SWI-Prolog (SWISH): No permission to modify static procedure `(=)/2'

Question

During some playing around with different list predicates in SWI-Prolog (SWISH), I was trying to check if an atom a was part of the list List1 which I defined in the program as List1 = [a,b,c,d].

I formulated my query as member(a, List1)., expecting something along the lines of a simple 'yes' (just as it shows in this youtube video at 59:25), but instead I got a warning

Singleton variables: [List1]

and an error

No permission to modify static procedure `(=)/2'

From what I understand from looking this up online, the warning is not that important here. I do not understand, however, why I get an error message while a is clearly a member of List1.

I tried this in two different ways:

1) By adding List1 = [a,b,c,d]. to the program and querying with member(a,List1). (which resulted in the error above);

2) By passing List1 = [a,b,c,d] directly to the interpreter and then using the same query ( member(a,List1). ), which resulted in an endless amount of results where a shifted positions in the Head of the list, like so:

List1 = [a|_1186]
List1 = [_1062, a|_1070]
List1 = [_1062, _1068, a|_1076]
List1 = [_1062, _1068, _1074, a|_1082]
List1 = [_1062, _1068, _1074, _1080, a|_1088]

Is this something about the specific Prolog version I am using, or am I missing something very simple?

EDIT

I was aware that a similar question was posed here , but I did not manage to fully understand the answer (nor the question) as it was immediately going about things as dynamic which I have not yet encountered in Prolog. I was looking for a more general, more 'high-level' answer which I have found by posing this question.

Answer 1

I defined in the program as List1 = [a,b,c,d].

This is not what it does. What it does is define a predicate =/2:

2 ?- write_canonical( (List1 = [a,b,c,d]) ).
=(_,[a,b,c,d])

^{(The ?-, or 2 ?- that you see there, is the interactive prompt of a Prolog system; SWI Prolog in my case. Whatever goes on that line after it is what I have typed; and then on the next line we see the system's response).}

Of course this tramples over the already existing built-in definition for = as the unification predicate. And hence the error which says precisely that. And yes, it is important.

To "define" a list in Prolog, we can define a predicate

8 ?- [user].
p([1,2,3,4]).

such that we can then query

9 ?- p(List1).
List1 = [1, 2, 3, 4].

and work further with List1,

10 ?- p(List1), member(A,List1).
List1 = [1, 2, 3, 4],
A = 1 ;
List1 = [1, 2, 3, 4],
A = 2 ;
List1 = [1, 2, 3, 4],
A = 3 ;
List1 = [1, 2, 3, 4],
A = 4.

We could also just directly specify the list as a sub-goal of our query,

11 ?- List1 = [1,2,3,4], member(A,List1).
List1 = [1, 2, 3, 4],
A = 1 ;
List1 = [1, 2, 3, 4],
A = 2 ;
List1 = [1, 2, 3, 4],
A = 3 ;
List1 = [1, 2, 3, 4],
A = 4.

making use of the predicate =/2, as opposed to redefining it, which is forbidden.

---

The above answers your 1). As for 2), you aren't telling us the whole truth. What you appear to have done, was to first make a query

12 ?- List1 = [a,b,c,d].
List1 = [a, b, c, d].

which is fine and dandy; and then make another query,

13 ?- member(a,List1).
List1 = [a|_G2181] ;
List1 = [_G2180, a|_G2184] ;
List1 = [_G2180, _G2183, a|_G2187] ;
List1 = [_G2180, _G2183, _G2186, a|_G2190] ;
List1 = [_G2180, _G2183, _G2186, _G2189, a|_G2193] .

Prolog prompt is not a REPL. We don't make definitions at it. We make queries.