our task is to write the following predicates in SWI-Prolog.

Your task is to write the following predicates in SWI-Prolog. The slash and number after the predicate name give the number of parameters for the predicate. These characters are NOT part of the predicate name. So, remove/3 means that remove is a predicate with 3 parameters.

1) remove/3. Removes from the list given by the first parameter, all elements appearing in the list given by the second parameter, and returns the result in the third parameter. Examples: 31 ?- remove([a, b, c, d, e, f], [a, d], X). X = [b, c, e, f] ; 32 ?- remove([a, b, c, d, e, f], [], X). X = [a, b, c, d, e, f] ; 33 ?- remove([a, b, c, d, e, f, a, a, d, e], [a, e], X). X = [b, c, d, f, d] ;

2) members_in_common/3. Given two lists through the first two parameters, it returns the list of common elements in the third parameter. Examples: 46 ?- members_in_common([a, b, c, d, a, b, a], [a, d, f, f, c, c, f], X). X = [a, c, d, a, a] Yes 47 ?- members_in_common([a, b, c, d, a, b, a], [p, q, r, s], X). X = [] Yes 48 ?- members_in_common([a, b, c, d, a, b, a], [], X). X = [] Yes

3) sum/2. Given two parameters, the first one being a list of numbers, return the sum of the numbers in the list in the second parameter. Examples: sum([3, 5, -7, 8], X). X = 9 sum([3], X). X = 3 4) min/2. Given two parameters, the first one being a list of numbers, return the minimum value in the list in the second parameter. min([3, 3, 8, 9, 2, 9, 2], X). X = 2 min([7], X). X = 7 None of the lists is a list of lists. The functions must all be independent. 

The only PROLOG predicate that you are allowed to use is member.