We consider the system Applicative_05__Ex5Folding. Alphabet: 0 : [] --> c add : [] --> a -> c -> c cons : [a * b] --> b fold : [a -> c -> c * c] --> b -> c mul : [] --> a -> c -> c nil : [] --> b plus : [c * c] --> c prod : [] --> b -> c s : [c] --> c sum : [] --> b -> c times : [c * c] --> c Rules: fold(f, x) nil => x fold(f, x) cons(y, z) => f y (fold(f, x) z) 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) sum => fold(add, 0) prod => fold(mul, 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: fold(F, X, nil) => X fold(F, X, cons(Y, Z)) => F Y fold(F, X, Z) 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) sum(X) => fold(/\x./\y.add(x, y), 0, X) prod(X) => fold(/\x./\y.mul(x, y), s(0), X) We thus obtain the following dependency pair problem (P_0, R_0, static, formative): Dependency Pairs P_0: 0] fold#(F, X, cons(Y, Z)) =#> fold#(F, X, Z) 1] plus#(s(X), Y) =#> plus#(X, Y) 2] times#(s(X), Y) =#> plus#(times(X, Y), Y) 3] times#(s(X), Y) =#> times#(X, Y) 4] sum#(X) =#> fold#(/\x./\y.add(x, y), 0, X) 5] prod#(X) =#> fold#(/\x./\y.mul(x, y), s(0), X) Rules R_0: fold(F, X, nil) => X fold(F, X, cons(Y, Z)) => F Y fold(F, X, Z) 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) sum(X) => fold(/\x./\y.add(x, y), 0, X) prod(X) => fold(/\x./\y.mul(x, y), s(0), 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 : 1 * 2 : 1 * 3 : 2, 3 * 4 : 0 * 5 : 0 This graph has the following strongly connected components: P_1: fold#(F, X, cons(Y, Z)) =#> fold#(F, X, Z) P_2: plus#(s(X), Y) =#> plus#(X, Y) P_3: times#(s(X), Y) =#> times#(X, 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: 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_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: 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_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: fold#(F, X, cons(Y, Z)) >? fold#(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 fold# = \G0y1y2.y2 Using this interpretation, the requirements translate to: [[fold#(_F0, _x1, cons(_x2, _x3))]] = 3 + 2x3 > x3 = [[fold#(_F0, _x1, _x3)]] 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.