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# FOL expression in Prolog

## FOL expression in Prolog

(OP)
Hi all, I'm trying to find a way to put the following first order logic expression into Prolog

#### CODE --> FOL

(p(0) or p(1)) and not (p(0) and p(1))

This means that it should respond in the following way to queries:

#### CODE --> Interpreter

?- p(0)
Yes.
?- p(1)
Yes.
?- p(0),p(1).
No. 

I tried to translate the logical expression:

#### CODE --> FOL

(p(0) or p(1)) and not (p(0) and p(1)) <=>
(not p(0) -> p(1)) and (p(0) -> not p(1)) <=>
p(0) <-> p(1) 

Using Clarks completion (that states that every definitional theory can be put in a logical program by giving the if-halves), I can obtain:

#### CODE --> Prolog

p(0) :- p(1).

Unfortunately, this resulting theory is only sound (it will not derive false information), but not complete (for example: p(1) cannot be derived). This is a consequence of Clarks theorem.

Does anybody know if there is a better solution? Thanks!
Replies continue below

### RE: FOL expression in Prolog

(OP)
I made a mistake in my post, and since I don't see an edit button I will post the correction here:

The derivation of the logical expression should have: p(0) <-> not p(1) as the last formula, and the Prolog code is: p(0) :- not p(1).

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