Question

I have a strange question with using negation in SWI Prolog. When I use the following code without a space after negation “\+”, it gives an error:

2 ?- assert((only_child(X) :- mother(M,X),\+(mother(M,_),\+ X==_))).
true.

3 ?- only_child(cathy).
ERROR: only_child/1: Undefined procedure: (\+)/2
ERROR:   However, there are definitions for:
ERROR:         (\+)/1
Exception: (7) \+(mother(melody, _G2102), \+cathy==_G2107) ? creep
4 ?-

However if I use space after negation, it works perfectly fine.

2 ?- assert((only_child(X) :- mother(M,X),\+ (mother(M,_),\+ X==_))).
true.

3 ?- only_child(cathy).
false.

4 ?-

But the following code works fine for both, using space after “\+” and without using space.

4 ?- \+ father(michael,cathy).
false.

5 ?- \+father(michael,cathy).
false.

6 ?-

Can anyone please explain this to me? Its really very puzzling. I will be really thankful.

Answer 1

The opening parenthesis is special when parsing Prolog. If there is no space between an identifier and an opening paren, the identifier is always treated as the name of a functor (even if it is not alphanumeric). Whatever is inside the parens is treated as a list of arguments for this functor.

However, if there is a space, the paren is treated in its normal, mathematical function of grouping expressions. The comma inside such an expression serves the role of a boolean and operator.

So:

"func(A,B)"  - OK - invoke `func` on parameters `A`, `B`    
"func (A,B)" - syntax error - interpreted as an identifier stuck together 
               with an AND expression    
"\+ (A,B)"   - OK - operator `\+` acting on `(A,B)` (A and B)    
"\+(A,B)"    - error - invoke `\+` with two parameters `A`, `B` 
               but \+ only takes one argument    
"\+(A)"      - OK - since `\+` takes one argument, it can 
               be invoked as a functor with one argument inside parens

The parser is also intelligent enough to split operators from identifiers (the first letter triggers a new token):

"\+ father(A,B)" - OK - invoke functor `father` with `A` and `B`, 
                   negate the result
"\+father(A,B)"  - still OK - the parser will stop reading the 
                   name of the operator when it encounters the 'f', 
                   so the result is same as above