We consider the system Applicative_05__Ex9Maps. Alphabet: cons : [d * e] --> e g : [] --> b map!fac62201 : [d -> d * e] --> e map!fac62202 : [d -> a -> d * a * e] --> e map!fac62203 : [b -> d -> c -> d * b * c * e] --> e Rules: map!fac62201(f, cons(x, y)) => cons(f x, map!fac62201(f, y)) map!fac62202(f, x, cons(y, z)) => cons(f y x, map!fac62202(f, x, z)) map!fac62203(f, g, x, cons(y, z)) => cons(f g y x, map!fac62203(f, g, x, z)) 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] map!fac62201#(F, cons(X, Y)) =#> map!fac62201#(F, Y) 1] map!fac62202#(F, X, cons(Y, Z)) =#> map!fac62202#(F, X, Z) 2] map!fac62203#(F, g, X, cons(Y, Z)) =#> map!fac62203#(F, g, X, Z) Rules R_0: map!fac62201(F, cons(X, Y)) => cons(F X, map!fac62201(F, Y)) map!fac62202(F, X, cons(Y, Z)) => cons(F Y X, map!fac62202(F, X, Z)) map!fac62203(F, g, X, cons(Y, Z)) => cons(F g Y X, map!fac62203(F, g, X, Z)) 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 : 1 * 2 : 2 This graph has the following strongly connected components: P_1: map!fac62201#(F, cons(X, Y)) =#> map!fac62201#(F, Y) P_2: map!fac62202#(F, X, cons(Y, Z)) =#> map!fac62202#(F, X, Z) P_3: map!fac62203#(F, g, X, cons(Y, Z)) =#> map!fac62203#(F, g, X, Z) 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. (P_3, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: map!fac62203#(F, g, X, cons(Y, Z)) >? map!fac62203#(F, g, X, Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + y1 g = 0 map!fac62203# = \G0y1y2y3.y3 Using this interpretation, the requirements translate to: [[map!fac62203#(_F0, g, _x1, cons(_x2, _x3))]] = 3 + x3 > x3 = [[map!fac62203#(_F0, g, _x1, _x3)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. (P_2, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: map!fac62202#(F, X, cons(Y, Z)) >? map!fac62202#(F, X, Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + 2y1 map!fac62202# = \G0y1y2.y2 Using this interpretation, the requirements translate to: [[map!fac62202#(_F0, _x1, cons(_x2, _x3))]] = 3 + 2x3 > x3 = [[map!fac62202#(_F0, _x1, _x3)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0). 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. (P_1, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: map!fac62201#(F, cons(X, Y)) >? map!fac62201#(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + 2y1 map!fac62201# = \G0y1.y1 Using this interpretation, the requirements translate to: [[map!fac62201#(_F0, cons(_x1, _x2))]] = 3 + 2x2 > x2 = [[map!fac62201#(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0). 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 +++ [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.