We consider the system AotoYamada_05__002. Alphabet: cons : [b * c] --> c false : [] --> a filter : [b -> a * c] --> c filtersub : [a * b -> a * c] --> c nil : [] --> c true : [] --> a Rules: filter(f, nil) => nil filter(f, cons(x, y)) => filtersub(f x, f, cons(x, y)) filtersub(true, f, cons(x, y)) => cons(x, filter(f, y)) filtersub(false, f, cons(x, y)) => filter(f, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 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] filter#(F, cons(X, Y)) =#> filtersub#(F X, F, cons(X, Y)) 1] filtersub#(true, F, cons(X, Y)) =#> filter#(F, Y) 2] filtersub#(false, F, cons(X, Y)) =#> filter#(F, Y) Rules R_0: filter(F, nil) => nil filter(F, cons(X, Y)) => filtersub(F X, F, cons(X, Y)) filtersub(true, F, cons(X, Y)) => cons(X, filter(F, Y)) filtersub(false, F, cons(X, Y)) => filter(F, Y) 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 will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: filter#(F, cons(X, Y)) >? filtersub#(F X, F, cons(X, Y)) filtersub#(true, F, cons(X, Y)) >? filter#(F, Y) filtersub#(false, F, cons(X, Y)) >? filter#(F, Y) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filtersub(F X, F, cons(X, Y)) filtersub(true, F, cons(X, Y)) >= cons(X, filter(F, Y)) filtersub(false, F, cons(X, Y)) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.1 + y1 false = 3 filter = \G0y1.y1 filter# = \G0y1.1 + y1 filtersub = \y0G1y2.y2 filtersub# = \y0G1y2.y2 nil = 0 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 > 1 + x2 = [[filtersub#(_F0 _x1, _F0, cons(_x1, _x2))]] [[filtersub#(true, _F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filter#(_F0, _x2)]] [[filtersub#(false, _F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filter#(_F0, _x2)]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filtersub(_F0 _x1, _F0, cons(_x1, _x2))]] [[filtersub(true, _F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filtersub(false, _F0, cons(_x1, _x2))]] = 1 + x2 >= x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of: filtersub#(true, F, cons(X, Y)) =#> filter#(F, Y) filtersub#(false, F, cons(X, Y)) =#> filter#(F, Y) 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 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. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [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.