We consider the system Applicative_first_order_05__hydra. Alphabet: 0 : [] --> a cons : [c * c] --> c copy : [a * c * c] --> c f : [c] --> c false : [] --> b filter : [c -> b * c] --> c filter2 : [b * c -> b * c * c] --> c map : [c -> c * c] --> c n : [] --> a nil : [] --> c s : [a] --> a true : [] --> b Rules: f(cons(nil, x)) => x f(cons(f(cons(nil, x)), y)) => copy(n, x, y) copy(0, x, y) => f(y) copy(s(x), y, z) => copy(x, y, cons(f(y), z)) map(g, nil) => nil map(g, cons(x, y)) => cons(g x, map(g, y)) filter(g, nil) => nil filter(g, cons(x, y)) => filter2(g x, g, x, y) filter2(true, g, x, y) => cons(x, filter(g, y)) filter2(false, g, x, y) => filter(g, 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] f#(cons(f(cons(nil, X)), Y)) =#> copy#(n, X, Y) 1] copy#(0, X, Y) =#> f#(Y) 2] copy#(s(X), Y, Z) =#> copy#(X, Y, cons(f(Y), Z)) 3] copy#(s(X), Y, Z) =#> f#(Y) 4] map#(F, cons(X, Y)) =#> map#(F, Y) 5] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 6] filter2#(true, F, X, Y) =#> filter#(F, Y) 7] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: f(cons(nil, X)) => X f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) copy(0, X, Y) => f(Y) copy(s(X), Y, Z) => copy(X, Y, cons(f(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). 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 : 0 * 2 : 1, 2, 3 * 3 : 0 * 4 : 4 * 5 : 6, 7 * 6 : 5 * 7 : 5 This graph has the following strongly connected components: P_1: copy#(s(X), Y, Z) =#> copy#(X, Y, cons(f(Y), Z)) P_2: map#(F, cons(X, Y)) =#> map#(F, Y) P_3: 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) and (P_3, 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) 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 [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) f(cons(nil, X)) >= X f(cons(f(cons(nil, X)), Y)) >= copy(n, X, Y) copy(0, X, Y) >= f(Y) copy(s(X), Y, Z) >= copy(X, Y, cons(f(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.2 + y1 copy = \y0y1y2.2y2 + 3y0 f = \y0.2y0 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.2 + y3 filter2# = \y0G1y2y3.2y3 filter# = \G0y1.2y1 map = \G0y1.y1 n = 0 nil = 0 s = \y0.3 + y0 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x2 > 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2 >= 2x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2 >= 2x2 = [[filter#(_F0, _x2)]] [[f(cons(nil, _x0))]] = 4 + 2x0 >= x0 = [[_x0]] [[f(cons(f(cons(nil, _x0)), _x1))]] = 4 + 2x1 >= 2x1 = [[copy(n, _x0, _x1)]] [[copy(0, _x0, _x1)]] = 9 + 2x1 >= 2x1 = [[f(_x1)]] [[copy(s(_x0), _x1, _x2)]] = 9 + 2x2 + 3x0 >= 4 + 2x2 + 3x0 = [[copy(_x0, _x1, cons(f(_x1), _x2))]] [[map(_F0, nil)]] = 0 >= 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))]] = 2 + x2 >= 2 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 >= 2 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x2 >= x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 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) and (P_4, R_0, static, formative) is finite. We consider the dependency pair problem (P_4, 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) 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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: map#(F, cons(X, Y)) >? map#(F, Y) f(cons(nil, X)) >= X f(cons(f(cons(nil, X)), Y)) >= copy(n, X, Y) copy(0, X, Y) >= f(Y) copy(s(X), Y, Z) >= copy(X, Y, cons(f(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.1 + y1 copy = \y0y1y2.y0 + y2 f = \y0.1 + y0 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + y3 map = \G0y1.2 + y1 map# = \G0y1.y1 n = 0 nil = 0 s = \y0.3 + y0 true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] [[f(cons(nil, _x0))]] = 2 + x0 >= x0 = [[_x0]] [[f(cons(f(cons(nil, _x0)), _x1))]] = 2 + x1 >= x1 = [[copy(n, _x0, _x1)]] [[copy(0, _x0, _x1)]] = 3 + x1 >= 1 + x1 = [[f(_x1)]] [[copy(s(_x0), _x1, _x2)]] = 3 + x0 + x2 >= 1 + x0 + x2 = [[copy(_x0, _x1, cons(f(_x1), _x2))]] [[map(_F0, nil)]] = 2 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x2 >= 3 + 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 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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: copy#(s(X), Y, Z) >? copy#(X, Y, cons(f(Y), Z)) f(cons(nil, X)) >= X f(cons(f(cons(nil, X)), Y)) >= copy(n, X, Y) copy(0, X, Y) >= f(Y) copy(s(X), Y, Z) >= copy(X, Y, cons(f(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.y1 copy = \y0y1y2.y0 + y2 copy# = \y0y1y2.y0 f = \y0.y0 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 map = \G0y1.0 n = 0 nil = 0 s = \y0.3 + y0 true = 3 Using this interpretation, the requirements translate to: [[copy#(s(_x0), _x1, _x2)]] = 3 + x0 > x0 = [[copy#(_x0, _x1, cons(f(_x1), _x2))]] [[f(cons(nil, _x0))]] = x0 >= x0 = [[_x0]] [[f(cons(f(cons(nil, _x0)), _x1))]] = x1 >= x1 = [[copy(n, _x0, _x1)]] [[copy(0, _x0, _x1)]] = 3 + x1 >= x1 = [[f(_x1)]] [[copy(s(_x0), _x1, _x2)]] = 3 + x0 + x2 >= x0 + x2 = [[copy(_x0, _x1, cons(f(_x1), _x2))]] [[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 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.