule.

max(X,Y,Z) :-
    (  X =< Y
    -> Z = Y
    ;  Z = X
    ).

Thanks to http://cs.union.edu/~striegnk/learn-prolog-now/html/node89.html