I'm new to using prolog and I'm trying things to get a text parsed and translated in this way:
?- go.
|: All boys run.
s(np(det(all),noun(boys)),vp(verb(run)))
S = all(_4326, boy(_4326)->run(_4326))
but I'm having an error: I run go. to get the translation, I get the following error:
go.
|: All boys run
**ERROR: Stream user_input:242:4 Syntax error: Operator expected**
Here is my code.
% Tokenizer
go1(Out):-
read(X),
name(X,L),
tokenize(L,Out).
tokenize([],[]):- !.
tokenize(L,[Word|Out]):- L\==[],
tokenize(L,Rest,WordChs),
name(Word,WordChs),
tokenize(Rest,Out).
tokenize([],[],[]):- !.
tokenize([46|_],[],[]):- !.
tokenize([32|T],T,[]):- !.
tokenize([H|T],Rest,[H|List]):-
tokenize(T,Rest,List).
% Main predicate
go :-
%go1(X),
readln(X, _, _, _, lowercase),
reset_gensym,
sentence(P, X, []),
translator(P, R),
write(R).
%Operator's definition
:- op(600,xfy,=>).
:- op(500,xfy,&).
%Grammar + Parser
sentence(sentence(NP, VP)) --> noun_phrase(Num, NP), verb_phrase(Num, VP).
%test cases
% All boys run
% All boys like all watermelons that contain some divine flavor
% Some boy eats some apple
% Some government conscripts some pacifist people
% All government that conscripts pacifist people are evil
noun_phrase(Num, np(DET, N)) --> determiner(Num, DET), noun(Num, N).
noun_phrase(Num, np(ADJ, N)) --> adjective(ADJ), noun(Num, N).
noun_phrase(Num, np(DET, ADJ, N)) --> determiner(Num, DET), adjective(ADJ), noun(Num, N).
noun_phrase(Num, np(DET, N, RCL)) --> determiner(Num, DET), noun(Num, N), relative_clause(Num, RCL).
noun_phrase(Num, np(N)) --> noun(Num, N).
verb_phrase(Num, vp(V)) --> verb(Num, itrans, V).
verb_phrase(Num, vp(V, NP)) --> verb(Num, trans, V), noun_phrase(_, NP).
verb_phrase(Num, vp(BV, ADJ)) --> beVerb(Num, BV), adjective(ADJ).
relative_clause(Num, rcl(REL, VP)) --> rel(REL), verb_phrase(Num,VP).
relative_clause(Num, rcl(REL, NP, V)) --> rel(REL), noun_phrase(Num,NP), verb(Num, itrans, V).
adjective(adj(good)) --> [good].
adjective(adj(evil)) --> [evil].
adjective(adj(big)) --> [big].
adjective(adj(divine)) --> [divine].
adjective(adj(pacifist)) --> [pacifist].
adjective(adj(nice)) --> [nice].
is_det(plural, all).
is_det(singular, a).
is_det(_, the).
is_det(_, some).
determiner(Num, determiner(D)) --> [D], {is_det(Num, D)}.
is_noun(plural, boys).
is_noun(singular, boy).
is_noun(plural, girls).
is_noun(singular, girl).
is_noun(plural, people).
is_noun(singular, person).
is_noun(plural, apples).
is_noun(singular, apple).
is_noun(plural, watermelons).
is_noun(singular, watermelon).
is_noun(plural, rabbits).
is_noun(plural, carrots).
is_noun(plural, evil).
is_noun(plural, governments).
is_noun(singular, government).
is_noun(singular, flavor).
noun(Num,noun(N)) --> [N], {is_noun(Num,N)}.
is_verb(plural, trans, like).
is_verb(singular, trans, likes).
is_verb(plural, _, eat).
is_verb(singular, _, eats).
is_verb(plural, itrans, run).
is_verb(singular, itrans, runs).
is_verb(plural, trans, contain).
is_verb(singular, trans, contains).
is_verb(plural, trans, conscript).
is_verb(singular, trans, conscripts).
verb(Num, Tp, v(V)) --> [V], {is_verb(Num, Tp, V)}.
beVerb(plural, bv(tastes)) --> [taste].
beVerb(plural, bv(iss)) --> [are].
beVerb(singular, bv(iss)) --> [iss].
beVerb(singular, bv(sounds)) --> [sounds].
rel(rel(that)) --> [that].
% Rules for translator
translator(sentence(NP,VP),F) :-
% genera un unico atomo de la X y lo unifica en Var
gensym(x,Var),
np(Var,NP,T),
vp(Var,T,VP,F).
% Translation of noun phrase
np(Var, np(DET,N), R) :-
n(Var, N, P1),
det(DET, P1, R).
np(Var, np(DET,ADJ,N), R) :-
adj(Var, ADJ, P1),
n(Var, N, P1, P2),
det(DET, P2, R).
np(Var, np(ADJ, N), R) :-
adj(Var, ADJ, P1),
n(Var, N, P1, R).
np(Var, np(DET, N, RCL), R) :-
n(Var, N, P1),
rcl(Var, RCL, P1, P2),
det(DET, P2, R).
np(Var, np(N), R) :-
n(Var, N, R).
rcl(Var, rcl(_, VP), T, R) :- vp(Var, T, VP, R).
% This predicate add an implication or conjuction between F1 and F2
q_join(all, V, F1, F2, all(V, F1 => F2)).
q_join(exists, V, F1, F2, exists(V, F1 & F2)).
% Translation of verb phrase
vp(Var, P, vp(V), R) :-
P =.. [Q, V1, F1],
v(Var, V, C),
q_join(Q, V1, F1, C, R).
vp(Var, P, vp(bv(BVERB), adj(ADJ)), R) :-
P =.. [Q, V1, F1],
C =.. [BVERB, Var, ADJ],
q_join(Q, V1, F1, C, R).
vp(Var, P, vp(V, NP), R) :-
P =.. [Q1, V1, F1],
gensym(x,X),
v(Var, X, V, C),
np(X, NP, T),
T =.. [Q2, V2, F2],
q_join(Q2, V2, F2, C, R1),
q_join(Q1, V1, F1, R1, R).
% Terminals symbols, they create functors
det(determiner(all), P, R) :-
P =.. [_, V, F],
R =.. [all, V, F],
!.
det(determiner(_), P, P).
n(Var, noun(N), exists(Var, P)) :- P =.. [N, Var].
n(Var, noun(N), A, exists(Var, (P & A))) :- P =.. [N, Var].
v(Var, v(V), P) :- P =.. [V, Var].
v(V1, Var, v(V), P) :- P =.. [V, V1, Var].
adj(Var,adj(ADJ),P) :- P =.. [ADJ, Var].
I have a function that uses rewrite to satisfy the Agda type checker. I thought that I had a reasonably good grasp of how to deal with the resulting "vertical bars" in proofs about such functions. And yet, I fail completely at dealing with these bars in my seemingly simple case.
Here are the imports and my function, step. The rewrites make Agda see that n is equal to n + 0 and that suc (acc + n) is equal to acc + suc n, respectively.
module Repro where
open import Relation.Binary.PropositionalEquality as P using (_≡_)
open import Data.Nat
open import Data.Nat.DivMod
open import Data.Nat.DivMod.Core
open import Data.Nat.Properties
open import Agda.Builtin.Nat using () renaming (mod-helper to modₕ)
step : (acc d n : ℕ) → modₕ acc (acc + n) d n ≤ acc + n
step zero d n rewrite P.sym (+-identityʳ n) = a[modₕ]n<n n (suc d) 0
step (suc acc) d n rewrite P.sym (+-suc acc n) = a[modₕ]n<n acc (suc d) (suc n)
Now for the proof, which pattern matches on acc, just like the function. Here's the zero case:
step-ok : ∀ (acc d n : ℕ) → step acc d n ≡ a[modₕ]n<n acc d n
step-ok zero d n with n | P.sym (+-identityʳ n)
step-ok zero d n | .(n + 0) | P.refl = ?
At this point, Agda tells me I'm not sure if there should be a case for the constructor P.refl, because I get stuck when trying to solve the following unification problems (inferred index ≟ expected index): w ≟ w + 0 [...]
I am also stuck in the second case, the suc acc case, albeit in a different way:
step-ok (suc acc) d n with suc (acc + n) | P.sym (+-suc acc n)
step-ok (suc acc) d n | .(acc + suc n) | P.refl = ?
Here, Agda says suc (acc + n) != w of type ℕ when checking that the type [...] of the generated with function is well-formed
Update after Sassa NF's response
I followed Sassa NF's advice and reformulated my function with P.subst instead of rewrite. I.e., I changed my right-hand side from being about n + 0 to being about n, instead of conversely changing the goal from being about n to being about n + 0:
step′ : (acc d n : ℕ) → modₕ acc (acc + n) d n ≤ acc + n
step′ zero d n = P.subst (λ # → modₕ 0 # d # ≤ #) (+-identityʳ n) (a[modₕ]n<n n (suc d) 0)
step′ (suc acc) d n = P.subst (λ # → modₕ (suc acc) # d n ≤ #) (+-suc acc n) (a[modₕ]n<n acc (suc d) (suc n))
During the proof, the P.subst in the function definition needs to be eliminated, which can be done with a with construct:
step-ok′ : ∀ (acc d n : ℕ) → step′ acc d n ≡ a[modₕ]n<n acc d n
step-ok′ zero d n with n + 0 | +-identityʳ n
... | .n | P.refl = P.refl
step-ok′ (suc acc) d n with acc + suc n | +-suc acc n
... | .(suc (acc + n)) | P.refl = P.refl
So, yay! I just finished my very first Agda proof involving a with.
Some progress on the original problem
My guess would be that my first issue is a unification issue during dependent pattern matching: there isn't any substitution that makes n identical to n + 0. More generally, in situations where one thing is a strict subterm of the other thing, I suppose that we may run into unification trouble. So, maybe using with to match n with n + 0 was asking for problems.
My second issue seems to be what the Agda language reference calls an ill-typed with-abstraction. According to the reference, this "happens when you abstract over a term that appears in the type of a subterm of the goal or argument types." The culprit seems to be the type of the goal's subterm a[modₕ]n<n (suc acc) d n, which is modₕ [...] ≤ (suc acc) + n, which contains the subterm I abstract over, (suc acc) + n.
It looks like this is usually resolved by additionally abstracting over the part of the goal that has the offending type. And, indeed, the following makes the error message go away:
step-ok (suc acc) d n with suc (acc + n) | P.sym (+-suc acc n) | a[modₕ]n<n (suc acc) d n
... | .(acc + suc n) | P.refl | rhs = {!!}
So far so good. Let's now introduce P.inspect to capture the rhs substitution:
step-ok (suc acc) d n with suc (acc + n) | P.sym (+-suc acc n) | a[modₕ]n<n (suc acc) d n | P.inspect (a[modₕ]n<n (suc acc) d) n
... | .(acc + suc n) | P.refl | rhs | P.[ rhs-eq ] = {!!}
Unfortunately, this leads to something like the original error: w != suc (acc + n) of type ℕ when checking that the type [...] of the generated with function is well-formed
One day later
Of course I'd run into the same ill-typed with-abstraction again! After all, the whole point of P.inspect is to preserve a[modₕ]n<n (suc acc) d n, so that it can construct the term a[modₕ]n<n (suc acc) d n ≡ rhs. However, preserved a[modₕ]n<n (suc acc) d n of course still has its preserved original type, modₕ [...] ≤ (suc acc) + n, whereas rhs has the modified type modₕ [...] ≤ acc + suc n. That's what's causing trouble now.
I guess one solution would be to use P.subst to change the type of the term we inspect. And, indeed, the following works, even though it is hilariously convoluted:
step-ok (suc acc) d n with suc (acc + n) | P.sym (+-suc acc n) | a[modₕ]n<n (suc acc) d n | P.inspect (λ n → P.subst (λ # → modₕ (suc acc) # d n ≤ #) (P.sym (+-suc acc n)) (a[modₕ]n<n (suc acc) d n)) n
... | .(acc + suc n) | P.refl | rhs | P.[ rhs-eq ] rewrite +-suc acc n = rhs-eq
So, yay again! I managed to fix my original second issue - basically by using P.subst in the proof instead of in the function definition. It seems, though, that using P.subst in the function definition as per Sassa NF's guidance is preferable, as it leads to much more concise code.
The unification issue is still a little mysterious to me, but on the positive side, I unexpectedly learned about the benefits of irrelevance on top of everything.
I'm accepting Sassa NF's response, as it put me on the right track towards a solution.
Your use of P.refl indicates some misunderstanding about the role of _≡_.
There is no magic in that type. It is just a dependent type with a single constructor. Proving that some x ≡ y resolves to P.refl does not tell Agda anything new about x and y: it only tells Agda that you managed to produce a witness of the type _≡_. This is the reason it cannot tell n and .(n + 0) are the same thing, or that suc (acc + n) is the same as .(acc + suc n). So both of the errors you see are really the same.
Now, what rewrite is for.
You cannot define C x ≡ C y for dependent type C _. C x and C y are different types. Equality is defined only for elements of the same type, so there is no way to even express the idea that an element of type C x is comparable to an element of type C y.
There is, however, an axiom of induction, which allows to produce elements of type C y, if you have an element of type C x and an element of type x ≡ y. Note there is no magic in the type _≡_ - that is, you can define your own type, and construct such a function, and Agda will be satisfied:
induction : {A : Set} {C : (x y : A) -> (x ≡ y) -> Set} (x y : A) (p : x ≡ y) ((x : A) -> C x x refl) -> C x y p
induction x .x refl f = f x
Or a simplified version that follows from the induction axiom:
transport : {A : Set} {C : A -> Set} (x y : A) (x ≡ y) (C x) -> C y
transport x .x refl cx = cx
What this means in practice, is that you get a proof for something - for example, A x ≡ A x, but then transport this proof along the equality x ≡ y to get a proof A x ≡ A y. This usually requires specifying the type explicitly, in this case {C = y -> A x ≡ A y}, and provide the x, the y and the C x. As such, it is a very cumbersome procedure, although the learners will benefit from doing these steps.
rewrite then is a syntactic mechanism that rewrites the types of the terms known before the rewrite, so that such transport is not needed after that. Because it is syntactic, it does interpret the type _≡_ in a special way (so if you define your own type, you need to tell Agda you are using a different type as equality). Rewriting types is not "telling" Agda that some types are equal. It just literally replaces occurrences of x in type signatures with y, so now you only need to construct things with y and refl.
Having said all that, you can see why it works for step. There rewrite P.sym ... literally replaced all occurrences of n with n + 0, including the return type of the function, so now it is modₕ acc (acc + (n + 0)) d (n + 0) ≤ acc + (n + 0). Then constructing a value of that type just works.
Then step-ok didn't work, because you only pattern-matched values. There is nothing to tell that n and (n + 0) are the same thing. But rewrite will. Or you could use a function like this transport.
I'm having a nasty problem with a formalisation of a theorem that uses a data type that have some constructors whose indices have list concatenation. When I try to use emacs mode to case split, Agda returns the following error message:
I'm not sure if there should be a case for the constructor
o-success, because I get stuck when trying to solve the following
unification problems (inferred index ≟ expected index):
e₁ o e'' , x₁ ++ x'' ++ y₁ ≟ e o e' , x ++ x' ++ y
suc (n₂ + n'') , x₁ ++ x'' ≟ m' , p''
when checking that the expression ? has type
suc (.n₁ + .n') == .m' × .x ++ .x' == p'
Since the code is has more than a small number of lines, I put it on the following gist:
https://gist.github.com/rodrigogribeiro/976b3d5cc82c970314c2
Any tip is appreciated.
Best,
There was a similar question.
However you want to unify xs1 ++ xs2 ++ xs3 with ys1 ++ ys2 ++ ys3, but _++_ is not a constructor — it's a function, and it's not injective. Consider this simplified example:
data Bar {A : Set} : List A -> Set where
bar : ∀ xs {ys} -> Bar (xs ++ ys)
ex : ∀ {A} {zs : List A} -> Bar zs -> Bar zs -> List A
ex (bar xs) b = {!!}
b is of type Bar (xs ++ .ys), but b is not necessarily equal to bar .xs, so you can't pattern-match like this. Here are two Bars, which have equal types but different values:
ok : ∃₂ λ (b1 b2 : Bar (tt ∷ [])) -> b1 ≢ b2
ok = bar [] , bar (tt ∷ []) , λ ()
This is because xs1 ++ xs2 ≡ ys1 ++ ys2 doesn't imply xs1 ≡ ys1 × xs2 ≡ ys2 in general.
But it's possible to generalize an index. You can use the technique described by Vitus at the link above, or you can use this simple combinator, which forgets the index:
generalize : ∀ {α β} {A : Set α} (B : A -> Set β) {x : A} -> B x -> ∃ B
generalize B y = , y
E.g.
ex : ∀ {A} {zs : List A} -> Bar zs -> Bar zs -> List A
ex {A} (bar xs) b with generalize Bar b
... | ._ , bar ys = xs ++ ys
After all, are you sure your lemma is true?
UPDATE
Some remarks first.
Your empty case states
empty : forall x -> G :: (emp , x) => (1 , x)
that the empty parser parses the whole string. It should be
empty : forall x -> G :: (emp , x) => (1 , [])
as in the paper.
Your definition of o-fail1 contains this part:
(n , fail ∷ o)
but fail fails everything, so it should be (n , fail ∷ []). With this representation you would probably need decidable equality on A to finish the lemma, and proofs would be dirty. Clean and idiomatic way to represent something, that can fail, is to wrap it in the Maybe monad, so here is my definition of _::_=>_:
data _::_=>_ {n} (G : Con n) : Foo n × List A -> Nat × Maybe (List A) -> Set where
empty : ∀ {x} -> G :: emp , x => 1 , just []
sym-success : ∀ {a x} -> G :: sym a , (a ∷ x) => 1 , just (a ∷ [])
sym-failure : ∀ {a b x} -> ¬ (a == b) -> G :: sym a , b ∷ x => 1 , nothing
var : ∀ {x m o} {v : Fin (suc n)}
-> G :: lookup v G , x => m , o -> G :: var v , x => suc m , o
o-success : ∀ {e e' x x' y n n'}
-> G :: e , x ++ x' ++ y => n , just x
-> G :: e' , x' ++ y => n' , just x'
-> G :: e o e' , x ++ x' ++ y => suc (n + n') , just (x ++ x')
o-fail1 : ∀ {e e' x x' y n}
-> G :: e , x ++ x' ++ y => n , nothing
-> G :: e o e' , x ++ x' ++ y => suc n , nothing
o-fail2 : ∀ {e e' x x' y n n'}
-> G :: e , x ++ x' ++ y => n , just x
-> G :: e' , x' ++ y => n' , nothing
-> G :: e o e' , x ++ x' ++ y => suc (n + n') , nothing
Here is the lemma:
postulate
cut : ∀ {α} {A : Set α} -> ∀ xs {ys zs : List A} -> xs ++ ys == xs ++ zs -> ys == zs
mutual
aux : ∀ {n} {G : Con n} {e e' z x x' y n n' m' p'}
-> z == x ++ x' ++ y
-> G :: e , z => n , just x
-> G :: e' , x' ++ y => n' , just x'
-> G :: e o e' , z => m' , p'
-> suc (n + n') == m' × just (x ++ x') == p'
aux {x = x} {x'} {n = n} {n'} r pr1 pr2 (o-success {x = x''} pr3 pr4) with x | n | lemma pr1 pr3
... | ._ | ._ | refl , refl rewrite cut x'' r with x' | n' | lemma pr2 pr4
... | ._ | ._ | refl , refl = refl , refl
aux ...
lemma : ∀ {n m m'} {G : Con n} {f x p p'}
-> G :: f , x => m , p -> G :: f , x => m' , p' -> m == m' × p == p'
lemma (o-success pr1 pr2) pr3 = aux refl pr1 pr2 pr3
lemma ...
The proof proceeds as follows:
We generalize the type of lemma's pr3 in an auxiliary function as in the Vitus' answer. Now it's possible to pattern-match on pr3.
We prove, that the first parser in lemma's pr3 (called also pr3 in the aux) produces the same output as pr1.
After some rewriting, we prove that the second parser in lemma's pr3 (called pr4 in the aux) produces the same output as pr2.
And since pr1 and pr3 produce the same output, and pr2 and pr4 produce the same output, o-success pr1 pr2 and o-success pr3 pr4 produce the same output, so we put refl , refl.
The code. I didn't prove the o-fail1 and o-fail2 cases, but they should be similar.
UPDATE
Amount of boilerplate can be reduced by
Fixing the definitions of the fail cases, which contain redundant information.
Returning Maybe (List A) instead of Nat × Maybe (List A). You can compute this Nat recursively, if needed.
Using the inspect idiom instead of auxiliary functions.
I don't think there is a simpler solution. The code.