We consider the system AotoYamada_05__Ex1SimplyTyped. Alphabet: 0 : [] --> a add : [a] --> a -> a cons : [b * c] --> c id : [] --> a -> a map : [b -> b * c] --> c nil : [] --> c s : [a] --> a Rules: id x => x add(0) => id add(s(x)) y => s(add(x) y) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: id(X) => X add(0, X) => id(X) add(s(X), Y) => s(add(X, Y)) Moreover, the system is orthogonal. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || proof of resources/system.trs || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 || || || Termination w.r.t. Q of the given QTRS could be proven: || || (0) QTRS || (1) QTRSRRRProof [EQUIVALENT] || (2) QTRS || (3) RisEmptyProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || id(%X) -> %X || add(0, %X) -> id(%X) || add(s(%X), %Y) -> s(add(%X, %Y)) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(0) = 2 || POL(add(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 || POL(id(x_1)) = 1 + x_1 || POL(s(x_1)) = 1 + x_1 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || id(%X) -> %X || add(0, %X) -> id(%X) || add(s(%X), %Y) -> s(add(%X, %Y)) || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (3) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. || ---------------------------------------- || || (4) || YES || 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]). In order to do so, we start by eta-expanding the system, which gives: id(X) => X add(0, X) => id(X) add(s(X), Y) => s(add(X, Y)) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] map#(F, cons(X, Y)) =#> map#(F, Y) Rules R_0: id(X) => X add(0, X) => id(X) add(s(X), Y) => s(add(X, Y)) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(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 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_0, 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.