We consider the system Applicative_first_order_05__#3.2. Alphabet: 0 : [] --> a cons : [c * d] --> d 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 pred : [a] --> a quot : [a * a] --> a s : [a] --> a true : [] --> b Rules: pred(s(x)) => x minus(x, 0) => x minus(x, s(y)) => pred(minus(x, y)) quot(0, s(x)) => 0 quot(s(x), s(y)) => s(quot(minus(x, y), s(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#(X, s(Y)) =#> pred#(minus(X, Y)) 1] minus#(X, s(Y)) =#> minus#(X, Y) 2] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 3] quot#(s(X), s(Y)) =#> minus#(X, 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: pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(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#(X, s(Y)) >? pred#(minus(X, Y)) minus#(X, s(Y)) >? minus#(X, Y) quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) quot#(s(X), s(Y)) >? minus#(X, 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) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) quot(0, s(X)) >= 0 quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(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 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.y3 filter# = \G0y1.y1 map = \G0y1.2y1 map# = \G0y1.y1 minus = \y0y1.y0 minus# = \y0y1.1 nil = 0 pred = \y0.y0 pred# = \y0.0 quot = \y0y1.0 quot# = \y0y1.3 + 2y1 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(_x0, s(_x1))]] = 1 > 0 = [[pred#(minus(_x0, _x1))]] [[minus#(_x0, s(_x1))]] = 1 >= 1 = [[minus#(_x0, _x1)]] [[quot#(s(_x0), s(_x1))]] = 3 + 2x1 >= 3 + 2x1 = [[quot#(minus(_x0, _x1), s(_x1))]] [[quot#(s(_x0), s(_x1))]] = 3 + 2x1 > 1 = [[minus#(_x0, _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > 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)]] [[pred(s(_x0))]] = x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_x1)))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 2 + 4x2 >= 1 + 4x2 = [[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_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of: minus#(X, s(Y)) =#> minus#(X, Y) quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(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#(X, s(Y)) >? minus#(X, Y) quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) quot(0, s(X)) >= 0 quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(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 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 pred = \y0.y0 quot = \y0y1.0 quot# = \y0y1.0 s = \y0.2y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(_x0, s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] [[quot#(s(_x0), s(_x1))]] = 0 >= 0 = [[quot#(minus(_x0, _x1), s(_x1))]] [[filter2#(true, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 3 > 0 = [[filter#(_F0, _x2)]] [[pred(s(_x0))]] = 2x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] [[quot(s(_x0), s(_x1))]] = 0 >= 0 = [[s(quot(minus(_x0, _x1), s(_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_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of: minus#(X, s(Y)) =#> minus#(X, Y) quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(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#(X, s(Y)) >? minus#(X, Y) quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) quot(0, s(X)) >= 0 quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(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 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 map = \G0y1.0 minus = \y0y1.y0 minus# = \y0y1.0 nil = 0 pred = \y0.y0 quot = \y0y1.y0 quot# = \y0y1.y0 s = \y0.1 + y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(_x0, s(_x1))]] = 0 >= 0 = [[minus#(_x0, _x1)]] [[quot#(s(_x0), s(_x1))]] = 1 + x0 > x0 = [[quot#(minus(_x0, _x1), s(_x1))]] [[pred(s(_x0))]] = 1 + x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] [[quot(s(_x0), s(_x1))]] = 1 + x0 >= 1 + x0 = [[s(quot(minus(_x0, _x1), s(_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_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of: minus#(X, s(Y)) =#> minus#(X, 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#(X, s(Y)) >? minus#(X, Y) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) quot(0, s(X)) >= 0 quot(s(X), s(Y)) >= s(quot(minus(X, Y), s(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 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 map = \G0y1.0 minus = \y0y1.y0 minus# = \y0y1.y1 nil = 0 pred = \y0.y0 quot = \y0y1.y0 s = \y0.1 + y0 true = 3 Using this interpretation, the requirements translate to: [[minus#(_x0, s(_x1))]] = 1 + x1 > x1 = [[minus#(_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))]] [[quot(0, s(_x0))]] = 0 >= 0 = [[0]] [[quot(s(_x0), s(_x1))]] = 1 + x0 >= 1 + x0 = [[s(quot(minus(_x0, _x1), s(_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 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. 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.