# 10 October 2006 Foundations of Logic and Constraint Programming 1 Unification ­An overview Need for Unification Ranked alfabeths and terms. Substitutions.

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10 October 2006 Foundations of Logic and Constraint Programming 1 Unification ­An overview Need for Unification Ranked alfabeths and terms. Substitutions Unifiers and mgus Martelli Montanari Algorithm

10 October 2006 Foundations of Logic and Constraint Programming 2 Why unification ? ­In propositional logic two clauses may resolve if they have two complementary literals. ­In the propositional case, two literals are the same if they have the same proposition (negated in exactly one of them). ­In first order logic, the situation is more complex, due to the existence of variables. We may assume that variables are universally quantified. ­In this case, to apply resolution, the proposition do not have to be the same in the complementary literals belonging to two clauses. ­Intuitively, since the variables are universally quantified and may be replaced by any “objects”, the predicate must refer to the same objects. For example A  P  P  B A  B q(X)  p(X,b)  P(a,Y)  r(Y) q(a)  r(b)

10 October 2006 Foundations of Logic and Constraint Programming 3 Term Universe ­Informally, the purpose of unification is to find what the common objects are. Such goal may be acomplished by purely syntactic means, through processing of the terms appearing in the common predicates. ­A term is composed of one or more symbols from Variable symbols (convention: starting with upper case) Alphabet is a finite set Σ of symbols (convention: starting with lower case) To every symbol a natural number  0 is assigned (its arity or rank). Σ (n) denotes the subset of Σ with arity n A symbol with arity 0 (belonging to Σ (0) ) is a constant. ­Given a set of variables V and a ranked alphabet of function symbols F, the term universe TU F,V (over F and V) is the smallest set T of terms such that a) V  T b) f  T, if f  F (0) c) f(t 1,t 2,...,t n )  T if f  F (n) and t 1,t 2,..., t n  T

10 October 2006 Foundations of Logic and Constraint Programming 4 Ground terms and Herbrand Universe ­The set of variables appearing in a term t is denoted by Var(t)  Var(t) :  set of variables in term t ­A term is ground if it contains no variables  t is ground :  Var(t) =  ­If F is a ranked alphabet of function symbols, then HU F, the Herbrand Universe over F, is the set of terms that can be formed without variables  HU F :  TU F,   Example: Given F = {a, f/1, g/2} it is HU F = = { a, f(a), g(a,a), g(a,f(a)), g(f(a),a), g(g(a,a),a), g(a, g(a,a)), f(f(a)), f(g(a,a)), g(f(a),f(a)), g(f(a),g(a,a)), g(g(a,a),f(a)), g(g(a,a),g(a,a)),... } ­A subterm s of t, is a term that is included in t  s = sub_term(t) :  s is a term that is a sub-string of t

10 October 2006 Foundations of Logic and Constraint Programming 5 Substitutions ­Informally a variable (universally quantified) may be substituted by a term (with some constraints). More formally, given V, a set of Variables X  V, a subset of variables to be substituted F, a ranked alphabet a subtitution θ :  X → TU F,V with x  θ(x) for any x  X. ­The following notation is commonly used for a substitution θ θ = {x 1 /t 1, x 2 /t 2,... x n /t n }, where 1.X = {x 1, x 2,... x n }, 2.θ(x i ) = t i for every x i  X ­It is also convenient to define  Empty substitution ε :  n = 0  θ is a ground substitution :  t 1, t 2,..., t n are ground terms.  θ is a pure variable substitution:  t 1, t 2,..., t n are variables.  θ is a renaming :  {t 1, t 2,..., t n }  {x 1, x 2,... x n }.

10 October 2006 Foundations of Logic and Constraint Programming 6 Substitutions ­For a subtitution θ it is also convenient to consider  The domain of the substitution is the set of variables to be substituted Dom(θ ) :  {x 1, x 2,... x n }.  The Range of the substitution is the set of terms for which the variable are substituted Range(θ ) :  {t 1, t 2,..., t n }  The Ran of the substitution is the set of variables appearing in these terms Ran(θ ) :  Var(Range(θ ) )  The variables of the substitution are those in its domain or in its ran. Var(θ ) :  Dom(θ)  Ran(θ )  The projection of a substitution to a subset of its domain Y  Dom(θ ) θ | Y :  {y/t | y/t  θ and y  Y)

10 October 2006 Foundations of Logic and Constraint Programming 7 Application of Substitutions ­A few notational conventions and definitions are commonly used when a substitution θ is applied  If x is a variable and x  Dom(θ), then xθ :  θ(x)  If x is a variable and x  Dom(θ), then xθ :  x  f(t 1, t 2,..., t n )θ :  f(t 1 θ, t 2 θ,..., t n θ )  t is an instance of s :  there is a substitution θ with sθ = t  s is more general than t :  t is an instance of s  t is a variant of s :  there is a renaming θ with sθ = t ­Lemma 2.5  t is a variant of s iff t is an instance of s and s is an instance of t.

10 October 2006 Foundations of Logic and Constraint Programming 8 Composition of Substitutions ­During a resolution proof, several substitutions are performed in sequence, in which variables are successively replaced by terms with other variables. The notion of composition captures the intuition of end result of these substitutions. Definition (2.6) ­ The composition of substitutions θ and σ is defined as (θ σ) x :  x (θ)σ for every variable x ­Lemma (2.3) Let θ = {x 1 /t 1, x 2 /t 2,... x n /t n }, σ = {y 1 /s 1, y 2 /s 2,... y m /s m }. Then θ σ can be constructed from the sequence x 1 /t 1 σ, x 2 /t 2 σ,... x n /t n σ,, y 1 /s 1, y 2 /s 2,... y m /s m 1.By removing all bindings x i /t i σ where x i = t i σ 2.By removing all bindings y j /s j where y j  {x 1, x 2,... x n } 3.By forming a substitution from the resulting sequence.

10 October 2006 Foundations of Logic and Constraint Programming 9 An Ordering of Substitutions ­A substitution restricts the values a variable may take, and the more so, when more substitutions are applied in sequence. This augmented restriction is captured by the ordering of substitutions. Definition (2.6) ­ Let θ and σ be substitutions. Then θ is more general than σ :  σ = θ for some substitution. (equivalently, σ is a specialisation of θ, or θ can be specialised to σ). ­Examples: ­θ = {x/y} is more general than σ = {x/a, y/a} since for = {y/a} it is σ = θ. Since, y = a, then θ = {x/yσ, y/a} = {x/a, y/a} = σ. ­θ = {x/y} is not more general than σ = {x/a}, since every such that θ = σ should contain a pair y/a, and that pair should appear in θ. x/a  {x/y}  y/a   y  Dom(σ ) = Dom (θ) Intuitively, σ = {x/a} can be specialised to {x/a, y/b} but θ = {x/y} cannot, so θ cannot be more general than σ.

10 October 2006 Foundations of Logic and Constraint Programming 10 Unifiers ­The resolution of two clauses does not require the occorrence of complementary literals that are equal (apart from the negation) but rather that the complementary literals are applicable to the same objects. Hence, the terms in the literals must be unifiable, i.e. equal after performing some substitution. Definition (2.9) ­Given substitution θ and terms s and t  θ is a unifier of s and t :  sθ = tθ.  s and t are unifiable :  sθ = tθ for some θ. Many unifiers may exist between two terms, but it is specially interesting (e.g. in resolution) to consider the most general unifier (mgu).  θ is a mgu of s and t :  sθ = tθ  s σ = t σ  σ = θ for some. ­The definition of unifiers for pairs of terms, may be extended for sets of pairs of terms.

10 October 2006 Foundations of Logic and Constraint Programming 11 Most General Unifiers The literals to be unified may have more than one argument. The following definitions are relevant in this case. -Let s 1, s 2,..., s n, t 1, t 2,..., t n, be terms; s i ≈ t i denote the (ordered ) pair (s i, t i ); and E = {s 1 ≈ t 1, s 2 ≈ t 2,... S n ≈ t n }. Then  θ is a unifier of E :  s i θ = t i θ for all i  1..n.  θ is an mgu of E :  σ is a unifier of E  σ = θ for some. ­Sets E and E’ are equivalent :  θ is a unifier of E  θ is a unifier of E’ ­E = {x 1 ≈ t 1, x 2 ≈ t 2,... x n ≈ t n } is solved :   x i, x j are pairwise distinct variables (1  i  j,  n)  no x i, occurs in t j (1  i, j,  n)

10 October 2006 Foundations of Logic and Constraint Programming 12 Most General Unifiers Lemma (2.15) If E = {x 1 ≈ t 1, x 2 ≈ t 2,... x n ≈ t n } is solved, then θ = {x 1 /t 1, x 2 /t 2,... x n /t n }, is an mgu of E. Proof: 1)x i θ = t i = t i θ 2)For every unifier σ of E, it is x i σ = t i σ. But t i = t i θ and so x i σ = x i θ σ; for every i  1..n. Additionally, x σ = x θ σ = x for every x  {x 1, x 2,..., x n }. Hence, σ = θ σ and θ is thus more general than σ. Once the unification of predicates and terms with more than one argument is defined, it is important to obtain an algorithm to perform the unification of two terms. In fact, it is easier to define an algorithm that unifies a set of pairs of terms – the Martelli – Montanari algorithm.

10 October 2006 Foundations of Logic and Constraint Programming 13 Martelli – Montanari algorithm Martelli-Montaneri algorithm (MM) Let E be a set of pairs of terms. As long as it is possible, choose nondeterministically a pair in E and perform the appropriate action: a)f(s 1,s 2,.., s n ) ≈ f(t 1,t 2,.., t n ) → replace, in E, by s 1 ≈ t 1, s 1 ≈ t 1,..., s n ≈ t n b)f(s 1,s 2,.., s n ) ≈ g(t 1,t 2,.., t n ) where f  g → halt with failure (clash) c)x ≈ x → delete the pair d)t ≈ x where t is not a variable → replace by x ≈ t e)x ≈ t where x  Var(t) and x occurs in some other pair → perform substitution { x / t } in those pairs f)x ≈ t where x  Var(t) and x  t → halt with failure (occurs check) The algorithm terminates when no action can be performed.

10 October 2006 Foundations of Logic and Constraint Programming 14 Martelli – Montanari algorithm Theorem 2.16 If the original set has a unifier, the algorithm MM successfully terminates and produces a solved set E´ that is equivalent to E; otherwise the algorithm terminates with failure. Proof Steps: 1.Prove that the algorithm terminates. 2.Prove that each action replaces the set of pairs by an equivalent one. 3.Prove that the algorithm terminates with failure, then the set of pairs at the moment of failure has no unifiers*. Given 2, the set E’ is equivalent to E, so E also has no unifiers 4.Prove that if the algorithm terminates successfully, the final set of pairs is solved**. By lemma 2.15, if the final set of pairs is solved, then it is an mgu.

10 October 2006 Foundations of Logic and Constraint Programming 15 Substitution Ordering ­Before going through each of the steps, the notion of well-founded orderings must be defined for relations on sets. ­Intuitively, even if the set is infinite, an well founded relation guarantees that the sets can be ordered so that they have an “initial” element. ­First some definitions on relations Definitions  R relation on set A:  R  A  A  R reflexive:  (a,a)  R for all a  A  R irreflexive:  (a,a)  R for all a  A  R antisymmetric:  (a,b)  R and (b,a)  R implies that a = b.  R transitive:  (a,b)  R and (b,c)  R implies that (a,c)  R.

10 October 2006 Foundations of Logic and Constraint Programming 16 Lexicographic Ordering ­The lexicographic ordering 1) is defined inductively on the set N n of n- tuples of natural numbers.  Base Clause (a 1 ) < 1 (b 1 ) :  a 1 < b 1  Inductive Clause (a 1, a 2,..., a n+1 ) < n+1 (b 1, b 2,..., b n+1 ) :  (a 1, a 2,..., a n ) < n+1 (b 1, b 2,..., b n ) ; or (a 1, a 2,..., a n ) = (b 1, b 2,..., b n ) and a n+1 < n b n+1 ­Examples: (5,3,9) < 3 (5,3,11), (1,4,8) < 3 (2,3,5). ­The lexicographic ordering (N n, < n ) is well-founded  Irreflexive partial ordering (irreflexible and transitive) with  Lowest element : (0,0,..., 0)

10 October 2006 Foundations of Logic and Constraint Programming 17 Well founded orderings ­Once defined the basic properties of relations some properties may be defined on the orderings they induce in particular sets.. Definitions  (A, R) is a (reflexive) partial ordering :  R is a reflexive, antisymmetric, and transitive relation on A  (A, R) is a irreflexive partial ordering :  R is a irreflexive and transitive relation on A  (A, R) is well-founded ordering :  R is a irreflexive partial ordering; and there is no infinite descending chain... a2 R a1 R a0. Examples Partial Orderings : (N,  ), (Z,  ), (  ({1,2,3,4}),  ) Irreflexive partial Orderings: (N, <), (Z, <), (  ({1,2,3,4}),  ) Well Founded:(N, <), (  ({1,2,3,4}),  ) Not Well Founded:(Z, <),

10 October 2006 Foundations of Logic and Constraint Programming 18 Algorithm MM terminates 1. MM terminates (1) ­Let the following denotations apply  Variable x solved in E :  x ≈ t  E, and this is the only occurrence of x in E  Uns(E) :  number of variables in E that are unsolved  lfun(E) :  number of occurrences of function symbols in the first components of pairs in E  card(E) :  number of pairs in E ­Let us consider tuple (Uns(E), lfun(E), card(E)). To prove termination of MM, it is sufficient to prove that each action in MM reduces this tuple. From the well- foundness of 3 tuples, the recursion must terminate (in the worst case, it cannot go below the lowest element).

10 October 2006 Foundations of Logic and Constraint Programming 19 Algorithm MM terminates 1. MM terminates (2) ­Let us consider tuple (u,l,c) = (Uns(E), lfun(E), card(E)) to show that the MM operations except b) and f), leading to failure, always lead to a lower tuple. a)f(s 1,s 2,.., s n ) ≈ f(t 1,t 2,.., t n ) → replace, in E, by s 1 ≈ t 1, s 2 ≈ t 2,..., s n ≈ t n (u,l,c) → (u-k, l-1, c+n-1) for some k  1..n c)x ≈ x → delete the pair (u,l,c) → (u-k, l, c-1) for some k  0..1 d)t ≈ x where t is not a variable → replace by x ≈ t (u,l,c) → (u-k 1, l-k 2, c) for some k 1  0..1 and k 2  1 e)x ≈ t where x  Var(t) and x occurs in some other pair → perform substitution { x / t } in those pairs (u,l,c) → (u-1, l+k, c) for some k  1

10 October 2006 Foundations of Logic and Constraint Programming 20 Algorithm MM is correct 2.MM replaces the set of pairs by an equivalents set a) f(s 1,s 2,.., s n ) ≈ f(t 1,t 2,.., t n ) → replace, in E, by s 1 ≈ t 1, s 1 ≈ t 1,..., s n ≈ t n c) x ≈ x → delete the pair d) t ≈ x where t is not a variable → replace by x ≈ t This is clearly true for actions a), c) and d) above. For action e) e) when x ≈ t where x  Var(t) and x occurs in some other pair → perform substitution {x/t } in those pairs consider E  {x ≈ t } and E{ x/t }  {x ≈ t } then θ is a unifier of E  {x ≈ t } iff(θ is a unifier of E) and xθ = tθ iff(θ is a unifier of E {x/t} ) and xθ = tθ iffθ is a unifier of E {x/t}  {x ≈ t }

10 October 2006 Foundations of Logic and Constraint Programming 21 Algorithm MM Result 3.If MM terminates with failure, then the set of pairs at the moment of failure has no unifiers In MM terminates with failure, the last action taken in E’ was either b) or d). b)f(s 1,s 2,.., s n ) ≈ g(t 1,t 2,.., t n ) where f  g → halt with failure (clash) f)x ≈ t where x  Var(t) and x  t → halt with failure (occurs check) Given the definition of unification, there must be no unifier of E’. Any substitution θ applied to f(s 1,s 2,.., s n ) θ would result in a term with functor f, not g, so no unifier θ exists. No substitution θ may contain a pair x/t where x  Var(t) and x  t, as would be needed to unify x and t.

10 October 2006 Foundations of Logic and Constraint Programming 22 Algorithm MM Result 4.If the algorithm terminates successfully, the final set of pairs is solved When algorithm MM terminates, with success, no left components of the pairs in E’ are functions (otherwise, actions a) or b) or d) would apply). a) f(s 1,s 2,.., s n ) ≈ f(t 1,t 2,.., t n ) → replace, in E, by s 1 ≈ t 1, s 1 ≈ t 1,..., s n ≈ t n d) t ≈ x where t is not a variable → replace by x ≈ t All pairs in E’ are in the form x ≈ t, where x is a variable. Furthermore t  x (otherwise action c) would apply) c) x ≈ x → delete the pair No pair other than x ≈ t contains variable x, otherwise action e) would apply e) x ≈ t where x  Var(t) and x occurs in some other pair → perform substitution { x / t } in those pairs Hence, E’ has the form E’ = {x 1 ≈ t 1, x 2 ≈ t 2,... x n ≈ t n }, where x i  Var(t j ), for 1  i  j  n, and is thus in solved form.

10 October 2006 Foundations of Logic and Constraint Programming 23 Examples ­Some non-trivial examples (from SICStus Prolog, where variables are written in upper case) | ?- f(X,b) = f(a,Y). X = a,Y = b ? ; no |?- f(X,f(b)) = f(g(a,Y),Y). X = g(a,f(b)), Y = f(b) ? ; no | ?- f(X,f(b,Z)) = f(g(a,Y),Y). X = g(a,f(b,Z)), Y = f(b,Z) ? ; no | ?- X = f(Y), Y = f(a). X = f(f(a)), Y = f(a) ? ; no | ?- X = f(Y,Z), g(a,Y) = g(Z,b). X = f(b,a),Y = b,Z = a ? ; no

10 October 2006 Foundations of Logic and Constraint Programming 24 Unifiers Size Can Be Exponential ­Example: | ?- f(X1) = f(g(X0,X0)). X1 = g(X0,X0) ? ; no | ?- f(X1,X2) = f(g(X0,X0),g(X1,X1)). X1 = g(X0,X0), X2 = g(g(X0,X0),g(X0,X0)) ? ; no | ?- f(X1,X2,X3) = f(g(X0,X0),g(X1,X1),g(X2,X2)). X1 = g(X0,X0), X2 = g(g(X0,X0),g(X0,X0)), X3 = g(g(g(X0,X0),g(X0,X0)),g(g(X0,X0),g(X0,X0))) ? ; no.... ­In this case, the “size” of variable X i, in terms of occurrences of X 0, is: size(X1) = 2 size(X2) = 2 2 = 4 size(X3) = 2 3 = 8... size(Xn) = 2 n

10 October 2006 Foundations of Logic and Constraint Programming 25 Strong MGUs Definition  Substitution θ is Idempotent :  θ θ = θ  :Mgu θ of terms s and t is strong :  σ is mgu of s and t  σ = θ σ Theorem 2.16 An mgu is strong iff it is idempotent. Proof :  Let θ be a strong mgu. Since θ is a strong mgu, for any unifier σ it is σ = θ σ. In particular, for unifier θ, we have θ = θ θ. Hence θ is idempotent.  Let θ be an idempotent mgu. Then, for any unifier σ there is some λ such that σ = θ λ. But since θ is idempotent it is σ = θ λ = (θ θ) λ = θ ( θ λ) = θ σ. Hence, θ is strong.

10 October 2006 Foundations of Logic and Constraint Programming 26 Strong MGUs Examples  The mgu θ = {x/y, y/x} of two identical terms is not strong. Indeed, θ θ = {x/y, y/x } ° {x/y, y/x} = {x/x, y/y, x/y, y/x} = ε  Substitution θ 1 = {x/u, z/f(u,v), y/v} is Idempotent { x/u, z/f(u,v), y/v } ° { x/u, z/f(u,v), y/v } = { x/u, z/f(u,v), y/v, x/u, z/f(u,v), y/v } = { x/u, z/f(u,v), y/v } = θ 1 Substitution θ 2 = {x/u, z/f(u,v), v/y} is not Idempotent { x/u, z/f(u,v), v/y } ° { x/u, z/f(u,v), v/y } = { x/u, z/f(u,y), v/y, x/u, z/f(u,v), y/v } = { x/u, z/f(u,y), y/v }  θ 2. ­The following lemma justifies a simple test to check whether a substitution mgu is idempotent. Lemma: A substitution θ is idempotent iff Dom(θ)  Ran (θ) 1 = 

10 October 2006 Foundations of Logic and Constraint Programming 27 Relevant MGUs Definition  Mgu θ of terms s and t is relevant :  Var(θ )  Var(s)  Var(t) Informaly, a relevant mgu does not introduce new variables. All mgus produced by the Martelli-Montanari algorithm are relevant. Moreover Theorem 2.22 (Relevance) Every idempotent mgu is relevant. Proof : in [Apt, 1997] Note that the converse is not true, i.e. Not all relevant mgus are idempotent. For example θ = {x/y, y/x}, which is a relevant mgu of terms s = f(x,y) and t = f(x,y), is not idempotent. As observed before, θ θ = ε, and ε is the mgu btained from the MM algorithm when presented with E = { f(x,y) ≈ f(x,y) }.

10 October 2006 Foundations of Logic and Constraint Programming 28 Properties of MGUs The following theorems can also be proved and are useful later (see [Apt97]) Lemma 2.23 (Equivalence): Let θ 1 be an mgu of a set of equations E. Then for every substitution θ 2, θ 2 is an mgu of E iff θ 2 = θ 1 ρ, where ρ is a renaming such that Var(ρ)  Var(θ 1 )  Var(θ 2 ). Lemma 2.24 (Iteration) : Let E 1 and E 2 be two sets of equations. Suppose that θ 1 is an mgu of E1, and θ 2 is an mgu of E 2 θ 1. Then θ 1 θ 2 is an mgu of E 1  E 2. Moreover, if E 1  E 2 is unifiable, then an mgu θ 1 of E 1 exists and for any mgu θ 1 of E 1 there also exists an mgu θ 2 of E 2 θ 1. Corollary 2.25 (Switching) : Let E 1 and E 2 be two sets of equations. Suppose that θ 1 is an mgu of E1, and θ 2 is an mgu of E 2 θ 1. Then E 2 is unifiable, and for every mgu θ 1 ’ of E 2 there exists an mgu θ 2 ’ of E 2 θ 1 ’ such that θ 1 ’ θ 2 ’ = θ 1 θ 2. Moreover, θ 2 ’ can be so chosen such that Var(θ 2 ’)  Var(E 1 )  Var(θ 1 ’)  Var(θ 1 θ 2 ).

10 October 2006 Foundations of Logic and Constraint Programming 29 Occurs Check in Prolog ­The occurs check is usually not implemented (by default) in Prolog, since  It makes unification significantly harder  It happens quite rarely ­Of course, unification behaves strangely when the occurs check is needed. For example, in SICStus, we may observe | ?- X = f(X). X = f(f(f(f(f(f(f(f(f(f(...)))))))))) ? ; no ­Even worse | ?- X = f(X). X = f(f(f(f(f(f(f(f(f(f(f(... % LOOP ­To avoid this incorrect beheviour, most Prolog systems provide the possibility of using correct, if innefficient unification.

10 October 2006 Foundations of Logic and Constraint Programming 30 Occurs Check in Prolog ­In SICStus, the explicit unification between two terms S and T can be expressed either as  S = T, the default mode, incorrect implementation of unification; or  unify_with_occurs_check(S,T), correct implementation, inneficient. ­Now, it is | ?- unify_with_occurs_check(X,f(X)). no | ?- unify_with_occurs_check(X,f(Y)). X = f(Y) ? ; no ­It is up to the user to select the appropriate form. For example, the test or an empty difference list L could be simply expressed as T-T. But if the list is not empty, and containts n elements, then it takes the form [a 1, a 2,..., a n | T] - T in which case the default unification (without occurs check) would not work properly, since it would try to unify T with [a 1, a 2,..., a n | T].

10 October 2006 Foundations of Logic and Constraint Programming 31 Occurs Check in Prolog ­This behaviour can be observed with the rev_diff/2 predicate to reverse a difference list rev_diff(L-L,L-L). % incorrect version rev_diff([H|T]-Z,X-W):- % according to diff_cat Z \== [H|T], % C=X, D=W if Y = [H|D] rev_diff(T-Z,X-[H|W]). leading to the following interaction | ?- rev_diff0([1,2|T]-T,L). L = [1,2,1,2,1,2,1,2,1|...]-[1,2,1,2,1,2,1,2,1|...], T = [1,2,1,2,1,2,1,2,1,2|...] ? ; L = [2,1|_A]-_A, T = [2,1|_A] ? ;no ­If the first clause is written instead as rev_diff(L-T,L-L):- unify_with_occurs_check(L,T). then the correct behaviour is observed | ?- rev_diff([1,2|T]-T,L). L = [2,1|_A]-_A, T = [2,1|_A] ? ;no

10 October 2006 Foundations of Logic and Constraint Programming 32 Exercises for the Week ­Solve exercises below from [Apt97], chapter 2  2.1 (pag. 21);  2.2 (pag. 21);  2.3 (pag. 22);  2.4 (pag. 24);  2.5 (pag. 26);  2.6 (pag. 33);  2.8 (pag. 36); ­Optionally  2.9 (pag. 37);  2.10 (pag. 39);

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