We consider the system Applicative_first_order_05__29. Alphabet: 0 : [] --> a ack : [a * a] --> 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 succ : [a] --> a true : [] --> b Rules: ack(0, x) => succ(x) ack(succ(x), y) => ack(x, succ(0)) ack(succ(x), succ(y)) => ack(x, ack(succ(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] ack#(succ(X), Y) =#> ack#(X, succ(0)) 1] ack#(succ(X), succ(Y)) =#> ack#(X, ack(succ(X), Y)) 2] ack#(succ(X), succ(Y)) =#> ack#(succ(X), Y) 3] map#(F, cons(X, Y)) =#> map#(F, Y) 4] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 5] filter2#(true, F, X, Y) =#> filter#(F, Y) 6] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: ack(0, X) => succ(X) ack(succ(X), Y) => ack(X, succ(0)) ack(succ(X), succ(Y)) => ack(X, ack(succ(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 : 0, 1, 2 * 1 : 0, 1, 2 * 2 : 0, 1, 2 * 3 : 3 * 4 : 5, 6 * 5 : 4 * 6 : 4 This graph has the following strongly connected components: P_1: ack#(succ(X), Y) =#> ack#(X, succ(0)) ack#(succ(X), succ(Y)) =#> ack#(X, ack(succ(X), Y)) ack#(succ(X), succ(Y)) =#> ack#(succ(X), 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) ack(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 ack = \y0y1.0 cons = \y0y1.1 + 2y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.1 + y3 filter# = \G0y1.y1 map = \G0y1.y1 nil = 0 succ = \y0.0 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] [[ack(0, _x0)]] = 0 >= 0 = [[succ(_x0)]] [[ack(succ(_x0), _x1)]] = 0 >= 0 = [[ack(_x0, succ(0))]] [[ack(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack(_x0, ack(succ(_x0), _x1))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[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 the dependency pair problem (P_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 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) and (P_4, R_0, static, formative) is finite. We consider the dependency pair problem (P_4, 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) 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) ack(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 ack = \y0y1.0 cons = \y0y1.1 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y3 map = \G0y1.2 + y1 map# = \G0y1.y1 nil = 1 succ = \y0.0 true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] [[ack(0, _x0)]] = 0 >= 0 = [[succ(_x0)]] [[ack(succ(_x0), _x1)]] = 0 >= 0 = [[ack(_x0, succ(0))]] [[ack(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack(_x0, ack(succ(_x0), _x1))]] [[map(_F0, nil)]] = 3 >= 1 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 2 >= 1 = [[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_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: ack#(succ(X), Y) >? ack#(X, succ(0)) ack#(succ(X), succ(Y)) >? ack#(X, ack(succ(X), Y)) ack#(succ(X), succ(Y)) >? ack#(succ(X), Y) ack(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(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: [[ack#(x_1, x_2)]] = ack#(x_1) [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) [[nil]] = _|_ We choose Lex = {ack, filter, filter2} and Mul = {0, @_{o -> o}, ack#, cons, false, map, succ, true}, and the following precedence: false > filter = filter2 > ack > succ > ack# > true > @_{o -> o} = map > cons > 0 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack#(succ(X)) > ack#(X) ack#(succ(X)) >= ack#(X) ack#(succ(X)) >= ack#(succ(X)) ack(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(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] ack#(succ(X)) > ack#(X) because [2], by definition 2] ack#*(succ(X)) >= ack#(X) because ack# in Mul and [3], by (Stat) 3] succ(X) > X because [4], by definition 4] succ*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] ack#(succ(X)) >= ack#(X) because ack# in Mul and [7], by (Fun) 7] succ(X) >= X because [8], by (Star) 8] succ*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] ack#(succ(X)) >= ack#(succ(X)) because ack# in Mul and [11], by (Fun) 11] succ(X) >= succ(X) because succ in Mul and [12], by (Fun) 12] X >= X by (Meta) 13] ack(0, X) >= succ(X) because [14], by (Star) 14] ack*(0, X) >= succ(X) because ack > succ and [15], by (Copy) 15] ack*(0, X) >= X because [16], by (Select) 16] X >= X by (Meta) 17] ack(succ(X), Y) >= ack(X, succ(0)) because [18], by (Star) 18] ack*(succ(X), Y) >= ack(X, succ(0)) because [3], [19] and [21], by (Stat) 19] ack*(succ(X), Y) >= X because [20], by (Select) 20] succ(X) >= X because [4], by (Star) 21] ack*(succ(X), Y) >= succ(0) because ack > succ and [22], by (Copy) 22] ack*(succ(X), Y) >= 0 because ack > 0, by (Copy) 23] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [24], by (Star) 24] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [25], [27] and [28], by (Stat) 25] succ(X) > X because [26], by definition 26] succ*(X) >= X because [12], by (Select) 27] ack*(succ(X), succ(Y)) >= X because [7], by (Select) 28] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [11], [29], [32] and [33], by (Stat) 29] succ(Y) > Y because [30], by definition 30] succ*(Y) >= Y because [31], by (Select) 31] Y >= Y by (Meta) 32] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [27], by (Copy) 33] ack*(succ(X), succ(Y)) >= Y because [34], by (Select) 34] succ(Y) >= Y because [30], by (Star) 35] map(F, _|_) >= _|_ by (Bot) 36] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [37], by (Star) 37] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [38] and [43], by (Copy) 38] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map = @_{o -> o}, map in Mul, [39] and [40], by (Stat) 39] F >= F by (Meta) 40] cons(X, Y) > X because [41], by definition 41] cons*(X, Y) >= X because [42], by (Select) 42] X >= X by (Meta) 43] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [39] and [44], by (Stat) 44] cons(X, Y) > Y because [45], by definition 45] cons*(X, Y) >= Y because [46], by (Select) 46] Y >= Y by (Meta) 47] filter(F, _|_) >= _|_ by (Bot) 48] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [49], by (Star) 49] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [50], [51], [54], [55], [56] and [60], 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, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [55] and [56], by (Copy) 55] filter*(F, cons(X, Y)) >= F because [50], by (Select) 56] filter*(F, cons(X, Y)) >= X because [57], by (Select) 57] cons(X, Y) >= X because [58], by (Star) 58] cons*(X, Y) >= X because [59], by (Select) 59] X >= X by (Meta) 60] filter*(F, cons(X, Y)) >= Y because [61], by (Select) 61] cons(X, Y) >= Y because [52], by (Star) 62] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [63], by (Star) 63] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [64] and [66], by (Copy) 64] filter2*(true, F, X, Y) >= X because [65], by (Select) 65] X >= X by (Meta) 66] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [67], [68], [69] and [70], by (Stat) 67] F >= F by (Meta) 68] Y >= Y by (Meta) 69] filter2*(true, F, X, Y) >= F because [67], by (Select) 70] filter2*(true, F, X, Y) >= Y because [68], by (Select) 71] filter2(false, F, X, Y) >= filter(F, Y) because [72], by (Star) 72] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [73], [74], [75] and [76], by (Stat) 73] F >= F by (Meta) 74] Y >= Y by (Meta) 75] filter2*(false, F, X, Y) >= F because [73], by (Select) 76] filter2*(false, F, X, Y) >= Y because [74], by (Select) 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_5, R_0, static, formative), where P_5 consists of: ack#(succ(X), succ(Y)) =#> ack#(X, ack(succ(X), Y)) ack#(succ(X), succ(Y)) =#> ack#(succ(X), 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: ack#(succ(X), succ(Y)) >? ack#(X, ack(succ(X), Y)) ack#(succ(X), succ(Y)) >? ack#(succ(X), Y) ack(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(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]] = _|_ [[ack#(x_1, x_2)]] = ack#(x_1) [[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 [[nil]] = _|_ We choose Lex = {ack} and Mul = {@_{o -> o}, ack#, false, map, succ, true}, and the following precedence: @_{o -> o} > false > map > ack > ack# > succ > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack#(succ(X)) > ack#(X) ack#(succ(X)) >= ack#(succ(X)) ack(_|_, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(_|_)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) map >= _|_ map >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ With these choices, we have: 1] ack#(succ(X)) > ack#(X) because [2], by definition 2] ack#*(succ(X)) >= ack#(X) because ack# in Mul and [3], by (Stat) 3] succ(X) > X because [4], by definition 4] succ*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] ack#(succ(X)) >= ack#(succ(X)) because ack# in Mul and [7], by (Fun) 7] succ(X) >= succ(X) because succ in Mul and [8], by (Fun) 8] X >= X by (Meta) 9] ack(_|_, X) >= succ(X) because [10], by (Star) 10] ack*(_|_, X) >= succ(X) because ack > succ and [11], by (Copy) 11] ack*(_|_, X) >= X because [12], by (Select) 12] X >= X by (Meta) 13] ack(succ(X), Y) >= ack(X, succ(_|_)) because [14], by (Star) 14] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [15], [18] and [20], by (Stat) 15] succ(X) > X because [16], by definition 16] succ*(X) >= X because [17], by (Select) 17] X >= X by (Meta) 18] ack*(succ(X), Y) >= X because [19], by (Select) 19] succ(X) >= X because [16], by (Star) 20] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [21], by (Copy) 21] ack*(succ(X), Y) >= _|_ by (Bot) 22] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [23], by (Star) 23] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [3], [24] and [26], by (Stat) 24] ack*(succ(X), succ(Y)) >= X because [25], by (Select) 25] succ(X) >= X because [4], by (Star) 26] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [7], [27], [30] and [31], by (Stat) 27] succ(Y) > Y because [28], by definition 28] succ*(Y) >= Y because [29], by (Select) 29] Y >= Y by (Meta) 30] ack*(succ(X), succ(Y)) >= succ(X) because [7], by (Select) 31] ack*(succ(X), succ(Y)) >= Y because [32], by (Select) 32] succ(Y) >= Y because [28], by (Star) 33] map >= _|_ by (Bot) 34] map >= _|_ by (Bot) 35] _|_ >= _|_ by (Bot) 36] _|_ >= _|_ by (Bot) 37] _|_ >= _|_ by (Bot) 38] _|_ >= _|_ by (Bot) 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: ack#(succ(X), succ(Y)) =#> ack#(succ(X), 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: ack#(succ(X), succ(Y)) >? ack#(succ(X), Y) ack(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(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]] = _|_ [[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 [[nil]] = _|_ We choose Lex = {ack} and Mul = {@_{o -> o}, ack#, false, map, succ, true}, and the following precedence: @_{o -> o} > ack > ack# > false > map > succ > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack#(succ(X), succ(Y)) > ack#(succ(X), Y) ack(_|_, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(_|_)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) map >= _|_ map >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ With these choices, we have: 1] ack#(succ(X), succ(Y)) > ack#(succ(X), Y) because [2], by definition 2] ack#*(succ(X), succ(Y)) >= ack#(succ(X), Y) because ack# in Mul, [3] and [5], by (Stat) 3] succ(X) >= succ(X) because succ in Mul and [4], by (Fun) 4] X >= X by (Meta) 5] succ(Y) > Y because [6], by definition 6] succ*(Y) >= Y because [7], by (Select) 7] Y >= Y by (Meta) 8] ack(_|_, X) >= succ(X) because [9], by (Star) 9] ack*(_|_, X) >= succ(X) because ack > succ and [10], by (Copy) 10] ack*(_|_, X) >= X because [11], by (Select) 11] X >= X by (Meta) 12] ack(succ(X), Y) >= ack(X, succ(_|_)) because [13], by (Star) 13] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [14], [17] and [19], by (Stat) 14] succ(X) > X because [15], by definition 15] succ*(X) >= X because [16], by (Select) 16] X >= X by (Meta) 17] ack*(succ(X), Y) >= X because [18], by (Select) 18] succ(X) >= X because [15], by (Star) 19] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [20], by (Copy) 20] ack*(succ(X), Y) >= _|_ by (Bot) 21] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [22], by (Star) 22] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [23], [25] and [27], by (Stat) 23] succ(X) > X because [24], by definition 24] succ*(X) >= X because [4], by (Select) 25] ack*(succ(X), succ(Y)) >= X because [26], by (Select) 26] succ(X) >= X because [24], by (Star) 27] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [3], [5], [28] and [29], by (Stat) 28] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [25], by (Copy) 29] ack*(succ(X), succ(Y)) >= Y because [30], by (Select) 30] succ(Y) >= Y because [6], by (Star) 31] map >= _|_ by (Bot) 32] map >= _|_ by (Bot) 33] _|_ >= _|_ by (Bot) 34] _|_ >= _|_ by (Bot) 35] _|_ >= _|_ by (Bot) 36] _|_ >= _|_ by (Bot) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, 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.