We consider the system Applicative_05__mapDivMinus. Alphabet: 0 : [] --> c cons : [a * b] --> b div : [c * c] --> c map : [a -> a * b] --> b minus : [c * c] --> c nil : [] --> b s : [c] --> c Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) div(0, s(x)) => 0 div(s(x), s(y)) => s(div(minus(x, y), s(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] minus#(s(X), s(Y)) =#> minus#(X, Y) 2] div#(s(X), s(Y)) =#> div#(minus(X, Y), s(Y)) 3] div#(s(X), s(Y)) =#> minus#(X, Y) Rules R_0: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) div(0, s(X)) => 0 div(s(X), s(Y)) => s(div(minus(X, Y), s(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 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 : 2, 3 * 3 : 1 This graph has the following strongly connected components: P_1: map#(F, cons(X, Y)) =#> map#(F, Y) P_2: minus#(s(X), s(Y)) =#> minus#(X, Y) P_3: div#(s(X), s(Y)) =#> div#(minus(X, Y), s(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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: div#(s(X), s(Y)) >? div#(minus(X, Y), s(Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) div(0, s(X)) >= 0 div(s(X), s(Y)) >= s(div(minus(X, Y), s(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 div = \y0y1.y0 div# = \y0y1.y0 map = \G0y1.y1 + 3y1G0(y1) + 3G0(y1) minus = \y0y1.y0 nil = 0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[div#(s(_x0), s(_x1))]] = 1 + x0 > x0 = [[div#(minus(_x0, _x1), s(_x1))]] [[map(_F0, nil)]] = 3F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x2 + 3x2F0(3 + x2) + 12F0(3 + x2) >= 3 + x2 + 3x2F0(x2) + 3F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 1 + x0 >= x0 = [[minus(_x0, _x1)]] [[div(0, s(_x0))]] = 0 >= 0 = [[0]] [[div(s(_x0), s(_x1))]] = 1 + x0 >= 1 + x0 = [[s(div(minus(_x0, _x1), s(_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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: minus#(s(X), s(Y)) >? minus#(X, Y) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) div(0, s(X)) >= 0 div(s(X), s(Y)) >= s(div(minus(X, Y), s(Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.0 div = \y0y1.2y0 map = \G0y1.0 minus = \y0y1.y0 minus# = \y0y1.y0 nil = 0 s = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[minus#(s(_x0), s(_x1))]] = 2 + 2x0 > x0 = [[minus#(_x0, _x1)]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 2 + 2x0 >= x0 = [[minus(_x0, _x1)]] [[div(0, s(_x0))]] = 0 >= 0 = [[0]] [[div(s(_x0), s(_x1))]] = 4 + 4x0 >= 2 + 4x0 = [[s(div(minus(_x0, _x1), s(_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 [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) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) div(0, s(X)) >= 0 div(s(X), s(Y)) >= s(div(minus(X, Y), s(Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.1 + 2y1 div = \y0y1.0 map = \G0y1.y1 map# = \G0y1.y1 minus = \y0y1.2y0 nil = 0 s = \y0.2y0 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[map#(_F0, _x2)]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[minus(_x0, 0)]] = 2x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 4x0 >= 2x0 = [[minus(_x0, _x1)]] [[div(0, s(_x0))]] = 0 >= 0 = [[0]] [[div(s(_x0), s(_x1))]] = 0 >= 0 = [[s(div(minus(_x0, _x1), s(_x1)))]] 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.