We consider the system Applicative_first_order_05__11. Alphabet: !facminus : [a * a] --> a !facplus : [a * a] --> a !factimes : [a * a] --> a 0 : [] --> a 1 : [] --> a 2 : [] --> a D : [a] --> a cons : [c * d] --> d constant : [] --> a div : [a * a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d ln : [a] --> a map : [c -> c * d] --> d minus : [a] --> a nil : [] --> d pow : [a * a] --> a t : [] --> a true : [] --> b Rules: D(t) => 1 D(constant) => 0 D(!facplus(x, y)) => !facplus(D(x), D(y)) D(!factimes(x, y)) => !facplus(!factimes(y, D(x)), !factimes(x, D(y))) D(!facminus(x, y)) => !facminus(D(x), D(y)) D(minus(x)) => minus(D(x)) D(div(x, y)) => !facminus(div(D(x), y), div(!factimes(x, D(y)), pow(y, 2))) D(ln(x)) => div(D(x), x) D(pow(x, y)) => !facplus(!factimes(!factimes(y, pow(x, !facminus(y, 1))), D(x)), !factimes(!factimes(pow(x, y), ln(x)), D(y))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(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] D#(!facplus(X, Y)) =#> D#(X) 1] D#(!facplus(X, Y)) =#> D#(Y) 2] D#(!factimes(X, Y)) =#> D#(X) 3] D#(!factimes(X, Y)) =#> D#(Y) 4] D#(!facminus(X, Y)) =#> D#(X) 5] D#(!facminus(X, Y)) =#> D#(Y) 6] D#(minus(X)) =#> D#(X) 7] D#(div(X, Y)) =#> D#(X) 8] D#(div(X, Y)) =#> D#(Y) 9] D#(ln(X)) =#> D#(X) 10] D#(pow(X, Y)) =#> D#(X) 11] D#(pow(X, Y)) =#> D#(Y) 12] map#(F, cons(X, Y)) =#> map#(F, Y) 13] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 14] filter2#(true, F, X, Y) =#> filter#(F, Y) 15] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: D(t) => 1 D(constant) => 0 D(!facplus(X, Y)) => !facplus(D(X), D(Y)) D(!factimes(X, Y)) => !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) D(!facminus(X, Y)) => !facminus(D(X), D(Y)) D(minus(X)) => minus(D(X)) D(div(X, Y)) => !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, 2))) D(ln(X)) => div(D(X), X) D(pow(X, Y)) => !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, 1))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(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). 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, 4, 5, 6, 7, 8, 9, 10, 11 * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 4 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 5 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 9 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 10 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 11 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 12 : 12 * 13 : 14, 15 * 14 : 13 * 15 : 13 This graph has the following strongly connected components: P_1: D#(!facplus(X, Y)) =#> D#(X) D#(!facplus(X, Y)) =#> D#(Y) D#(!factimes(X, Y)) =#> D#(X) D#(!factimes(X, Y)) =#> D#(Y) D#(!facminus(X, Y)) =#> D#(X) D#(!facminus(X, Y)) =#> D#(Y) D#(minus(X)) =#> D#(X) D#(div(X, Y)) =#> D#(X) D#(div(X, Y)) =#> D#(Y) D#(ln(X)) =#> D#(X) D#(pow(X, Y)) =#> D#(X) D#(pow(X, Y)) =#> D#(Y) P_2: map#(F, cons(X, Y)) =#> map#(F, Y) P_3: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, 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: filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) D(t) >= 1 D(constant) >= 0 D(!facplus(X, Y)) >= !facplus(D(X), D(Y)) D(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) D(!facminus(X, Y)) >= !facminus(D(X), D(Y)) D(minus(X)) >= minus(D(X)) D(div(X, Y)) >= !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, 2))) D(ln(X)) >= div(D(X), X) D(pow(X, Y)) >= !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, 1))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !facminus = \y0y1.0 !facplus = \y0y1.0 !factimes = \y0y1.3 0 = 0 1 = 0 2 = 0 D = \y0.3 cons = \y0y1.1 + y1 constant = 3 div = \y0y1.0 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.1 + 2y3 filter# = \G0y1.2y1 ln = \y0.3 map = \G0y1.1 + 2y1 minus = \y0.0 nil = 0 pow = \y0y1.3 t = 3 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 2 + 2x2 > 1 + 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] [[D(t)]] = 3 >= 0 = [[1]] [[D(constant)]] = 3 >= 0 = [[0]] [[D(!facplus(_x0, _x1))]] = 3 >= 0 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 3 >= 0 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 3 >= 0 = [[!facminus(D(_x0), D(_x1))]] [[D(minus(_x0))]] = 3 >= 0 = [[minus(D(_x0))]] [[D(div(_x0, _x1))]] = 3 >= 0 = [[!facminus(div(D(_x0), _x1), div(!factimes(_x0, D(_x1)), pow(_x1, 2)))]] [[D(ln(_x0))]] = 3 >= 0 = [[div(D(_x0), _x0)]] [[D(pow(_x0, _x1))]] = 3 >= 0 = [[!facplus(!factimes(!factimes(_x1, pow(_x0, !facminus(_x1, 1))), D(_x0)), !factimes(!factimes(pow(_x0, _x1), ln(_x0)), D(_x1)))]] [[map(_F0, nil)]] = 1 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + 2x2 >= 2 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + 2x2 >= 2x2 = [[filter(_F0, _x2)]] 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: map#(F, cons(X, Y)) >? map#(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + 2y1 map# = \G0y1.y1 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x2 > x2 = [[map#(_F0, _x2)]] 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: D#(!facplus(X, Y)) >? D#(X) D#(!facplus(X, Y)) >? D#(Y) D#(!factimes(X, Y)) >? D#(X) D#(!factimes(X, Y)) >? D#(Y) D#(!facminus(X, Y)) >? D#(X) D#(!facminus(X, Y)) >? D#(Y) D#(minus(X)) >? D#(X) D#(div(X, Y)) >? D#(X) D#(div(X, Y)) >? D#(Y) D#(ln(X)) >? D#(X) D#(pow(X, Y)) >? D#(X) D#(pow(X, Y)) >? D#(Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !facminus = \y0y1.3 + y0 + y1 !facplus = \y0y1.3 + y0 + y1 !factimes = \y0y1.3 + y0 + y1 D# = \y0.y0 div = \y0y1.3 + y0 + y1 ln = \y0.3 + y0 minus = \y0.3 + y0 pow = \y0y1.3 + y0 + y1 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[D#(_x0)]] [[D#(!facplus(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[D#(_x1)]] [[D#(!factimes(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[D#(_x0)]] [[D#(!factimes(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[D#(_x1)]] [[D#(minus(_x0))]] = 3 + x0 > x0 = [[D#(_x0)]] [[D#(div(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[D#(_x0)]] [[D#(div(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[D#(_x1)]] [[D#(ln(_x0))]] = 3 + x0 > x0 = [[D#(_x0)]] [[D#(pow(_x0, _x1))]] = 3 + x0 + x1 > x0 = [[D#(_x0)]] [[D#(pow(_x0, _x1))]] = 3 + x0 + x1 > x1 = [[D#(_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.