Notes on computational linguistics.pdf - UCLA Department of ...

Stabler - Lx 185/209 2003
Loading the axioms of people.pl, displayed on page 13 above, prolog will use these rules to establish consequences
of the theory. We can ask prolog to prove that Frege is a mathematician by typing
mathematician(frege). at the prolog prompt. Prolog will respond with yes. We can also use a variable
to ask prolog what things X are mathematicians (or computers). If the loaded axioms do not entail that any X
is a mathematician (or computer), prolog will say: no. If something can be proven to be a mathematician (or
computer), prolog will show what it is. And after receiving one answer, we can ask for other answers by typing
asemi-colon:
| ?- mathematician(X).
X = frege ? ;
X = hilbert ? ;
X = turing ? ;
X = montague ? ;
no
| ?- mathematician(fido).
no
Prolog establishes these claims just by finding substitutions of terms for the variable X which make the goal
identical to an axiom. So, in effect, a variable in a goal is existentially quantified. The goal ?-p(X) is, in effect,
a request to prove that there is some X such that p(X). 5
| ?- mathematician(X),linguist(X),piano_player(X).
X = montague ? ;
no
| ?-
We can display a proof as follows, this time showing the clause and bindings used at each step:
goal theory workspace
?-human(X) , Γ ⊢ ?-human(X)
?-human(X) , human(X ′ ):-mathematician(X ′ ) ⊢ ?-mathematician(X) {X ′ ↦ X}
?-human(X) , mathematician(frege) ⊢ {X ↦ frege}
5More precisely, the prolog proof is a refutation of ¬(∀X) p(X), and this is equivalent to an existential claim: ¬(∀X) p(X) ≡
(∃X)¬p(X).
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