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 [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) 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 + 2y1 constant = 3 div = \y0y1.0 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 ln = \y0.3 map = \G0y1.1 + 3y1 map# = \G0y1.y1 minus = \y0.0 nil = 0 pow = \y0y1.3 t = 3 true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[map#(_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))]] = 4 + 6x2 >= 3 + 6x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 1 + 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 >= x2 = [[filter(_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 [Kop12, Thm. 7.16]. 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) 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) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[1]] = _|_ [[2]] = _|_ [[cons(x_1, x_2)]] = _|_ [[filter(x_1, x_2)]] = _|_ [[filter2(x_1, x_2, x_3, x_4)]] = _|_ [[map(x_1, x_2)]] = map(x_1) [[nil]] = _|_ We choose Lex = {} and Mul = {!facminus, !facplus, !factimes, @_{o -> o}, D, D#, constant, div, false, ln, map, minus, pow, t, true}, and the following precedence: @_{o -> o} > D > !facplus > false > ln > D# > !facminus > !factimes > map > constant > pow > minus > div > t > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 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) D(t) >= _|_ D(constant) >= _|_ 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, _|_))) D(ln(X)) >= div(D(X), X) D(pow(X, Y)) >= !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) map(F) >= _|_ map(F) >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ With these choices, we have: 1] D#(!facplus(X, Y)) > D#(X) because [2], by definition 2] D#*(!facplus(X, Y)) >= D#(X) because [3], by (Select) 3] !facplus(X, Y) >= D#(X) because [4], by (Star) 4] !facplus*(X, Y) >= D#(X) because !facplus > D# and [5], by (Copy) 5] !facplus*(X, Y) >= X because [6], by (Select) 6] X >= X by (Meta) 7] D#(!facplus(X, Y)) >= D#(Y) because D# in Mul and [8], by (Fun) 8] !facplus(X, Y) >= Y because [9], by (Star) 9] !facplus*(X, Y) >= Y because [10], by (Select) 10] Y >= Y by (Meta) 11] D#(!factimes(X, Y)) >= D#(X) because D# in Mul and [12], by (Fun) 12] !factimes(X, Y) >= X because [13], by (Star) 13] !factimes*(X, Y) >= X because [14], by (Select) 14] X >= X by (Meta) 15] D#(!factimes(X, Y)) >= D#(Y) because D# in Mul and [16], by (Fun) 16] !factimes(X, Y) >= Y because [17], by (Star) 17] !factimes*(X, Y) >= Y because [18], by (Select) 18] Y >= Y by (Meta) 19] D#(!facminus(X, Y)) >= D#(X) because D# in Mul and [20], by (Fun) 20] !facminus(X, Y) >= X because [21], by (Star) 21] !facminus*(X, Y) >= X because [22], by (Select) 22] X >= X by (Meta) 23] D#(!facminus(X, Y)) >= D#(Y) because D# in Mul and [24], by (Fun) 24] !facminus(X, Y) >= Y because [25], by (Star) 25] !facminus*(X, Y) >= Y because [26], by (Select) 26] Y >= Y by (Meta) 27] D#(minus(X)) >= D#(X) because D# in Mul and [28], by (Fun) 28] minus(X) >= X because [29], by (Star) 29] minus*(X) >= X because [30], by (Select) 30] X >= X by (Meta) 31] D#(div(X, Y)) >= D#(X) because D# in Mul and [32], by (Fun) 32] div(X, Y) >= X because [33], by (Star) 33] div*(X, Y) >= X because [34], by (Select) 34] X >= X by (Meta) 35] D#(div(X, Y)) >= D#(Y) because D# in Mul and [36], by (Fun) 36] div(X, Y) >= Y because [37], by (Star) 37] div*(X, Y) >= Y because [38], by (Select) 38] Y >= Y by (Meta) 39] D#(ln(X)) >= D#(X) because D# in Mul and [40], by (Fun) 40] ln(X) >= X because [41], by (Star) 41] ln*(X) >= X because [42], by (Select) 42] X >= X by (Meta) 43] D#(pow(X, Y)) >= D#(X) because D# in Mul and [44], by (Fun) 44] pow(X, Y) >= X because [45], by (Star) 45] pow*(X, Y) >= X because [46], by (Select) 46] X >= X by (Meta) 47] D#(pow(X, Y)) >= D#(Y) because D# in Mul and [48], by (Fun) 48] pow(X, Y) >= Y because [49], by (Star) 49] pow*(X, Y) >= Y because [50], by (Select) 50] Y >= Y by (Meta) 51] D(t) >= _|_ by (Bot) 52] D(constant) >= _|_ by (Bot) 53] D(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because [54], by (Star) 54] D*(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because D > !facplus, [55] and [57], by (Copy) 55] D*(!facplus(X, Y)) >= D(X) because D in Mul and [56], by (Stat) 56] !facplus(X, Y) > X because [5], by definition 57] D*(!facplus(X, Y)) >= D(Y) because D in Mul and [58], by (Stat) 58] !facplus(X, Y) > Y because [59], by definition 59] !facplus*(X, Y) >= Y because [10], by (Select) 60] D(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because [61], by (Star) 61] D*(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because D > !facplus, [62] and [67], by (Copy) 62] D*(!factimes(X, Y)) >= !factimes(Y, D(X)) because D > !factimes, [63] and [64], by (Copy) 63] D*(!factimes(X, Y)) >= Y because [16], by (Select) 64] D*(!factimes(X, Y)) >= D(X) because D in Mul and [65], by (Stat) 65] !factimes(X, Y) > X because [66], by definition 66] !factimes*(X, Y) >= X because [14], by (Select) 67] D*(!factimes(X, Y)) >= !factimes(X, D(Y)) because D > !factimes, [68] and [69], by (Copy) 68] D*(!factimes(X, Y)) >= X because [12], by (Select) 69] D*(!factimes(X, Y)) >= D(Y) because D in Mul and [70], by (Stat) 70] !factimes(X, Y) > Y because [71], by definition 71] !factimes*(X, Y) >= Y because [18], by (Select) 72] D(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because [73], by (Star) 73] D*(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because D > !facminus, [74] and [77], by (Copy) 74] D*(!facminus(X, Y)) >= D(X) because D in Mul and [75], by (Stat) 75] !facminus(X, Y) > X because [76], by definition 76] !facminus*(X, Y) >= X because [22], by (Select) 77] D*(!facminus(X, Y)) >= D(Y) because D in Mul and [78], by (Stat) 78] !facminus(X, Y) > Y because [79], by definition 79] !facminus*(X, Y) >= Y because [26], by (Select) 80] D(minus(X)) >= minus(D(X)) because [81], by (Star) 81] D*(minus(X)) >= minus(D(X)) because D > minus and [82], by (Copy) 82] D*(minus(X)) >= D(X) because D in Mul and [83], by (Stat) 83] minus(X) > X because [84], by definition 84] minus*(X) >= X because [30], by (Select) 85] D(div(X, Y)) >= !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, _|_))) because [86], by (Star) 86] D*(div(X, Y)) >= !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, _|_))) because D > !facminus, [87] and [92], by (Copy) 87] D*(div(X, Y)) >= div(D(X), Y) because D > div, [88] and [91], by (Copy) 88] D*(div(X, Y)) >= D(X) because D in Mul and [89], by (Stat) 89] div(X, Y) > X because [90], by definition 90] div*(X, Y) >= X because [34], by (Select) 91] D*(div(X, Y)) >= Y because [36], by (Select) 92] D*(div(X, Y)) >= div(!factimes(X, D(Y)), pow(Y, _|_)) because D > div, [93] and [98], by (Copy) 93] D*(div(X, Y)) >= !factimes(X, D(Y)) because D > !factimes, [94] and [95], by (Copy) 94] D*(div(X, Y)) >= X because [32], by (Select) 95] D*(div(X, Y)) >= D(Y) because D in Mul and [96], by (Stat) 96] div(X, Y) > Y because [97], by definition 97] div*(X, Y) >= Y because [38], by (Select) 98] D*(div(X, Y)) >= pow(Y, _|_) because D > pow, [91] and [99], by (Copy) 99] D*(div(X, Y)) >= _|_ by (Bot) 100] D(ln(X)) >= div(D(X), X) because [101], by (Star) 101] D*(ln(X)) >= div(D(X), X) because D > div, [102] and [105], by (Copy) 102] D*(ln(X)) >= D(X) because D in Mul and [103], by (Stat) 103] ln(X) > X because [104], by definition 104] ln*(X) >= X because [42], by (Select) 105] D*(ln(X)) >= X because [40], by (Select) 106] D(pow(X, Y)) >= !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) because [107], by (Star) 107] D*(pow(X, Y)) >= !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) because D > !facplus, [108] and [118], by (Copy) 108] D*(pow(X, Y)) >= !factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)) because D > !factimes, [109] and [115], by (Copy) 109] D*(pow(X, Y)) >= !factimes(Y, pow(X, !facminus(Y, _|_))) because D > !factimes, [110] and [111], by (Copy) 110] D*(pow(X, Y)) >= Y because [48], by (Select) 111] D*(pow(X, Y)) >= pow(X, !facminus(Y, _|_)) because D > pow, [112] and [113], by (Copy) 112] D*(pow(X, Y)) >= X because [44], by (Select) 113] D*(pow(X, Y)) >= !facminus(Y, _|_) because D > !facminus, [110] and [114], by (Copy) 114] D*(pow(X, Y)) >= _|_ by (Bot) 115] D*(pow(X, Y)) >= D(X) because D in Mul and [116], by (Stat) 116] pow(X, Y) > X because [117], by definition 117] pow*(X, Y) >= X because [46], by (Select) 118] D*(pow(X, Y)) >= !factimes(!factimes(pow(X, Y), ln(X)), D(Y)) because D > !factimes, [119] and [122], by (Copy) 119] D*(pow(X, Y)) >= !factimes(pow(X, Y), ln(X)) because D > !factimes, [120] and [121], by (Copy) 120] D*(pow(X, Y)) >= pow(X, Y) because D > pow, [112] and [110], by (Copy) 121] D*(pow(X, Y)) >= ln(X) because D > ln and [112], by (Copy) 122] D*(pow(X, Y)) >= D(Y) because D in Mul and [123], by (Stat) 123] pow(X, Y) > Y because [124], by definition 124] pow*(X, Y) >= Y because [50], by (Select) 125] map(F) >= _|_ by (Bot) 126] map(F) >= _|_ by (Bot) 127] _|_ >= _|_ by (Bot) 128] _|_ >= _|_ by (Bot) 129] _|_ >= _|_ by (Bot) 130] _|_ >= _|_ by (Bot) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of: 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) Thus, the original system is terminating if (P_4, R_0, static, formative) is finite. We consider the dependency pair problem (P_4, 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: 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) 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) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[1]] = _|_ [[2]] = _|_ [[cons(x_1, x_2)]] = x_2 [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4) [[nil]] = _|_ We choose Lex = {} and Mul = {!facminus, !facplus, !factimes, @_{o -> o}, D, D#, constant, div, false, filter, filter2, ln, map, minus, pow, t, true}, and the following precedence: D# > constant > false > t > filter = filter2 > D = minus > pow > !factimes > !facminus > div > @_{o -> o} > !facplus > true > ln > map Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 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) D(t) >= _|_ D(constant) >= _|_ 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, _|_))) D(ln(X)) >= div(D(X), X) D(pow(X, Y)) >= !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) map(F, _|_) >= _|_ map(F, X) >= map(F, X) filter(F, _|_) >= _|_ filter(F, X) >= filter2(F, X) filter2(F, X) >= filter(F, X) filter2(F, X) >= filter(F, X) With these choices, we have: 1] D#(!facplus(X, Y)) >= D#(Y) because D# in Mul and [2], by (Fun) 2] !facplus(X, Y) >= Y because [3], by (Star) 3] !facplus*(X, Y) >= Y because [4], by (Select) 4] Y >= Y by (Meta) 5] D#(!factimes(X, Y)) > D#(X) because [6], by definition 6] D#*(!factimes(X, Y)) >= D#(X) because D# in Mul and [7], by (Stat) 7] !factimes(X, Y) > X because [8], by definition 8] !factimes*(X, Y) >= X because [9], by (Select) 9] X >= X by (Meta) 10] D#(!factimes(X, Y)) >= D#(Y) because D# in Mul and [11], by (Fun) 11] !factimes(X, Y) >= Y because [12], by (Star) 12] !factimes*(X, Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] D#(!facminus(X, Y)) >= D#(X) because D# in Mul and [15], by (Fun) 15] !facminus(X, Y) >= X because [16], by (Star) 16] !facminus*(X, Y) >= X because [17], by (Select) 17] X >= X by (Meta) 18] D#(!facminus(X, Y)) >= D#(Y) because [19], by (Star) 19] D#*(!facminus(X, Y)) >= D#(Y) because D# in Mul and [20], by (Stat) 20] !facminus(X, Y) > Y because [21], by definition 21] !facminus*(X, Y) >= Y because [22], by (Select) 22] Y >= Y by (Meta) 23] D#(minus(X)) >= D#(X) because D# in Mul and [24], by (Fun) 24] minus(X) >= X because [25], by (Star) 25] minus*(X) >= X because [26], by (Select) 26] X >= X by (Meta) 27] D#(div(X, Y)) >= D#(X) because [28], by (Star) 28] D#*(div(X, Y)) >= D#(X) because D# in Mul and [29], by (Stat) 29] div(X, Y) > X because [30], by definition 30] div*(X, Y) >= X because [31], by (Select) 31] X >= X by (Meta) 32] D#(div(X, Y)) >= D#(Y) because D# in Mul and [33], by (Fun) 33] div(X, Y) >= Y because [34], by (Star) 34] div*(X, Y) >= Y because [35], by (Select) 35] Y >= Y by (Meta) 36] D#(ln(X)) >= D#(X) because [37], by (Star) 37] D#*(ln(X)) >= D#(X) because D# in Mul and [38], by (Stat) 38] ln(X) > X because [39], by definition 39] ln*(X) >= X because [40], by (Select) 40] X >= X by (Meta) 41] D#(pow(X, Y)) >= D#(X) because D# in Mul and [42], by (Fun) 42] pow(X, Y) >= X because [43], by (Star) 43] pow*(X, Y) >= X because [44], by (Select) 44] X >= X by (Meta) 45] D#(pow(X, Y)) >= D#(Y) because D# in Mul and [46], by (Fun) 46] pow(X, Y) >= Y because [47], by (Star) 47] pow*(X, Y) >= Y because [48], by (Select) 48] Y >= Y by (Meta) 49] D(t) >= _|_ by (Bot) 50] D(constant) >= _|_ by (Bot) 51] D(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because [52], by (Star) 52] D*(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because D > !facplus, [53] and [57], by (Copy) 53] D*(!facplus(X, Y)) >= D(X) because D in Mul and [54], by (Stat) 54] !facplus(X, Y) > X because [55], by definition 55] !facplus*(X, Y) >= X because [56], by (Select) 56] X >= X by (Meta) 57] D*(!facplus(X, Y)) >= D(Y) because D in Mul and [58], by (Stat) 58] !facplus(X, Y) > Y because [59], by definition 59] !facplus*(X, Y) >= Y because [4], by (Select) 60] D(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because [61], by (Star) 61] D*(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because D > !facplus, [62] and [65], by (Copy) 62] D*(!factimes(X, Y)) >= !factimes(Y, D(X)) because D > !factimes, [63] and [64], by (Copy) 63] D*(!factimes(X, Y)) >= Y because [11], by (Select) 64] D*(!factimes(X, Y)) >= D(X) because D in Mul and [7], by (Stat) 65] D*(!factimes(X, Y)) >= !factimes(X, D(Y)) because D > !factimes, [66] and [68], by (Copy) 66] D*(!factimes(X, Y)) >= X because [67], by (Select) 67] !factimes(X, Y) >= X because [8], by (Star) 68] D*(!factimes(X, Y)) >= D(Y) because D in Mul and [69], by (Stat) 69] !factimes(X, Y) > Y because [70], by definition 70] !factimes*(X, Y) >= Y because [13], by (Select) 71] D(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because [72], by (Star) 72] D*(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because D > !facminus, [73] and [76], by (Copy) 73] D*(!facminus(X, Y)) >= D(X) because D in Mul and [74], by (Stat) 74] !facminus(X, Y) > X because [75], by definition 75] !facminus*(X, Y) >= X because [17], by (Select) 76] D*(!facminus(X, Y)) >= D(Y) because D in Mul and [20], by (Stat) 77] D(minus(X)) >= minus(D(X)) because D = minus, D in Mul and [78], by (Fun) 78] minus(X) >= D(X) because minus = D, minus in Mul and [79], by (Fun) 79] X >= X by (Meta) 80] D(div(X, Y)) >= !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, _|_))) because [81], by (Star) 81] D*(div(X, Y)) >= !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, _|_))) because D > !facminus, [82] and [85], by (Copy) 82] D*(div(X, Y)) >= div(D(X), Y) because D > div, [83] and [84], by (Copy) 83] D*(div(X, Y)) >= D(X) because D in Mul and [29], by (Stat) 84] D*(div(X, Y)) >= Y because [33], by (Select) 85] D*(div(X, Y)) >= div(!factimes(X, D(Y)), pow(Y, _|_)) because D > div, [86] and [92], by (Copy) 86] D*(div(X, Y)) >= !factimes(X, D(Y)) because D > !factimes, [87] and [89], by (Copy) 87] D*(div(X, Y)) >= X because [88], by (Select) 88] div(X, Y) >= X because [30], by (Star) 89] D*(div(X, Y)) >= D(Y) because D in Mul and [90], by (Stat) 90] div(X, Y) > Y because [91], by definition 91] div*(X, Y) >= Y because [35], by (Select) 92] D*(div(X, Y)) >= pow(Y, _|_) because D > pow, [84] and [93], by (Copy) 93] D*(div(X, Y)) >= _|_ by (Bot) 94] D(ln(X)) >= div(D(X), X) because [95], by (Star) 95] D*(ln(X)) >= div(D(X), X) because D > div, [96] and [97], by (Copy) 96] D*(ln(X)) >= D(X) because D in Mul and [38], by (Stat) 97] D*(ln(X)) >= X because [98], by (Select) 98] ln(X) >= X because [39], by (Star) 99] D(pow(X, Y)) >= !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) because [100], by (Star) 100] D*(pow(X, Y)) >= !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) because D > !facplus, [101] and [111], by (Copy) 101] D*(pow(X, Y)) >= !factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)) because D > !factimes, [102] and [108], by (Copy) 102] D*(pow(X, Y)) >= !factimes(Y, pow(X, !facminus(Y, _|_))) because D > !factimes, [103] and [104], by (Copy) 103] D*(pow(X, Y)) >= Y because [46], by (Select) 104] D*(pow(X, Y)) >= pow(X, !facminus(Y, _|_)) because D > pow, [105] and [106], by (Copy) 105] D*(pow(X, Y)) >= X because [42], by (Select) 106] D*(pow(X, Y)) >= !facminus(Y, _|_) because D > !facminus, [103] and [107], by (Copy) 107] D*(pow(X, Y)) >= _|_ by (Bot) 108] D*(pow(X, Y)) >= D(X) because D in Mul and [109], by (Stat) 109] pow(X, Y) > X because [110], by definition 110] pow*(X, Y) >= X because [44], by (Select) 111] D*(pow(X, Y)) >= !factimes(!factimes(pow(X, Y), ln(X)), D(Y)) because D > !factimes, [112] and [115], by (Copy) 112] D*(pow(X, Y)) >= !factimes(pow(X, Y), ln(X)) because D > !factimes, [113] and [114], by (Copy) 113] D*(pow(X, Y)) >= pow(X, Y) because D > pow, [105] and [103], by (Copy) 114] D*(pow(X, Y)) >= ln(X) because D > ln and [105], by (Copy) 115] D*(pow(X, Y)) >= D(Y) because D in Mul and [116], by (Stat) 116] pow(X, Y) > Y because [117], by definition 117] pow*(X, Y) >= Y because [48], by (Select) 118] map(F, _|_) >= _|_ by (Bot) 119] map(F, X) >= map(F, X) because map in Mul, [120] and [121], by (Fun) 120] F >= F by (Meta) 121] X >= X by (Meta) 122] filter(F, _|_) >= _|_ by (Bot) 123] filter(F, X) >= filter2(F, X) because filter = filter2, filter in Mul, [124] and [125], by (Fun) 124] F >= F by (Meta) 125] X >= X by (Meta) 126] filter2(F, X) >= filter(F, X) because filter2 = filter, filter2 in Mul, [127] and [128], by (Fun) 127] F >= F by (Meta) 128] X >= X by (Meta) 129] filter2(F, X) >= filter(F, X) because filter2 = filter, filter2 in Mul, [130] and [131], by (Fun) 130] F >= F by (Meta) 131] X >= X by (Meta) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_0, static, formative) by (P_5, R_0, static, formative), where P_5 consists of: D#(!facplus(X, Y)) =#> D#(Y) 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) Thus, the original system is terminating if (P_5, R_0, static, formative) is finite. We consider the dependency pair problem (P_5, 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: D#(!facplus(X, Y)) >? D#(Y) 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) 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.y0 + y1 !facplus = \y0y1.y1 !factimes = \y0y1.y1 0 = 0 1 = 0 2 = 0 D = \y0.3y0 D# = \y0.2y0 cons = \y0y1.0 constant = 3 div = \y0y1.y0 + 3y1 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 ln = \y0.1 + 3y0 map = \G0y1.0 minus = \y0.y0 nil = 0 pow = \y0y1.y0 + y1 t = 3 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 2x1 >= 2x1 = [[D#(_x1)]] [[D#(!factimes(_x0, _x1))]] = 2x1 >= 2x1 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[D#(_x1)]] [[D#(minus(_x0))]] = 2x0 >= 2x0 = [[D#(_x0)]] [[D#(div(_x0, _x1))]] = 2x0 + 6x1 >= 2x0 = [[D#(_x0)]] [[D#(div(_x0, _x1))]] = 2x0 + 6x1 >= 2x1 = [[D#(_x1)]] [[D#(ln(_x0))]] = 2 + 6x0 > 2x0 = [[D#(_x0)]] [[D#(pow(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 = [[D#(_x0)]] [[D#(pow(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[D#(_x1)]] [[D(t)]] = 9 >= 0 = [[1]] [[D(constant)]] = 9 >= 0 = [[0]] [[D(!facplus(_x0, _x1))]] = 3x1 >= 3x1 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 3x1 >= 3x1 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[!facminus(D(_x0), D(_x1))]] [[D(minus(_x0))]] = 3x0 >= 3x0 = [[minus(D(_x0))]] [[D(div(_x0, _x1))]] = 3x0 + 9x1 >= 3x0 + 9x1 = [[!facminus(div(D(_x0), _x1), div(!factimes(_x0, D(_x1)), pow(_x1, 2)))]] [[D(ln(_x0))]] = 3 + 9x0 >= 6x0 = [[div(D(_x0), _x0)]] [[D(pow(_x0, _x1))]] = 3x0 + 3x1 >= 3x1 = [[!facplus(!factimes(!factimes(_x1, pow(_x0, !facminus(_x1, 1))), D(_x0)), !factimes(!factimes(pow(_x0, _x1), ln(_x0)), D(_x1)))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, static, formative) by (P_6, R_0, static, formative), where P_6 consists of: D#(!facplus(X, Y)) =#> D#(Y) 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#(pow(X, Y)) =#> D#(X) D#(pow(X, Y)) =#> D#(Y) Thus, the original system is terminating if (P_6, R_0, static, formative) is finite. We consider the dependency pair problem (P_6, 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: D#(!facplus(X, Y)) >? D#(Y) 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#(pow(X, Y)) >? D#(X) D#(pow(X, Y)) >? D#(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.y0 + y1 !facplus = \y0y1.y1 !factimes = \y0y1.y1 0 = 0 1 = 0 2 = 0 D = \y0.3y0 D# = \y0.2y0 cons = \y0y1.0 constant = 3 div = \y0y1.3 + y0 + 3y1 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 ln = \y0.1 + 2y0 map = \G0y1.0 minus = \y0.y0 nil = 0 pow = \y0y1.1 + y0 + y1 t = 3 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 2x1 >= 2x1 = [[D#(_x1)]] [[D#(!factimes(_x0, _x1))]] = 2x1 >= 2x1 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[D#(_x1)]] [[D#(minus(_x0))]] = 2x0 >= 2x0 = [[D#(_x0)]] [[D#(div(_x0, _x1))]] = 6 + 2x0 + 6x1 > 2x0 = [[D#(_x0)]] [[D#(div(_x0, _x1))]] = 6 + 2x0 + 6x1 > 2x1 = [[D#(_x1)]] [[D#(pow(_x0, _x1))]] = 2 + 2x0 + 2x1 > 2x0 = [[D#(_x0)]] [[D#(pow(_x0, _x1))]] = 2 + 2x0 + 2x1 > 2x1 = [[D#(_x1)]] [[D(t)]] = 9 >= 0 = [[1]] [[D(constant)]] = 9 >= 0 = [[0]] [[D(!facplus(_x0, _x1))]] = 3x1 >= 3x1 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 3x1 >= 3x1 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 3x0 + 3x1 >= 3x0 + 3x1 = [[!facminus(D(_x0), D(_x1))]] [[D(minus(_x0))]] = 3x0 >= 3x0 = [[minus(D(_x0))]] [[D(div(_x0, _x1))]] = 9 + 3x0 + 9x1 >= 9 + 3x0 + 9x1 = [[!facminus(div(D(_x0), _x1), div(!factimes(_x0, D(_x1)), pow(_x1, 2)))]] [[D(ln(_x0))]] = 3 + 6x0 >= 3 + 6x0 = [[div(D(_x0), _x0)]] [[D(pow(_x0, _x1))]] = 3 + 3x0 + 3x1 >= 3x1 = [[!facplus(!factimes(!factimes(_x1, pow(_x0, !facminus(_x1, 1))), D(_x0)), !factimes(!factimes(pow(_x0, _x1), ln(_x0)), D(_x1)))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_0, static, formative) by (P_7, R_0, static, formative), where P_7 consists of: D#(!facplus(X, Y)) =#> D#(Y) D#(!factimes(X, Y)) =#> D#(Y) D#(!facminus(X, Y)) =#> D#(X) D#(!facminus(X, Y)) =#> D#(Y) D#(minus(X)) =#> D#(X) Thus, the original system is terminating if (P_7, R_0, static, formative) is finite. We consider the dependency pair problem (P_7, 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: D#(!facplus(X, Y)) >? D#(Y) D#(!factimes(X, Y)) >? D#(Y) D#(!facminus(X, Y)) >? D#(X) D#(!facminus(X, Y)) >? D#(Y) D#(minus(X)) >? D#(X) 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.y0 + y1 !facplus = \y0y1.y1 !factimes = \y0y1.1 + y1 0 = 0 1 = 0 2 = 0 D = \y0.2y0 D# = \y0.2y0 cons = \y0y1.0 constant = 3 div = \y0y1.0 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 ln = \y0.1 map = \G0y1.0 minus = \y0.y0 nil = 0 pow = \y0y1.1 + y1 t = 3 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 2x1 >= 2x1 = [[D#(_x1)]] [[D#(!factimes(_x0, _x1))]] = 2 + 2x1 > 2x1 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 2x0 + 2x1 >= 2x1 = [[D#(_x1)]] [[D#(minus(_x0))]] = 2x0 >= 2x0 = [[D#(_x0)]] [[D(t)]] = 6 >= 0 = [[1]] [[D(constant)]] = 6 >= 0 = [[0]] [[D(!facplus(_x0, _x1))]] = 2x1 >= 2x1 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 2 + 2x1 >= 1 + 2x1 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!facminus(D(_x0), D(_x1))]] [[D(minus(_x0))]] = 2x0 >= 2x0 = [[minus(D(_x0))]] [[D(div(_x0, _x1))]] = 0 >= 0 = [[!facminus(div(D(_x0), _x1), div(!factimes(_x0, D(_x1)), pow(_x1, 2)))]] [[D(ln(_x0))]] = 2 >= 0 = [[div(D(_x0), _x0)]] [[D(pow(_x0, _x1))]] = 2 + 2x1 >= 1 + 2x1 = [[!facplus(!factimes(!factimes(_x1, pow(_x0, !facminus(_x1, 1))), D(_x0)), !factimes(!factimes(pow(_x0, _x1), ln(_x0)), D(_x1)))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_7, R_0, static, formative) by (P_8, R_0, static, formative), where P_8 consists of: D#(!facplus(X, Y)) =#> D#(Y) D#(!facminus(X, Y)) =#> D#(X) D#(!facminus(X, Y)) =#> D#(Y) D#(minus(X)) =#> D#(X) Thus, the original system is terminating if (P_8, R_0, static, formative) is finite. We consider the dependency pair problem (P_8, 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: D#(!facplus(X, Y)) >? D#(Y) D#(!facminus(X, Y)) >? D#(X) D#(!facminus(X, Y)) >? D#(Y) D#(minus(X)) >? D#(X) 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 = \y0y1.1 + y0 + y1 !facplus = \y0y1.y1 !factimes = \y0y1.0 0 = 0 1 = 0 2 = 0 D = \y0.1 + 2y0 D# = \y0.y0 cons = \y0y1.0 constant = 3 div = \y0y1.0 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 ln = \y0.0 map = \G0y1.2 minus = \y0.1 + 2y0 nil = 0 pow = \y0y1.0 t = 3 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = x1 >= x1 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 1 + x0 + x1 > x0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 1 + x0 + x1 > x1 = [[D#(_x1)]] [[D#(minus(_x0))]] = 1 + 2x0 > x0 = [[D#(_x0)]] [[D(t)]] = 7 >= 0 = [[1]] [[D(constant)]] = 7 >= 0 = [[0]] [[D(!facplus(_x0, _x1))]] = 1 + 2x1 >= 1 + 2x1 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 1 >= 0 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 3 + 2x0 + 2x1 >= 3 + 2x0 + 2x1 = [[!facminus(D(_x0), D(_x1))]] [[D(minus(_x0))]] = 3 + 4x0 >= 3 + 4x0 = [[minus(D(_x0))]] [[D(div(_x0, _x1))]] = 1 >= 1 = [[!facminus(div(D(_x0), _x1), div(!factimes(_x0, D(_x1)), pow(_x1, 2)))]] [[D(ln(_x0))]] = 1 >= 0 = [[div(D(_x0), _x0)]] [[D(pow(_x0, _x1))]] = 1 >= 0 = [[!facplus(!factimes(!factimes(_x1, pow(_x0, !facminus(_x1, 1))), D(_x0)), !factimes(!factimes(pow(_x0, _x1), ln(_x0)), D(_x1)))]] [[map(_F0, nil)]] = 2 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 2 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_8, R_0, static, formative) by (P_9, R_0, static, formative), where P_9 consists of: D#(!facplus(X, Y)) =#> D#(Y) Thus, the original system is terminating if (P_9, R_0, static, formative) is finite. We consider the dependency pair problem (P_9, 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: D#(!facplus(X, Y)) >? D#(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 = \y0y1.1 + 2y1 !factimes = \y0y1.0 0 = 0 1 = 0 2 = 0 D = \y0.1 + 2y0 D# = \y0.y0 cons = \y0y1.0 constant = 3 div = \y0y1.0 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 ln = \y0.0 map = \G0y1.0 minus = \y0.0 nil = 0 pow = \y0y1.0 t = 3 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 1 + 2x1 > x1 = [[D#(_x1)]] [[D(t)]] = 7 >= 0 = [[1]] [[D(constant)]] = 7 >= 0 = [[0]] [[D(!facplus(_x0, _x1))]] = 3 + 4x1 >= 3 + 4x1 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 1 >= 1 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 1 >= 0 = [[!facminus(D(_x0), D(_x1))]] [[D(minus(_x0))]] = 1 >= 0 = [[minus(D(_x0))]] [[D(div(_x0, _x1))]] = 1 >= 0 = [[!facminus(div(D(_x0), _x1), div(!factimes(_x0, D(_x1)), pow(_x1, 2)))]] [[D(ln(_x0))]] = 1 >= 0 = [[div(D(_x0), _x0)]] [[D(pow(_x0, _x1))]] = 1 >= 1 = [[!facplus(!factimes(!factimes(_x1, pow(_x0, !facminus(_x1, 1))), D(_x0)), !factimes(!factimes(pow(_x0, _x1), ln(_x0)), D(_x1)))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_9, 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.