We consider the system Applicative_first_order_05__#3.18. Alphabet: 0 : [] --> a cons : [c * d] --> d double : [a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d minus : [a * a] --> a nil : [] --> d plus : [a * a] --> a s : [a] --> a true : [] --> b Rules: minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) double(0) => 0 double(s(x)) => s(s(double(x))) plus(0, x) => x plus(s(x), y) => s(plus(x, y)) plus(s(x), y) => plus(x, s(y)) plus(s(x), y) => s(plus(minus(x, y), double(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] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] double#(s(X)) =#> double#(X) 2] plus#(s(X), Y) =#> plus#(X, Y) 3] plus#(s(X), Y) =#> plus#(X, s(Y)) 4] plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y)) 5] plus#(s(X), Y) =#> minus#(X, Y) 6] plus#(s(X), Y) =#> double#(Y) 7] map#(F, cons(X, Y)) =#> map#(F, Y) 8] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 9] filter2#(true, F, X, Y) =#> filter#(F, Y) 10] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) double(0) => 0 double(s(X)) => s(s(double(X))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) plus(s(X), Y) => plus(X, s(Y)) plus(s(X), Y) => s(plus(minus(X, Y), double(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 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: minus#(s(X), s(Y)) >? minus#(X, Y) double#(s(X)) >? double#(X) plus#(s(X), Y) >? plus#(X, Y) plus#(s(X), Y) >? plus#(X, s(Y)) plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) plus#(s(X), Y) >? minus#(X, Y) plus#(s(X), Y) >? double#(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) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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 cons = \y0y1.0 double = \y0.0 double# = \y0.0 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 filter2# = \y0G1y2y3.0 filter# = \G0y1.0 map = \G0y1.2G0(0) + 2G0(y1) map# = \G0y1.0 minus = \y0y1.y0 minus# = \y0y1.0 nil = 0 plus = \y0y1.y1 plus# = \y0y1.3 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] [[double#(s(_x0))]] = 0 >= 0 = [[double#(_x0)]] [[plus#(s(_x0), _x1)]] = 3 >= 3 = [[plus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 3 >= 3 = [[plus#(_x0, s(_x1))]] [[plus#(s(_x0), _x1)]] = 3 >= 3 = [[plus#(minus(_x0, _x1), double(_x1))]] [[plus#(s(_x0), _x1)]] = 3 > 0 = [[minus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 3 > 0 = [[double#(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[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)]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] [[double(0)]] = 0 >= 0 = [[0]] [[double(s(_x0))]] = 0 >= 0 = [[s(s(double(_x0)))]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] [[plus(s(_x0), _x1)]] = x1 >= x1 = [[plus(_x0, s(_x1))]] [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(minus(_x0, _x1), double(_x1)))]] [[map(_F0, nil)]] = 4F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 4F0(0) >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[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_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of: minus#(s(X), s(Y)) =#> minus#(X, Y) double#(s(X)) =#> double#(X) plus#(s(X), Y) =#> plus#(X, Y) plus#(s(X), Y) =#> plus#(X, s(Y)) plus#(s(X), Y) =#> plus#(minus(X, Y), double(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) 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: minus#(s(X), s(Y)) >? minus#(X, Y) double#(s(X)) >? double#(X) plus#(s(X), Y) >? plus#(X, Y) plus#(s(X), Y) >? plus#(X, s(Y)) plus#(s(X), Y) >? plus#(minus(X, Y), double(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) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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 cons = \y0y1.1 + 2y1 double = \y0.0 double# = \y0.0 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.2G1(0) + y3G1(y3) filter# = \G0y1.G0(0) + y1G0(y1) map = \G0y1.y1 map# = \G0y1.y1 minus = \y0y1.y0 minus# = \y0y1.0 nil = 0 plus = \y0y1.2y1 plus# = \y0y1.0 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] [[double#(s(_x0))]] = 0 >= 0 = [[double#(_x0)]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, s(_x1))]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(minus(_x0, _x1), double(_x1))]] [[map#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = F0(0) + F0(1 + 2x2) + 2x2F0(1 + 2x2) >= 2F0(0) + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 2F0(0) + x2F0(x2) >= F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2F0(0) + x2F0(x2) >= F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] [[double(0)]] = 0 >= 0 = [[0]] [[double(s(_x0))]] = 0 >= 0 = [[s(s(double(_x0)))]] [[plus(0, _x0)]] = 2x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 2x1 >= 2x1 = [[s(plus(_x0, _x1))]] [[plus(s(_x0), _x1)]] = 2x1 >= 2x1 = [[plus(_x0, s(_x1))]] [[plus(s(_x0), _x1)]] = 2x1 >= 0 = [[s(plus(minus(_x0, _x1), double(_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_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of: minus#(s(X), s(Y)) =#> minus#(X, Y) double#(s(X)) =#> double#(X) plus#(s(X), Y) =#> plus#(X, Y) plus#(s(X), Y) =#> plus#(X, s(Y)) plus#(s(X), Y) =#> plus#(minus(X, Y), double(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_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: minus#(s(X), s(Y)) >? minus#(X, Y) double#(s(X)) >? double#(X) plus#(s(X), Y) >? plus#(X, Y) plus#(s(X), Y) >? plus#(X, s(Y)) plus#(s(X), Y) >? plus#(minus(X, Y), double(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) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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 cons = \y0y1.1 + 2y1 double = \y0.0 double# = \y0.0 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.2y3 filter# = \G0y1.y1 map = \G0y1.y1 minus = \y0y1.y0 minus# = \y0y1.0 nil = 0 plus = \y0y1.y1 plus# = \y0y1.0 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] [[double#(s(_x0))]] = 0 >= 0 = [[double#(_x0)]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, s(_x1))]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(minus(_x0, _x1), double(_x1))]] [[filter#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2 >= x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2 >= x2 = [[filter#(_F0, _x2)]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] [[double(0)]] = 0 >= 0 = [[0]] [[double(s(_x0))]] = 0 >= 0 = [[s(s(double(_x0)))]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] [[plus(s(_x0), _x1)]] = x1 >= x1 = [[plus(_x0, s(_x1))]] [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(minus(_x0, _x1), double(_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_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of: minus#(s(X), s(Y)) =#> minus#(X, Y) double#(s(X)) =#> double#(X) plus#(s(X), Y) =#> plus#(X, Y) plus#(s(X), Y) =#> plus#(X, s(Y)) plus#(s(X), Y) =#> plus#(minus(X, Y), double(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_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: minus#(s(X), s(Y)) >? minus#(X, Y) double#(s(X)) >? double#(X) plus#(s(X), Y) >? plus#(X, Y) plus#(s(X), Y) >? plus#(X, s(Y)) plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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 cons = \y0y1.0 double = \y0.0 double# = \y0.0 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 filter2# = \y0G1y2y3.3 filter# = \G0y1.0 map = \G0y1.0 minus = \y0y1.y0 minus# = \y0y1.0 nil = 0 plus = \y0y1.y1 plus# = \y0y1.0 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(s(_x0), s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] [[double#(s(_x0))]] = 0 >= 0 = [[double#(_x0)]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, s(_x1))]] [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(minus(_x0, _x1), double(_x1))]] [[filter2#(true, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = x0 >= x0 = [[minus(_x0, _x1)]] [[double(0)]] = 0 >= 0 = [[0]] [[double(s(_x0))]] = 0 >= 0 = [[s(s(double(_x0)))]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x1 >= x1 = [[s(plus(_x0, _x1))]] [[plus(s(_x0), _x1)]] = x1 >= x1 = [[plus(_x0, s(_x1))]] [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(minus(_x0, _x1), double(_x1)))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[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_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of: minus#(s(X), s(Y)) =#> minus#(X, Y) double#(s(X)) =#> double#(X) plus#(s(X), Y) =#> plus#(X, Y) plus#(s(X), Y) =#> plus#(X, s(Y)) plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y)) Thus, the original system is terminating if (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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: minus#(s(X), s(Y)) >? minus#(X, Y) double#(s(X)) >? double#(X) plus#(s(X), Y) >? plus#(X, Y) plus#(s(X), Y) >? plus#(X, s(Y)) plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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)]] = cons(x_2) [[filter(x_1, x_2)]] = x_2 [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4) [[minus(x_1, x_2)]] = x_1 [[nil]] = _|_ We choose Lex = {plus, plus#} and Mul = {@_{o -> o}, cons, double, double#, false, filter2, map, minus#, s, true}, and the following precedence: @_{o -> o} > double# > false > map > cons = filter2 > minus# > plus > plus# > double > s > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: minus#(s(X), s(Y)) >= minus#(X, Y) double#(s(X)) >= double#(X) plus#(s(X), Y) >= plus#(X, Y) plus#(s(X), Y) > plus#(X, s(Y)) plus#(s(X), Y) >= plus#(X, double(Y)) X >= X s(X) >= X double(_|_) >= _|_ double(s(X)) >= s(s(double(X))) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(X, double(Y))) map(F, _|_) >= _|_ map(F, cons(X)) >= cons(map(F, X)) _|_ >= _|_ cons(X) >= filter2(X) filter2(X) >= cons(X) filter2(X) >= X With these choices, we have: 1] minus#(s(X), s(Y)) >= minus#(X, Y) because minus# in Mul, [2] and [5], by (Fun) 2] s(X) >= X because [3], by (Star) 3] s*(X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] s(Y) >= Y because [6], by (Star) 6] s*(Y) >= Y because [7], by (Select) 7] Y >= Y by (Meta) 8] double#(s(X)) >= double#(X) because double# in Mul and [9], by (Fun) 9] s(X) >= X because [10], by (Star) 10] s*(X) >= X because [11], by (Select) 11] X >= X by (Meta) 12] plus#(s(X), Y) >= plus#(X, Y) because [13] and [16], by (Fun) 13] s(X) >= X because [14], by (Star) 14] s*(X) >= X because [15], by (Select) 15] X >= X by (Meta) 16] Y >= Y by (Meta) 17] plus#(s(X), Y) > plus#(X, s(Y)) because [18], by definition 18] plus#*(s(X), Y) >= plus#(X, s(Y)) because [19], [22] and [24], by (Stat) 19] s(X) > X because [20], by definition 20] s*(X) >= X because [21], by (Select) 21] X >= X by (Meta) 22] plus#*(s(X), Y) >= X because [23], by (Select) 23] s(X) >= X because [20], by (Star) 24] plus#*(s(X), Y) >= s(Y) because plus# > s and [25], by (Copy) 25] plus#*(s(X), Y) >= Y because [26], by (Select) 26] Y >= Y by (Meta) 27] plus#(s(X), Y) >= plus#(X, double(Y)) because [28], by (Star) 28] plus#*(s(X), Y) >= plus#(X, double(Y)) because [29], [32] and [34], by (Stat) 29] s(X) > X because [30], by definition 30] s*(X) >= X because [31], by (Select) 31] X >= X by (Meta) 32] plus#*(s(X), Y) >= X because [33], by (Select) 33] s(X) >= X because [30], by (Star) 34] plus#*(s(X), Y) >= double(Y) because plus# > double and [35], by (Copy) 35] plus#*(s(X), Y) >= Y because [36], by (Select) 36] Y >= Y by (Meta) 37] X >= X by (Meta) 38] s(X) >= X because [3], by (Star) 39] double(_|_) >= _|_ by (Bot) 40] double(s(X)) >= s(s(double(X))) because [41], by (Star) 41] double*(s(X)) >= s(s(double(X))) because double > s and [42], by (Copy) 42] double*(s(X)) >= s(double(X)) because double > s and [43], by (Copy) 43] double*(s(X)) >= double(X) because double in Mul and [44], by (Stat) 44] s(X) > X because [45], by definition 45] s*(X) >= X because [11], by (Select) 46] plus(_|_, X) >= X because [47], by (Star) 47] plus*(_|_, X) >= X because [48], by (Select) 48] X >= X by (Meta) 49] plus(s(X), Y) >= s(plus(X, Y)) because [50], by (Star) 50] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [51], by (Copy) 51] plus*(s(X), Y) >= plus(X, Y) because [52], [54] and [55], by (Stat) 52] s(X) > X because [53], by definition 53] s*(X) >= X because [15], by (Select) 54] plus*(s(X), Y) >= X because [13], by (Select) 55] plus*(s(X), Y) >= Y because [16], by (Select) 56] plus(s(X), Y) >= plus(X, s(Y)) because [57], by (Star) 57] plus*(s(X), Y) >= plus(X, s(Y)) because [19], [58] and [59], by (Stat) 58] plus*(s(X), Y) >= X because [23], by (Select) 59] plus*(s(X), Y) >= s(Y) because plus > s and [60], by (Copy) 60] plus*(s(X), Y) >= Y because [26], by (Select) 61] plus(s(X), Y) >= s(plus(X, double(Y))) because [62], by (Star) 62] plus*(s(X), Y) >= s(plus(X, double(Y))) because plus > s and [63], by (Copy) 63] plus*(s(X), Y) >= plus(X, double(Y)) because [29], [64] and [65], by (Stat) 64] plus*(s(X), Y) >= X because [33], by (Select) 65] plus*(s(X), Y) >= double(Y) because plus > double and [66], by (Copy) 66] plus*(s(X), Y) >= Y because [36], by (Select) 67] map(F, _|_) >= _|_ by (Bot) 68] map(F, cons(X)) >= cons(map(F, X)) because [69], by (Star) 69] map*(F, cons(X)) >= cons(map(F, X)) because map > cons and [70], by (Copy) 70] map*(F, cons(X)) >= map(F, X) because map in Mul, [71] and [72], by (Stat) 71] F >= F by (Meta) 72] cons(X) > X because [73], by definition 73] cons*(X) >= X because [74], by (Select) 74] X >= X by (Meta) 75] _|_ >= _|_ by (Bot) 76] cons(X) >= filter2(X) because cons = filter2, cons in Mul and [77], by (Fun) 77] X >= X by (Meta) 78] filter2(X) >= cons(X) because filter2 = cons, filter2 in Mul and [79], by (Fun) 79] X >= X by (Meta) 80] filter2(X) >= X because [81], by (Star) 81] filter2*(X) >= X because [82], by (Select) 82] X >= X by (Meta) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_0, static, formative) by (P_5, R_0, static, formative), where P_5 consists of: minus#(s(X), s(Y)) =#> minus#(X, Y) double#(s(X)) =#> double#(X) plus#(s(X), Y) =#> plus#(X, Y) plus#(s(X), Y) =#> plus#(minus(X, Y), double(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: minus#(s(X), s(Y)) >? minus#(X, Y) double#(s(X)) >? double#(X) plus#(s(X), Y) >? plus#(X, Y) plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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 [[minus(x_1, x_2)]] = minus(x_1) [[minus#(x_1, x_2)]] = x_1 [[nil]] = _|_ We choose Lex = {plus, plus#} and Mul = {@_{o -> o}, double, double#, false, map, minus, s, true}, and the following precedence: @_{o -> o} > double# > false > map > plus > true > plus# > double > s > minus Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: s(X) >= X double#(s(X)) >= double#(X) plus#(s(X), Y) > plus#(X, Y) plus#(s(X), Y) >= plus#(minus(X), double(Y)) minus(X) >= X minus(s(X)) >= minus(X) double(_|_) >= _|_ double(s(X)) >= s(s(double(X))) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X), double(Y))) map >= _|_ map >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ With these choices, we have: 1] s(X) >= X because [2], by (Star) 2] s*(X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] double#(s(X)) >= double#(X) because double# in Mul and [5], by (Fun) 5] s(X) >= X because [6], by (Star) 6] s*(X) >= X because [7], by (Select) 7] X >= X by (Meta) 8] plus#(s(X), Y) > plus#(X, Y) because [9], by definition 9] plus#*(s(X), Y) >= plus#(X, Y) because [10], [13] and [15], by (Stat) 10] s(X) > X because [11], by definition 11] s*(X) >= X because [12], by (Select) 12] X >= X by (Meta) 13] plus#*(s(X), Y) >= X because [14], by (Select) 14] s(X) >= X because [11], by (Star) 15] plus#*(s(X), Y) >= Y because [16], by (Select) 16] Y >= Y by (Meta) 17] plus#(s(X), Y) >= plus#(minus(X), double(Y)) because [18], by (Star) 18] plus#*(s(X), Y) >= plus#(minus(X), double(Y)) because [19], [23] and [25], by (Stat) 19] s(X) > minus(X) because [20], by definition 20] s*(X) >= minus(X) because s > minus and [21], by (Copy) 21] s*(X) >= X because [22], by (Select) 22] X >= X by (Meta) 23] plus#*(s(X), Y) >= minus(X) because [24], by (Select) 24] s(X) >= minus(X) because [20], by (Star) 25] plus#*(s(X), Y) >= double(Y) because plus# > double and [26], by (Copy) 26] plus#*(s(X), Y) >= Y because [27], by (Select) 27] Y >= Y by (Meta) 28] minus(X) >= X because [29], by (Star) 29] minus*(X) >= X because [30], by (Select) 30] X >= X by (Meta) 31] minus(s(X)) >= minus(X) because [32], by (Star) 32] minus*(s(X)) >= minus(X) because [33], by (Select) 33] s(X) >= minus(X) because [34], by (Star) 34] s*(X) >= minus(X) because s > minus and [35], by (Copy) 35] s*(X) >= X because [3], by (Select) 36] double(_|_) >= _|_ by (Bot) 37] double(s(X)) >= s(s(double(X))) because [38], by (Star) 38] double*(s(X)) >= s(s(double(X))) because double > s and [39], by (Copy) 39] double*(s(X)) >= s(double(X)) because double > s and [40], by (Copy) 40] double*(s(X)) >= double(X) because double in Mul and [41], by (Stat) 41] s(X) > X because [42], by definition 42] s*(X) >= X because [7], by (Select) 43] plus(_|_, X) >= X because [44], by (Star) 44] plus*(_|_, X) >= X because [45], by (Select) 45] X >= X by (Meta) 46] plus(s(X), Y) >= s(plus(X, Y)) because [47], by (Star) 47] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [48], by (Copy) 48] plus*(s(X), Y) >= plus(X, Y) because [10], [49] and [50], by (Stat) 49] plus*(s(X), Y) >= X because [14], by (Select) 50] plus*(s(X), Y) >= Y because [16], by (Select) 51] plus(s(X), Y) >= plus(X, s(Y)) because [52], by (Star) 52] plus*(s(X), Y) >= plus(X, s(Y)) because [53], [56] and [58], by (Stat) 53] s(X) > X because [54], by definition 54] s*(X) >= X because [55], by (Select) 55] X >= X by (Meta) 56] plus*(s(X), Y) >= X because [57], by (Select) 57] s(X) >= X because [54], by (Star) 58] plus*(s(X), Y) >= s(Y) because plus > s and [59], by (Copy) 59] plus*(s(X), Y) >= Y because [60], by (Select) 60] Y >= Y by (Meta) 61] plus(s(X), Y) >= s(plus(minus(X), double(Y))) because [62], by (Star) 62] plus*(s(X), Y) >= s(plus(minus(X), double(Y))) because plus > s and [63], by (Copy) 63] plus*(s(X), Y) >= plus(minus(X), double(Y)) because [19], [64] and [65], by (Stat) 64] plus*(s(X), Y) >= minus(X) because [24], by (Select) 65] plus*(s(X), Y) >= double(Y) because plus > double and [66], by (Copy) 66] plus*(s(X), Y) >= Y because [27], by (Select) 67] map >= _|_ by (Bot) 68] map >= _|_ by (Bot) 69] _|_ >= _|_ by (Bot) 70] _|_ >= _|_ by (Bot) 71] _|_ >= _|_ by (Bot) 72] _|_ >= _|_ 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: minus#(s(X), s(Y)) =#> minus#(X, Y) double#(s(X)) =#> double#(X) plus#(s(X), Y) =#> plus#(minus(X, Y), double(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: minus#(s(X), s(Y)) >? minus#(X, Y) double#(s(X)) >? double#(X) plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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)]] = x_2 [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4) [[minus(x_1, x_2)]] = minus(x_1) [[minus#(x_1, x_2)]] = minus#(x_1) [[nil]] = _|_ [[plus#(x_1, x_2)]] = plus#(x_1) We choose Lex = {double#, plus} and Mul = {@_{o -> o}, double, false, filter, filter2, map, minus, minus#, plus#, s, true}, and the following precedence: @_{o -> o} > double# > false > filter = filter2 > map > plus > double > plus# > minus# = s > minus > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: minus#(s(X)) >= minus#(X) double#(s(X)) > double#(X) plus#(s(X)) >= plus#(minus(X)) minus(X) >= X minus(s(X)) >= minus(X) double(_|_) >= _|_ double(s(X)) >= s(s(double(X))) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X), double(Y))) map(F, _|_) >= _|_ map(F, X) >= map(F, X) filter(F, _|_) >= _|_ filter(F, X) >= filter2(F, X) filter2(F, X) >= filter(F, X) filter2(F, X) >= filter(F, X) With these choices, we have: 1] minus#(s(X)) >= minus#(X) because [2], by (Star) 2] minus#*(s(X)) >= minus#(X) because [3], by (Select) 3] s(X) >= minus#(X) because s = minus#, s in Mul and [4], by (Fun) 4] X >= X by (Meta) 5] double#(s(X)) > double#(X) because [6], by definition 6] double#*(s(X)) >= double#(X) because [7] and [10], by (Stat) 7] s(X) > X because [8], by definition 8] s*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] double#*(s(X)) >= X because [11], by (Select) 11] s(X) >= X because [8], by (Star) 12] plus#(s(X)) >= plus#(minus(X)) because plus# in Mul and [13], by (Fun) 13] s(X) >= minus(X) because [14], by (Star) 14] s*(X) >= minus(X) because s > minus and [15], by (Copy) 15] s*(X) >= X because [16], by (Select) 16] X >= X by (Meta) 17] minus(X) >= X because [18], by (Star) 18] minus*(X) >= X because [19], by (Select) 19] X >= X by (Meta) 20] minus(s(X)) >= minus(X) because [21], by (Star) 21] minus*(s(X)) >= minus(X) because [22], by (Select) 22] s(X) >= minus(X) because [23], by (Star) 23] s*(X) >= minus(X) because s > minus and [24], by (Copy) 24] s*(X) >= X because [4], by (Select) 25] double(_|_) >= _|_ by (Bot) 26] double(s(X)) >= s(s(double(X))) because [27], by (Star) 27] double*(s(X)) >= s(s(double(X))) because double > s and [28], by (Copy) 28] double*(s(X)) >= s(double(X)) because double > s and [29], by (Copy) 29] double*(s(X)) >= double(X) because double in Mul and [7], by (Stat) 30] plus(_|_, X) >= X because [31], by (Star) 31] plus*(_|_, X) >= X because [32], by (Select) 32] X >= X by (Meta) 33] plus(s(X), Y) >= s(plus(X, Y)) because [34], by (Star) 34] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [35], by (Copy) 35] plus*(s(X), Y) >= plus(X, Y) because [36], [39] and [41], by (Stat) 36] s(X) > X because [37], by definition 37] s*(X) >= X because [38], by (Select) 38] X >= X by (Meta) 39] plus*(s(X), Y) >= X because [40], by (Select) 40] s(X) >= X because [37], by (Star) 41] plus*(s(X), Y) >= Y because [42], by (Select) 42] Y >= Y by (Meta) 43] plus(s(X), Y) >= plus(X, s(Y)) because [44], by (Star) 44] plus*(s(X), Y) >= plus(X, s(Y)) because [45], [48] and [50], by (Stat) 45] s(X) > X because [46], by definition 46] s*(X) >= X because [47], by (Select) 47] X >= X by (Meta) 48] plus*(s(X), Y) >= X because [49], by (Select) 49] s(X) >= X because [46], by (Star) 50] plus*(s(X), Y) >= s(Y) because plus > s and [51], by (Copy) 51] plus*(s(X), Y) >= Y because [52], by (Select) 52] Y >= Y by (Meta) 53] plus(s(X), Y) >= s(plus(minus(X), double(Y))) because [54], by (Star) 54] plus*(s(X), Y) >= s(plus(minus(X), double(Y))) because plus > s and [55], by (Copy) 55] plus*(s(X), Y) >= plus(minus(X), double(Y)) because [56], [58] and [61], by (Stat) 56] s(X) > minus(X) because [57], by definition 57] s*(X) >= minus(X) because s > minus and [15], by (Copy) 58] plus*(s(X), Y) >= minus(X) because plus > minus and [59], by (Copy) 59] plus*(s(X), Y) >= X because [60], by (Select) 60] s(X) >= X because [15], by (Star) 61] plus*(s(X), Y) >= double(Y) because plus > double and [62], by (Copy) 62] plus*(s(X), Y) >= Y because [63], by (Select) 63] Y >= Y by (Meta) 64] map(F, _|_) >= _|_ by (Bot) 65] map(F, X) >= map(F, X) because map in Mul, [66] and [67], by (Fun) 66] F >= F by (Meta) 67] X >= X by (Meta) 68] filter(F, _|_) >= _|_ by (Bot) 69] filter(F, X) >= filter2(F, X) because filter = filter2, filter in Mul, [70] and [71], by (Fun) 70] F >= F by (Meta) 71] X >= X by (Meta) 72] filter2(F, X) >= filter(F, X) because filter2 = filter, filter2 in Mul, [73] and [74], by (Fun) 73] F >= F by (Meta) 74] X >= X by (Meta) 75] filter2(F, X) >= filter(F, X) because filter2 = filter, filter2 in Mul, [76] and [77], by (Fun) 76] F >= F by (Meta) 77] X >= X by (Meta) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_0, static, formative) by (P_7, R_0, static, formative), where P_7 consists of: minus#(s(X), s(Y)) =#> minus#(X, Y) plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y)) Thus, the original system is terminating if (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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: minus#(s(X), s(Y)) >? minus#(X, Y) plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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]] = _|_ [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) [[minus(x_1, x_2)]] = minus(x_1) [[nil]] = _|_ [[plus#(x_1, x_2)]] = plus#(x_1) We choose Lex = {@_{o -> o}, filter, filter2, plus} and Mul = {cons, double, false, map, minus, minus#, plus#, s, true}, and the following precedence: false > map > @_{o -> o} = filter = filter2 > cons > minus# > plus > double > plus# > s > minus > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: minus#(s(X), s(Y)) > minus#(X, Y) plus#(s(X)) >= plus#(minus(X)) minus(X) >= X minus(s(X)) >= minus(X) double(_|_) >= _|_ double(s(X)) >= s(s(double(X))) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X), double(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] minus#(s(X), s(Y)) > minus#(X, Y) because [2], by definition 2] minus#*(s(X), s(Y)) >= minus#(X, Y) because minus# in Mul, [3] and [6], by (Stat) 3] s(X) > X because [4], by definition 4] s*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] s(Y) >= Y because [7], by (Star) 7] s*(Y) >= Y because [8], by (Select) 8] Y >= Y by (Meta) 9] plus#(s(X)) >= plus#(minus(X)) because plus# in Mul and [10], by (Fun) 10] s(X) >= minus(X) because [11], by (Star) 11] s*(X) >= minus(X) because s > minus and [12], by (Copy) 12] s*(X) >= X because [13], by (Select) 13] X >= X by (Meta) 14] minus(X) >= X because [15], by (Star) 15] minus*(X) >= X because [16], by (Select) 16] X >= X by (Meta) 17] minus(s(X)) >= minus(X) because [18], by (Star) 18] minus*(s(X)) >= minus(X) because minus in Mul and [3], by (Stat) 19] double(_|_) >= _|_ by (Bot) 20] double(s(X)) >= s(s(double(X))) because [21], by (Star) 21] double*(s(X)) >= s(s(double(X))) because double > s and [22], by (Copy) 22] double*(s(X)) >= s(double(X)) because double > s and [23], by (Copy) 23] double*(s(X)) >= double(X) because double in Mul and [24], by (Stat) 24] s(X) > X because [25], by definition 25] s*(X) >= X because [26], by (Select) 26] X >= X by (Meta) 27] plus(_|_, X) >= X because [28], by (Star) 28] plus*(_|_, X) >= X because [29], by (Select) 29] X >= X by (Meta) 30] plus(s(X), Y) >= s(plus(X, Y)) because [31], by (Star) 31] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [32], by (Copy) 32] plus*(s(X), Y) >= plus(X, Y) because [33], [36] and [38], by (Stat) 33] s(X) > X because [34], by definition 34] s*(X) >= X because [35], by (Select) 35] X >= X by (Meta) 36] plus*(s(X), Y) >= X because [37], by (Select) 37] s(X) >= X because [34], by (Star) 38] plus*(s(X), Y) >= Y because [39], by (Select) 39] Y >= Y by (Meta) 40] plus(s(X), Y) >= plus(X, s(Y)) because [41], by (Star) 41] plus*(s(X), Y) >= plus(X, s(Y)) because [42], [45] and [47], by (Stat) 42] s(X) > X because [43], by definition 43] s*(X) >= X because [44], by (Select) 44] X >= X by (Meta) 45] plus*(s(X), Y) >= X because [46], by (Select) 46] s(X) >= X because [43], by (Star) 47] plus*(s(X), Y) >= s(Y) because plus > s and [48], by (Copy) 48] plus*(s(X), Y) >= Y because [49], by (Select) 49] Y >= Y by (Meta) 50] plus(s(X), Y) >= s(plus(minus(X), double(Y))) because [51], by (Star) 51] plus*(s(X), Y) >= s(plus(minus(X), double(Y))) because plus > s and [52], by (Copy) 52] plus*(s(X), Y) >= plus(minus(X), double(Y)) because [53], [55] and [56], by (Stat) 53] s(X) > minus(X) because [54], by definition 54] s*(X) >= minus(X) because s > minus and [12], by (Copy) 55] plus*(s(X), Y) >= minus(X) because [10], by (Select) 56] plus*(s(X), Y) >= double(Y) because plus > double and [57], by (Copy) 57] plus*(s(X), Y) >= Y because [58], by (Select) 58] Y >= Y by (Meta) 59] map(F, _|_) >= _|_ by (Bot) 60] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [61], by (Star) 61] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [62] and [69], by (Copy) 62] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [63] and [65], by (Copy) 63] map*(F, cons(X, Y)) >= F because [64], by (Select) 64] F >= F by (Meta) 65] map*(F, cons(X, Y)) >= X because [66], by (Select) 66] cons(X, Y) >= X because [67], by (Star) 67] cons*(X, Y) >= X because [68], by (Select) 68] X >= X by (Meta) 69] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [70] and [71], by (Stat) 70] F >= F by (Meta) 71] cons(X, Y) > Y because [72], by definition 72] cons*(X, Y) >= Y because [73], by (Select) 73] Y >= Y by (Meta) 74] filter(F, _|_) >= _|_ by (Bot) 75] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [76], by (Star) 76] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [77], [78], [81], [85], [86] and [88], by (Stat) 77] F >= F by (Meta) 78] cons(X, Y) > Y because [79], by definition 79] cons*(X, Y) >= Y because [80], by (Select) 80] Y >= Y by (Meta) 81] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter = @_{o -> o}, [77], [82], [85] and [86], by (Stat) 82] cons(X, Y) > X because [83], by definition 83] cons*(X, Y) >= X because [84], by (Select) 84] X >= X by (Meta) 85] filter*(F, cons(X, Y)) >= F because [77], by (Select) 86] filter*(F, cons(X, Y)) >= X because [87], by (Select) 87] cons(X, Y) >= X because [83], by (Star) 88] filter*(F, cons(X, Y)) >= Y because [89], by (Select) 89] cons(X, Y) >= Y because [79], by (Star) 90] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [91], by (Star) 91] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [92] and [94], by (Copy) 92] filter2*(true, F, X, Y) >= X because [93], by (Select) 93] X >= X by (Meta) 94] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [95], [96], [97] and [98], by (Stat) 95] F >= F by (Meta) 96] Y >= Y by (Meta) 97] filter2*(true, F, X, Y) >= F because [95], by (Select) 98] filter2*(true, F, X, Y) >= Y because [96], by (Select) 99] filter2(false, F, X, Y) >= filter(F, Y) because [100], by (Star) 100] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [101], [102], [103] and [104], by (Stat) 101] F >= F by (Meta) 102] Y >= Y by (Meta) 103] filter2*(false, F, X, Y) >= F because [101], by (Select) 104] filter2*(false, F, X, Y) >= Y because [102], by (Select) 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: plus#(s(X), Y) =#> plus#(minus(X, Y), double(Y)) Thus, the original system is terminating if (P_8, R_0, static, formative) is finite. We consider the dependency pair problem (P_8, 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: plus#(s(X), Y) >? plus#(minus(X, Y), double(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) double(0) >= 0 double(s(X)) >= s(s(double(X))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X, Y), double(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 [[minus(x_1, x_2)]] = minus(x_1) [[nil]] = _|_ [[plus#(x_1, x_2)]] = plus#(x_1) We choose Lex = {plus} and Mul = {@_{o -> o}, double, false, map, minus, plus#, s, true}, and the following precedence: @_{o -> o} > false > map > plus > double > s > minus > plus# > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: plus#(s(X)) > plus#(minus(X)) minus(X) >= X minus(s(X)) >= minus(X) double(_|_) >= _|_ double(s(X)) >= s(s(double(X))) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) plus(s(X), Y) >= plus(X, s(Y)) plus(s(X), Y) >= s(plus(minus(X), double(Y))) map >= _|_ map >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ With these choices, we have: 1] plus#(s(X)) > plus#(minus(X)) because [2], by definition 2] plus#*(s(X)) >= plus#(minus(X)) because [3], by (Select) 3] s(X) >= plus#(minus(X)) because [4], by (Star) 4] s*(X) >= plus#(minus(X)) because s > plus# and [5], by (Copy) 5] s*(X) >= minus(X) because s > minus and [6], by (Copy) 6] s*(X) >= X because [7], by (Select) 7] X >= X by (Meta) 8] minus(X) >= X because [9], by (Star) 9] minus*(X) >= X because [10], by (Select) 10] X >= X by (Meta) 11] minus(s(X)) >= minus(X) because [12], by (Star) 12] minus*(s(X)) >= minus(X) because minus in Mul and [13], by (Stat) 13] s(X) > X because [14], by definition 14] s*(X) >= X because [15], by (Select) 15] X >= X by (Meta) 16] double(_|_) >= _|_ by (Bot) 17] double(s(X)) >= s(s(double(X))) because [18], by (Star) 18] double*(s(X)) >= s(s(double(X))) because double > s and [19], by (Copy) 19] double*(s(X)) >= s(double(X)) because double > s and [20], by (Copy) 20] double*(s(X)) >= double(X) because double in Mul and [21], by (Stat) 21] s(X) > X because [22], by definition 22] s*(X) >= X because [23], by (Select) 23] X >= X by (Meta) 24] plus(_|_, X) >= X because [25], by (Star) 25] plus*(_|_, X) >= X because [26], by (Select) 26] X >= X by (Meta) 27] plus(s(X), Y) >= s(plus(X, Y)) because [28], by (Star) 28] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [29], by (Copy) 29] plus*(s(X), Y) >= plus(X, Y) because [30], [33] and [35], by (Stat) 30] s(X) > X because [31], by definition 31] s*(X) >= X because [32], by (Select) 32] X >= X by (Meta) 33] plus*(s(X), Y) >= X because [34], by (Select) 34] s(X) >= X because [31], by (Star) 35] plus*(s(X), Y) >= Y because [36], by (Select) 36] Y >= Y by (Meta) 37] plus(s(X), Y) >= plus(X, s(Y)) because [38], by (Star) 38] plus*(s(X), Y) >= plus(X, s(Y)) because [39], [42] and [44], by (Stat) 39] s(X) > X because [40], by definition 40] s*(X) >= X because [41], by (Select) 41] X >= X by (Meta) 42] plus*(s(X), Y) >= X because [43], by (Select) 43] s(X) >= X because [40], by (Star) 44] plus*(s(X), Y) >= s(Y) because plus > s and [45], by (Copy) 45] plus*(s(X), Y) >= Y because [46], by (Select) 46] Y >= Y by (Meta) 47] plus(s(X), Y) >= s(plus(minus(X), double(Y))) because [48], by (Star) 48] plus*(s(X), Y) >= s(plus(minus(X), double(Y))) because plus > s and [49], by (Copy) 49] plus*(s(X), Y) >= plus(minus(X), double(Y)) because [50], [51] and [53], by (Stat) 50] s(X) > minus(X) because [5], by definition 51] plus*(s(X), Y) >= minus(X) because [52], by (Select) 52] s(X) >= minus(X) because [5], by (Star) 53] plus*(s(X), Y) >= double(Y) because plus > double and [54], by (Copy) 54] plus*(s(X), Y) >= Y because [55], by (Select) 55] Y >= Y by (Meta) 56] map >= _|_ by (Bot) 57] map >= _|_ by (Bot) 58] _|_ >= _|_ by (Bot) 59] _|_ >= _|_ by (Bot) 60] _|_ >= _|_ by (Bot) 61] _|_ >= _|_ by (Bot) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_8, 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.