We consider the system AotoYamada_05__014. Alphabet: 0 : [] --> b cons : [b * a] --> a double : [] --> a -> a inc : [] --> a -> a map : [b -> b] --> a -> a nil : [] --> a plus : [b] --> b -> b s : [b] --> b times : [b] --> b -> b Rules: plus(0) x => x plus(s(x)) y => s(plus(x) y) times(0) x => 0 times(s(x)) y => plus(times(x) y) y map(f) nil => nil map(f) cons(x, y) => cons(f x, map(f) y) inc => map(plus(s(0))) double => map(times(s(s(0)))) 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]). In order to do so, we start by eta-expanding the system, which gives: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(times(X, Y), Y) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) inc(X) => map(/\x.plus(s(0), x), X) double(X) => map(/\x.times(s(s(0)), x), X) We thus obtain the following dependency pair problem (P_0, R_0, static, formative): Dependency Pairs P_0: 0] plus#(s(X), Y) =#> plus#(X, Y) 1] times#(s(X), Y) =#> plus#(times(X, Y), Y) 2] times#(s(X), Y) =#> times#(X, Y) 3] map#(F, cons(X, Y)) =#> map#(F, Y) 4] inc#(X) =#> map#(/\x.plus(s(0), x), X) 5] inc#(X) =#> plus#(s(0), Y) 6] double#(X) =#> map#(/\x.times(s(s(0)), x), X) 7] double#(X) =#> times#(s(s(0)), Y) Rules R_0: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(times(X, Y), Y) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) inc(X) => map(/\x.plus(s(0), x), X) double(X) => map(/\x.times(s(s(0)), x), 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 : 0 * 2 : 1, 2 * 3 : 3 * 4 : 3 * 5 : 0 * 6 : 3 * 7 : 1, 2 This graph has the following strongly connected components: P_1: plus#(s(X), Y) =#> plus#(X, Y) P_2: times#(s(X), Y) =#> times#(X, Y) P_3: map#(F, cons(X, Y)) =#> map#(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 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#(F, cons(X, Y)) >? map#(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# = \G0y1.y1 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x2 > x2 = [[map#(_F0, _x2)]] 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: times#(s(X), Y) >? times#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: s = \y0.1 + 2y0 times# = \y0y1.y0 Using this interpretation, the requirements translate to: [[times#(s(_x0), _x1)]] = 1 + 2x0 > x0 = [[times#(_x0, _x1)]] 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: plus#(s(X), Y) >? plus#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: plus# = \y0y1.y0 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 1 + 2x0 > x0 = [[plus#(_x0, _x1)]] 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.