[Next] [Up] [Previous] [Contents]
Next: Running Prolog Up: Tutorial Previous: Definitions   Contents

# Functions

Now let's take on a harder problem: defining capture-avoiding substitution. We'll define substitution as a function since we usually want to run it forward'' (i.e., given a term and a substitution, construct the result) rather than reverse''.

To declare a function write (in tutorial.apl):

func subst(tm,[(id,tm)]) = tm.

This expresses that is a function symbol taking two arguments: a term and a list of identifier-term pairs (i.e., the subsitution), and producing a term. Now let's define cases. Clauses for function definitions are of the form , where is the function being defined, the are the inputs for which the clause defines a value, is the value, and is a subgoal that must be solved for the clause to apply. For example
subst(var(A),S)     = T                           :- mem((A,T),S).
subst(var(A),S)     = var A                       :- not(mem(A,_),S).
subst(app(T1,T2),S) = app(subst(T1,S),subst(T2,S)).
subst(lam(x\T),S)   = lam(x\subst(T,S))           :- x # S.

The first clause asserts that if is bound to a term in then is the result. The second, that if is not so bound, then is the result. The third clause just propagates past an application. The final clause propagates past a , provided the bound variable is not mentioned in . Since appears only in an abstraction, it is always possible to satisfy this constraint by renaming (although at times renaming may not be necessary). In fact, this is what Prolog does by default.

Is this definition really correct? Let's try some examples.

?- X = subst(var(a), [(a,var(b))]).
Yes.
X = var b
?- X = subst(lam(a\var(a)), [(a,var(b))]).
Yes.
X = lam (a\var(a))
?- X = subst(lam(b\var(a)), [(a,var(b))]).
Yes.
X = lam (b1\var(b))
;
No.

Note that in the final case, the bound variable has been freshened to . Remember that when a name is used in an abstraction, we lose control over its specific value since the abstraction really represents an equivalence class of -equivalent terms. The last example also verifies that a simple form of variable capture is actually impossible. That is, it's not the case that happens to get the right answer the first time it succeeds; in fact, no alternative wrong answers are possible.

Now this definition of has some disadvantages: for example the first two clauses overlap and may repeat the membership test . We can make this more efficient using Prolog's if-then-else'' construct:

subst(var A,S) = T :- mem ((A,V),S) -> T = V | T = var(A).

Similarly, we can collapse the other two clauses into one completely general case using disjunction:
subst(T,A) = T' :- ( T = var(A),
if mem((A,V),S)
then T = V
else T = var(A)
; T = app(T1,T2),
T' = app(subst(T1,S),subst(T2,S)))
; T = lam(x\T1),
x # S,
T' = lam(x\subst(T1,S)))


Another alternative would be to replace the call to with a substitution-specific function:

func lookup(id,[(id,exp)]) = exp.
lookup(A,[]) = var A.
lookup(A,[(A,V)|S]) = V.
lookup(A,[(B,V)|S]) = lookup(A,S).
...
subst(var A,S) = lookup(A,S).


Both this and the earlier definition have the potential disadvantage that if duplicate bindings are present in a substitution (e.g., ), then or can succeed with multiple possible answers. We therefore regard substitutions to be well-formed only if there are no such duplicates.

[Next] [Up] [Previous] [Contents]
Next: Running Prolog Up: Tutorial Previous: Definitions   Contents
James Cheney 2003-10-23