We consider the system Applicative_first_order_05__#3.16. Alphabet: 0 : [] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d plus : [a * a] --> a s : [a] --> a times : [a * a] --> a true : [] --> b Rules: times(x, 0) => 0 times(x, s(y)) => plus(times(x, y), x) plus(x, 0) => x plus(0, x) => x plus(x, s(y)) => s(plus(x, y)) plus(s(x), y) => s(plus(x, 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] times#(X, s(Y)) =#> plus#(times(X, Y), X) 1] times#(X, s(Y)) =#> times#(X, Y) 2] plus#(X, s(Y)) =#> plus#(X, Y) 3] plus#(s(X), Y) =#> plus#(X, Y) 4] map#(F, cons(X, Y)) =#> map#(F, Y) 5] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 6] filter2#(true, F, X, Y) =#> filter#(F, Y) 7] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(0, X) => X plus(X, s(Y)) => s(plus(X, Y)) plus(s(X), Y) => s(plus(X, 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 : 2, 3 * 1 : 0, 1 * 2 : 2, 3 * 3 : 2, 3 * 4 : 4 * 5 : 6, 7 * 6 : 5 * 7 : 5 This graph has the following strongly connected components: P_1: times#(X, s(Y)) =#> times#(X, Y) P_2: plus#(X, s(Y)) =#> plus#(X, Y) plus#(s(X), Y) =#> plus#(X, Y) P_3: map#(F, cons(X, Y)) =#> map#(F, Y) P_4: 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), (P_3, R_0, m, f) and (P_4, 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), (P_3, R_0, static, formative) and (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: 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) times(X, 0) >= 0 times(X, s(Y)) >= plus(times(X, Y), X) plus(X, 0) >= X plus(0, X) >= X plus(X, s(Y)) >= s(plus(X, Y)) plus(s(X), Y) >= s(plus(X, 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]] = _|_ [[filter(x_1, x_2)]] = filter(x_2, x_1) [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_3, x_1) [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_4, x_2, x_3, x_1) [[filter#(x_1, x_2)]] = filter#(x_2, x_1) [[nil]] = _|_ We choose Lex = {filter, filter2, filter2#, filter#} and Mul = {@_{o -> o}, cons, false, map, plus, s, times, true}, and the following precedence: filter2# = filter# > filter = filter2 > map > cons > @_{o -> o} > true > false > times > plus > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: filter#(F, cons(X, Y)) >= filter2#(@_{o -> o}(F, X), F, X, Y) filter2#(true, F, X, Y) >= filter#(F, Y) filter2#(false, F, X, Y) > filter#(F, Y) times(X, _|_) >= _|_ times(X, s(Y)) >= plus(times(X, Y), X) plus(X, _|_) >= X plus(_|_, X) >= X plus(X, s(Y)) >= s(plus(X, Y)) plus(s(X), Y) >= s(plus(X, Y)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) With these choices, we have: 1] filter#(F, cons(X, Y)) >= filter2#(@_{o -> o}(F, X), F, X, Y) because [2], by (Star) 2] filter#*(F, cons(X, Y)) >= filter2#(@_{o -> o}(F, X), F, X, Y) because filter# = filter2#, [3], [6], [7], [9] and [13], by (Stat) 3] cons(X, Y) > Y because [4], by definition 4] cons*(X, Y) >= Y because [5], by (Select) 5] Y >= Y by (Meta) 6] filter#*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter# > @_{o -> o}, [7] and [9], by (Copy) 7] filter#*(F, cons(X, Y)) >= F because [8], by (Select) 8] F >= F by (Meta) 9] filter#*(F, cons(X, Y)) >= X because [10], by (Select) 10] cons(X, Y) >= X because [11], by (Star) 11] cons*(X, Y) >= X because [12], by (Select) 12] X >= X by (Meta) 13] filter#*(F, cons(X, Y)) >= Y because [14], by (Select) 14] cons(X, Y) >= Y because [4], by (Star) 15] filter2#(true, F, X, Y) >= filter#(F, Y) because [16], by (Star) 16] filter2#*(true, F, X, Y) >= filter#(F, Y) because filter2# = filter#, [17], [18], [19] and [20], by (Stat) 17] F >= F by (Meta) 18] Y >= Y by (Meta) 19] filter2#*(true, F, X, Y) >= F because [17], by (Select) 20] filter2#*(true, F, X, Y) >= Y because [18], by (Select) 21] filter2#(false, F, X, Y) > filter#(F, Y) because [22], by definition 22] filter2#*(false, F, X, Y) >= filter#(F, Y) because filter2# = filter#, [23], [24], [25] and [26], by (Stat) 23] F >= F by (Meta) 24] Y >= Y by (Meta) 25] filter2#*(false, F, X, Y) >= F because [23], by (Select) 26] filter2#*(false, F, X, Y) >= Y because [24], by (Select) 27] times(X, _|_) >= _|_ by (Bot) 28] times(X, s(Y)) >= plus(times(X, Y), X) because [29], by (Star) 29] times*(X, s(Y)) >= plus(times(X, Y), X) because times > plus, [30] and [35], by (Copy) 30] times*(X, s(Y)) >= times(X, Y) because times in Mul, [31] and [32], by (Stat) 31] X >= X by (Meta) 32] s(Y) > Y because [33], by definition 33] s*(Y) >= Y because [34], by (Select) 34] Y >= Y by (Meta) 35] times*(X, s(Y)) >= X because [31], by (Select) 36] plus(X, _|_) >= X because [37], by (Star) 37] plus*(X, _|_) >= X because [38], by (Select) 38] X >= X by (Meta) 39] plus(_|_, X) >= X because [40], by (Star) 40] plus*(_|_, X) >= X because [41], by (Select) 41] X >= X by (Meta) 42] plus(X, s(Y)) >= s(plus(X, Y)) because [43], by (Star) 43] plus*(X, s(Y)) >= s(plus(X, Y)) because plus > s and [44], by (Copy) 44] plus*(X, s(Y)) >= plus(X, Y) because plus in Mul, [45] and [46], by (Stat) 45] X >= X by (Meta) 46] s(Y) > Y because [47], by definition 47] s*(Y) >= Y because [48], by (Select) 48] Y >= Y by (Meta) 49] plus(s(X), Y) >= s(plus(X, Y)) because [50], by (Star) 50] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [51], by (Copy) 51] plus*(s(X), Y) >= plus(X, Y) because plus in Mul, [52] and [55], by (Stat) 52] s(X) > X because [53], by definition 53] s*(X) >= X because [54], by (Select) 54] X >= X by (Meta) 55] Y >= Y by (Meta) 56] map(F, _|_) >= _|_ by (Bot) 57] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [58], by (Star) 58] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [59] and [66], by (Copy) 59] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [60] and [62], by (Copy) 60] map*(F, cons(X, Y)) >= F because [61], by (Select) 61] F >= F by (Meta) 62] map*(F, cons(X, Y)) >= X because [63], by (Select) 63] cons(X, Y) >= X because [64], by (Star) 64] cons*(X, Y) >= X because [65], by (Select) 65] X >= X by (Meta) 66] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [67] and [68], by (Stat) 67] F >= F by (Meta) 68] cons(X, Y) > Y because [69], by definition 69] cons*(X, Y) >= Y because [70], by (Select) 70] Y >= Y by (Meta) 71] filter(F, _|_) >= _|_ by (Bot) 72] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [73], by (Star) 73] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [3], [74], [75], [76] and [77], by (Stat) 74] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [75] and [76], by (Copy) 75] filter*(F, cons(X, Y)) >= F because [8], by (Select) 76] filter*(F, cons(X, Y)) >= X because [10], by (Select) 77] filter*(F, cons(X, Y)) >= Y because [14], by (Select) 78] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [79], by (Star) 79] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [80] and [82], by (Copy) 80] filter2*(true, F, X, Y) >= X because [81], by (Select) 81] X >= X by (Meta) 82] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [17], [18], [83] and [84], by (Stat) 83] filter2*(true, F, X, Y) >= F because [17], by (Select) 84] filter2*(true, F, X, Y) >= Y because [18], by (Select) 85] filter2(false, F, X, Y) >= filter(F, Y) because [86], by (Star) 86] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [23], [24], [87] and [88], by (Stat) 87] filter2*(false, F, X, Y) >= F because [23], by (Select) 88] filter2*(false, F, X, Y) >= Y because [24], by (Select) 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: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative), (P_3, R_0, static, formative) and (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: filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter2#(true, F, X, Y) >? filter#(F, Y) times(X, 0) >= 0 times(X, s(Y)) >= plus(times(X, Y), X) plus(X, 0) >= X plus(0, X) >= X plus(X, s(Y)) >= s(plus(X, Y)) plus(s(X), Y) >= s(plus(X, 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]] = _|_ [[@_{o -> o}(x_1, x_2)]] = @_{o -> o}(x_2, x_1) [[filter(x_1, x_2)]] = filter(x_2, x_1) [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_3, x_1) [[filter2#(x_1, x_2, x_3, x_4)]] = filter2#(x_3, x_4) [[filter#(x_1, x_2)]] = x_2 [[nil]] = _|_ We choose Lex = {@_{o -> o}, filter, filter2} and Mul = {cons, false, filter2#, map, plus, s, times, true}, and the following precedence: map > false > @_{o -> o} = filter = filter2 > cons > times > plus > filter2# > s > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: cons(X, Y) >= filter2#(X, Y) filter2#(X, Y) > Y times(X, _|_) >= _|_ times(X, s(Y)) >= plus(times(X, Y), X) plus(X, _|_) >= X plus(_|_, X) >= X plus(X, s(Y)) >= s(plus(X, Y)) plus(s(X), Y) >= s(plus(X, Y)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) With these choices, we have: 1] cons(X, Y) >= filter2#(X, Y) because [2], by (Star) 2] cons*(X, Y) >= filter2#(X, Y) because cons > filter2#, [3] and [5], by (Copy) 3] cons*(X, Y) >= X because [4], by (Select) 4] X >= X by (Meta) 5] cons*(X, Y) >= Y because [6], by (Select) 6] Y >= Y by (Meta) 7] filter2#(X, Y) > Y because [8], by definition 8] filter2#*(X, Y) >= Y because [9], by (Select) 9] Y >= Y by (Meta) 10] times(X, _|_) >= _|_ by (Bot) 11] times(X, s(Y)) >= plus(times(X, Y), X) because [12], by (Star) 12] times*(X, s(Y)) >= plus(times(X, Y), X) because times > plus, [13] and [18], by (Copy) 13] times*(X, s(Y)) >= times(X, Y) because times in Mul, [14] and [15], by (Stat) 14] X >= X by (Meta) 15] s(Y) > Y because [16], by definition 16] s*(Y) >= Y because [17], by (Select) 17] Y >= Y by (Meta) 18] times*(X, s(Y)) >= X because [14], by (Select) 19] plus(X, _|_) >= X because [20], by (Star) 20] plus*(X, _|_) >= X because [21], by (Select) 21] X >= X by (Meta) 22] plus(_|_, X) >= X because [23], by (Star) 23] plus*(_|_, X) >= X because [24], by (Select) 24] X >= X by (Meta) 25] plus(X, s(Y)) >= s(plus(X, Y)) because [26], by (Star) 26] plus*(X, s(Y)) >= s(plus(X, Y)) because plus > s and [27], by (Copy) 27] plus*(X, s(Y)) >= plus(X, Y) because plus in Mul, [28] and [29], by (Stat) 28] X >= X by (Meta) 29] s(Y) > Y because [30], by definition 30] s*(Y) >= Y because [31], by (Select) 31] Y >= Y by (Meta) 32] plus(s(X), Y) >= s(plus(X, Y)) because [33], by (Star) 33] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [34], by (Copy) 34] plus*(s(X), Y) >= plus(X, Y) because plus in Mul, [35] and [38], by (Stat) 35] s(X) > X because [36], by definition 36] s*(X) >= X because [37], by (Select) 37] X >= X by (Meta) 38] Y >= Y by (Meta) 39] map(F, _|_) >= _|_ by (Bot) 40] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [41], by (Star) 41] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [42] and [49], by (Copy) 42] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [43] and [45], by (Copy) 43] map*(F, cons(X, Y)) >= F because [44], by (Select) 44] F >= F by (Meta) 45] map*(F, cons(X, Y)) >= X because [46], by (Select) 46] cons(X, Y) >= X because [47], by (Star) 47] cons*(X, Y) >= X because [48], by (Select) 48] X >= X by (Meta) 49] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [50] and [51], by (Stat) 50] F >= F by (Meta) 51] cons(X, Y) > Y because [52], by definition 52] cons*(X, Y) >= Y because [53], by (Select) 53] Y >= Y by (Meta) 54] filter(F, _|_) >= _|_ by (Bot) 55] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [56], by (Star) 56] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [57], [58], [61], [63] and [65], by (Stat) 57] cons(X, Y) > Y because [5], by definition 58] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter = @_{o -> o}, [59], [61] and [63], by (Stat) 59] cons(X, Y) > X because [60], by definition 60] cons*(X, Y) >= X because [4], by (Select) 61] filter*(F, cons(X, Y)) >= F because [62], by (Select) 62] F >= F by (Meta) 63] filter*(F, cons(X, Y)) >= X because [64], by (Select) 64] cons(X, Y) >= X because [60], by (Star) 65] filter*(F, cons(X, Y)) >= Y because [66], by (Select) 66] cons(X, Y) >= Y because [5], by (Star) 67] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [68], by (Star) 68] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [69] and [71], by (Copy) 69] filter2*(true, F, X, Y) >= X because [70], by (Select) 70] X >= X by (Meta) 71] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [72], [73], [74] and [75], by (Stat) 72] F >= F by (Meta) 73] Y >= Y by (Meta) 74] filter2*(true, F, X, Y) >= F because [72], by (Select) 75] filter2*(true, F, X, Y) >= Y because [73], by (Select) 76] filter2(false, F, X, Y) >= filter(F, Y) because [77], by (Star) 77] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [78], [79], [80] and [81], by (Stat) 78] F >= F by (Meta) 79] Y >= Y by (Meta) 80] filter2*(false, F, X, Y) >= F because [78], by (Select) 81] filter2*(false, F, X, Y) >= Y because [79], by (Select) 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: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative), (P_3, R_0, static, formative) and (P_6, R_0, static, formative) is finite. We consider the dependency pair problem (P_6, 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 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 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: 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_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#(X, s(Y)) >? plus#(X, Y) 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 + y1 s = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[plus#(_x0, s(_x1))]] = 3 + x0 + 2x1 > x0 + x1 = [[plus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 3 + x1 + 2x0 > x0 + x1 = [[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: times#(X, s(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.y1 Using this interpretation, the requirements translate to: [[times#(_x0, s(_x1))]] = 1 + 2x1 > x1 = [[times#(_x0, _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.