We consider the system Applicative_first_order_05__02. Alphabet: !facdot : [a * a] --> a 1 : [] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d i : [a] --> a map : [c -> c * d] --> d nil : [] --> d true : [] --> b Rules: !facdot(1, x) => x !facdot(x, 1) => x !facdot(i(x), x) => 1 !facdot(x, i(x)) => 1 !facdot(i(x), !facdot(x, y)) => y !facdot(x, !facdot(i(x), y)) => y i(1) => 1 i(i(x)) => 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, all): Dependency Pairs P_0: 0] map#(F, cons(X, Y)) =#> map#(F, Y) 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 3] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: !facdot(1, X) => X !facdot(X, 1) => X !facdot(i(X), X) => 1 !facdot(X, i(X)) => 1 !facdot(i(X), !facdot(X, Y)) => Y !facdot(X, !facdot(i(X), Y)) => Y i(1) => 1 i(i(X)) => 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, all) is finite. We consider the dependency pair problem (P_0, R_0, static, all). 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, 3 * 2 : 1 * 3 : 1 This graph has the following strongly connected components: P_1: map#(F, cons(X, Y)) =#> map#(F, Y) P_2: 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) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, static, all) and (P_2, R_0, static, all) is finite. We consider the dependency pair problem (P_2, R_0, static, all). 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) !facdot(1, X) >= X !facdot(X, 1) >= X !facdot(i(X), X) >= 1 !facdot(X, i(X)) >= 1 !facdot(i(X), !facdot(X, Y)) >= Y !facdot(X, !facdot(i(X), Y)) >= Y i(1) >= 1 i(i(X)) >= 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: !facdot = \y0y1.3 + 3y0 + 3y1 1 = 0 cons = \y0y1.1 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.1 + y3 filter# = \G0y1.y1 i = \y0.3 + 3y0 map = \G0y1.y1 nil = 0 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] [[!facdot(1, _x0)]] = 3 + 3x0 >= x0 = [[_x0]] [[!facdot(_x0, 1)]] = 3 + 3x0 >= x0 = [[_x0]] [[!facdot(i(_x0), _x0)]] = 12 + 12x0 >= 0 = [[1]] [[!facdot(_x0, i(_x0))]] = 12 + 12x0 >= 0 = [[1]] [[!facdot(i(_x0), !facdot(_x0, _x1))]] = 21 + 9x1 + 18x0 >= x1 = [[_x1]] [[!facdot(_x0, !facdot(i(_x0), _x1))]] = 39 + 9x1 + 30x0 >= x1 = [[_x1]] [[i(1)]] = 3 >= 0 = [[1]] [[i(i(_x0))]] = 12 + 9x0 >= x0 = [[_x0]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[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_2, R_0, static, all) by (P_3, R_0, static, all), where P_3 consists of: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) Thus, the original system is terminating if each of (P_1, R_0, static, all) and (P_3, R_0, static, all) is finite. We consider the dependency pair problem (P_3, R_0, static, all). 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 (P_1, R_0, static, all) is finite. We consider the dependency pair problem (P_1, R_0, static, all). 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) !facdot(1, X) >= X !facdot(X, 1) >= X !facdot(i(X), X) >= 1 !facdot(X, i(X)) >= 1 !facdot(i(X), !facdot(X, Y)) >= Y !facdot(X, !facdot(i(X), Y)) >= Y i(1) >= 1 i(i(X)) >= 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: !facdot = \y0y1.3 + 3y0 + 3y1 1 = 0 cons = \y0y1.1 + 2y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 i = \y0.3 + 3y0 map = \G0y1.1 + 3y1 map# = \G0y1.y1 nil = 0 true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[map#(_F0, _x2)]] [[!facdot(1, _x0)]] = 3 + 3x0 >= x0 = [[_x0]] [[!facdot(_x0, 1)]] = 3 + 3x0 >= x0 = [[_x0]] [[!facdot(i(_x0), _x0)]] = 12 + 12x0 >= 0 = [[1]] [[!facdot(_x0, i(_x0))]] = 12 + 12x0 >= 0 = [[1]] [[!facdot(i(_x0), !facdot(_x0, _x1))]] = 21 + 9x1 + 18x0 >= x1 = [[_x1]] [[!facdot(_x0, !facdot(i(_x0), _x1))]] = 39 + 9x1 + 30x0 >= x1 = [[_x1]] [[i(1)]] = 3 >= 0 = [[1]] [[i(i(_x0))]] = 12 + 9x0 >= x0 = [[_x0]] [[map(_F0, nil)]] = 1 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 4 + 6x2 >= 3 + 6x2 = [[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 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.