We consider the system AotoYamada_05__009. Alphabet: and : [c * c] --> c cons : [a * b] --> b false : [] --> c forall : [a -> c * b] --> c forsome : [a -> c * b] --> c nil : [] --> b or : [c * c] --> c true : [] --> c Rules: and(true, true) => true and(true, false) => false and(false, true) => false and(false, false) => false or(true, true) => true or(true, false) => true or(false, true) => true or(false, false) => false forall(f, nil) => true forall(f, cons(x, y)) => and(f x, forall(f, y)) forsome(f, nil) => false forsome(f, cons(x, y)) => or(f x, forsome(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] forall#(F, cons(X, Y)) =#> and#(F X, forall(F, Y)) 1] forall#(F, cons(X, Y)) =#> forall#(F, Y) 2] forsome#(F, cons(X, Y)) =#> or#(F X, forsome(F, Y)) 3] forsome#(F, cons(X, Y)) =#> forsome#(F, Y) Rules R_0: and(true, true) => true and(true, false) => false and(false, true) => false and(false, false) => false or(true, true) => true or(true, false) => true or(false, true) => true or(false, false) => false forall(F, nil) => true forall(F, cons(X, Y)) => and(F X, forall(F, Y)) forsome(F, nil) => false forsome(F, cons(X, Y)) => or(F X, forsome(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). This combination (P_0, R_0) has no formative rules! We will name the empty set of rules:R_1. By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, static, formative) by (P_0, R_1, static, formative). Thus, the original system is terminating if (P_0, R_1, static, formative) is finite. We consider the dependency pair problem (P_0, R_1, 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: forall#(F, cons(X, Y)) >? and#(F X, forall(F, Y)) forall#(F, cons(X, Y)) >? forall#(F, Y) forsome#(F, cons(X, Y)) >? or#(F X, forsome(F, Y)) forsome#(F, cons(X, Y)) >? forsome#(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and# = \y0y1.0 cons = \y0y1.3 + 2y1 forall = \G0y1.0 forall# = \G0y1.3 + y1 + y1G0(y1) forsome = \G0y1.0 forsome# = \G0y1.3 + y1 + y1G0(y1) or# = \y0y1.0 Using this interpretation, the requirements translate to: [[forall#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 0 = [[and#(_F0 _x1, forall(_F0, _x2))]] [[forall#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 3 + x2 + x2F0(x2) = [[forall#(_F0, _x2)]] [[forsome#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 0 = [[or#(_F0 _x1, forsome(_F0, _x2))]] [[forsome#(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 2x2F0(3 + 2x2) + 3F0(3 + 2x2) > 3 + x2 + x2F0(x2) = [[forsome#(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_1) by ({}, R_1). 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.