We consider the system Applicative_first_order_05__13. Alphabet: !facplus : [a * a] --> a !factimes : [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 true : [] --> b Rules: !factimes(x, !facplus(y, z)) => !facplus(!factimes(x, y), !factimes(x, z)) !factimes(!facplus(x, y), z) => !facplus(!factimes(z, x), !factimes(z, y)) !factimes(!factimes(x, y), z) => !factimes(x, !factimes(y, z)) !facplus(!facplus(x, y), z) => !facplus(x, !facplus(y, z)) 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] !factimes#(X, !facplus(Y, Z)) =#> !facplus#(!factimes(X, Y), !factimes(X, Z)) 1] !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y) 2] !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z) 3] !factimes#(!facplus(X, Y), Z) =#> !facplus#(!factimes(Z, X), !factimes(Z, Y)) 4] !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X) 5] !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y) 6] !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z)) 7] !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z) 8] !facplus#(!facplus(X, Y), Z) =#> !facplus#(X, !facplus(Y, Z)) 9] !facplus#(!facplus(X, Y), Z) =#> !facplus#(Y, Z) 10] map#(F, cons(X, Y)) =#> map#(F, Y) 11] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 12] filter2#(true, F, X, Y) =#> filter#(F, Y) 13] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) => !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) => !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) => !facplus(X, !facplus(Y, Z)) 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 ::= !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) => !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) => !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) => !facplus(X, !facplus(Y, Z)) 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: !factimes#(X, !facplus(Y, Z)) >? !facplus#(!factimes(X, Y), !factimes(X, Z)) !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) >? !facplus#(!factimes(Z, X), !factimes(Z, Y)) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) >? !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) >? !factimes#(Y, Z) !facplus#(!facplus(X, Y), Z) >? !facplus#(X, !facplus(Y, Z)) !facplus#(!facplus(X, Y), Z) >? !facplus#(Y, Z) 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) !factimes(X, !facplus(Y, Z)) >= !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) >= !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) >= !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) >= !facplus(X, !facplus(Y, Z)) 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: !facplus = \y0y1.0 !facplus# = \y0y1.2y1 !factimes = \y0y1.0 !factimes# = \y0y1.3 cons = \y0y1.1 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.0 filter# = \G0y1.0 map = \G0y1.y1 map# = \G0y1.y1 true = 3 Using this interpretation, the requirements translate to: [[!factimes#(_x0, !facplus(_x1, _x2))]] = 3 > 0 = [[!facplus#(!factimes(_x0, _x1), !factimes(_x0, _x2))]] [[!factimes#(_x0, !facplus(_x1, _x2))]] = 3 >= 3 = [[!factimes#(_x0, _x1)]] [[!factimes#(_x0, !facplus(_x1, _x2))]] = 3 >= 3 = [[!factimes#(_x0, _x2)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 3 > 0 = [[!facplus#(!factimes(_x2, _x0), !factimes(_x2, _x1))]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 3 >= 3 = [[!factimes#(_x2, _x0)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 3 >= 3 = [[!factimes#(_x2, _x1)]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 3 >= 3 = [[!factimes#(_x0, !factimes(_x1, _x2))]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 3 >= 3 = [[!factimes#(_x1, _x2)]] [[!facplus#(!facplus(_x0, _x1), _x2)]] = 2x2 >= 0 = [[!facplus#(_x0, !facplus(_x1, _x2))]] [[!facplus#(!facplus(_x0, _x1), _x2)]] = 2x2 >= 2x2 = [[!facplus#(_x1, _x2)]] [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] [[!factimes(_x0, !facplus(_x1, _x2))]] = 0 >= 0 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] [[!factimes(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] [[!factimes(!factimes(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes(_x0, !factimes(_x1, _x2))]] [[!facplus(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(_x0, !facplus(_x1, _x2))]] [[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_0, R_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of: !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z) !facplus#(!facplus(X, Y), Z) =#> !facplus#(X, !facplus(Y, Z)) !facplus#(!facplus(X, Y), Z) =#> !facplus#(Y, Z) 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: !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) >? !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) >? !factimes#(Y, Z) !facplus#(!facplus(X, Y), Z) >? !facplus#(X, !facplus(Y, Z)) !facplus#(!facplus(X, Y), Z) >? !facplus#(Y, Z) 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) !factimes(X, !facplus(Y, Z)) >= !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) >= !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) >= !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) >= !facplus(X, !facplus(Y, Z)) 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: !facplus = \y0y1.0 !facplus# = \y0y1.0 !factimes = \y0y1.0 !factimes# = \y0y1.0 cons = \y0y1.1 + 2y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.1 + 2y3 filter# = \G0y1.2y1 map = \G0y1.y1 true = 3 Using this interpretation, the requirements translate to: [[!factimes#(_x0, !facplus(_x1, _x2))]] = 0 >= 0 = [[!factimes#(_x0, _x1)]] [[!factimes#(_x0, !facplus(_x1, _x2))]] = 0 >= 0 = [[!factimes#(_x0, _x2)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x2, _x0)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x2, _x1)]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x0, !factimes(_x1, _x2))]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x1, _x2)]] [[!facplus#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus#(_x0, !facplus(_x1, _x2))]] [[!facplus#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus#(_x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2 + 4x2 > 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)]] [[!factimes(_x0, !facplus(_x1, _x2))]] = 0 >= 0 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] [[!factimes(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] [[!factimes(!factimes(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes(_x0, !factimes(_x1, _x2))]] [[!facplus(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(_x0, !facplus(_x1, _x2))]] [[map(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[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_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of: !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z) !facplus#(!facplus(X, Y), Z) =#> !facplus#(X, !facplus(Y, Z)) !facplus#(!facplus(X, Y), Z) =#> !facplus#(Y, Z) 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 ::= !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) => !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) => !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) => !facplus(X, !facplus(Y, Z)) 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: !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) >? !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) >? !factimes#(Y, Z) !facplus#(!facplus(X, Y), Z) >? !facplus#(X, !facplus(Y, Z)) !facplus#(!facplus(X, Y), Z) >? !facplus#(Y, Z) !factimes(X, !facplus(Y, Z)) >= !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) >= !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) >= !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) >= !facplus(X, !facplus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !facplus = \y0y1.2 + y1 + 2y0 !facplus# = \y0y1.y0 !factimes = \y0y1.y1 + 2y0 + 3y0y1 !factimes# = \y0y1.0 Using this interpretation, the requirements translate to: [[!factimes#(_x0, !facplus(_x1, _x2))]] = 0 >= 0 = [[!factimes#(_x0, _x1)]] [[!factimes#(_x0, !facplus(_x1, _x2))]] = 0 >= 0 = [[!factimes#(_x0, _x2)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x2, _x0)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x2, _x1)]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x0, !factimes(_x1, _x2))]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 0 >= 0 = [[!factimes#(_x1, _x2)]] [[!facplus#(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 > x0 = [[!facplus#(_x0, !facplus(_x1, _x2))]] [[!facplus#(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 > x1 = [[!facplus#(_x1, _x2)]] [[!factimes(_x0, !facplus(_x1, _x2))]] = 2 + x2 + 2x1 + 3x0x2 + 6x0x1 + 8x0 >= 2 + x2 + 2x1 + 3x0x2 + 6x0 + 6x0x1 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] [[!factimes(!facplus(_x0, _x1), _x2)]] = 4 + 2x1 + 3x1x2 + 4x0 + 6x0x2 + 7x2 >= 2 + x1 + 2x0 + 3x1x2 + 6x0x2 + 6x2 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] [[!factimes(!factimes(_x0, _x1), _x2)]] = x2 + 2x1 + 3x1x2 + 4x0 + 6x0x1 + 6x0x2 + 9x0x1x2 >= x2 + 2x0 + 2x1 + 3x0x2 + 3x1x2 + 6x0x1 + 9x0x1x2 = [[!factimes(_x0, !factimes(_x1, _x2))]] [[!facplus(!facplus(_x0, _x1), _x2)]] = 6 + x2 + 2x1 + 4x0 >= 4 + x2 + 2x0 + 2x1 = [[!facplus(_x0, !facplus(_x1, _x2))]] 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: !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z) 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: !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) >? !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) >? !factimes#(Y, Z) !factimes(X, !facplus(Y, Z)) >= !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) >= !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) >= !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) >= !facplus(X, !facplus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !facplus = \y0y1.2 + y1 + 2y0 !factimes = \y0y1.y0 + y1 + y0y1 !factimes# = \y0y1.2y0 + 2y0y1 + 2y1 Using this interpretation, the requirements translate to: [[!factimes#(_x0, !facplus(_x1, _x2))]] = 4 + 2x0x2 + 2x2 + 4x0x1 + 4x1 + 6x0 > 2x0 + 2x0x1 + 2x1 = [[!factimes#(_x0, _x1)]] [[!factimes#(_x0, !facplus(_x1, _x2))]] = 4 + 2x0x2 + 2x2 + 4x0x1 + 4x1 + 6x0 > 2x0 + 2x0x2 + 2x2 = [[!factimes#(_x0, _x2)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 4 + 2x1 + 2x1x2 + 4x0 + 4x0x2 + 6x2 > 2x0 + 2x0x2 + 2x2 = [[!factimes#(_x2, _x0)]] [[!factimes#(!facplus(_x0, _x1), _x2)]] = 4 + 2x1 + 2x1x2 + 4x0 + 4x0x2 + 6x2 > 2x1 + 2x1x2 + 2x2 = [[!factimes#(_x2, _x1)]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 2x0 + 2x0x1 + 2x0x1x2 + 2x0x2 + 2x1 + 2x1x2 + 2x2 >= 2x0 + 2x0x1 + 2x0x1x2 + 2x0x2 + 2x1 + 2x1x2 + 2x2 = [[!factimes#(_x0, !factimes(_x1, _x2))]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 2x0 + 2x0x1 + 2x0x1x2 + 2x0x2 + 2x1 + 2x1x2 + 2x2 >= 2x1 + 2x1x2 + 2x2 = [[!factimes#(_x1, _x2)]] [[!factimes(_x0, !facplus(_x1, _x2))]] = 2 + x2 + 2x0x1 + 2x1 + 3x0 + x0x2 >= 2 + x2 + 2x0x1 + 2x1 + 3x0 + x0x2 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] [[!factimes(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 + 2x0x2 + 3x2 + x1x2 >= 2 + x1 + 2x0 + 2x0x2 + 3x2 + x1x2 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] [[!factimes(!factimes(_x0, _x1), _x2)]] = x0 + x1 + x2 + x0x1 + x0x1x2 + x0x2 + x1x2 >= x0 + x1 + x2 + x0x1 + x0x1x2 + x0x2 + x1x2 = [[!factimes(_x0, !factimes(_x1, _x2))]] [[!facplus(!facplus(_x0, _x1), _x2)]] = 6 + x2 + 2x1 + 4x0 >= 4 + x2 + 2x0 + 2x1 = [[!facplus(_x0, !facplus(_x1, _x2))]] 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: !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z) 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). The formative rules of (P_4, R_2) are R_3 ::= !factimes(!factimes(X, Y), Z) => !factimes(X, !factimes(Y, Z)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_4, R_2, static, formative) by (P_4, R_3, static, formative). Thus, the original system is terminating if (P_4, R_3, static, formative) is finite. We consider the dependency pair problem (P_4, 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: !factimes#(!factimes(X, Y), Z) >? !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) >? !factimes#(Y, Z) !factimes(!factimes(X, Y), Z) >= !factimes(X, !factimes(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !factimes = \y0y1.1 + y0 + y1 !factimes# = \y0y1.y1 + 2y0 Using this interpretation, the requirements translate to: [[!factimes#(!factimes(_x0, _x1), _x2)]] = 2 + x2 + 2x0 + 2x1 > 1 + x1 + x2 + 2x0 = [[!factimes#(_x0, !factimes(_x1, _x2))]] [[!factimes#(!factimes(_x0, _x1), _x2)]] = 2 + x2 + 2x0 + 2x1 > x2 + 2x1 = [[!factimes#(_x1, _x2)]] [[!factimes(!factimes(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[!factimes(_x0, !factimes(_x1, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, 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.