We consider the system Applicative_first_order_05__perfect2. Alphabet: 0 : [] --> a cons : [c * d] --> d f : [a * a * a * a] --> b false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d if : [b * b * b] --> b le : [a * a] --> b map : [c -> c * d] --> d minus : [a * a] --> a nil : [] --> d perfectp : [a] --> b s : [a] --> a true : [] --> b Rules: minus(0, x) => 0 minus(s(x), 0) => s(x) minus(s(x), s(y)) => minus(x, y) le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) if(true, x, y) => x if(false, x, y) => y perfectp(0) => false perfectp(s(x)) => f(x, s(0), s(x), s(x)) f(0, x, 0, y) => true f(0, x, s(y), z) => false f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) 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 [FuhKop19]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] le#(s(X), s(Y)) =#> le#(X, Y) 2] perfectp#(s(X)) =#> f#(X, s(0), s(X), s(X)) 3] f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) 4] f#(s(X), 0, Y, Z) =#> minus#(Y, s(X)) 5] f#(s(X), s(Y), Z, U) =#> if#(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 6] f#(s(X), s(Y), Z, U) =#> le#(X, Y) 7] f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) 8] f#(s(X), s(Y), Z, U) =#> minus#(Y, X) 9] f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) 10] map#(F, cons(X, Y)) =#> map#(F, Y) 11] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 12] filter2#(true, F, X, Y) =#> filter#(F, Y) 13] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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, computable, formative) is finite. We consider the dependency pair problem (P_0, R_0, computable, 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 : 5, 6, 7, 8, 9 * 3 : 3, 4, 5, 6, 7, 8, 9 * 4 : 0 * 5 : * 6 : 1 * 7 : 3, 4, 5, 6, 7, 8, 9 * 8 : 0 * 9 : 3, 4, 5, 6, 7, 8, 9 * 10 : 10 * 11 : 12, 13 * 12 : 11 * 13 : 11 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: le#(s(X), s(Y)) =#> le#(X, Y) P_3: f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) 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, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative) and (P_5, R_0, computable, formative) is finite. We consider the dependency pair problem (P_5, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(filter2#) = 4 nu(filter#) = 2 Thus, we can orient the dependency pairs as follows: nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_5, R_0, computable, f) by (P_6, R_0, computable, f), where P_6 contains: 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, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative) and (P_6, R_0, computable, formative) is finite. We consider the dependency pair problem (P_6, R_0, computable, 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, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative) and (P_4, R_0, computable, formative) is finite. We consider the dependency pair problem (P_4, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(map#) = 2 Thus, we can orient the dependency pairs as follows: nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_4, R_0, computable, f) by ({}, R_0, computable, f). 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, computable, formative), (P_2, R_0, computable, formative) and (P_3, R_0, computable, formative) is finite. We consider the dependency pair problem (P_3, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(f#) = 1 Thus, we can orient the dependency pairs as follows: nu(f#(s(X), 0, Y, Z)) = s(X) |> X = nu(f#(X, Z, minus(Y, s(X)), Z)) nu(f#(s(X), s(Y), Z, U)) = s(X) = s(X) = nu(f#(s(X), minus(Y, X), Z, U)) nu(f#(s(X), s(Y), Z, U)) = s(X) |> X = nu(f#(X, U, Z, U)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, R_0, computable, f) by (P_7, R_0, computable, f), where P_7 contains: f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative) and (P_7, R_0, computable, formative) is finite. We consider the dependency pair problem (P_7, R_0, computable, formative). The formative rules of (P_7, R_0) are R_1 ::= minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_7, R_0, computable, formative) by (P_7, R_1, computable, formative). Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative) and (P_7, R_1, computable, formative) is finite. We consider the dependency pair problem (P_7, R_1, computable, 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: f#(s(X), s(Y), Z, U) >? f#(s(X), minus(Y, X), Z, U) minus(0, X) >= 0 minus(s(X), 0) >= s(X) 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: 0 = 0 f# = \y0y1y2y3.y1 minus = \y0y1.y0 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[f#(s(_x0), s(_x1), _x2, _x3)]] = 1 + 2x1 > x1 = [[f#(s(_x0), minus(_x1, _x0), _x2, _x3)]] [[minus(0, _x0)]] = 0 >= 0 = [[0]] [[minus(s(_x0), 0)]] = 1 + 2x0 >= 1 + 2x0 = [[s(_x0)]] [[minus(s(_x0), s(_x1))]] = 1 + 2x0 >= x0 = [[minus(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_7, R_1) by ({}, R_1). 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, computable, formative) and (P_2, R_0, computable, formative) is finite. We consider the dependency pair problem (P_2, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(le#) = 1 Thus, we can orient the dependency pairs as follows: nu(le#(s(X), s(Y))) = s(X) |> X = nu(le#(X, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by ({}, R_0, computable, f). 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, computable, formative) is finite. We consider the dependency pair problem (P_1, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, f). 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 +++ [FuhKop19] C. Fuhs, and C. Kop. A static higher-order dependency pair framework. In Proceedings of ESOP 2019, 2019. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [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.