We consider the system Applicative_first_order_05__#3.25. Alphabet: cons : [c * d] --> d f : [a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d g : [a] --> a h : [a] --> a map : [c -> c * d] --> d nil : [] --> d true : [] --> b Rules: f(g(x)) => g(f(f(x))) f(h(x)) => h(g(x)) map(i, nil) => nil map(i, cons(x, y)) => cons(i x, map(i, y)) filter(i, nil) => nil filter(i, cons(x, y)) => filter2(i x, i, x, y) filter2(true, i, x, y) => cons(x, filter(i, y)) filter2(false, i, x, y) => filter(i, 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#(g(X)) =#> f#(f(X)) 1] f#(g(X)) =#> f#(X) 2] map#(F, cons(X, Y)) =#> map#(F, Y) 3] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 4] filter2#(true, F, X, Y) =#> filter#(F, Y) 5] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: f(g(X)) => g(f(f(X))) f(h(X)) => h(g(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, 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 : 0, 1 * 2 : 2 * 3 : 4, 5 * 4 : 3 * 5 : 3 This graph has the following strongly connected components: P_1: f#(g(X)) =#> f#(f(X)) f#(g(X)) =#> f#(X) 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 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 [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_3, R_0, static, f) by (P_4, R_0, static, f), where P_4 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, 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 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 [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by ({}, R_0, static, 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, static, formative) is finite. We consider the dependency pair problem (P_1, R_0, static, formative). The formative rules of (P_1, R_0) are R_1 ::= f(g(X)) => g(f(f(X))) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, static, formative) by (P_1, R_1, static, formative). Thus, the original system is terminating if (P_1, R_1, static, formative) is finite. We consider the dependency pair problem (P_1, R_1, 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: f#(g(X)) >? f#(f(X)) f#(g(X)) >? f#(X) f(g(X)) >= g(f(f(X))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0.y0 f# = \y0.3y0 g = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[f#(g(_x0))]] = 9 + 9x0 > 3x0 = [[f#(f(_x0))]] [[f#(g(_x0))]] = 9 + 9x0 > 3x0 = [[f#(_x0)]] [[f(g(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[g(f(f(_x0)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1). 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.