We consider the system Applicative_first_order_05__08. Alphabet: !facminus : [a * a] --> a !facplus : [a * a] --> a !factimes : [a * a] --> a 0 : [] --> a 1 : [] --> a D : [a] --> a cons : [c * d] --> d constant : [] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d 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)) 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] map#(F, cons(X, Y)) =#> map#(F, Y) 7] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 8] filter2#(true, F, X, Y) =#> filter#(F, Y) 9] 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)) 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 ::= 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)) 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: 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) 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) 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)) 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: !facminus = \y0y1.0 !facplus = \y0y1.0 !factimes = \y0y1.0 D = \y0.0 D# = \y0.0 cons = \y0y1.2 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.2 + 2y3 filter2# = \y0G1y2y3.2y3G1(y3) + 2G1(y3) filter# = \G0y1.2y1G0(y1) map = \G0y1.y1 map# = \G0y1.y1 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!facplus(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[D#(!factimes(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!factimes(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 2 + x2 > x2 = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2x2F0(2 + x2) + 4F0(2 + x2) >= 2x2F0(x2) + 2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2F0(x2) + 2F0(x2) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2F0(x2) + 2F0(x2) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] [[D(!facplus(_x0, _x1))]] = 0 >= 0 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 0 >= 0 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 0 >= 0 = [[!facminus(D(_x0), D(_x1))]] [[map(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 4 + 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 >= 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_0, R_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of: 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) 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) 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: 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) 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(!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)) 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: !facminus = \y0y1.0 !facplus = \y0y1.0 !factimes = \y0y1.0 D = \y0.0 D# = \y0.0 cons = \y0y1.2 + y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.2 + y3 filter2# = \y0G1y2y3.2y3 + y3G1(y3) filter# = \G0y1.2y1 + y1G0(y1) map = \G0y1.y1 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!facplus(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[D#(!factimes(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!factimes(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 2F0(2 + x2) + x2F0(2 + x2) > 2x2 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2 + x2F0(x2) >= 2x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2 + x2F0(x2) >= 2x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[D(!facplus(_x0, _x1))]] = 0 >= 0 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 0 >= 0 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 0 >= 0 = [[!facminus(D(_x0), D(_x1))]] [[map(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 >= 2 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x2 >= 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_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of: 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) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(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). The formative rules of (P_2, R_1) are R_2 ::= 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)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_1, static, formative) by (P_2, R_2, static, formative). Thus, the original system is terminating if (P_2, R_2, static, formative) is finite. We consider the dependency pair problem (P_2, 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: 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) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) 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)) 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 = \y0y1.0 D = \y0.0 D# = \y0.0 false = 3 filter2# = \y0G1y2y3.3 filter# = \G0y1.0 true = 3 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!facplus(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[D#(!factimes(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!factimes(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[D#(!facminus(_x0, _x1))]] = 0 >= 0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 0 >= 0 = [[D#(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[D(!facplus(_x0, _x1))]] = 0 >= 0 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 0 >= 0 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 0 >= 0 = [[!facminus(D(_x0), D(_x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_2, static, formative) by (P_3, R_2, static, formative), where P_3 consists of: 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) 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: 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(!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)) 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]. We choose Lex = {} and Mul = {!facminus, !facplus, !factimes, D, D#}, and the following precedence: D > !facplus > !factimes > D# > !facminus With these choices, we have: 1] D#(!facplus(X, Y)) >= D#(X) because D# in Mul and [2], by (Fun) 2] !facplus(X, Y) >= X because [3], by (Star) 3] !facplus*(X, Y) >= X because [4], by (Select) 4] X >= X by (Meta) 5] D#(!facplus(X, Y)) >= D#(Y) because [6], by (Star) 6] D#*(!facplus(X, Y)) >= D#(Y) because [7], by (Select) 7] !facplus(X, Y) >= D#(Y) because [8], by (Star) 8] !facplus*(X, Y) >= D#(Y) because !facplus > D# and [9], by (Copy) 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 [16], by definition 16] D#*(!factimes(X, Y)) >= D#(Y) because D# in Mul and [17], by (Stat) 17] !factimes(X, Y) > Y because [18], by definition 18] !factimes*(X, Y) >= Y because [19], by (Select) 19] Y >= Y by (Meta) 20] D#(!facminus(X, Y)) >= D#(X) because [21], by (Star) 21] D#*(!facminus(X, Y)) >= D#(X) because D# in Mul and [22], by (Stat) 22] !facminus(X, Y) > X because [23], by definition 23] !facminus*(X, Y) >= X because [24], by (Select) 24] X >= X by (Meta) 25] D#(!facminus(X, Y)) >= D#(Y) because D# in Mul and [26], by (Fun) 26] !facminus(X, Y) >= Y because [27], by (Star) 27] !facminus*(X, Y) >= Y because [28], by (Select) 28] Y >= Y by (Meta) 29] D(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because [30], by (Star) 30] D*(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because D > !facplus, [31] and [34], by (Copy) 31] D*(!facplus(X, Y)) >= D(X) because D in Mul and [32], by (Stat) 32] !facplus(X, Y) > X because [33], by definition 33] !facplus*(X, Y) >= X because [4], by (Select) 34] D*(!facplus(X, Y)) >= D(Y) because D in Mul and [35], by (Stat) 35] !facplus(X, Y) > Y because [9], by definition 36] D(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because [37], by (Star) 37] D*(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because D > !facplus, [38] and [44], by (Copy) 38] D*(!factimes(X, Y)) >= !factimes(Y, D(X)) because D > !factimes, [39] and [41], by (Copy) 39] D*(!factimes(X, Y)) >= Y because [40], by (Select) 40] !factimes(X, Y) >= Y because [18], by (Star) 41] D*(!factimes(X, Y)) >= D(X) because D in Mul and [42], by (Stat) 42] !factimes(X, Y) > X because [43], by definition 43] !factimes*(X, Y) >= X because [14], by (Select) 44] D*(!factimes(X, Y)) >= !factimes(X, D(Y)) because D > !factimes, [45] and [46], by (Copy) 45] D*(!factimes(X, Y)) >= X because [12], by (Select) 46] D*(!factimes(X, Y)) >= D(Y) because D in Mul and [17], by (Stat) 47] D(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because [48], by (Star) 48] D*(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because D > !facminus, [49] and [50], by (Copy) 49] D*(!facminus(X, Y)) >= D(X) because D in Mul and [22], by (Stat) 50] D*(!facminus(X, Y)) >= D(Y) because D in Mul and [51], by (Stat) 51] !facminus(X, Y) > Y because [52], by definition 52] !facminus*(X, Y) >= Y because [28], 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: D#(!facplus(X, Y)) =#> D#(X) D#(!facplus(X, Y)) =#> D#(Y) D#(!factimes(X, Y)) =#> D#(X) D#(!facminus(X, Y)) =#> D#(X) D#(!facminus(X, Y)) =#> D#(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: D#(!facplus(X, Y)) >? D#(X) D#(!facplus(X, Y)) >? D#(Y) D#(!factimes(X, Y)) >? D#(X) D#(!facminus(X, Y)) >? D#(X) D#(!facminus(X, Y)) >? D#(Y) 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)) 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]. We choose Lex = {} and Mul = {!facminus, !facplus, !factimes, D, D#}, and the following precedence: D > !facplus > !factimes > !facminus > D# With these choices, we have: 1] D#(!facplus(X, Y)) >= D#(X) because D# in Mul and [2], by (Fun) 2] !facplus(X, Y) >= X because [3], by (Star) 3] !facplus*(X, Y) >= X because [4], by (Select) 4] X >= X by (Meta) 5] D#(!facplus(X, Y)) > D#(Y) because [6], by definition 6] D#*(!facplus(X, Y)) >= D#(Y) because [7], by (Select) 7] !facplus(X, Y) >= D#(Y) because [8], by (Star) 8] !facplus*(X, Y) >= D#(Y) because !facplus > D# and [9], by (Copy) 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#(!facminus(X, Y)) >= D#(X) because [16], by (Star) 16] D#*(!facminus(X, Y)) >= D#(X) because D# in Mul and [17], by (Stat) 17] !facminus(X, Y) > X because [18], by definition 18] !facminus*(X, Y) >= X because [19], by (Select) 19] X >= X by (Meta) 20] D#(!facminus(X, Y)) >= D#(Y) because [21], by (Star) 21] D#*(!facminus(X, Y)) >= D#(Y) because [22], by (Select) 22] !facminus(X, Y) >= D#(Y) because [23], by (Star) 23] !facminus*(X, Y) >= D#(Y) because !facminus > D# and [24], by (Copy) 24] !facminus*(X, Y) >= Y because [25], by (Select) 25] Y >= Y by (Meta) 26] D(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because [27], by (Star) 27] D*(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because D > !facplus, [28] and [31], by (Copy) 28] D*(!facplus(X, Y)) >= D(X) because D in Mul and [29], by (Stat) 29] !facplus(X, Y) > X because [30], by definition 30] !facplus*(X, Y) >= X because [4], by (Select) 31] D*(!facplus(X, Y)) >= D(Y) because D in Mul and [32], by (Stat) 32] !facplus(X, Y) > Y because [9], by definition 33] D(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because [34], by (Star) 34] D*(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) because D > !facplus, [35] and [43], by (Copy) 35] D*(!factimes(X, Y)) >= !factimes(Y, D(X)) because D > !factimes, [36] and [40], by (Copy) 36] D*(!factimes(X, Y)) >= Y because [37], by (Select) 37] !factimes(X, Y) >= Y because [38], by (Star) 38] !factimes*(X, Y) >= Y because [39], by (Select) 39] Y >= Y by (Meta) 40] D*(!factimes(X, Y)) >= D(X) because D in Mul and [41], by (Stat) 41] !factimes(X, Y) > X because [42], by definition 42] !factimes*(X, Y) >= X because [14], by (Select) 43] D*(!factimes(X, Y)) >= !factimes(X, D(Y)) because D > !factimes, [44] and [45], by (Copy) 44] D*(!factimes(X, Y)) >= X because [12], by (Select) 45] D*(!factimes(X, Y)) >= D(Y) because D in Mul and [46], by (Stat) 46] !factimes(X, Y) > Y because [47], by definition 47] !factimes*(X, Y) >= Y because [39], by (Select) 48] D(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because [49], by (Star) 49] D*(!facminus(X, Y)) >= !facminus(D(X), D(Y)) because D > !facminus, [50] and [51], by (Copy) 50] D*(!facminus(X, Y)) >= D(X) because D in Mul and [17], by (Stat) 51] D*(!facminus(X, Y)) >= D(Y) because D in Mul and [52], by (Stat) 52] !facminus(X, Y) > Y because [24], by definition 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: D#(!facplus(X, Y)) =#> D#(X) D#(!factimes(X, Y)) =#> D#(X) D#(!facminus(X, Y)) =#> D#(X) D#(!facminus(X, Y)) =#> D#(Y) 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: D#(!facplus(X, Y)) >? D#(X) D#(!factimes(X, Y)) >? D#(X) D#(!facminus(X, Y)) >? D#(X) D#(!facminus(X, Y)) >? D#(Y) 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)) 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.y0 !factimes = \y0y1.2y0 + 2y1 D = \y0.2y0 D# = \y0.y0 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = x0 >= x0 = [[D#(_x0)]] [[D#(!factimes(_x0, _x1))]] = 2x0 + 2x1 >= x0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 1 + x0 + x1 > x0 = [[D#(_x0)]] [[D#(!facminus(_x0, _x1))]] = 1 + x0 + x1 > x1 = [[D#(_x1)]] [[D(!facplus(_x0, _x1))]] = 2x0 >= 2x0 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 4x0 + 4x1 >= 2x1 + 4x0 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] [[D(!facminus(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[!facminus(D(_x0), D(_x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_2, static, formative) by (P_6, R_2, static, formative), where P_6 consists of: D#(!facplus(X, Y)) =#> D#(X) D#(!factimes(X, Y)) =#> D#(X) Thus, the original system is terminating if (P_6, R_2, static, formative) is finite. We consider the dependency pair problem (P_6, R_2, static, formative). The formative rules of (P_6, R_2) are R_3 ::= D(!facplus(X, Y)) => !facplus(D(X), D(Y)) D(!factimes(X, Y)) => !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_6, R_2, static, formative) by (P_6, R_3, static, formative). Thus, the original system is terminating if (P_6, R_3, static, formative) is finite. We consider the dependency pair problem (P_6, R_3, 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#(!factimes(X, Y)) >? D#(X) D(!facplus(X, Y)) >= !facplus(D(X), D(Y)) D(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !facplus = \y0y1.2 + y0 !factimes = \y0y1.3 + y1 + 2y0 D = \y0.2 + 2y0 D# = \y0.y0 Using this interpretation, the requirements translate to: [[D#(!facplus(_x0, _x1))]] = 2 + x0 > x0 = [[D#(_x0)]] [[D#(!factimes(_x0, _x1))]] = 3 + x1 + 2x0 > x0 = [[D#(_x0)]] [[D(!facplus(_x0, _x1))]] = 6 + 2x0 >= 4 + 2x0 = [[!facplus(D(_x0), D(_x1))]] [[D(!factimes(_x0, _x1))]] = 8 + 2x1 + 4x0 >= 7 + 2x0 + 2x1 = [[!facplus(!factimes(_x1, D(_x0)), !factimes(_x0, D(_x1)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, R_3) by ({}, R_3). 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.