different behavior of predicates in prolog

Question

I'm trying to understand the different behavior of predicate in and outside of a rule. when I write:

test(1,2).

in the console I get "predicate not found". but when I write

S=[test(1,2,3)],member(test(1,A,B),S).

prolog answers although the predicate test is still not defined. why is that ?

Answer 1

In your second example, test(1,2,3) is just a structure. In a conventional language, something like [test(1,2,3)] would cause the language to go and evaluate some function . Python for instance:

def test(x, y, z):
   return 3

Then [test(1,2,3)] would reduce to [3] .

This is not how Prolog works. Prolog does not have functions, it has relations , and relations do not have a pre-defined data flow into and out of them. To give a concrete example, consider nth1/3 : if you call nth1(3, [a,b,c], X) then Prolog gives you back X = c , but you can also call nth1(X, [a,b,c], c) and Prolog gives you back X = 3 . You can also call nth1(X, [a,b,c], Y) and then Prolog will give you back three solutions: X = 1, Y = a , X = 2, Y = b and X = 3, Y = c .

I like this example a lot because it shows you that nth1/3 is doing all the work of Python's enumerate() and index() and __getitem__() . But it also shows you why Prolog cannot just evaluate predicates and replace them with what they "return."

That still leaves you with some ambiguity about what exactly [test(1,2,3)] means in Prolog. The answer is that it is a list containing Prolog terms. There's just one term in there, a functor or structure test(1,2,3) . Your second example shows that you are able to unify that structure with another structure. This is a very powerful thing that Prolog lets you do. It's part of what makes Prolog a homoiconic language, meaning a language that has a built-in understanding of its own constituents. For the time being, you can think of it as analogous to Javascript objects or Python dictionaries or C structures. It just happens to have the same shape as the predicate definitions. I consider this fact both a source of great power and great confusion for beginners, so take some heart! It is not intuitive, especially coming from other languages.

Answer 2

In your first example, test is a predicate of arity 2 which you try to prove and that does not exist.

In your second example, test is a functor of arity 3. That does not define any test predicate. It is just a function whose inputs are the values 1 , 2 , and 3 , and whose value is test(1,2,3) . So, test(1,2,3) is the value of the function test/3 . A constant like a may also be viewed as a function, it has no input, and its value is a .

Predicates define relations between functions.