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Tutorial on the Elpi programming language

Author: Enrico Tassi

This little tutorial does not talk about Coq, but rather focuses on Elpi as a programming language. It assumes no previous knowledge of Prolog, λProlog or Elpi. Coq is used as an environment for stepping trough the tutorial one paragraph at a time. The text between lp:{{ and }} is Elpi code, while the rest are Coq directives to drive the Elpi interpreter.

Logic programming

Elpi is a dialect of λProlog enriched with constraints. We start by introducing the first order fragment of λProlog, i.e. the terms will not contain binders. Later we cover terms with binders and constraints.

Our first program is called tutorial. We begin by declaring the signature of our terms. Here we declare that person is a type, and that mallory, bob and alice are terms of that type.

Elpi Program tutorial lp:{{

kind person  type.
type mallory, bob, alice  person.

}}.

An Elpi program is made of rules that declare when predicates hold and that are accumulated one after the other. Rules are also called clauses in Prolog's slang, so we may use both terms interchangeably.

The next code snippet accumulates on top of the current tutorial program a predicate declaration for age and three rules representing our knowledge about our terms.

Elpi Accumulate lp:{{

pred age o:person, o:int.

age mallory 23.
age bob 23.
age alice 20.

}}.

The predicate age has two arguments, the former is a person while the latter is an integer. The label o: (standing for output) is a mode declaration, which we will explain later (ignore it for now).

Note

int is the built-in data type of integers

Integers come with usual arithmetic operators, see the calc built-in.

In order to run our program we have to write a query, i.e. a predicate expression containing variables such as:

age alice A


The execution of the program is expected to assign a value to A, which represents the age of alice.

Important

Syntactic conventions:

• variables are identifiers starting with a capital letter, eg A, B, FooBar, Foo_bar, X1
• constants (for individuals or predicates) are identifiers starting with a lowercase letter, eg foo, bar, this_that, camelCase, dash-allowed, qmark_too?, arrows->and.dots.as<-well

A query can be composed of many predicate expressions separated by , that stands for conjunction: we want to get an answer to all the predicate expressions.

Elpi Query lp:{{

age alice A, coq.say "The age of alice is" A

}}.constant get-option has no declared type.
[elpi.typecheck,elpi]The age of alice is 20Query assignments:A = 20

coq.say is a built-in predicate provided by Coq-Elpi which prints its arguments. You can look at the output buffer of Coq to see the value for A or hover or toggle the little bubble after }}. if you are reading the tutorial with a web browser.

Note

string is a built-in data type

Strings are delimited by double quotes and \ is the escape symbol.

The predicate age represents a relation (in contrast to a function) and it computes both ways: we can ask Elpi which person P is 23 years old.

Elpi Query lp:{{

age P 23, coq.say P "is 23 years old"

}}.mallory is 23 years oldQuery assignments:P = mallory

Unification

Operationally the query age P 23 is unified with each and every rule present in the program starting from the first one.

Unification compares two terms structurally and eventually assigns variables. For example for the first rule of the program we obtain the following unification problem:

age P 23 = age mallory 23


This problem can be simplified into smaller unification problems following the structure of the terms:

age = age
P = mallory
23 = 23


The first and last are trivial, while the second one can be satisfied by assigning mallory to P. All equations are solved, hence unification succeeds.

Since the first part of the query is succesful the rest of the query is run: the value of P is printed as well as the "is 23 years old" string.

Note

= is a regular predicate

The query age P 23 can be also written as follows:

A = 23, age P A, Msg = "is 23 years old", coq.say P Msg


Let's try a query harder to solve!

Elpi Query lp:{{

age P 20, coq.say P "is 20 years old"

}}.alice is 20 years oldQuery assignments:P = alice

This time the unification problem for the first rule in the program is:

age P 20 = age mallory 23


that is simplified to:

age = age
P = mallory
20 = 23


The second equation can be solved by assigning mallory to P, but the third one has no solution, so unification fails.

Backtracking

When failure occurs all assignements are undone (i.e. P is unset again) and the next rule in the program is tried. This operation is called backtracking.

The unification problem for the next rule is:

age P 20 = age bob 23


This one also fails. The unification problem for the last rule is:

age P 20 = age alice 20


This one works, and the assigment P = alice is kept as the result of the first part of the query. Then P is printed and the program ends.

An even harder query is the following one where we ask for two distinct individuals to have the same age.

Elpi Query lp:{{

age P A, age Q A, not(P = Q),
coq.say P "and" Q "are" A "years old"

}}.mallory and bob are 23 years oldQuery assignments:A = 23P = malloryQ = bob

This example shows that backtracking is global. The first solution for age P A and age Q A picks P and Q to be the same individual mallory, but then not(P = Q) fails and forces the last choice that was made to be reconsidered, so Q becomes bob.

Look at the output of the following code to better understand how backtracking works.

Elpi Query lp:{{

age P A, coq.say "I picked P =" P,
age Q A, coq.say "I picked Q =" Q,
not(P = Q),
coq.say "the last choice worked!",
coq.say P "and" Q "are" A "years old"

}}.I picked P = malloryI picked Q = malloryI picked Q = bobthe last choice worked!mallory and bob are 23 years oldQuery assignments:A = 23P = malloryQ = bob

Note

not is a black hole

The not(P) predicate tries to solve the query P: it fails if P succeeds, and succeeds if P fails. In any case no trace is left of the computation for P. E.g. not(X = 1, 2 < 1) suceeds, but the assignment for X is undone. See also the section about the foundations of λProlog.

Facts and conditional rules

The rules we have seen so far are facts: they always hold. In general rules can only be applied if some condition holds. Conditions are also called premises, we may use the two terms interchangeably.

Here we add to our program a rule that defines what older P Q means in terms of the age of P and Q.

Note

:- separates the head of a rule from the premises

Elpi Accumulate lp:{{

pred older o:person, o:person.
older P Q :- age P N, age Q M, N > M.

}}.

The rule reads: P is older than Q if N is the age of P and M is the age of Q and N is greater than M.

Let's run a query using older:

Elpi Query lp:{{

older bob X,
coq.say "bob is older than" X

}}.bob is older than aliceQuery assignments:X = alice

The query older bob X is unified with the head of the program rule older P Q assigning P = bob and X = Q. Then three new queries are run:

age bob N
age Q M
N > M


The former assigns N = 23, the second one first sets Q = mallory and M = 23. This makes the last query to fail, since 23 > 23 is false. Hence the second query is run again and again until Q is set to alice and M to 20.

Variables in the query are said to be existentially quantified because Elpi will try to find one possible value for them.

Conversely, the variables used in rules are universally quantified in the front of the rule. This means that the same program rule can be used multiple times, and each time the variables are fresh.

In the following example the variable P in older P Q :- ... once takes bob and another time takes mallory.

Elpi Query lp:{{

older bob X, older mallory X,
coq.say "both bob and mallory are older than" X

}}.both bob and mallory are older than aliceQuery assignments:X = alice

Terms with binders

So far the syntax of terms is based on constants (eg age or mallory) and variables (eg X).

λProlog adds another term constructor: λ-abstraction (written x\ ...).

Note

the variable name before the \ can be a capital

Given that it is explicitly bound Elpi needs not to guess if it is a global symbol or a rule variable (that required the convention of using capitals for variables in the first place).

λ-abstraction

Functions built using λ-abstraction can be applied to arguments and honor the usual β-reduction rule (the argument is substituted for the bound variable).

In the following example F 23 reads, once the β-reduction is performed, age alice 23.

Elpi Query lp:{{

F = (x\ age alice x),
coq.say "F =" F,
coq.say "F 20 =" (F 20),
coq.say "F 23 =" (F 23)

}}.F = c0 \ age alice c0F 20 = age alice 20F 23 = age alice 23Query assignments:  F = c0 \
age alice c0

Let's now write the "hello world" of λProlog: an interpreter and type checker for the simply typed λ-calculus. We call this program stlc.

We start by declaring that term is a type and that app and fun are constructors of that type.

Elpi Program stlc lp:{{

kind  term  type.

type  app   term -> term -> term.
type  fun   (term -> term) -> term.

}}.

The constructor app takes two terms while fun only one (of functional type).

Note that

• there is no constructor for variables, we will use the notion of bound variable of λProlog in order to represent variables
• fun takes a function as subterm, i.e. something we can build using the λ-abstraction x\ ...

As a consequence, the identity function λx.x is written like this:

fun (x\ x)


while the first projection λx.λy.x is written:

fun (x\ fun (y\ x))


Another consequence of this approach is that there is no such thing as a free variable in our representation of the λ-calculus. Variables are only available under the λ-abstraction of the programming language, that gives them a well defined scope and substitution operation (β-reduction).

This approach is called HOAS.

We can now implement weak head reduction, that is we stop reducing when the term is a fun or a global constant (potentially applied). If the term is app (fun F) A then we compute the reduct F A. Note that F is a λProlog function, so passing an argument to it implements the substitution of the actual argument for the bound variable.

We first give a type and a mode for our predicate whd. It reads "whd takes a term in input and gives a term in output". We will explain what input means precisely later, for now just think about it as a comment.

Elpi Accumulate lp:{{

pred whd i:term, o:term.

% when the head "Hd" of an "app" (lication) is a
% "fun" we substitute and continue
whd (app Hd Arg) Reduct :- whd Hd (fun F), !,
whd (F Arg) Reduct.

% otherwise a term X is already in normal form.
whd X Reduct :- Reduct = X.

}}.

Recall that, due to backtracking, all rules are potentially used. Here whenever the first premise of the first rule succeeds we want the second rule to be skipped, since we found a redex.

The premises of a rule are run in order and the ! operator discards all other rules following the current one. Said otherwise it commits to the currently chosen rule for the current query (but leaves all rules available for subsequent queries, they are not erased from the program). So, as soon as whd Hd (fun F) succeeds we discard the second rule.

Elpi Query lp:{{

I = (fun x\x),
whd I T, coq.say "λx.x ~>" T,
whd (app I I) T1, coq.say "(λx.x) (λx.x) ~>" T1

}}.λx.x ~> fun c0 \ c0(λx.x) (λx.x) ~> fun c0 \ c0Query assignments:I = fun c0 \ c0T = fun c0 \ c0T1 = fun c0 \ c0

Another little test using global constants:

Elpi Accumulate lp:{{

type foo, bar term.

}}.

Elpi Query lp:{{

Fst = fun (x\ fun y\ x),
T = app (app Fst foo) bar,
whd T T1, coq.say "(Fst foo bar) ~>" T1,
S = app foo bar,
whd S S1, coq.say "(foo bar) ~>" S1

}}.(Fst foo bar) ~> foo(foo bar) ~> app foo barQuery assignments:Fst = fun c0 \ fun c1 \ c0S = app foo barS1 = app foo barT = app (app (fun c0 \ fun c1 \ c0) foo) barT1 = foo

A last test with a lambda term that has no weak head normal form:

Elpi Bound Steps 1000. (* Let's be cautious *)
Elpi Query lp:{{

Delta = fun (x\ app x x),
Omega = app Delta Delta,
whd Omega Hummm, coq.say "not going to happen"

}}.elpi run out of steps (1000)
Elpi Bound Steps 0.

pi x\  and =>

We have seen how to implement subtitution using the binders of λProlog. More often than not we need to move under binders rather than remove them by substituting some term in place of the bound variable.

In order to move under a binder and inspect the body of a function λProlog provides the pi quantifier and the => connective.

A good showcase for these features is to implement a type checker for the simply typed lambda calculus. See also the Wikipedia page on the simply typed lambda calculus.

We start by defining the data type of simple types. We then declare a new predicate of (for "type of") and finally we provide two rules, one for each term constructor.

Elpi Accumulate lp:{{

kind  ty   type.           % the data type of types
type  arr  ty -> ty -> ty. % our type constructor

pred of i:term, o:ty. % the type checking algorithm

% for the app node we ensure the head is a function from
% A to B, and that the argument is of type A
of (app Hd Arg) B :-
of Hd (arr A B), of Arg A.

% for lambda, instead of using a context (a list) of bound
% variables we use pi and => , explained below
of (fun F) (arr A B) :-
pi x\ of x A => of (F x) B.

}}.

The pi name\ code syntax is reserved, as well as rule => code.

Operationally pi x\ code introduces a fresh constant c for x and then runs code. Operationally rule => code adds rule to the program and runs code. Such extra rule is said to be hypothetical. Both the constant for x and rule are removed once code terminates.

Important

hypothetical rules are added at the top of the program

Hypothetical rules hence take precedence over static rules, since they are tried first.

Note that in this last example the hypothetical rule is going to be of c A for a fixed A and a fresh constant c. The variable A is fixed but not assigned yet, meaning that c has a type, and only one, but we may not know it yet.

Now let's assign a type to λx.λy.x:

Elpi Query lp:{{

of (fun (x\ fun y\ x)) Ty, coq.say "The type of λx.λy.x is:" Ty

}}.The type of λx.λy.x is: arr X0 (arr X1 X0)Query assignments:Ty = arr X0 (arr X1 X0)

Let's run this example step by step:

The rule for fun is used:

• arrow A1 B1 is assigned to Ty by unification
• the pi x\  quantifier creates a fresh constant c1 to play the role of x
• the => connective adds the rule of c1 A1 the program
• the new query of (fun y\ c1) B1 is run.

Again, the rule for fun is used (since its variables are universally quantified, we use A2, B2... this time):

• arrow A2 B2 is assigned to B1 by unification
• the pi x\  quantifier creates a fresh constant c2 to play the role of x
• the => connective adds the rule of c2 A2 the program
• the new query of c1 B2 is run.

The (hypotetical) rule of c1 A1 is used:

• unification assigns A1 to B2

The value of Ty is hence arr A1 (arr A2 A1), a good type for λx.λy.x (the first argument and the output have the same type A1).

What about the term λx.(x x) ?

Elpi Query lp:{{

Delta = fun (x\ app x x),
(of Delta Ty ; coq.say "Error:" Delta "has no type")

}}.Error: fun c0 \ app c0 c0 has no typeQuery assignments:Delta = fun c0 \ app c0 c0Ty = X0

The ; infix operator stands for disjunction. Since we see the message of failed: the term fun (x\ app x x) is not well typed.

First, the rule for elpi:fun is selected:

• arrow A1 B1 is assigned to Ty by unification
• the pi x\  quantifier creates a fresh constant c1 to play the role of x
• the => connective adds the rule of c1 A1 the program
• the new query of (app c1 c1) B1 is run.

Then it's the turn of typing the application:

• the query of c1 (arr A2 B2) assignes to A1 the value arr A2 B2. This means that the hypothetical rule is now of c1 (arr A2 B2).

• the query of c1 A2 fails because the unification

of c1 A2 = of c1 (arr A2 B2)


has no solution, in particular the sub problem A2 = (arr A2 B2) fails the so called occur check.

Logical foundations

This section tries to link, informally, λProlog with logic, assuming the reader has some familiarity with first order intuitionistic logic and proof theory. The reader which is not familiar with that can probably skip this section, although section functional-style contains some explanations about the scope of variables which are based on the logical foundations of the language.

The semantics of a λProlog program is given by interpreting it in terms of logical formulas and proof search in intuitionistic logic.

A rule

p A B :- q A C, r C B.


has to be understood as a formula

\begin{equation*} \forall A~B~C, (\mathrm{q}~A~C \wedge \mathrm{r}~C~B) \rightarrow \mathrm{p}~A~B \end{equation*}

A query is a goal that is proved by backchaining rules. For example p 3 X is solved by unifying it with the conclusion of the formula above (that sets A to 3) and generating two new goals, q 3 C and r C B. Note that C is an argument to both q and r and acts as a link: if solving q fixes C then the query for r sees that. Similarly for B, that is identified with X, and is hence a link from the solution of r to the solution of p.

A rule like:

of (fun F) (arr A B) :-
pi x\ of x A => of (F x) B.


\begin{equation*} \forall F~A~B, (\forall x, \mathrm{of}~x~A \rightarrow \mathrm{of}~(F~x)~B) \rightarrow \mathrm{of}~(\mathrm{fun}~F)~(\mathrm{arr}~A~B) \end{equation*}

where $$F$$ stands for a function. Alternatively, using the inference rule notation typically used for type systems:

\begin{equation*} \frac{\Gamma, \mathrm{of}~x~A \vdash \mathrm{of}~(F~x)~B \quad x~\mathrm{fresh}}{\Gamma \vdash \mathrm{of}~(\mathrm{fun}~F)~(\mathrm{arr}~A~B)} \end{equation*}

Hence, x and of x A are available only temporarily to prove of (F x) B and this is also why A cannot change during this sub proof (A is quantified once and forall outside).

Each program execution is a proof (tree) of the query and is made of the program rules seen as proof rules or axioms.

As we hinted before negation is a black hole, indeed the usual definition of $$\neg A$$ as $$A \to \bot$$ is the one of a function with no output (see also the the Wikipedia page on the Curry-Howard correspondence).

Modes and constraints

Elpi extends λProlog with syntactic constraints and rules to manipulate the store of constraints.

Syntactic constraints are goals suspended on a variable which are resumed as soon as that variable gets instantiated. While suspended they are kept in a store which can be manipulated by dedicated rules.

A companion facility is the declaration of modes. The argument of a predicate can be marked as input to avoid instantiating the goal when it is unified with the head of a rule (an input argument is matched, rather than unified).

Elpi Program peano lp:{{

kind nat type.
type z nat.
type s nat -> nat.

}}.

Elpi Query lp:{{

add (s (s z)) (s z) R, coq.say "2 + 1 =" R

}}.2 + 1 = s (s (s z))Query assignments:R = s (s (s z))

Unfortunately the relation does not work well when the first argument is a variable. Depending on the order of the rules for add Elpi can either diverge or pick z as a value for X (that may not be what one wants)

Elpi Bound Steps 100.
Elpi Query lp:{{ add X (s z) Y }}.elpi run out of steps (100)
Elpi Bound Steps 0.

Indeed the first rule for add can be applied forever. If one exchanges the two rules in the program, then Elpi terminates picking z for X.

We can use the mode directive in order to match arguments marked as i: against the patterns in the head of rules, rather than unifying them.

Elpi Program peano2 lp:{{

kind nat type.
type z nat.
type s nat -> nat.

pred sum i:nat, i:nat, o:nat.

sum (s X) Y (s Z) :- sum X Y Z.
sum z X X.
sum _ _ _ :-
coq.error "nothing matched but for this catch all clause!".

}}.

Elpi Query lp:{{ sum X (s z) Y }}.nothing matched but for this catch all clause!

The query fails because no rule first argument matches X.

Instead of failing we can suspend goals and turn them into syntactic constraints

Elpi Program peano3 lp:{{

kind nat type.
type z nat.
type s nat -> nat.

pred sum i:nat, i:nat, o:nat.

sum (s X) Y (s Z) :- sum X Y Z.
sum z X X.
sum X Y Z :-
% the head of the rule always unifies with the query, so
% we double check X is a variable (we could also be
% here because the other rules failed)
var X,
% then we declare the constraint and schedule its resumption
% on the assignment of X
declare_constraint (sum X Y Z) [X].

}}.

Elpi Query lp:{{ sum X (s z) Z }}.Query assignments:X = X0Z = X1Syntactic constraints:
sum X0 (s z) X1  /* suspended on X0 */

Syntactic constraints are resumed when the variable they are suspended on is assigned:

Elpi Query lp:{{

sum X (s z) Z, X = z,
coq.say "The result is:" Z

}}.The result is: s zQuery assignments:X = zZ = s z

Here a couple more examples. Keep in mind that:

• resumption can cause failure
• recall that ; stands for disjunction
Elpi Query lp:{{ sum X (s z) (s (s z)),  X = z }}.The elpi tactic/command peano3 failed without giving a
specific error message. Please report this
inconvenience to the authors of the program.
Elpi Query lp:{{ sum X (s z) (s (s z)), (X = z ; X = s z) }}.Query assignments:X = s z

In this example the computation suspends, then makes progess, then suspends again...

Elpi Query lp:{{

sum X (s z) Y,
print_constraints, coq.say "Currently Y =" Y,
X = s Z,
print_constraints, coq.say "Currently Y =" Y,
Z = z,
coq.say "Finally Y =" Y

}}.sum X0 (s z) X1  /* suspended on X0 */Currently Y = X1sum X2 (s z) X3  /* suspended on X2 */Currently Y = s X3Finally Y = s (s z)Query assignments:X = s zY = s (s z)Z = z

Sometimes the set of syntactic constraints becomes unsatisfiable and we would like to be able to fail early.

Elpi Accumulate lp:{{

pred even i:nat.
pred odd  i:nat.

even z.
even (s X) :- odd X.
odd (s X) :- even X.

odd X :- var X, declare_constraint (odd X) [X].
even X :- var X, declare_constraint (even X) [X].

}}.

Elpi Query lp:{{ even (s X), odd (s X) }}. (* hum, not nice *)Query assignments:X = X0Syntactic constraints:
even X0  /* suspended on X0 */ odd X0  /* suspended on X0 */

Constraint Handling Rules

A constraint (handling) rule can see the store of syntactic constraints as a whole, remove constraints and/or create new goals:

Elpi Accumulate lp:{{

constraint even odd {
% if two distinct, conflicting, constraints about the same X
% are part of the constraint store
rule (even X) (odd X) <=>
% generate the following goal
(coq.say X "can't be even and odd at the same time", fail).
}

}}.

Elpi Query lp:{{ even (s X), odd (s X) }}.X0 can't be even and odd at the same timeThe elpi tactic/command peano3 failed without giving a
specific error message. Please report this
inconvenience to the authors of the program.

Note

fail is a predicate with no solution

Functional style

Elpi is a relational language, not a functional one. Still some features typical of functional programming are available, with some caveats.

Spilling (relation composition)

Chaining "relations" can be painful, especially when they look like functions. Here we use std.append and std.rev to build a palindrome:

Elpi Program function lp:{{

pred make-palindrome i:list A, o:list A.

make-palindrome L Result :-
std.rev L TMP,
std.append L TMP Result.

}}.

Elpi Query lp:{{

make-palindrome [1,2,3] A

}}.Query assignments:A = [1, 2, 3, 3, 2, 1]

Note

variables (capital letters) can be used in types in order to describe ML-like polymorphism.

Note

list A is a built-in data type

The empty list is written [], while the cons constructor is written [Hd | Tail]. Iterated cons can be written [ E1, E2 | Tail ] and | Tail can be omitted if the list is nil terminated.

The make-palindrome predicate has to use a temporary variable just to pass the output of a function as the input to another function.

Spilling is a syntactic elaboration which does that for you. Expressions between { and } are said to be spilled out and placed just before the predicate that contains them.

Elpi Accumulate lp:{{

pred make-palindrome2 i:list A, o:list A.

make-palindrome2 L Result :-
std.append L {std.rev L} Result.

}}.

Elpi Query lp:{{

make-palindrome2 [1,2,3] A

}}.Query assignments:A = [1, 2, 3, 3, 2, 1]

The two versions of make-palindrome are equivalent. Actually the latter is elaborated into the former.

APIs for built-in data

Functions about built-in data types are available via the calc predicate or its infix version is. Example:

Elpi Query lp:{{

calc ( "result " ^ "=" ) X,
Y is 3 + 2,
coq.say X Y

}}.result = 5Query assignments:X = result =Y = 5

The calc predicate works nicely with spilling:

Elpi Query lp:{{ coq.say "result =" {calc (2 + 3)} }}.result = 5Query assignments:Spilled_1 = 5

Allocation of variables

The language let's one use λ-abstraction also to write anonymous rules but one has to be wary of where variables are bound (allocated really).

In our example we use the higher order predicate std.map:

pred std.map i:list A, i:(A -> B -> prop), o:list B.


Note

prop is the type of predicates

The actual type of the std.map symbol is:

type std.map list A -> (A -> B -> prop) -> list B -> prop.


The pred directive complements a type declaration for predicates (the trailing -> prop is implicit) with a mode declaration for each argument.

The type of the second argument of std.map is the one of a predicate relating A with B.

Let's try to call std.map passing an anonymous rule (as we would do in a functional language by passing an anonymous function):

Elpi Accumulate lp:{{

pred bad i:list int, o:list int.

std.map L (x\ r\ TMP is x + 1, r = TMP) Result.

pred good i:list int, o:list int.
good L Result :-
std.map L good.aux Result.
good.aux X R :- TMP is X + 1, R = TMP.

pred good2 i:list int, o:list int.
good2 L Result :-
std.map L (x\ r\ sigma TMP\ TMP is x + 1, r = TMP) Result.

}}.

Elpi Query lp:{{

good [1,2,3] R2,
good2 [1,2,3] R3

}}.Query assignments:R1 = X0R2 = [2, 3, 4]R3 = [2, 3, 4]

The problem with bad is that TMP is fresh each time the rule is used, but not every time the anonymous rule passed to map is used. Technically TMP is quantified (allocated) where L and Result are.

There are two ways to quantify TMP correctly, that is inside the anonymous predicate. One is to actually name the predicate. Another one is to use the sigma x\  quantifier to allocate TMP at every call. We recommend to name the auxiliary predicate.

Tip

predicates whose name ends in .aux don't trigger a missing type declaration warning

One last way to skin the cat is to use => as follows. It gives us the occasion to clarify further the scope of variables.

Elpi Accumulate lp:{{

pred good3 i:list int, o:list int.
good3 L Result :-
pi aux\
(pi TMP X R\ aux X R :- TMP is X + 1, R = TMP) =>
std.map L aux Result.

}}.

Elpi Query lp:{{

good3 [1,2,3] R

}}.Query assignments:R = [2, 3, 4]

In this case the auxiliary predicate aux is only visible inside good3. What is interesting to remark is that the quantifications are explicit in the hypothetical rule, and they indicate clearly that each and every time aux is used TMP, X and R are fresh.

The pi x\  quantifier is dual to sigma x\ : since here it occurs negatively it has the same meaning. That is, the hypothetical rule could be written pi X R\ aux X R :- sigma TMP\ TMP is X + 1, R = TMP.

Tip

pi x\  and sigma x\  can quantify on a bunch of variables at once

That is, pi x y\ ... is equivalent to pi x\ pi y\ ... and sigma x y\ ... is equivalent to sigma x\ sigma y\ ....

Tip

=> can load more than one clause at once

It is sufficient to put a list on the left hand side, eg [ rule1, rule2 ] => code. Moreover one can synthesize a rule before loading it, eg:

Rules = [ one-more-rule | ExtraRules ], Rules => code


The last remark worth making is that bound variables are intimately related to universal quantification, while unification variables are related to existential quantification. It goes without saying that the following two formulas are not equivalent and while the former is trivial the latter is in general false:

\begin{align*} \forall x, \exists Y, Y = x\\ \exists Y, \forall x, Y = x \end{align*}

Let's run these two corresponding queries:

Elpi Query lp:{{ pi x\ sigma Y\ Y = x, coq.say "Y =" Y }}.Y = c0
Elpi Query lp:{{ sigma Y\ pi x\ Y = x, coq.say "Y =" Y }}.The elpi tactic/command function failed without giving
a specific error message. Please report this
inconvenience to the authors of the program.

Another way to put it: x is in the scope of Y only in the first formula since it is quantified before it. Hence x can be assigned to Y in that case, but not in the second query, where it is quantified after.

More in general, λProlog tracks the bound variables that are in scope of each unification variable. There are only two ways to put a bound variable in the scope:

• quantify the unification variable under the bound one (first formula)

• pass the bound variable to the unification variable explicitly: in this case the unification variable needs to have a functional type. Indeed $$\exists Y, \forall x, (Y x) = x$$ has a solution: Y can be the identity function.

Elpi Query lp:{{ sigma Y\ pi x\ Y x = x, coq.say "Y =" Y }}.Y = c1 \ c1

If we look again at the rule for type checking λ-abstraction:

of (fun F) (arr A B) :-
pi x\ of x A => of (F x) B.


we can see that the only unification variable that sees the fresh x is F, because we pass x to F explicitly (recall all unification variables such as F, A, B are quantified upfront, before the pi x\ ). Indeed when we write:

\begin{equation*} \frac{\Gamma, x : A \vdash f : B}{\Gamma \vdash \lambda x.f : A \rightarrow B} \end{equation*}

on paper, the variable denoted by x being bound there can only occur in $$f$$, not in $$\Gamma$$ or $$B$$ for example (although a different variable could be named the same, hence the usual freshness side conditions which are not really necessary using HOAS).

Remark that in the premise the variable $$x$$ is still bound, this time not by a λ-abstraction but by the context $$\Gamma, x : A$$. In λProlog the context is the set of hypothetical rules and pi\  -quantified variables and is implicitly handled by the runtime of the programming language.

A slogan to keep in mind is that:

Important

There is no such thing as a free variable!

Indeed the variable bound by the λ-abstraction (of our data) is replaced by a fresh variable bound by the context (of our program). This is called binder mobility. See also the paper Mechanized metatheory revisited by Dale Miller which is an excellent introduction to these concepts.

Debugging

A common λProlog idiom is to have a debug rule lying around. The :if attribute can be used to make the rule conditionally interpreted (only if the given debug variable is set).

Elpi Debug "DEBUG_MYPRED".
Elpi Program debug lp:{{

pred mypred i:int.

:if "DEBUG_MYPRED"
mypred X :-
coq.say "calling mypred on " X, fail.

mypred 0 :- coq.say "ok".
mypred M :- N is M - 1, mypred N.

}}.
Elpi Query lp:{{ mypred 3 }}.calling mypred on  3calling mypred on  2calling mypred on  1calling mypred on  0ok

As a slightly more sophisticated debugging feature, the Elpi interpreter provides tracing facilities.

Elpi Trace.

Elpi Query stlc lp:{{ % We run the query in the stlc program

of (fun (x\ fun y\ x)) Ty, coq.say Ty

}}.run 1 {{{
rid:0 step:1 gid:5 user:curgoal = ,
of (fun c0 \ fun c1 \ c0) X0 , coq.say X0
rid:0 step:1 gid:5 user:rule = and
rid:0 step:1 gid:5 user:subgoal = 6
rid:0 step:1 gid:6 user:newgoal = of (fun c0 \ fun c1 \ c0) X0
rid:0 step:1 gid:5 user:subgoal = 7
rid:0 step:1 gid:7 user:newgoal = coq.say X0
rid:0 step:1 gid:5 user:rule:and = success
}}} ->  (0.000s)run 2 {{{
rid:0 step:2 gid:6 user:curgoal = of
of (fun c0 \ fun c1 \ c0) X0
rid:0 step:2 gid:6 user:rule = backchain
rid:0 step:2 gid:6 user:rule:backchain:candidates = File "(stdin)", line 15, column 3, character 438:
}}} ->  (0.000s)select 3 {{{
rid:0 step:2 gid:6 user:rule:backchain:try = File "(stdin)", line 15, column 3, character 438:
(of (fun A0) (arr A1 A2)) :- (
pi (c0 \
(of c0 A1 => of (A0 c0) A2))).
rid:0 step:2 gid:0 user:assign = A0 := c0 \
fun c1 \ c0
rid:0 step:2 gid:0 user:assign = X0 := arr X1 X2
rid:0 step:2 gid:6 user:subgoal = 8
rid:0 step:2 gid:8 user:newgoal = pi c0 \ of c0 X1 => of (fun c1 \ c0) X2
rid:0 step:2 gid:8 user:rule:backchain = success
}}} ->  (0.002s)run 3 {{{
rid:0 step:3 gid:8 user:curgoal = pi
pi c0 \ of c0 X1 => of (fun c1 \ c0) X2
rid:0 step:3 gid:8 user:rule = pi
rid:0 step:3 gid:8 user:subgoal = 9
rid:0 step:3 gid:9 user:newgoal = of c0 X1 => of (fun c1 \ c0) X2
rid:0 step:3 gid:9 user:rule:pi = success
}}} ->  (0.000s)run 4 {{{
rid:0 step:4 gid:9 user:curgoal = =>
of c0 X1 => of (fun c1 \ c0) X2
rid:0 step:4 gid:9 user:rule = implication
rid:0 step:4 gid:9 user:subgoal = 10
rid:0 step:4 gid:10 user:newgoal = of (fun c1 \ c0) X2
rid:0 step:4 gid:10 user:rule:implication = success
}}} ->  (0.000s)run 5 {{{
rid:0 step:5 gid:10 user:curgoal = of
of (fun c1 \ c0) X2
rid:0 step:5 gid:10 user:rule = backchain
rid:0 step:5 gid:10 user:rule:backchain:candidates = File "(stdin)", line 15, column 3, character 438:
}}} ->  (0.000s)select 4 {{{
rid:0 step:5 gid:10 user:rule:backchain:try = File "(stdin)", line 15, column 3, character 438:
(of (fun A0) (arr A1 A2)) :- (
pi (c0 \
(of c0 A1 => of (A0 c0) A2))).
rid:0 step:5 gid:0 user:assign = A0 := c1 \
c0
rid:0 step:5 gid:0 user:assign = X2 := arr X3 X4
rid:0 step:5 gid:10 user:subgoal = 11
rid:0 step:5 gid:11 user:newgoal = pi c1 \ of c1 X3 => of c0 X4
rid:0 step:5 gid:11 user:rule:backchain = success
}}} ->  (0.000s)run 6 {{{
rid:0 step:6 gid:11 user:curgoal = pi
pi c1 \ of c1 X3 => of c0 X4
rid:0 step:6 gid:11 user:rule = pi
rid:0 step:6 gid:11 user:subgoal = 12
rid:0 step:6 gid:12 user:newgoal = of c1 X3 => of c0 X4
rid:0 step:6 gid:12 user:rule:pi = success
}}} ->  (0.000s)run 7 {{{
rid:0 step:7 gid:12 user:curgoal = =>
of c1 X3 => of c0 X4
rid:0 step:7 gid:12 user:rule = implication
rid:0 step:7 gid:12 user:subgoal = 13
rid:0 step:7 gid:13 user:newgoal = of c0 X4
rid:0 step:7 gid:13 user:rule:implication = success
}}} ->  (0.000s)run 8 {{{
rid:0 step:8 gid:13 user:curgoal = of
of c0 X4
rid:0 step:8 gid:13 user:rule = backchain
rid:0 step:8 gid:13 user:rule:backchain:candidates = File "(context step_id:4)", line 1, column 0, character 0:
}}} ->  (0.000s)select 5 {{{
rid:0 step:8 gid:13 user:rule:backchain:try = File "(context step_id:4)", line 1, column 0, character 0:
(of c0 X1) :- .
rid:0 step:8 gid:0 user:assign = X1 := X4
rid:0 step:8 gid:13 user:rule:backchain = success
}}} ->  (0.000s)run 9 {{{
rid:0 step:9 gid:7 user:curgoal = coq.say
coq.say (arr X4 (arr X3 X4))
rid:0 step:9 gid:7 user:rule = builtin
rid:0 step:9 gid:7 user:rule:builtin:name = coq.say
arr X4 (arr X3 X4)rid:0 step:9 gid:7 user:rule:builtin = success
}}} ->  (0.000s)Query assignments:Ty = arr X4 (arr X3 X4)

Elpi Query stlc lp:{{

of (fun (x\ app x x)) Ty, coq.say Ty

}}.run 1 {{{
rid:1 step:1 gid:14 user:curgoal = ,
of (fun c0 \ app c0 c0) X0 , coq.say X0
rid:1 step:1 gid:14 user:rule = and
rid:1 step:1 gid:14 user:subgoal = 15
rid:1 step:1 gid:15 user:newgoal = of (fun c0 \ app c0 c0) X0
rid:1 step:1 gid:14 user:subgoal = 16
rid:1 step:1 gid:16 user:newgoal = coq.say X0
rid:1 step:1 gid:14 user:rule:and = success
}}} ->  (0.001s)run 2 {{{
rid:1 step:2 gid:15 user:curgoal = of
of (fun c0 \ app c0 c0) X0
rid:1 step:2 gid:15 user:rule = backchain
rid:1 step:2 gid:15 user:rule:backchain:candidates = File "(stdin)", line 15, column 3, character 438:
}}} ->  (0.000s)select 3 {{{
rid:1 step:2 gid:15 user:rule:backchain:try = File "(stdin)", line 15, column 3, character 438:
(of (fun A0) (arr A1 A2)) :- (
pi (c0 \
(of c0 A1 => of (A0 c0) A2))).
rid:1 step:2 gid:0 user:assign = A0 := c0 \
app c0 c0
rid:1 step:2 gid:0 user:assign = X0 := arr X1 X2
rid:1 step:2 gid:15 user:subgoal = 17
rid:1 step:2 gid:17 user:newgoal = pi c0 \ of c0 X1 => of (app c0 c0) X2
rid:1 step:2 gid:17 user:rule:backchain = success
}}} ->  (0.000s)run 3 {{{
rid:1 step:3 gid:17 user:curgoal = pi
pi c0 \ of c0 X1 => of (app c0 c0) X2
rid:1 step:3 gid:17 user:rule = pi
rid:1 step:3 gid:17 user:subgoal = 18
rid:1 step:3 gid:18 user:newgoal = of c0 X1 => of (app c0 c0) X2
rid:1 step:3 gid:18 user:rule:pi = success
}}} ->  (0.000s)run 4 {{{
rid:1 step:4 gid:18 user:curgoal = =>
of c0 X1 => of (app c0 c0) X2
rid:1 step:4 gid:18 user:rule = implication
rid:1 step:4 gid:18 user:subgoal = 19
rid:1 step:4 gid:19 user:newgoal = of (app c0 c0) X2
rid:1 step:4 gid:19 user:rule:implication = success
}}} ->  (0.000s)run 5 {{{
rid:1 step:5 gid:19 user:curgoal = of
of (app c0 c0) X2
rid:1 step:5 gid:19 user:rule = backchain
rid:1 step:5 gid:19 user:rule:backchain:candidates = File "(stdin)", line 10, column 3, character 273:
}}} ->  (0.001s)select 4 {{{
rid:1 step:5 gid:19 user:rule:backchain:try = File "(stdin)", line 10, column 3, character 273:
(of (app A0 A1) A2) :- (
of A0 (arr A3 A2)),
(of A1 A3).
rid:1 step:5 gid:0 user:assign = A0 := c0
rid:1 step:5 gid:0 user:assign = A1 := c0
rid:1 step:5 gid:0 user:assign = A2 := X2
rid:1 step:5 gid:19 user:subgoal = 20
rid:1 step:5 gid:20 user:newgoal = of c0 (arr X3^1 X2)
rid:1 step:5 gid:20 user:subgoal = 21
rid:1 step:5 gid:21 user:newgoal = of c0 X3^1
rid:1 step:5 gid:20 user:rule:backchain = success
}}} ->  (0.000s)run 6 {{{
rid:1 step:6 gid:20 user:curgoal = of
of c0 (arr X3^1 X2)
rid:1 step:6 gid:20 user:rule = backchain
rid:1 step:6 gid:20 user:rule:backchain:candidates = File "(context step_id:4)", line 1, column 0, character 0:
}}} ->  (0.000s)select 5 {{{
rid:1 step:6 gid:20 user:rule:backchain:try = File "(context step_id:4)", line 1, column 0, character 0:
(of c0 X1) :- .
rid:1 step:6 gid:0 user:assign:expand = X3^1 := X4 c0
rid:1 step:6 gid:0 user:assign:restrict = 0 X4 c0 := c0 \
.X5
rid:1 step:6 gid:0 user:assign = X1 := arr X5 X2
rid:1 step:6 gid:20 user:rule:backchain = success
}}} ->  (0.000s)run 7 {{{
rid:1 step:7 gid:21 user:curgoal = of
of c0 X5
rid:1 step:7 gid:21 user:rule = backchain
rid:1 step:7 gid:21 user:rule:backchain:candidates = File "(context step_id:4)", line 1, column 0, character 0:
}}} ->  (0.000s)select 6 {{{
rid:1 step:7 gid:21 user:rule:backchain:try = File "(context step_id:4)", line 1, column 0, character 0:
(of c0 (arr X5 X2)) :- .
rid:1 step:7 gid:21 user:backchain:fail-to = unify X5 with arr X5 X2
}}} ->  (0.000s)select 7 {{{
rid:1 step:7 gid:21 user:rule:backchain = fail
}}} ->  (0.000s)The elpi tactic/command stlc failed without giving a
specific error message. Please report this
inconvenience to the authors of the program.

The trace can be limited to a range of steps. Look at the numbers run HERE {{{.

Elpi Trace 6 8.
Elpi Query stlc lp:{{

of (fun (x\ fun y\ x)) Ty, coq.say Ty

}}.run 6 {{{
rid:2 step:6 gid:28 user:curgoal = pi
pi c1 \ of c1 X0 => of c0 X1
rid:2 step:6 gid:28 user:rule = pi
rid:2 step:6 gid:28 user:subgoal = 29
rid:2 step:6 gid:29 user:newgoal = of c1 X0 => of c0 X1
rid:2 step:6 gid:29 user:rule:pi = success
}}} ->  (0.000s)run 7 {{{
rid:2 step:7 gid:29 user:curgoal = =>
of c1 X0 => of c0 X1
rid:2 step:7 gid:29 user:rule = implication
rid:2 step:7 gid:29 user:subgoal = 30
rid:2 step:7 gid:30 user:newgoal = of c0 X1
rid:2 step:7 gid:30 user:rule:implication = success
}}} ->  (0.000s)run 8 {{{
rid:2 step:8 gid:30 user:curgoal = of
of c0 X1
rid:2 step:8 gid:30 user:rule = backchain
rid:2 step:8 gid:30 user:rule:backchain:candidates = File "(context step_id:4)", line 1, column 0, character 0:
}}} ->  (0.000s)select 5 {{{
rid:2 step:8 gid:30 user:rule:backchain:try = File "(context step_id:4)", line 1, column 0, character 0:
(of c0 X2) :- .
rid:2 step:8 gid:0 user:assign = X2 := X1
rid:2 step:8 gid:30 user:rule:backchain = success
}}} ->  (0.000s)arr X1 (arr X0 X1)Query assignments:Ty = arr X1 (arr X0 X1)

The trace can be limited to a (list of) predicates as follows:

Elpi Trace "of".
Elpi Query stlc lp:{{

of (fun (x\ fun y\ x)) Ty, coq.say Ty

}}.run 2 {{{
rid:3 step:2 gid:32 user:curgoal = of
of (fun c0 \ fun c1 \ c0) X0
rid:3 step:2 gid:32 user:rule = backchain
rid:3 step:2 gid:32 user:rule:backchain:candidates = File "(stdin)", line 15, column 3, character 438:
}}} ->  (0.000s)select 3 {{{
rid:3 step:2 gid:32 user:rule:backchain:try = File "(stdin)", line 15, column 3, character 438:
(of (fun A0) (arr A1 A2)) :- (
pi (c0 \
(of c0 A1 => of (A0 c0) A2))).
rid:3 step:2 gid:0 user:assign = A0 := c0 \
fun c1 \ c0
rid:3 step:2 gid:0 user:assign = X0 := arr X1 X2
rid:3 step:2 gid:32 user:subgoal = 34
rid:3 step:2 gid:34 user:newgoal = pi c0 \ of c0 X1 => of (fun c1 \ c0) X2
rid:3 step:2 gid:34 user:rule:backchain = success
}}} ->  (0.000s)run 5 {{{
rid:3 step:5 gid:36 user:curgoal = of
of (fun c1 \ c0) X2
rid:3 step:5 gid:36 user:rule = backchain
rid:3 step:5 gid:36 user:rule:backchain:candidates = File "(stdin)", line 15, column 3, character 438:
}}} ->  (0.000s)select 4 {{{
rid:3 step:5 gid:36 user:rule:backchain:try = File "(stdin)", line 15, column 3, character 438:
(of (fun A0) (arr A1 A2)) :- (
pi (c0 \
(of c0 A1 => of (A0 c0) A2))).
rid:3 step:5 gid:0 user:assign = A0 := c1 \
c0
rid:3 step:5 gid:0 user:assign = X2 := arr X3 X4
rid:3 step:5 gid:36 user:subgoal = 37
rid:3 step:5 gid:37 user:newgoal = pi c1 \ of c1 X3 => of c0 X4
rid:3 step:5 gid:37 user:rule:backchain = success
}}} ->  (0.001s)run 8 {{{
rid:3 step:8 gid:39 user:curgoal = of
of c0 X4
rid:3 step:8 gid:39 user:rule = backchain
rid:3 step:8 gid:39 user:rule:backchain:candidates = File "(context step_id:4)", line 1, column 0, character 0:
}}} ->  (0.000s)select 5 {{{
rid:3 step:8 gid:39 user:rule:backchain:try = File "(context step_id:4)", line 1, column 0, character 0:
(of c0 X1) :- .
rid:3 step:8 gid:0 user:assign = X1 := X4
rid:3 step:8 gid:39 user:rule:backchain = success
}}} ->  (0.000s)arr X4 (arr X3 X4)Query assignments:Ty = arr X4 (arr X3 X4)

One can combine the range of steps with the predicate:

Elpi Trace 6 8 "of".
Elpi Query stlc lp:{{

of (fun (x\ fun y\ x)) Ty, coq.say Ty

}}.run 8 {{{
rid:4 step:8 gid:48 user:curgoal = of
of c0 X0
rid:4 step:8 gid:48 user:rule = backchain
rid:4 step:8 gid:48 user:rule:backchain:candidates = File "(context step_id:4)", line 1, column 0, character 0:
}}} ->  (0.000s)select 5 {{{
rid:4 step:8 gid:48 user:rule:backchain:try = File "(context step_id:4)", line 1, column 0, character 0:
(of c0 X1) :- .
rid:4 step:8 gid:0 user:assign = X1 := X0
rid:4 step:8 gid:48 user:rule:backchain = success
}}} ->  (0.000s)arr X0 (arr X2 X0)Query assignments:Ty = arr X0 (arr X2 X0)

To switch traces off:

Elpi Trace Off.

Given that programs are not written in a single place, but rather obtained by accumulating code, Elpi is able to print a (full) program to an html file as follows. The obtained file provides a facility to filter rules by their predicate.

Elpi Print stlc "stlc.html".

Look at the generated page and type of in the filter.

Finally, one can bound the number of backchaining steps performed by the interpreter:

Elpi Query lp:{{ 0 = 0, 1 = 1 }}.
Elpi Bound Steps 1.
Elpi Query lp:{{ 0 = 0, 1 = 1 }}. (* it needs 2 steps! *)elpi run out of steps (1)
Elpi Bound Steps 0. (* Go back to no bound *)

Common pitfalls

Well, no programming language is perfect.

Precedence of , and =>

The precedence of , and => can be surprising

Elpi Query stlc lp:{{

pi x\
of x A => of x B, of x C

}}.The elpi tactic/command stlc failed without giving a
specific error message. Please report this
inconvenience to the authors of the program.

Elpi Query stlc lp:{{

pi x\
of x A => (of x B, of x C) % both goals see of x A

}}.Query assignments:A = X0B = X0C = X0

Backtracking

Backtracking can lead to weird execution traces. The std.do! predicate should be used to write non-backtracking code.

pred not-a-backtracking-one.
not-a-backtracking-one :- condition, !, std.do! [
step,
(generate, test),
step,
].


In the example above once condition holds we start a sequence of steps which we will not reconsider. Locally, backtracking is still available, e.g. between generate and test. See also the std.spy-do! predicate which prints each and every step, and the std.spy one which can be used to spy on a single one.

Unification variables v.s. Imparative variables

Unification variables sit in between variables in imperative programming and functional programming. In imperative programming a variable can hold a value, and that value can change over time via assignment. In functional languages variables always hold a value, and that value never changes. In logic programming a unification variable can either be unset (no value) or set to a value that never changes. Backtracking goes back in time, it is not visible to the program.

As a result of this, code like

pred bad-example.
bad-example :- X is 1 + 2, X is 4 + 5.


fails, because X cannot be at the same time 3 and 9. Initially X is unset, then it is set to 3, and finally the programmer is asserting that 3 (the value hold by X) is equal to 9, not changing the value carried by X. Unification, and hence the = pradicate, plays two roles. When X is unset, X = v sets the variable. When X is set to u, X = v checks if the value of X is equal to u: it is equivalent to u = v.