How to get the second factor from an equation in Prolog?

Question

Prolog

Well... How do you say equations? It is known that a certain number X is the result of multiplying two prime numbers. At first I tried
143 = prime(P) * prime(_)
But since there are 2 unknowns, I tried
143 = prime(P) * 13on the idea, Prolog should take 143 / 13 = 11 from here and check if the number 11 is prime. But it
143 = P * 13doesn't even work!
Here is the code: *poke*
PS {
Prime checker working
I don't think *(P, 13) is better than P * 13
SWI-Prolog
}

Answer 1

I didn’t write from scratch, I spied part of the code on the net (because “I haven’t picked up checkers for a long time”), but the idea is quite simple: you need to figure out the principles of inference in the prologue, in the sense that candidates must be generated by some process .
By reading The Art of Prolog
The idea is really simple, you need to write "generators" that will give candidates for the solution. In the code below, these are lists of prime numbers less than the given one, then the candidate is just a prime number from the list, pulled out by the standard member function
. The output is done like this

?- factorization(143,P1,P2).
1113
P1 = 11,
P2 = 13 .

?- factorization(25,P1,P2).
55
P1 = P2, P2 = 5 .

Solution code

divisible(X,Y) :- 0 is X mod Y, !.

divisible(X,Y) :- X > Y+1, divisible(X, Y+1).

is_prime(2) :- true,!.
is_prime(X) :- X < 2,!,false.
is_prime(X) :- not(divisible(X, 2)).

factorization(X,P1,P2) :- prime_list(2,X,L), member(P1,L), member(P2,L), is_prime(P1), is_prime(P2), X is P1 * P2.

prime_list(A,B,L) :- A =< 2, !, p_list(2,B,L).
prime_list(A,B,L) :- A1 is (A // 2) * 2 + 1, p_list(A1,B,L).

p_list(A,B,[]) :- A > B, !.
p_list(A,B,[A|L]) :- is_prime(A), !, next(A,A1), p_list(A1,B,L). 
p_list(A,B,L) :-  next(A,A1), p_list(A1,B,L). 
next(2,3) :- !.  
next(A,A1) :- A1 is A + 2.