We consider the system prefixsum. Alphabet: !plus : [nat * nat] --> nat cons : [nat * list] --> list map : [nat -> nat * list] --> list nil : [] --> list ps : [list] --> list Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) ps(nil) => nil ps(cons(x, y)) => cons(x, ps(map(/\z.!plus(x, z), 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] map#(F, cons(X, Y)) =#> map#(F, Y) 1] ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y)) 2] ps#(cons(X, Y)) =#> map#(/\x.!plus(X, x), Y) Rules R_0: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) ps(nil) => nil ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), 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). The formative rules of (P_0, R_0) are R_1 ::= map(F, cons(X, Y)) => cons(F X, map(F, Y)) ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) 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: map#(F, cons(X, Y)) >? map#(F, Y) ps#(cons(X, Y)) >? ps#(map(/\x.!plus(X, x), Y)) ps#(cons(X, Y)) >? map#(/\x.!plus(X, x), Y) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.0 cons = \y0y1.0 map = \G0y1.2G0(0) + 2G0(y1) map# = \G0y1.0 ps = \y0.0 ps# = \y0.3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] [[ps#(cons(_x0, _x1))]] = 3 >= 3 = [[ps#(map(/\x.!plus(_x0, x), _x1))]] [[ps#(cons(_x0, _x1))]] = 3 > 0 = [[map#(/\x.!plus(_x0, x), _x1)]] [[map(_F0, cons(_x1, _x2))]] = 4F0(0) >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[ps(cons(_x0, _x1))]] = 0 >= 0 = [[cons(_x0, ps(map(/\x.!plus(_x0, x), _x1)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of: map#(F, cons(X, Y)) =#> map#(F, Y) ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y)) Thus, the original system is terminating if (P_1, R_1, static, formative) is finite. We consider the dependency pair problem (P_1, 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: map#(F, cons(X, Y)) >? map#(F, Y) ps#(cons(X, Y)) >? ps#(map(/\x.!plus(X, x), Y)) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.0 cons = \y0y1.1 + y1 map = \G0y1.y1 map# = \G0y1.y1 ps = \y0.y0 ps# = \y0.0 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] [[ps#(cons(_x0, _x1))]] = 0 >= 0 = [[ps#(map(/\x.!plus(_x0, x), _x1))]] [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[ps(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[cons(_x0, ps(map(/\x.!plus(_x0, x), _x1)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of: ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y)) Thus, the original system is terminating if (P_2, R_1, static, formative) is finite. We consider the dependency pair problem (P_2, 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: ps#(cons(X, Y)) >? ps#(map(/\x.!plus(X, x), Y)) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.0 cons = \y0y1.1 + y1 map = \G0y1.y1 ps = \y0.y0 ps# = \y0.2y0 Using this interpretation, the requirements translate to: [[ps#(cons(_x0, _x1))]] = 2 + 2x1 > 2x1 = [[ps#(map(/\x.!plus(_x0, x), _x1))]] [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[ps(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[cons(_x0, ps(map(/\x.!plus(_x0, x), _x1)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, 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.