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). The formative rules of (P_0, R_0) are R_1 ::= 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, cons(X, Y)) => cons(F X, map(F, Y)) 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) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, static, formative) by (P_0, R_1, static, formative). Thus, the original system is terminating if (P_0, R_1, static, formative) is finite. We consider the dependency pair problem (P_0, R_1, 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) map#(F, cons(X, Y)) >? map#(F, Y) 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, cons(X, Y)) >= cons(F X, map(F, Y)) 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 = \y0y1.2 ack# = \y0y1.0 cons = \y0y1.2 + 2y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.2 + 2y3 filter2# = \y0G1y2y3.y3 filter# = \G0y1.y1 map = \G0y1.y1 map# = \G0y1.0 succ = \y0.0 true = 3 Using this interpretation, the requirements translate to: [[ack#(succ(_x0), _x1)]] = 0 >= 0 = [[ack#(_x0, succ(0))]] [[ack#(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack#(_x0, ack(succ(_x0), _x1))]] [[ack#(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack#(succ(_x0), _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2 + 2x2 > x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = x2 >= x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = x2 >= x2 = [[filter#(_F0, _x2)]] [[ack(0, _x0)]] = 2 >= 0 = [[succ(_x0)]] [[ack(succ(_x0), _x1)]] = 2 >= 2 = [[ack(_x0, succ(0))]] [[ack(succ(_x0), succ(_x1))]] = 2 >= 2 = [[ack(_x0, ack(succ(_x0), _x1))]] [[map(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 2 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 2 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + 2x2 >= 2 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + 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_0, R_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of: 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) map#(F, cons(X, Y)) =#> map#(F, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if (P_1, R_1, static, formative) is finite. We consider the dependency pair problem (P_1, R_1, 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) map#(F, cons(X, Y)) >? map#(F, 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, cons(X, Y)) >= cons(F X, map(F, Y)) 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 = \y0y1.2 ack# = \y0y1.0 cons = \y0y1.0 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 filter2# = \y0G1y2y3.3 filter# = \G0y1.0 map = \G0y1.0 map# = \G0y1.0 succ = \y0.0 true = 3 Using this interpretation, the requirements translate to: [[ack#(succ(_x0), _x1)]] = 0 >= 0 = [[ack#(_x0, succ(0))]] [[ack#(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack#(_x0, ack(succ(_x0), _x1))]] [[ack#(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack#(succ(_x0), _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[ack(0, _x0)]] = 2 >= 0 = [[succ(_x0)]] [[ack(succ(_x0), _x1)]] = 2 >= 2 = [[ack(_x0, succ(0))]] [[ack(succ(_x0), succ(_x1))]] = 2 >= 2 = [[ack(_x0, ack(succ(_x0), _x1))]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[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_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of: 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) map#(F, cons(X, Y)) =#> map#(F, Y) Thus, the original system is terminating if (P_2, R_1, static, formative) is finite. We consider the dependency pair problem (P_2, R_1, 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) 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, cons(X, Y)) >= cons(F X, map(F, Y)) 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 ack# = \y0y1.0 cons = \y0y1.1 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y3 map = \G0y1.y1 map# = \G0y1.y1 succ = \y0.0 true = 3 Using this interpretation, the requirements translate to: [[ack#(succ(_x0), _x1)]] = 0 >= 0 = [[ack#(_x0, succ(0))]] [[ack#(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack#(_x0, ack(succ(_x0), _x1))]] [[ack#(succ(_x0), succ(_x1))]] = 0 >= 0 = [[ack#(succ(_x0), _x1)]] [[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, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[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 the dependency pair problem (P_2, R_1, static, formative) by (P_3, R_1, static, formative), where P_3 consists of: 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) Thus, the original system is terminating if (P_3, R_1, static, formative) is finite. We consider the dependency pair problem (P_3, R_1, static, formative). The formative rules of (P_3, R_1) are R_2 ::= ack(0, X) => succ(X) ack(succ(X), Y) => ack(X, succ(0)) ack(succ(X), succ(Y)) => ack(X, ack(succ(X), Y)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_1, static, formative) by (P_3, R_2, static, formative). Thus, the original system is terminating if (P_3, R_2, static, formative) is finite. We consider the dependency pair problem (P_3, R_2, 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)) 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]] = _|_ We choose Lex = {ack, ack#} and Mul = {succ}, and the following precedence: ack# > ack > succ Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack#(succ(X), Y) >= ack#(X, succ(_|_)) ack#(succ(X), succ(Y)) >= ack#(X, ack(succ(X), Y)) 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)) With these choices, we have: 1] ack#(succ(X), Y) >= ack#(X, succ(_|_)) because [2], by (Star) 2] ack#*(succ(X), Y) >= ack#(X, succ(_|_)) because [3], [6] and [8], 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), Y) >= X because [7], by (Select) 7] succ(X) >= X because [4], by (Star) 8] ack#*(succ(X), Y) >= succ(_|_) because ack# > succ and [9], by (Copy) 9] ack#*(succ(X), Y) >= _|_ by (Bot) 10] ack#(succ(X), succ(Y)) >= ack#(X, ack(succ(X), Y)) because [11], by (Star) 11] ack#*(succ(X), succ(Y)) >= ack#(X, ack(succ(X), Y)) because [12], [15] and [17], by (Stat) 12] succ(X) > X because [13], by definition 13] succ*(X) >= X because [14], by (Select) 14] X >= X by (Meta) 15] ack#*(succ(X), succ(Y)) >= X because [16], by (Select) 16] succ(X) >= X because [13], by (Star) 17] ack#*(succ(X), succ(Y)) >= ack(succ(X), Y) because ack# > ack, [18] and [21], by (Copy) 18] ack#*(succ(X), succ(Y)) >= succ(X) because [19], by (Select) 19] succ(X) >= succ(X) because succ in Mul and [20], by (Fun) 20] X >= X by (Meta) 21] ack#*(succ(X), succ(Y)) >= Y because [22], by (Select) 22] succ(Y) >= Y because [23], by (Star) 23] succ*(Y) >= Y because [24], by (Select) 24] Y >= Y by (Meta) 25] ack#(succ(X), succ(Y)) > ack#(succ(X), Y) because [26], by definition 26] ack#*(succ(X), succ(Y)) >= ack#(succ(X), Y) because [27], [28], [18] and [21], by (Stat) 27] succ(X) >= succ(X) because succ in Mul and [20], by (Fun) 28] succ(Y) > Y because [29], by definition 29] succ*(Y) >= Y because [24], by (Select) 30] ack(_|_, X) >= succ(X) because [31], by (Star) 31] ack*(_|_, X) >= succ(X) because ack > succ and [32], by (Copy) 32] ack*(_|_, X) >= X because [33], by (Select) 33] X >= X by (Meta) 34] ack(succ(X), Y) >= ack(X, succ(_|_)) because [35], by (Star) 35] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [3], [36] and [37], by (Stat) 36] ack*(succ(X), Y) >= X because [7], by (Select) 37] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [38], by (Copy) 38] ack*(succ(X), Y) >= _|_ by (Bot) 39] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [40], by (Star) 40] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [12], [41] and [42], by (Stat) 41] ack*(succ(X), succ(Y)) >= X because [16], by (Select) 42] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [27], [28], [43] and [44], by (Stat) 43] ack*(succ(X), succ(Y)) >= succ(X) because ack > succ and [41], by (Copy) 44] ack*(succ(X), succ(Y)) >= Y because [22], by (Select) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_2, static, formative) by (P_4, R_2, static, formative), where P_4 consists of: ack#(succ(X), Y) =#> ack#(X, succ(0)) ack#(succ(X), succ(Y)) =#> ack#(X, ack(succ(X), Y)) Thus, the original system is terminating if (P_4, R_2, static, formative) is finite. We consider the dependency pair problem (P_4, R_2, 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(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), 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) We choose Lex = {ack} and Mul = {0, ack#, succ}, and the following precedence: ack > 0 > succ > ack# 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(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) With these choices, we have: 1] ack#(succ(X)) >= ack#(X) because ack# in Mul and [2], by (Fun) 2] succ(X) >= X because [3], by (Star) 3] succ*(X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] ack#(succ(X)) > ack#(X) because [6], by definition 6] ack#*(succ(X)) >= ack#(X) because ack# in Mul and [7], by (Stat) 7] succ(X) > X because [8], by definition 8] succ*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] ack(0, X) >= succ(X) because [11], by (Star) 11] ack*(0, X) >= succ(X) because ack > succ and [12], by (Copy) 12] ack*(0, X) >= X because [13], by (Select) 13] X >= X by (Meta) 14] ack(succ(X), Y) >= ack(X, succ(0)) because [15], by (Star) 15] ack*(succ(X), Y) >= ack(X, succ(0)) because [16], [18] and [19], by (Stat) 16] succ(X) > X because [17], by definition 17] succ*(X) >= X because [4], by (Select) 18] ack*(succ(X), Y) >= X because [2], by (Select) 19] ack*(succ(X), Y) >= succ(0) because ack > succ and [20], by (Copy) 20] ack*(succ(X), Y) >= 0 because ack > 0, by (Copy) 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 [7], [23] and [25], by (Stat) 23] ack*(succ(X), succ(Y)) >= X because [24], by (Select) 24] succ(X) >= X because [8], by (Star) 25] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [26], [28], [31] and [32], by (Stat) 26] succ(X) >= succ(X) because succ in Mul and [27], by (Fun) 27] X >= X by (Meta) 28] succ(Y) > Y because [29], by definition 29] succ*(Y) >= Y because [30], by (Select) 30] Y >= Y by (Meta) 31] ack*(succ(X), succ(Y)) >= succ(X) because [26], by (Select) 32] ack*(succ(X), succ(Y)) >= Y because [33], by (Select) 33] succ(Y) >= Y because [29], by (Star) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_2, static, formative) by (P_5, R_2, static, formative), where P_5 consists of: ack#(succ(X), Y) =#> ack#(X, succ(0)) Thus, the original system is terminating if (P_5, R_2, static, formative) is finite. We consider the dependency pair problem (P_5, R_2, 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(0, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(0)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), 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) We choose Lex = {ack} and Mul = {ack#, succ}, and the following precedence: ack > succ > ack# 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(_|_, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(_|_)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) With these choices, we have: 1] ack#(succ(X)) > ack#(X) because [2], by definition 2] ack#*(succ(X)) >= ack#(X) because [3], by (Select) 3] succ(X) >= ack#(X) because [4], by (Star) 4] succ*(X) >= ack#(X) because succ > ack# and [5], by (Copy) 5] succ*(X) >= X because [6], by (Select) 6] X >= X by (Meta) 7] ack(_|_, X) >= succ(X) because [8], by (Star) 8] ack*(_|_, X) >= succ(X) because ack > succ and [9], by (Copy) 9] ack*(_|_, X) >= X because [10], by (Select) 10] X >= X by (Meta) 11] ack(succ(X), Y) >= ack(X, succ(_|_)) because [12], by (Star) 12] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [13], [14] and [16], by (Stat) 13] succ(X) > X because [5], by definition 14] ack*(succ(X), Y) >= X because [15], by (Select) 15] succ(X) >= X because [5], by (Star) 16] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [17], by (Copy) 17] ack*(succ(X), Y) >= _|_ by (Bot) 18] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [19], by (Star) 19] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [20], [23] and [25], by (Stat) 20] succ(X) > X because [21], by definition 21] succ*(X) >= X because [22], by (Select) 22] X >= X by (Meta) 23] ack*(succ(X), succ(Y)) >= X because [24], by (Select) 24] succ(X) >= X because [21], by (Star) 25] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [26], [28], [31] and [32], by (Stat) 26] succ(X) >= succ(X) because succ in Mul and [27], by (Fun) 27] X >= X by (Meta) 28] succ(Y) > Y because [29], by definition 29] succ*(Y) >= Y because [30], by (Select) 30] Y >= Y by (Meta) 31] ack*(succ(X), succ(Y)) >= succ(X) because [26], by (Select) 32] ack*(succ(X), succ(Y)) >= Y because [33], by (Select) 33] succ(Y) >= Y because [29], by (Star) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_5, R_2) by ({}, R_2). 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.