We consider the system AotoYamada_05__004. Alphabet: 0 : [] --> b cons : [a * b] --> b nil : [] --> b plus : [b * b] --> b s : [b] --> b sumwith : [a -> b * b] --> b Rules: plus(0, x) => x plus(s(x), y) => s(plus(x, y)) sumwith(f, nil) => nil sumwith(f, cons(x, y)) => plus(f x, sumwith(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] plus#(s(X), Y) =#> plus#(X, Y) 1] sumwith#(F, cons(X, Y)) =#> plus#(F X, sumwith(F, Y)) 2] sumwith#(F, cons(X, Y)) =#> sumwith#(F, Y) Rules R_0: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) sumwith(F, nil) => nil sumwith(F, cons(X, Y)) => plus(F X, sumwith(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: plus#(s(X), Y) >? plus#(X, Y) sumwith#(F, cons(X, Y)) >? plus#(F X, sumwith(F, Y)) sumwith#(F, cons(X, Y)) >? sumwith#(F, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) sumwith(F, nil) >= nil sumwith(F, cons(X, Y)) >= plus(F X, sumwith(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.3 + y0 nil = 0 plus = \y0y1.2y1 plus# = \y0y1.0 s = \y0.0 sumwith = \G0y1.0 sumwith# = \G0y1.3 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[sumwith#(_F0, cons(_x1, _x2))]] = 3 > 0 = [[plus#(_F0 _x1, sumwith(_F0, _x2))]] [[sumwith#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[sumwith#(_F0, _x2)]] [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 2x1 >= 0 = [[s(plus(_x0, _x1))]] [[sumwith(_F0, nil)]] = 0 >= 0 = [[nil]] [[sumwith(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[plus(_F0 _x1, sumwith(_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: plus#(s(X), Y) =#> plus#(X, Y) sumwith#(F, cons(X, Y)) =#> sumwith#(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 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: plus#(s(X), Y) >? plus#(X, Y) sumwith#(F, cons(X, Y)) >? sumwith#(F, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) sumwith(F, nil) >= nil sumwith(F, cons(X, Y)) >= plus(F X, sumwith(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.3 + y1 nil = 0 plus = \y0y1.y1 plus# = \y0y1.0 s = \y0.0 sumwith = \G0y1.0 sumwith# = \G0y1.y1 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[sumwith#(_F0, cons(_x1, _x2))]] = 3 + x2 > x2 = [[sumwith#(_F0, _x2)]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(_x0, _x1))]] [[sumwith(_F0, nil)]] = 0 >= 0 = [[nil]] [[sumwith(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[plus(_F0 _x1, sumwith(_F0, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of: plus#(s(X), Y) =#> plus#(X, Y) Thus, the original system is terminating if (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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: plus#(s(X), Y) >? plus#(X, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) sumwith(F, nil) >= nil sumwith(F, cons(X, Y)) >= plus(F X, sumwith(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.3 + y0 + 2y1 nil = 0 plus = \y0y1.y0 + 2y1 plus# = \y0y1.y0 s = \y0.2 + y0 sumwith = \G0y1.y1G0(y1) Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 2 + x0 > x0 = [[plus#(_x0, _x1)]] [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 2 + x0 + 2x1 >= 2 + x0 + 2x1 = [[s(plus(_x0, _x1))]] [[sumwith(_F0, nil)]] = 0 >= 0 = [[nil]] [[sumwith(_F0, cons(_x1, _x2))]] = 2x2F0(3 + x1 + 2x2) + 3F0(3 + x1 + 2x2) + x1F0(3 + x1 + 2x2) >= F0(x1) + 2x2F0(x2) = [[plus(_F0 _x1, sumwith(_F0, _x2))]] 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. 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.