We consider the system Applicative_first_order_05__#3.6. Alphabet: 0 : [] --> b cons : [c * d] --> d false : [] --> a filter : [c -> a * d] --> d filter2 : [a * c -> a * c * d] --> d gcd : [b * b] --> b if!fac6220gcd : [a * b * b] --> b le : [b * b] --> a map : [c -> c * d] --> d minus : [b * b] --> b nil : [] --> d pred : [b] --> b s : [b] --> b true : [] --> a Rules: le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) pred(s(x)) => x minus(x, 0) => x minus(x, s(y)) => pred(minus(x, y)) gcd(0, x) => x gcd(s(x), 0) => s(x) gcd(s(x), s(y)) => if!fac6220gcd(le(y, x), s(x), s(y)) if!fac6220gcd(true, s(x), s(y)) => gcd(minus(x, y), s(y)) if!fac6220gcd(false, s(x), s(y)) => gcd(minus(y, x), s(x)) 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] le#(s(X), s(Y)) =#> le#(X, Y) 1] minus#(X, s(Y)) =#> pred#(minus(X, Y)) 2] minus#(X, s(Y)) =#> minus#(X, Y) 3] gcd#(s(X), s(Y)) =#> if!fac6220gcd#(le(Y, X), s(X), s(Y)) 4] gcd#(s(X), s(Y)) =#> le#(Y, X) 5] if!fac6220gcd#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) 6] if!fac6220gcd#(true, s(X), s(Y)) =#> minus#(X, Y) 7] if!fac6220gcd#(false, s(X), s(Y)) =#> gcd#(minus(Y, X), s(X)) 8] if!fac6220gcd#(false, s(X), s(Y)) =#> minus#(Y, X) 9] map#(F, cons(X, Y)) =#> map#(F, Y) 10] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 11] filter2#(true, F, X, Y) =#> filter#(F, Y) 12] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) gcd(0, X) => X gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if!fac6220gcd(le(Y, X), s(X), s(Y)) if!fac6220gcd(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if!fac6220gcd(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 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, 2 * 3 : 5, 6, 7, 8 * 4 : 0 * 5 : 3, 4 * 6 : 1, 2 * 7 : 3, 4 * 8 : 1, 2 * 9 : 9 * 10 : 11, 12 * 11 : 10 * 12 : 10 This graph has the following strongly connected components: P_1: le#(s(X), s(Y)) =#> le#(X, Y) P_2: minus#(X, s(Y)) =#> minus#(X, Y) P_3: gcd#(s(X), s(Y)) =#> if!fac6220gcd#(le(Y, X), s(X), s(Y)) if!fac6220gcd#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) if!fac6220gcd#(false, s(X), s(Y)) =#> gcd#(minus(Y, X), s(X)) P_4: map#(F, cons(X, Y)) =#> map#(F, Y) P_5: 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), (P_4, R_0, m, f) and (P_5, 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), (P_4, 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) filter2#(false, F, X, Y) >? filter#(F, Y) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) gcd(0, X) >= X gcd(s(X), 0) >= s(X) gcd(s(X), s(Y)) >= if!fac6220gcd(le(Y, X), s(X), s(Y)) if!fac6220gcd(true, s(X), s(Y)) >= gcd(minus(X, Y), s(Y)) if!fac6220gcd(false, s(X), s(Y)) >= gcd(minus(Y, X), s(X)) 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 = 3 cons = \y0y1.3 + y1 + 2y0 false = 0 filter = \G0y1.2y1 + 3y1G0(y1) filter2 = \y0G1y2y3.2 + 2y0 + 2y2 + 2y3 + 2G1(y3) + 3y3G1(y3) filter2# = \y0G1y2y3.2y0 + 2G1(y3) + 3y3G1(y3) filter# = \G0y1.3y1G0(y1) gcd = \y0y1.2y0 + 2y1 if!fac6220gcd = \y0y1y2.2y1 + 2y2 le = \y0y1.1 map = \G0y1.y1 + 3y1G0(y1) minus = \y0y1.y0 nil = 0 pred = \y0.y0 s = \y0.y0 true = 1 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) >= 2F0(x1) + 2F0(x2) + 3x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 2 + 2F0(x2) + 3x2F0(x2) > 3x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2F0(x2) + 3x2F0(x2) >= 3x2F0(x2) = [[filter#(_F0, _x2)]] [[le(0, _x0)]] = 1 >= 1 = [[true]] [[le(s(_x0), 0)]] = 1 >= 0 = [[false]] [[le(s(_x0), s(_x1))]] = 1 >= 1 = [[le(_x0, _x1)]] [[pred(s(_x0))]] = x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] [[gcd(0, _x0)]] = 6 + 2x0 >= x0 = [[_x0]] [[gcd(s(_x0), 0)]] = 6 + 2x0 >= x0 = [[s(_x0)]] [[gcd(s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[if!fac6220gcd(le(_x1, _x0), s(_x0), s(_x1))]] [[if!fac6220gcd(true, s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[gcd(minus(_x0, _x1), s(_x1))]] [[if!fac6220gcd(false, s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[gcd(minus(_x1, _x0), s(_x0))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x2 + 2x1 + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) >= 3 + x2 + 2F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) >= 2 + 2x1 + 2x2 + 2F0(x1) + 2F0(x2) + 3x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 4 + 2x1 + 2x2 + 2F0(x2) + 3x2F0(x2) >= 3 + 2x1 + 2x2 + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 + 2F0(x2) + 3x2F0(x2) >= 2x2 + 3x2F0(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_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) filter2#(false, 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), (P_4, 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 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#(false, F, X, Y) >? filter#(F, Y) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) gcd(0, X) >= X gcd(s(X), 0) >= s(X) gcd(s(X), s(Y)) >= if!fac6220gcd(le(Y, X), s(X), s(Y)) if!fac6220gcd(true, s(X), s(Y)) >= gcd(minus(X, Y), s(Y)) if!fac6220gcd(false, s(X), s(Y)) >= gcd(minus(Y, X), s(X)) 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 = 3 cons = \y0y1.3 + y0 + y1 false = 2 filter = \G0y1.y1 filter2 = \y0G1y2y3.3 + y2 + y3 filter2# = \y0G1y2y3.2y0 + 2G1(y3) + y3G1(y3) filter# = \G0y1.3 + G0(y1) + y1G0(y1) gcd = \y0y1.y1 + 2y0 if!fac6220gcd = \y0y1y2.y1 + y2 le = \y0y1.2 map = \G0y1.y1 + y1G0(y1) minus = \y0y1.y0 nil = 0 pred = \y0.y0 s = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 3 + 4F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) > 2F0(x1) + 2F0(x2) + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 4 + 2F0(x2) + x2F0(x2) > 3 + F0(x2) + x2F0(x2) = [[filter#(_F0, _x2)]] [[le(0, _x0)]] = 2 >= 0 = [[true]] [[le(s(_x0), 0)]] = 2 >= 2 = [[false]] [[le(s(_x0), s(_x1))]] = 2 >= 2 = [[le(_x0, _x1)]] [[pred(s(_x0))]] = 2x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] [[gcd(0, _x0)]] = 6 + x0 >= x0 = [[_x0]] [[gcd(s(_x0), 0)]] = 3 + 4x0 >= 2x0 = [[s(_x0)]] [[gcd(s(_x0), s(_x1))]] = 2x1 + 4x0 >= 2x0 + 2x1 = [[if!fac6220gcd(le(_x1, _x0), s(_x0), s(_x1))]] [[if!fac6220gcd(true, s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[gcd(minus(_x0, _x1), s(_x1))]] [[if!fac6220gcd(false, s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[gcd(minus(_x1, _x0), s(_x0))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 3F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) >= 3 + x2 + F0(x1) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 >= 3 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 3 + x1 + x2 >= 3 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 3 + x1 + x2 >= x2 = [[filter(_F0, _x2)]] 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. 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 with usable rules [Kop12, Thm. 7.44]. (P_4, 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_4, 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), (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]. The usable rules of (P_3, R_0) are: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: gcd#(s(X), s(Y)) >? if!fac6220gcd#(le(Y, X), s(X), s(Y)) if!fac6220gcd#(true, s(X), s(Y)) >? gcd#(minus(X, Y), s(Y)) if!fac6220gcd#(false, s(X), s(Y)) >? gcd#(minus(Y, X), s(X)) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 false = 2 gcd# = \y0y1.2y1 + 3y0 if!fac6220gcd# = \y0y1y2.2y0 + 2y1 + 2y2 le = \y0y1.y1 minus = \y0y1.y0 pred = \y0.y0 s = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[gcd#(s(_x0), s(_x1))]] = 4x1 + 6x0 >= 4x1 + 6x0 = [[if!fac6220gcd#(le(_x1, _x0), s(_x0), s(_x1))]] [[if!fac6220gcd#(true, s(_x0), s(_x1))]] = 4x0 + 4x1 >= 3x0 + 4x1 = [[gcd#(minus(_x0, _x1), s(_x1))]] [[if!fac6220gcd#(false, s(_x0), s(_x1))]] = 4 + 4x0 + 4x1 > 3x1 + 4x0 = [[gcd#(minus(_x1, _x0), s(_x0))]] [[le(0, _x0)]] = x0 >= 0 = [[true]] [[le(s(_x0), 0)]] = 3 >= 2 = [[false]] [[le(s(_x0), s(_x1))]] = 2x1 >= x1 = [[le(_x0, _x1)]] [[pred(s(_x0))]] = 2x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] 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_7, R_0, static, formative), where P_7 consists of: gcd#(s(X), s(Y)) =#> if!fac6220gcd#(le(Y, X), s(X), s(Y)) if!fac6220gcd#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (P_7, R_0, static, formative) is finite. We consider the dependency pair problem (P_7, R_0, static, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_7, R_0) are: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: gcd#(s(X), s(Y)) >? if!fac6220gcd#(le(Y, X), s(X), s(Y)) if!fac6220gcd#(true, s(X), s(Y)) >? gcd#(minus(X, Y), s(Y)) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 false = 0 gcd# = \y0y1.y0 if!fac6220gcd# = \y0y1y2.y1 le = \y0y1.0 minus = \y0y1.y0 pred = \y0.y0 s = \y0.1 + y0 true = 0 Using this interpretation, the requirements translate to: [[gcd#(s(_x0), s(_x1))]] = 1 + x0 >= 1 + x0 = [[if!fac6220gcd#(le(_x1, _x0), s(_x0), s(_x1))]] [[if!fac6220gcd#(true, s(_x0), s(_x1))]] = 1 + x0 > x0 = [[gcd#(minus(_x0, _x1), s(_x1))]] [[le(0, _x0)]] = 0 >= 0 = [[true]] [[le(s(_x0), 0)]] = 0 >= 0 = [[false]] [[le(s(_x0), s(_x1))]] = 0 >= 0 = [[le(_x0, _x1)]] [[pred(s(_x0))]] = 1 + x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_7, R_0, static, formative) by (P_8, R_0, static, formative), where P_8 consists of: gcd#(s(X), s(Y)) =#> if!fac6220gcd#(le(Y, X), s(X), s(Y)) Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (P_8, R_0, static, formative) is finite. We consider the dependency pair problem (P_8, 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 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: minus#(X, s(Y)) >? minus#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: minus# = \y0y1.y1 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[minus#(_x0, s(_x1))]] = 1 + 2x1 > x1 = [[minus#(_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: le#(s(X), s(Y)) >? le#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: le# = \y0y1.y0 + y1 s = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[le#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[le#(_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.