/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: !faccolon : [] --> b -> b -> b !facplus : [] --> b -> b -> b a : [] --> b cons : [] --> d -> e -> e false : [] --> c filter : [] --> d -> c -> e -> e filter2 : [] --> c -> d -> c -> d -> e -> e g : [] --> b -> a -> b map : [] --> d -> d -> e -> e nil : [] --> e true : [] --> c Rules: !faccolon (!faccolon x y) z => !faccolon x (!faccolon y z) !faccolon (!facplus x y) z => !facplus (!faccolon x z) (!faccolon y z) !faccolon x (!facplus y (f z)) => !faccolon (g x z) (!facplus y a) map f nil => nil map f (cons x y) => cons (f x) (map f y) filter f nil => nil filter f (cons x y) => filter2 (f x) f x y filter2 true f x y => cons x (filter f y) filter2 false f x y => filter f y Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: Alphabet: !faccolon : [b * b] --> b !facplus : [b * b] --> b a : [] --> b cons : [d * e] --> e false : [] --> c filter : [d -> c * e] --> e filter2 : [c * d -> c * d * e] --> e g : [b * a] --> b map : [d -> d * e] --> e nil : [] --> e true : [] --> c ~AP1 : [a -> b * a] --> b Rules: !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(g(X, Z), !facplus(Y, a)) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) !faccolon(X, !facplus(Y, g(Z, U))) => !faccolon(g(X, U), !facplus(Y, a)) ~AP1(F, X) => F X We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]). We thus obtain the following dependency pair problem (P_0, R_0, static, formative): Dependency Pairs P_0: 0] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z)) 1] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z) 2] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z) 3] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z) 4] !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(g(X, Z), !facplus(Y, a)) 5] map#(F, cons(X, Y)) =#> map#(F, Y) 6] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 7] filter2#(true, F, X, Y) =#> filter#(F, Y) 8] filter2#(false, F, X, Y) =#> filter#(F, Y) 9] !faccolon#(X, !facplus(Y, g(Z, U))) =#> !faccolon#(g(X, U), !facplus(Y, a)) Rules R_0: !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(g(X, Z), !facplus(Y, a)) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) !faccolon(X, !facplus(Y, g(Z, U))) => !faccolon(g(X, U), !facplus(Y, a)) ~AP1(F, X) => F X Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. We consider the dependency pair problem (P_0, R_0, static, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0, 1, 2, 3, 4, 9 * 1 : 0, 1, 2, 3, 4, 9 * 2 : 0, 1, 2, 3, 4, 9 * 3 : 0, 1, 2, 3, 4, 9 * 4 : * 5 : 5 * 6 : 7, 8 * 7 : 6 * 8 : 6 * 9 : This graph has the following strongly connected components: P_1: !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z)) !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z) P_2: map#(F, cons(X, Y)) =#> map#(F, Y) P_3: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f) and (P_3, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (P_3, R_0, static, formative) is finite. We consider the dependency pair problem (P_3, R_0, static, formative). We apply the subterm criterion with the following projection function: nu(filter2#) = 4 nu(filter#) = 2 Thus, we can orient the dependency pairs as follows: nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_3, R_0, static, f) by (P_4, R_0, static, f), where P_4 contains: filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (P_4, R_0, static, formative) is finite. We consider the dependency pair problem (P_4, R_0, static, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. We consider the dependency pair problem (P_2, R_0, static, formative). We apply the subterm criterion with the following projection function: nu(map#) = 2 Thus, we can orient the dependency pairs as follows: nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by ({}, R_0, static, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. We consider the dependency pair problem (P_1, R_0, static, formative). We apply the subterm criterion with the following projection function: nu(!faccolon#) = 1 Thus, we can orient the dependency pairs as follows: nu(!faccolon#(!faccolon(X, Y), Z)) = !faccolon(X, Y) |> X = nu(!faccolon#(X, !faccolon(Y, Z))) nu(!faccolon#(!faccolon(X, Y), Z)) = !faccolon(X, Y) |> Y = nu(!faccolon#(Y, Z)) nu(!faccolon#(!facplus(X, Y), Z)) = !facplus(X, Y) |> X = nu(!faccolon#(X, Z)) nu(!faccolon#(!facplus(X, Y), Z)) = !facplus(X, Y) |> Y = nu(!faccolon#(Y, Z)) By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop11] C. Kop. Simplifying Algebraic Functional Systems. In Proceedings of CAI 2011, volume 6742 of LNCS. 201--215, Springer, 2011. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.