We consider the system AotoYamada_05__019. Alphabet: comp : [a -> a * a -> a] --> a -> a twice : [a -> a] --> a -> a Rules: comp(f, g) x => f (g x) twice(f) => comp(f, f) 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 dynamic dependency pairs. After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] comp(F, G) X =#> F(G X) 1] comp(F, G) X =#> G(X) 2] twice(F) X =#> comp(F, F) X 3] twice#(F) =#> comp#(F, F) Rules R_0: comp(F, G) X => F (G X) twice(F) => comp(F, F) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, 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, 2, 3 * 1 : 0, 1, 2, 3 * 2 : 0, 1 * 3 : This graph has the following strongly connected components: P_1: comp(F, G) X =#> F(G X) comp(F, G) X =#> G(X) twice(F) X =#> comp(F, F) X 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). Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). This combination (P_1, 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_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: comp(F, G, X) >? F(G X) comp(F, G, X) >? G(X) twice(F, X) >? comp(F, F, X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( comp(F, G, X) ) = #argfun-comp#(F (G X), G X) pi( twice(F, X) ) = #argfun-twice#(#argfun-comp#(F (F X), F X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: #argfun-comp# = \y0y1.3 + max(y0, y1) #argfun-twice# = \y0.3 + y0 comp = \G0G1y2.0 twice = \G0y1.0 Using this interpretation, the requirements translate to: [[#argfun-comp#(_F0 (_F1 _x2), _F1 _x2)]] = 3 + max(x2, F0(max(x2, F1(x2))), F1(x2)) > F0(max(x2, F1(x2))) = [[_F0(_F1 _x2)]] [[#argfun-comp#(_F0 (_F1 _x2), _F1 _x2)]] = 3 + max(x2, F0(max(x2, F1(x2))), F1(x2)) > F1(x2) = [[_F1(_x2)]] [[#argfun-twice#(#argfun-comp#(_F0 (_F0 _x1), _F0 _x1))]] = 6 + max(x1, F0(x1), F0(max(x1, F0(x1)))) > 3 + max(x1, F0(x1), F0(max(x1, F0(x1)))) = [[#argfun-comp#(_F0 (_F0 _x1), _F0 _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, 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.