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 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 : 1 * 2 : 2, 3, 4, 5, 6 * 3 : 2, 3, 4, 5, 6 * 4 : 2, 3, 4, 5, 6 * 5 : 0 * 6 : 1 * 7 : 7 * 8 : 9, 10 * 9 : 8 * 10 : 8 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: double#(s(X)) =#> double#(X) P_3: 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)) 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) 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 + y1 double = \y0.0 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + y3 filter2# = \y0G1y2y3.y3 filter# = \G0y1.y1 map = \G0y1.1 + y1 minus = \y0y1.y0 nil = 0 plus = \y0y1.y1 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = x2 >= x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = x2 >= x2 = [[filter#(_F0, _x2)]] [[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)]] = 1 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + x2 >= 1 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + 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_5, R_0, static, formative) by (P_6, R_0, static, formative), where P_6 consists of: filter2#(true, F, X, Y) =#> filter#(F, 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : 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), (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: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) double(0) => 0 double(s(X)) => s(s(double(X))) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: 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))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 double = \y0.2y0 minus = \y0y1.y0 plus# = \y0y1.y0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 1 + x0 > x0 = [[plus#(_x0, _x1)]] [[plus#(s(_x0), _x1)]] = 1 + x0 > x0 = [[plus#(_x0, s(_x1))]] [[plus#(s(_x0), _x1)]] = 1 + x0 > x0 = [[plus#(minus(_x0, _x1), double(_x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 1 + x0 >= x0 = [[minus(_x0, _x1)]] [[double(0)]] = 0 >= 0 = [[0]] [[double(s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[s(s(double(_x0)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, 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) 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: double#(s(X)) >? double#(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: double# = \y0.y0 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[double#(s(_x0))]] = 1 + 2x0 > x0 = [[double#(_x0)]] 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: minus#(s(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.y0 + y1 s = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[minus#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + 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_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.